MFE

Deeper Understanding, Faster Calc: SOA MFEand CAS Exam 3F

Yufeng Guo

July 14, 2009

Contents

Introduction ix

9 Parity and other option relationships 19.1 Put-call parity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

9.1.1 Option on stocks . . . . . . . . . . . . . . . . . . . . . . . 19.1.2 Options on currencies . . . . . . . . . . . . . . . . . . . . 119.1.3 Options on bonds . . . . . . . . . . . . . . . . . . . . . . . 129.1.4 Generalized parity and exchange options . . . . . . . . . . 139.1.5 Comparing options with respect to style, maturity, and

strike . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

10 Binomial option pricing: I 3510.1 One-period binomial model: simple examples . . . . . . . . . . . 3510.2 General one-period binomial model . . . . . . . . . . . . . . . . . 36

10.2.1 Two or more binomial trees . . . . . . . . . . . . . . . . . 4910.2.2 Options on stock index . . . . . . . . . . . . . . . . . . . 6410.2.3 Options on currency . . . . . . . . . . . . . . . . . . . . . 6710.2.4 Options on futures contracts . . . . . . . . . . . . . . . . 71

11 Binomial option pricing: II 7911.1 Understanding early exercise . . . . . . . . . . . . . . . . . . . . 7911.2 Understanding risk-neutral probability . . . . . . . . . . . . . . . 80

11.2.1 Pricing an option using real probabilities . . . . . . . . . 8111.2.2 Binomial tree and lognormality . . . . . . . . . . . . . . . 8811.2.3 Estimate stock volatility . . . . . . . . . . . . . . . . . . . 91

11.3 Stocks paying discrete dividends . . . . . . . . . . . . . . . . . . 9511.3.1 Problems with discrete dividend tree . . . . . . . . . . . . 9611.3.2 Binomial tree using prepaid forward . . . . . . . . . . . . 98

12 Black-Scholes 10512.1 Introduction to the Black-Scholes formula . . . . . . . . . . . . . 105

12.1.1 Call and put option price . . . . . . . . . . . . . . . . . . 10512.1.2 When is the Black-Scholes formula valid? . . . . . . . . . 107

12.2 Applying the formula to other assets . . . . . . . . . . . . . . . . 107

iii

Yufeng Guo, Fall 09 MFE, actuary88.com

iv CONTENTS

12.2.1 Black-Scholes formula in terms of prepaid forward price . 10712.2.2 Options on stocks with discrete dividends . . . . . . . . . 10812.2.3 Options on currencies . . . . . . . . . . . . . . . . . . . . 10812.2.4 Options on futures . . . . . . . . . . . . . . . . . . . . . . 110

12.3 Option the Greeks . . . . . . . . . . . . . . . . . . . . . . . . . . 11012.3.1 Delta . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11112.3.2 Gamma . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11112.3.3 Vega . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11212.3.4 Theta . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11212.3.5 Rho . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11212.3.6 Psi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11212.3.7 Greek measures for a portfolio . . . . . . . . . . . . . . . 11212.3.8 Option elasticity and volatility . . . . . . . . . . . . . . . 11312.3.9 Option risk premium and Sharp ratio . . . . . . . . . . . 11412.3.10Elasticity and risk premium of a portfolio . . . . . . . . . 115

12.4 Profit diagrams before maturity . . . . . . . . . . . . . . . . . . . 11512.4.1 Holding period profit . . . . . . . . . . . . . . . . . . . . . 11512.4.2 Calendar spread . . . . . . . . . . . . . . . . . . . . . . . 118

12.5 Implied volatility . . . . . . . . . . . . . . . . . . . . . . . . . . . 11912.5.1 Calculate the implied volatility . . . . . . . . . . . . . . . 11912.5.2 Volatility skew . . . . . . . . . . . . . . . . . . . . . . . . 12012.5.3 Using implied volatility . . . . . . . . . . . . . . . . . . . 121

12.6 Perpetual American options . . . . . . . . . . . . . . . . . . . . . 12112.6.1 Perpetual calls and puts . . . . . . . . . . . . . . . . . . . 12112.6.2 Barrier present values . . . . . . . . . . . . . . . . . . . . 125

13 Market-making and delta-hedging 12913.1 Delta hedging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12913.2 Examples of Delta hedging . . . . . . . . . . . . . . . . . . . . . 12913.3 Textbook Table 13.2 . . . . . . . . . . . . . . . . . . . . . . . . . 13813.4 Textbook Table 13.3 . . . . . . . . . . . . . . . . . . . . . . . . . 14013.5 Mathematics of Delta hedging . . . . . . . . . . . . . . . . . . . . 141

13.5.1 Delta-Gamma-Theta approximation . . . . . . . . . . . . 14113.5.2 Understanding the market maker’s profit . . . . . . . . . 142

14 Exotic options: I 14514.1 Asian option (i.e. average options) . . . . . . . . . . . . . . . . . 145

14.1.1 Characteristics . . . . . . . . . . . . . . . . . . . . . . . . 14514.1.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 14614.1.3 Geometric average . . . . . . . . . . . . . . . . . . . . . . 14614.1.4 Payoff at maturity T . . . . . . . . . . . . . . . . . . . . . 147

14.2 Barrier option . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14714.2.1 Knock-in option . . . . . . . . . . . . . . . . . . . . . . . 14714.2.2 Knock-out option . . . . . . . . . . . . . . . . . . . . . . . 14714.2.3 Rebate option . . . . . . . . . . . . . . . . . . . . . . . . . 148


CONTENTS v

14.2.4 Barrier parity . . . . . . . . . . . . . . . . . . . . . . . . . 14814.2.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

14.3 Compound option . . . . . . . . . . . . . . . . . . . . . . . . . . 14914.4 Gap option . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

14.4.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . 15114.4.2 Pricing formula . . . . . . . . . . . . . . . . . . . . . . . . 15114.4.3 How to memorize the pricing formula . . . . . . . . . . . 151

14.5 Exchange option . . . . . . . . . . . . . . . . . . . . . . . . . . . 152

18 Lognormal distribution 15518.1 Normal distribution . . . . . . . . . . . . . . . . . . . . . . . . . 15518.2 Lognormal distribution . . . . . . . . . . . . . . . . . . . . . . . . 15618.3 Lognormal model of stock prices . . . . . . . . . . . . . . . . . . 15618.4 Lognormal probability calculation . . . . . . . . . . . . . . . . . . 157

18.4.1 Lognormal confidence interval . . . . . . . . . . . . . . . . 15818.4.2 Conditional expected prices . . . . . . . . . . . . . . . . . 16218.4.3 Black-Scholes formula . . . . . . . . . . . . . . . . . . . . 162

18.5 Estimating the parameters of a lognormal distribution . . . . . . 163

18.6 How are asset prices distributed . . . . . . . . . . . . . . . . . . . 16518.6.1 Histogram . . . . . . . . . . . . . . . . . . . . . . . . . . . 16518.6.2 Normal probability plots . . . . . . . . . . . . . . . . . . . 166

18.7 Sample problems . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

19 Monte Carlo valuation 17319.1 Example 1 Estimate E (ez) . . . . . . . . . . . . . . . . . . . . . 17319.2 Example 2 Estimate π . . . . . . . . . . . . . . . . . . . . . . . . 17719.3 Example 3 Estimate the price of European call or put options . . 18019.4 Example 4 Arithmetic and geometric options . . . . . . . . . . . 18419.5 Efficient Monte Carlo valuation . . . . . . . . . . . . . . . . . . . 193

19.5.1 Control variance method . . . . . . . . . . . . . . . . . . . 19319.6 Antithetic variate method . . . . . . . . . . . . . . . . . . . . . . 19719.7 Stratified sampling . . . . . . . . . . . . . . . . . . . . . . . . . . 199

19.7.1 Importance sampling . . . . . . . . . . . . . . . . . . . . . 19919.8 Sample problems . . . . . . . . . . . . . . . . . . . . . . . . . . . 199

20 Brownian motion and Ito’s Lemma 20520.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205

20.1.1 Big picture . . . . . . . . . . . . . . . . . . . . . . . . . . 20620.2 Brownian motion . . . . . . . . . . . . . . . . . . . . . . . . . . . 207

20.2.1 Stochastic process . . . . . . . . . . . . . . . . . . . . . . 20720.2.2 Definition of Brownian motion . . . . . . . . . . . . . . . 20720.2.3 Martingale . . . . . . . . . . . . . . . . . . . . . . . . . . 20920.2.4 Properties of Brownian motion . . . . . . . . . . . . . . . 21020.2.5 Arithmetic Brownian motion and Geometric Brownian

motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214


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20.2.6 Ornstein-Uhlenbeck process . . . . . . . . . . . . . . . . . 21520.3 Definition of the stochastic calculus . . . . . . . . . . . . . . . . . 21620.4 Properties of the stochastic calculus . . . . . . . . . . . . . . . . 22220.5 Ito’s lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223

20.5.1 Multiplication rules . . . . . . . . . . . . . . . . . . . . . 22320.5.2 Ito’s lemma . . . . . . . . . . . . . . . . . . . . . . . . . . 223

20.6 Geometric Brownian motion revisited . . . . . . . . . . . . . . . 22520.6.1 Relative importance of drift and noise term . . . . . . . . 22520.6.2 Correlated Ito processes . . . . . . . . . . . . . . . . . . . 226

20.7 Sharpe ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23020.8 Risk neutral process . . . . . . . . . . . . . . . . . . . . . . . . . 23320.9 Valuing a claim on Sa . . . . . . . . . . . . . . . . . . . . . . . . 233

20.9.1 Process followed by Sa . . . . . . . . . . . . . . . . . . . . 23320.9.2 Formula for SA (t) and E

£SA (t)

¤. . . . . . . . . . . . . . 234

20.9.3 Expected return of a claim on SA (t) . . . . . . . . . . . . 23520.9.4 Specific examples . . . . . . . . . . . . . . . . . . . . . . . 235

21 Black-Scholes equation 24521.1 Differential equations and valuation under certainty . . . . . . . 245

21.1.1 Valuation equation . . . . . . . . . . . . . . . . . . . . . . 24521.1.2 Bonds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24621.1.3 Dividend paying stock . . . . . . . . . . . . . . . . . . . . 246

21.2 Black-Scholes equation . . . . . . . . . . . . . . . . . . . . . . . . 24621.2.1 How to derive Black-Scholes equation . . . . . . . . . . . 24621.2.2 Verifying the formula for a derivative . . . . . . . . . . . . 24721.2.3 Black-Scholes equation and equilibrium returns . . . . . . 250

21.3 Risk-neutral pricing . . . . . . . . . . . . . . . . . . . . . . . . . 252

22 Exotic options: II 25322.1 All-or-nothing options . . . . . . . . . . . . . . . . . . . . . . . . 253

23 Volatility 255

24 Interest rate models 25724.1 Market-making and bond pricing . . . . . . . . . . . . . . . . . . 257

24.1.1 Review of duration and convexity . . . . . . . . . . . . . . 25724.1.2 Interest rate is not so simple . . . . . . . . . . . . . . . . 26424.1.3 Impossible bond pricing model . . . . . . . . . . . . . . . 26524.1.4 Equilibrium equation for bonds . . . . . . . . . . . . . . . 26824.1.5 Delta-Gamma approximation for bonds . . . . . . . . . . 270

24.2 Equilibrium short-rate bond price models . . . . . . . . . . . . . 27124.2.1 Arithmetic Brownian motion (i.e. Merton model) . . . . . 27124.2.2 Rendleman-Bartter model . . . . . . . . . . . . . . . . . . 27224.2.3 Vasicek model . . . . . . . . . . . . . . . . . . . . . . . . 27224.2.4 CIR model . . . . . . . . . . . . . . . . . . . . . . . . . . 274

24.3 Bond options, caps, and the Black model . . . . . . . . . . . . . 275


CONTENTS vii

24.3.1 Black formula . . . . . . . . . . . . . . . . . . . . . . . . . 27524.3.2 Interest rate caplet . . . . . . . . . . . . . . . . . . . . . . 279

24.4 Binomial interest rate model . . . . . . . . . . . . . . . . . . . . 27924.5 Black-Derman-Toy model . . . . . . . . . . . . . . . . . . . . . . 283


Introduction

This study guide is for SOA MFE and CAS Exam 3F. Before you start, makesure you have the following items:

1. Derivatives Markets, the 2nd edition.

2. Errata of Derivatives Markets. You can download the errata at http://www.kellogg.northwestern.edu/faculty/mcdonald/htm/typos2e_01.html.

3. Download the syllabus from the SOA or CAS website.

4. Download the sample MFE problems and solutions from the SOA website.

5. Download the recent SOA MFE and CAS Exam 3 problems.

ix


Chapter 9

Parity and other optionrelationships

9.1 Put-call parity

9.1.1 Option on stocks

Notation

t = 0 Current date (date when an option is sold or bought)T Option expiration date (maturity date)S0 Current price of the underlying assetST The price of the underlying asset at the option expiration dateK Strike price or exercise price

CEur (K,T ) Premium of a European call option with strike price K and T years to expirationCEur (K, 0) Premium of a European call option on the expiration datePEur (K,T ) Premium of a European put option with strike price K and T years to expirationPEur (K, 0) Premium of a European put option on the expiration date

r The continuously compounded annual risk-free interest rateδ The continuously compounded annual dividend rate

F0,T Delivery price in a forward contract expiring in T

Put-call parity

The textbook gives the following formula

CEur (K,T )− PEur (K,T ) = PV0,T (F0,T −K) = e−rT (F0,T −K) (9.1)

The textbook explains the intuition behind Equation 9.1. If we set theforward price F0,T as the common strike price for both the call and the put

1


2 CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS

(i.e. K = F0,T ) , then CEur (K,T ) − PEur (K,T ) = PV0,T (F0,T −K) = 0.Buying a call and selling a put with K = F0,T synthetically creates a forwardcontract and the premium for a forward contract is zero.However, often I find that Equation 9.1 is not intuitive at all. In fact, it’s

annoyingly complex and hard to memorize. So I like to rewrite 9.1 as follows:

CEur (K,T ) + PV (K) = PEur (K,T ) + S0 (9.2)

Later I will explain the intuition behind 9.2. First, let’s prove 9.2. The proofis extremely important.Suppose at time zero we have two portfolios:

• Portfolio #1 consists of a European call option on a stock and PV (K) =Ke−rT . PV (K) is the present value of the strike price K and r is thecontinuously compounded risk free interest rate per year.

• Portfolio #2 consists of a European put option on the stock and one shareof the stock with current price S0.

• Both the call and put have the same underlying stock, the same strike priceK, and the same expiration date T . Notice that at time zero Portfolio #1is worthy CEur (K,T ) +PV (K); Portfolio #2 is worth PEur (K,T ) +S0.

Since it’s difficult to compare the value of Portfolio #1 and the value ofPortfolio #2 at time zero, let’s compare them at the expiration date T . We’llsoon see that the two portfolios have the same value at T .

Payoff of Portfolio #1 at the expiration date TIf ST < K If ST ≥ K

Call payoff is max (0, ST −K) 0 ST −KPayoff of PV (K) K KTotal K ST

If you have PV (K) at t = 0, you’ll have K at T .

Payoff of Portfolio #2 at the expiration date TIf ST < K If ST ≥ K

Put payoff max (0,K − ST ) K − ST 0Payoff of S0 ST STTotal K ST

If you have one stock worth S0 at t = 0, you’ll have one stock at T worthST .You see that Portfolio #1 and Portfolio #2 have identical payoffs ofmax (K,ST )

at T . If ST < K, both portfolios are worth the strike price K; if ST ≥ K, both


9.1. PUT-CALL PARITY 3

portfolios are worth the stock price ST . Since Portfolio #1 and #2 are worththe same at T , to avoid arbitrage, they must be worth the same at any timeprior to T . Otherwise, anyone can make free money by buying the lower pricedportfolio and selling the higher priced one. So Portfolio #1 and #2 are worththe same at time zero. Equation 9.2 holds.

I recommend that from this point now you throw Equation 9.1 away and useEquation 9.2 instead.How to memorize Equation 9.2:

Tip 9.1.1. Many candidates have trouble memorizing Equation 9.2. For exam-ple, it’s very easy to write a wrong formula CEur (K,T ) + S0 = PEur (K,T ) +PV (K). To memorize Equation 9.2, notice that for a call to work, a call mustgo hand in hand with the strike price K. When exercising a call option, yougive the call seller both the call certificate and the strike price K. In return, thecall seller gives you one stock. Similarly, for a put to work, a put must go handin hand with one stock. When exercising a put, you must give the put seller boththe put certificate and one stock. In return, the put seller gives you the strikeprice K.

Tip 9.1.2. Another technique that helps me memorize Equation 9.2 is thephrase “Check (CK) Please (PS).” At expiration T , CEur (K, 0)+K = PEur (K, 0)+ST . Discounting this equation to time zero, we get: CEur (K,T ) + PV (K) =PEur (K,T ) + S0.

Example 9.1.1. The price of a 6-month 30-strike European call option is 12.22.The stock price is 35. The continuously compounded risk-free interest rate is 8%per year. What”s the price of a 6-month 30-strike European put option on thesame stock?

Solution.

K = 30 T = 6/12 = 0.5 S0 = 35 r = 0.08 CEur = 12.22CEur + PV (K) = PEur + S0→ 12.22 + 30e−0.08(0.5) = PEur + 35, PEur = 6. 043

Parity if the stock pays discrete dividends

If the stock pays discrete dividend, the parity formula is

CEur (K,T ) + PV (K) = PEur (K,T ) + S0 − PV (Div) (9.3)

This is why we need to subtract the term PV (Div). At expiration, CEur (K,T )+K = PEur (K,T ) + ST . If we discount ST from T to time zero, we’ll get



S0−PV (Div). If you have one stock worth S0 at time zero, then during [0, T ],you’ll receive dividend payments. Then at T , you not only have one share ofstock, you also have the accumulated value of the dividend. To get exactly onestock at T , you need to have S0 − PV (Div) at time zero.Discounting this equation back to time zero, we get Equation 9.3Please note that Equation 9.3 assumes that both the timing and the amount

of each dividend are 100% known in advance.

Example 9.1.2. The price of a 9-month 95-strike European call option is 19.24.The stock price is 100. The stock pays dividend of $1 in 3 months and $2 in 6months. The continuously compounded risk-free interest rate is 10% per year.What”s the price of a 9-month 95-strike European put option on the same stock?

Solution.

T = 9/12 = 0.75 K = 95 S0 = 100 r = 0.1CEur = 19.24PV (Div) = e−0.1(3/12) + 2e−0.1(6/12) = 2. 877 8

CEur (K,T ) + PV (K) = PEur (K,T ) + S0 − PV (Div)19.24 + 95e−0.1(0.75) = PEur (K,T ) + 100− 2. 877 8, PEur (K,T ) = 10. 25

Example 9.1.3. The price of a 9-month 83-strike European put option is 13.78.The stock price is 75. The stock pays dividend of $1 in 3 months, $2 in 6 months,$3 in 9 months, and $4 in 12 months. The continuously compounded risk-freeinterest rate is 6% per year. What”s the price of a 9-month 95-strike Europeancall option on the same stock?

Solution.

T = 9/12 = 0.75 K = 83 S0 = 75 r = 0.06PEur = 13.78PV (Div) = e−0.06(3/12) + 2e−0.06(6/12) + 3e−0.06(9/12) = 5. 794

CEur (K,T ) + PV (K) = PEur (K,T ) + S0 − PV (Div)CEur (K,T ) + 83e−0.06(0.75) = 13.78 + 75− 5. 794 CEur (K,T ) = 3. 64

Tip 9.1.3. When calculating PV (Div) in Equation 9.3, discard any dividendpaid after the option expiration date. In this example, the $4 is paid in 12months, which is after the expiration date of the option. This dividend is ignoredwhen we use Equation 9.3.



Parity if the stock pays continuous dividends

If the stock pays dividends at a continuously compounded rate of δ per year,the parity formula is:

CEur (K,T ) + PV (K) = PEur (K,T ) + S0e−δT (9.4)

At expiration, CEur (K,T )+K = PEur (K,T )+ST . If we discount ST fromT to time zero, we’ll get S0e−δT . If you have e−δT share of a stock, by investingdividends and buying additional stock, you’ll have exactly one stock at T . Thisconcept is called tailing. Refer to Derivatives Markets Section 5.2 about tailing.

Example 9.1.4. The price of a 6-month 90-strike European put option is 5.54.The stock price is 110. The stock pays dividend at a continuously compoundedrate of 2% per year. The continuously compounded risk-free interest rate is 6%per year. What”s the price of a 6-month 90-strike European call option on thesame stock?

Solution.

T = 6/12 = 0.5 K = 90 S0 = 110 r = 0.06 δ = 0.02PEur = 5.54CEur (K,T ) + PV (K) = PEur (K,T ) + S0e

−δT

CEur (K,T ) + 90e−0.06(0.5) = 5.54 + 110e−0.02(0.5) CEur (K,T ) = 27. 11

Example 9.1.5. The price of a 3-month 40-strike European call option is 6.57.The stock price is 44. The stock pays dividend at a continuously compoundedrate of 5% per year. The continuously compounded risk-free interest rate is 8%per year. What”s the price of a 3-month 40-strike European put option on thesame stock?

Solution.

T = 3/12 = 0.25 K = 40 S0 = 44 r = 0.08 δ = 0.08CEur = 6.57CEur (K,T ) + PV (K) = PEur (K,T ) + S0e

−δT

6.57 + 40e−0.08(0.25) = PEur (K,T ) + 44e−0.05(0.25) PEur (K,T ) = 2. 32

Synthetic stock

Rearranging Equation 9.3, we get:

S0 = CEur (K,T )− PEur (K,T ) + PV (K) + PV (Div) (9.5)

To understand the meaning of Equation 9.5, notice



Symbol Meaning at t = 0+CEur (K,T ) buy a K-strike call expiring in T−PEur (K,T ) sell a K-strike put expiring in T+PV (K) buy a zero-coupon bond that pays K at T+PV (Div) buy a zero-coupon bond that pays Div at T

Next, notice CEur (K,T ) − PEur (K,T ) + PV (K) is worth ST . At T , youalways receive K from the zero-coupon bond seller. However, the call and putvalues depend on whether ST ≥ K.

• If ST ≥ K, the sold put expires worthless; you exercise the call, paying Kand receiving one stock.

• If on the other hand, ST ≤ K, the purchased call expires worthless and thesold put is exercised against you. You pay the put holder K and receiveone stock from him.

• Either way, if at time zero you buy a call, sell a put, and invest PV (K)in a zero-coupon bond, then at T you are guaranteed to have one stock.

Once you understand CEur (K,T )−PEur (K,T )+PV (K) is worth one stockat T , the meaning of Equation 9.5 is obvious: if you buy one stock at time zero,then at time T , you’ll have one stock worth ST . In addition, you’ll receive thefuture value of the dividends due to owning a stock.

Example 9.1.6. The price of a 9-month 52-strike European call option on anon dividend-paying stock is 33.4420. The price of a 9-month 52-strike Europeanput option on the same stock is 15.1538. The continuously compounded risk-freeinterest rate is 6% per year. How can you synthetically create one stock at timezero? What’s the price of this synthetically created stock at time zero?

Solution.

The parity for a non-dividend paying stock isCEur (K,T ) + PV (K) = PEur (K,T ) + S0Rearranging this equation, we get:S0 = CEur (K,T ) + PV (K)− PEur (K,T )

To synthetically create the ownership of one stock, you need to do the fol-lowing at time zero:

• Buy a 9-month 52-strike European call option

• Sell a 9-month 52-strike European put option

• Invest 52e−0.06(0.75) = 49. 711 9 in an account earning risk-free interestrate (i.e. buying a zero coupon bond that pays 52 in 9 months)

The current price of this synthetically created stock is:S0 = 33.4420 + 49. 711 9− 15.1538 = 68



Synthetic T-bill

Rearranging Equation 9.3, we get:

PV (Div) + PV (K) = S0 + PEur (K,T )− CEur (K,T ) (9.6)

According to Equation 9.6, buying one stock, buying a K-strike put, andselling a K-strike call synthetically creates a zero coupon bond with a presentvalue equal to PV (Div) + PV (K).

Creating synthetic T-bill by buying the stock, buying a put, and selling acall is called a conversion. If we short the stock, buy a call, and sell a put, wecreate a short position on T-bill. This is called a reverse conversion.

Example 9.1.7. Your mother-in-law desperately wants to borrow $1000 fromyou for one year. She’s willing to pay you 50% interest rate for using yourmoney for one year. You really want to take her offer and earn 50% interest.However, state anti-usury laws prohibits any lender from charging an interestrate equal to or greater than 50%. Since you happen to know the put-call parity,you decide to synthetically create a loan and circumvent the state anti-usurylaw. Explain how you can synthetically create a loan and earn 50% interest.

You want to lend $1000 at time zero and receive $1000 (1.5) = 1500 at T = 1.To achieve this, at time zero you can

• have your mother-in-law sell you an asset that’s worth $1000

• have your mother-in-law sell you a 1500-strike, 1-year to expiration putoption on the asset

• sell your mother-in-law a 1500-strike, 1-year to expiration call option onthe asset.

Let’s see what’s happens at T = 1.

• If ST ≥ 1500, you exercise the put and sell the asset to your mother-in-lawfor 1500

• if ST ≤ 1500, your mother-in-law exercises the call and buys the assetfrom you for 1500.

The net effect is that you give your mother-in-law $1000 at time zero andreceive $1500 from her at T = 1.In this example, you have a long position on the synthetically created T-

bill. This is an example of conversion. In contrast, your mother-in-law has ashort position in the synthetically created T-bill. This is an example of reverseconversion.



Synthetic call option

Rearranging Equation 9.3, we get:CEur (K,T ) = PEur (K,T ) + S0 − PV (Div)− PV (K)

Example 9.1.8. The price of a 6-month 75-strike European put option on adividend-paying stock is 8.06. The stock price is 80. The continuously com-pounded risk-free interest rate is 5% per year. The continuously compoundeddividend rate is 3% per year. Explain how you can create a synthetic 6-month75-strike European call option on the stock. Calculate the premium for such asynthetic call option.

Solution.

The put-call parity for a stock paying continuous dividend isCEur (K,T ) + PV (K) = PEur (K,T ) + S0e

−δT

Rearranging this equation, we get:CEur (K,T ) = PEur (K,T ) + S0e

−δT − PV (K)

= 8.06 + 80e−0.03(0.5) − 75e−0.05(0.5) = 13. 72This is how to synthetically create a 6-month 75-strike European callsymbol at t = 0+PEur (K,T ) buy a 6-month 75-strike European put+S0e

−δT buy e−0.03(0.5) = 0.985 share of stock−PV (K) sell a bond that pays 75 in 6 months

Let’s see why a call is synthetically created. 0.985 share of stock at t = 0will grow into one stock if you reinvest the dividend and buy additional share ofstock. If at T = 0.5 the stock price is greater than 75 (i.e. ST > 75), then twothings happen: your purchased put expires worthless; the bond matures andyou need to pay the bond holder K = 75. So the net effect is that if ST > 75then at T you pay K = 75 and own one stock. This is the same as if ST > 75you exercise the call, paying K = 75 and receiving one stock.If ST ≤ 75, then two things will happen. You exercise the put, selling your

stock for K = 75; the bond matures and you pay the bond holder K = 75. Sothe net effect is that if ST ≤ 75 you net cash flow is zero and you don’t own astock. This is the same as if ST ≤ 75 you do nothing and let your call expireworthless.

Example 9.1.9. The price of a 9-month 110-strike European put option ona dividend-paying stock is 42.81. The stock price is 100. The continuouslycompounded risk-free interest rate is 8% per year. The stock will pay $1 dividendin 3 months and $1 dividend in 6 months. Explain how you can create a synthetic9-month 110-strike European call option on the stock. Calculate the premiumfor such a synthetic call option.



CEur (K,T ) = PEur (K,T ) + S0 − PV (Div)− PV (K)= 42.81+100− e−0.08(0.25)− e−0.08(0.5)− 110e−0.08(0.75) = 37. 27

How to synthetically create a 9-month 110-strike European callsymbol at t = 0+PEur (K,T ) buy a 9-month 110-strike European put+S0 buy one share of stock paying 100−PV (Div) sell Bond #1 that pays $1 in 3 months and $1 in 6 months−PV (K) sell Bond #2 that pays $110 in 6 months

Let’s see why a synthetic call is created.If at expiration date ST > 110

• You’ll receive $1 dividend at the end of Month 3 and $1 dividend at theend of Month 6. You use the dividends to pay off Bond #1

• Your purchased put expires worthless

• You have one stock at T = 0.75

• Your pay K = 110 at T = 0.75

The net effect is that you’ll pay K = 110 at T = 0.75 and own one stock.This is the same as owning a call option and ST > 110.If at expiration date ST ≤ 110

• You’ll receive $1 dividend at the end of Month 3 and $1 dividend at theend of Month 6. You use the dividends to pay off Bond #1

• You have one stock at T = 0.75

• You exercise your put, surrendering one stock and receiving K = 110

• Your pay K = 110 at T = 0.75

The net effect is that you have zero cash left and don’t own one stock. Thisis the same as owning a call option and ST ≤ 110.

Synthetic put option

Rearranging Equation 9.3, we get:PEur (K,T ) = CEur (K,T )− S0 + PV (Div) + PV (K)

If the stock pays continuous dividend, then

PEur (K,T ) = CEur (K,T ) + PV (K)− S0e−δT



Example 9.1.10. The price of a 6-month 45-strike European call option on adividend-paying stock is 18.62. The stock price is 50. The continuously com-pounded risk-free interest rate is 6% per year. The stock will pay $1 dividend in3 months. Explain how you can create a 6-month 45-strike European put optionon the stock. Calculate the premium for such a synthetic put option.

Solution.

PEur (K,T ) = CEur (K,T )− S0 + PV (Div) + PV (K)= 18.62− 50 + e−0.06(0.25) + 45e−0.06(0.5) = 13. 28

How to create a synthetic 6-month 45-strike European put option

symbol at t = 0+CEur (K,T ) buy a 6-month 45-strike European call−S0 short sell one share receiving 50 and invest in a savings account+PV (Div) buy Bond #1 that pays $1 in 3 months+PV (K) sell Bond #2 that pays $45 in 6 months

This is why CEur (K,T )− S0 + PV (Div) + PV (K) behaves like a put.If ST ≥ 45

• (1) You will receive K = 45 at T = 0.5 from Bond #2

• (2) You exercise the call at T = 0.5, paying K = 45 which you get from(1) and receiving one stock

• (3) at T = 0.5 you give the stock you get from (2) to the broker fromwhom you borrowed the stock for short sale

• (4) Bond #1 pays you $1 dividend at the end of Month 3. After receivingthis dividend, you Immediately pay this dividend to the original owner ofthe stock you sold short

(2)+(3) will close out your short position on the stockThe net effect is that if ST ≥ 45 you keep the proceeds from the short sale,

which you can use to buy a stock. This is "keeping your asset (i.e. proceeds fromthe short sale)." This is the same as if you own a 6-month 45-strike Europeanput and ST ≥ 45. If you own a 6-month 45-strike European put and ST ≥ 45,you let the put option expire worthless and you still own a stock.If ST < 45

• (1) You will receive K = 45 at T = 0.5 from Bond #2

• (2) You let the call expire worthless

• (3) At T = 0.5 you buy a stock from the open market using the proceedsfrom the short sale; you give the stock to the broker from whom youborrowed the stock for short sale



• (4) Bond #1 pays you $1 dividend at the end of Month 3. After receivingthis dividend, you Immediately pay this dividend to the original owner ofthe stock you sold short

(3)+(4) will close out your short position on the stock.The net effect is that if ST ≥ 45 you receive K = 45 and you spent the

proceeds from the short sale. This is "giving up an asset (proceeds from theshort sale) and getting the strike price K."

9.1.2 Options on currencies

The put-call parity when currencies are underlying is

CEur (K,T ) + PV (K) = x0e−r€ + PEur (K,T ) (9.7)

In Equation 9.7, the underlying asset is 1 euro. The call holder has theright, at T , to buy the underlying (i.e. 1 euro) by paying a fixed dollar amountK. The premium of this call option is CEur (K,T ) dollars. Similarly, the putholder has the right, at T , to sell the underlying (i.e. 1 euro) for a fixed dollaramount K. The premium of this put option is PEur (K,T ) dollars. x0 is theprice, in dollars, of buying the underlying (i.e. 1 euro) at time zero. r€ is thecontinuously compounded euro interest rate per year earned by the underlying(i.e. 1 euro).To understand the term x0e

−r€ in 9.7, notice that to have 1€ at T , you needto have e−r€ € at time zero. Since the cost of buying 1€ at time zero is x0dollars, the cost of buying e−r€ € at time zero is x0e−r€ dollars.

Equation 9.7 is very similar to Equation 9.4. If you set S0 = x0 and δ = r€,Equation 9.4 becomes 9.7. This shouldn’t surprise us. Both S0 and x0 refer tothe price of an underlying asset at time zero. S0 is the dollar price of a stock;x0 is the dollar price of 1 euro. Both δ and r€ measure the continuous rate ofreward earned by an underlying asset.

Tip 9.1.4. How to memorize Equation 9.7. Just memorize Equation 9.4. Next,set S0 = x0 and δ = r€.

Tip 9.1.5. When applying Equation 9.7, remember thatK, CEur (K,T ), PEur (K,T ),and x0 are in U.S. dollars. To remember this, assume that you are living in theU.S. (i.e. US dollar is your home currency); that your goal is to either buy orsell the underlying asset (i.e.1€) with a fixed dollar amount. Also rememberthat r€ is the euro interest rate on euro money.

Example 9.1.11. The current exchange rate is 1€= 1.33 US dollars. Thedollar-denominated 6-month to expiration $1.2-strike European call option onone euro has a premium $0.1736. The continuously compounded risk-free in-terest rate on dollars is 6% per year. The continuously compounded risk-free



interest rate on euros is 4% per year. Calculate the premium for the dollar-denominated 6-month to expiration $1.2-strike European put option on one euro.

Solution.

CEur (K,T ) + PV (K) = PEur (K,T ) + S0e−δT

K = 1.2 T = 0.5 CEur (K,T ) = $0.1736r = 0.06 δ = 0.04 S0 = 1.330.1736 + 1.2e−0.06(0.5) = PEur (K,T ) + 1.33e−0.04(0.5)

PEur (K,T ) = 0.03 447

Example 9.1.12. The current exchange rate is $1 = 0.78€. The dollar-denominated9-month to expiration $1-strike European put option on one euro has a premium$0.0733. The continuously compounded risk-free interest rate on dollars is 7%per year. The continuously compounded risk-free interest rate on euros is 3% peryear. Calculate the premium for the dollar-denominated 9-month to expiration$1-strike European call option on one euro.

CEur (K,T ) + PV (K) = PEur (K,T ) + S0e−δT

K = 1 T = 0.5 PEur (K,T ) = 0.0733

r = 0.07 δ = 0.03 S0 =1

0.78= 1. 282 05

CEur (K,T )+e−0.07(0.75) = 0.0733+1

0.78e−0.03(0.75) CEur (K,T ) = $0.3780

9.1.3 Options on bonds

This is an option contract where the underlying asset is a bond. Other thanhaving different underlying assets, a stock option and a bond option have nomajor differences. You buy a call option on stock if you think the stock pricewill go up; you buy a put option if you think the stock price will drop. Similarly,you buy a call option on a bond if you think that the market interest rate willgo down (hence the price of a bond will go up); you buy a put option on a bondif you think that the market interest rate will go up (hence the price of the bondwill go down).The coupon payments in a bond are like discrete dividends in a stock. Using

Equation 9.3, we get the

CEur (K,T ) + PV (K) = PEur (K,T ) +B0 − PV (Coupon) (9.8)

Please note that in Equation 9.8, in the term PV (Coupon), "coupon" refersto the coupon payments during [0, T ]. In other words, only the coupons madeduring the life of the options are used in Equation 9.8. Coupons made after Tare ignored.



Example 9.1.13. A 10-year $1,000 par bond pays 8% annual coupons. Theyield of the bond is equal to the continuously compounded risk-free rate of 6%per year. A 15-month $1,000-strike European call option on the bond has apremium $180. Calculate the premium for a 15-month $1,000-strike Europeanput option on the bond.

Solution.

The annual effective interest rate is i = e0.06 − 1 = 6. 184%

B0 = 1000 (0.08) a10|+1000v10 = 1000 (0.08)×1− 1.06184

−10

0.06184+1000

¡1.06184−10

¢=

1132. 50We need to be careful about calculating PV (Coupon). The bond matures

in 10 years. There are 10 annual coupons made at t = 1, 2, ..., 10. However,since the option expires at T = 15/12 = 1. 25, only the coupon paid at t = 1 isused in Equation 9.8.

PV (Coupon) = e−0.06 (1000) (0.08) = 75. 34CEur (K,T ) + PV (K) = PEur (K,T ) +B0 − PV (Coupon)180 + 1000e−0.06(1.25) = PEur (K,T ) + 1132. 50− 75. 34PEur (K,T ) = 50. 59

9.1.4 Generalized parity and exchange options

General definition of call and put

Most times, the strike price is a constant and we can easily tell whether anoption is a call or a put. For example, an option that gives you the privilege ofbuying one Google stock in 1 year for $35 is a call option; an option that givesyou the privilege of selling one Google stock in 1 year for $35 is a put option.However, occasionally, the strike price is not a constant and it’s hard to

determine whether an option is a call or put. For example, you purchase anoption that gives you the privilege of receiving one Google stock by surrenderingone Microsoft stock in 1 year. If at T = 1, the price of a Google stock is higherthan that of a Micros0ft stock (i.e. SGoogleT >.SMicrosoft

T ), it’s advantageous foryou to exercise the option. To exercise the call, you buy one Microsoft stockfrom the open market for SMicrosoft

T , give it to the option writer. In return,the option writer gives you one Google stock worth SGoogleT . Your payoff isSGoogleT −.SMicrosoft

T . If, on the other hand, SGoogleT ≤.SMicrosoftT , you let your

option expires worthless and your payoff is zero. Is this option a call or a put?

It turns out that this option can be labeled as either a call or a put. Ifyou view the Google stock as the underlying asset and the Microsoft stock asthe strike asset, then it’s a call option. This option gives you the privilege ofbuying, at T = 1, one Google stock by paying one Microsoft stock. If you view



the Microsoft stock as the underlying asset and the Google stock as the strikeasset, then it’s a put option. This option gives you the privilege of selling, atT = 1, one Microsoft stock for the price of one Google stock.

Definition 9.1.1. An option gives the option holder the privilege, at T , ofsurrendering an asset AT and receiving an asset BT (we denote the option asAT → BT ). This is a call option if we view B as the underlying asset and A asthe strike asset (the option holder has the privilege of buying BT by paying AT ).This is a put option if we view A as the underlying asset and B as the strikeasset (the option holder has the privilege of selling AT and receiving BT ).

Please note that the payoff of Option AT → BT is max (0, BT −AT ).

Example 9.1.14. An option gives the option holder the privilege, at T = 0.25(i.e. 3 months later), of buying €1 with $1.25. Explain why this option can beviewed (perhaps annoyingly) as either a call or a put.

Solution.

This option is $1.25→€1. The option holder has the privilege, at T = 0.25,of surrendering $1.25 and receiving €1 (i.e. give $1.25 and get €1).This is a call option if we view €1 as the underlying asset. The option holder

has the privilege of buying €1 by paying $1.25.This is also a put option if we view $1.25 as the underlying asset. The option

holder has the privilege of selling $1.25 for €1 (i.e. selling $1 for €1

1.25= €

0.8).

Example 9.1.15. An option gives the option holder the privilege, at T = 0.25,of buying one Microsoft stock for $35. Explain why this option can be viewed(perhaps annoyingly) as either a call or a put.

This option is $35→ 1Microsoft stock (give $35 and get 1 Microsoft stock).If we view the Microsoft stock as the underlying asset, this is a call option. Theoption holder has the privilege of buying one Microsoft stock by paying $35.This is also a put option if we view $35 as the underlying. The option holderhas the privilege of selling $35 for the price of one Microsoft stock.

Example 9.1.16. An option gives the option holder the privilege, at T = 0.25,of selling one Microsoft stock for $35. Explain why this option can be viewed aseither a call or a put.

This option is 1Microsoft stock→ $35 (give 1 Microsoft stock and get $35).If we view the Microsoft stock as the underlying asset, this is a put option. Theoption holder has the privilege of selling one Microsoft stock for $35. This isalso a call option if we view $35 as the underlying. The option holder has theprivilege of buying $35 by paying one Microsoft stock.



Generalized put and call parity

(AT → BT )0 + PV (AT ) = (BT → AT )0 + PV (BT ) (9.9)

In Equation 9.9, (AT → BT )0 is the premium paid at t = 0 for the privilegeof giving AT and getting BT at T . Similarly, (BT → AT )0 is the premium paidat t = 0 for the privilege of giving BT and getting AT at T . PV (AT ) is thepresent value of AT . If you have PV (AT ) at time zero and invest it from timezero to T , you’ll have exactly AT . Similarly, if you have PV (BT ) at time zeroand invest it from time zero to T , you’ll have exactly BT .This is the proof of Equation 9.9. Suppose we have two portfolios. Portfolio

#1 consists a European option AT → BT and the present value of the asset AT .Portfolio #2 consists a European option BT → AT and the present value of theasset BT .Payoff of Portfolio #1 at T

If AT ≥ BT If AT < BT

Option AT → BT 0 BT −AT

PV (AT ) AT AT

Total AT BT

Payoff of Portfolio #2 at TIf AT ≥ BT If AT < BT

Option BT → AT AT −BT 0PV (BT ) BT BT

Total AT BT

Portfolio #1 and Portfolio #2 have the identical payoff at T . To avoidarbitrage, these two portfolios must cost us the same to set up at any time priorto T . The cost of setting up Portfolio #1 is (AT → BT )0 + PV (AT ). The costof setting up Portfolio #2 is (BT → AT )0+PV (BT ). Hence Equation 9.9 holds.

Example 9.1.17. Stock A currently sells for $30 per share. It doesn’t payany dividend. Stock B currently sells for $50 per share. It pays dividend at acontinuously compounded rate of 5% per year. The continuously compoundedrisk-free interest rate is 6% per year. A European option gives the option holderthe right to surrender one share of Stock B and receive one share of Stock A atthe end of Year 1. This option currently sells for $8.54. Calculate the premiumfor another European option that gives the option holder the right to surrenderone Stock A and receive one Stock B at the end of Year 1.

(AT → BT )0 + PV (AT ) = (BT → AT )0 + PV (BT )(AT → BT )0 + 30 = 8.54 + 50e

−0.05

(AT → BT )0 = 26. 1Please note that the risk free interest rate 6% is not needed for solving the

problem. In addition, you don’t need to decide whether to call the option "give



Stock B and receive Stock A" or "give Stock A and receive Stock B" as a call orput. Equation 9.9 holds no matter you call the option AT → BT or BT → AT

a call or put.

Example 9.1.18. Stock A currently sells for $55 per share. It pays dividend of$1.2 at the end of each quarter. Stock B currently sells for $72 per share. It paysdividend at a continuously compounded rate of 8% per year. The continuouslycompounded risk-free interest rate is 6% per year. A European option gives theoption holder the right to surrender one share of Stock A and receive one shareof Stock B at the end of Year 1. This option currently sells for $27.64. Calculatethe premium for another European option that gives the option holder the rightto surrender one Stock B and receive one Stock A at the end of Year 1.

(AT → BT )0 + PV (AT ) = (BT → AT )0 + PV (BT )PV (AT ) = A0 − 1.2a4|i = 55− 1.2a4|ii is the effective interest rate per quarter.i = e0.25r − 1 = e0.25(0.06) − 1 = 1. 511 3%PV (AT ) = 55− 1.2a4|i = 55− 1.2

µ1− 1.01 511 3−40.01 511 3

¶= 50. 38

PV (BT ) = B0e−δBT = 72e−0.08 = 66. 46

27.64 + 50. 38 = (BT → AT )0 + 66. 46(BT → AT )0 = 11. 56

Currency options

Example 9.1.19. Let’s go through the textbook example. Suppose that a 1-yeardollar-denominated call option on €1 with the strike price $0.92 is $0.00337.The current exchange rate is €1 = $0.9. What’s the premium for a 1-year

euro-denominated put option on $1 with strike price €1

0.92= €1. 087?

First, let’s walk through the vocabulary. The phrase "dollar-denominatedoption" means that both the strike price and the option premium are expressedin U.S. dollars. Similarly, the phrase "euro-denominated option" means thatboth the strike price and the option premium are expressed in euros.Next, let’s summarize the information using symbols. "dollar-denominated

call option on €1 with the strike price $0.92" is $0.92 →€1. The premium forthis option is $0.00337. This is represented by ($0.92→ €1)0 =$0.0337.

"Euro-denominated put option on $1 with strike price €1

0.92= €1. 087" can

be represented by $1→€ 1

0.92. The premium for this option is

µ$1→ €

1

0.92

¶0

Now the solution should be simple.µ$1→ €

1

0.92

¶0

=1

0.92

µ$1× 0.92→ €

1

0.92× 0.92

¶0

=1

0.92($0.92→ €1)0 =

1

0.92×$0.0337



Since the exchange rate is €1 = $0.9 or $1 =€1

0.9, we have:µ

$1→ €1

0.92

¶0

=1

0.92×$0.0337 = 1

0.92× 0.0337×€ 1

0.9= €0.04 07



General formula:The current exchange rate is €1 = $x0 or $1=€

1

x0

A dollar-denominated $K strike call on €1 has a premium of $am

( $K → €1)0 = $a or ( $K → €1)$0 = a

A euro-denominated €1

Kstrike put on $1 has a premium of €b

mµ$1→ €

1

K

¶0

=€b orµ$1→ €

1

K

¶€0

= b

It then follows:

µ$1→ €

1

K

¶€0

× x0 =

µ$1→ €

1

K

¶$0

=1

K× ( $K → €1)$0 (9.10)

In Equation 9.10,µ$1→ €

1

K

¶€0

is the euro-cost of "give $1, get €1

K." Since

the exchange rate is €1 = $x0,µ$1→ €

1

K

¶€0

× x0 is the dollar cost of "give

$1, get1

K." Similarly, ( $K → €1)$0 is the dollar-cost of "give $K, get €1."

Equation 9.10 essentially says that the dollar cost of "give $1, get1

K" is

1

Kof

the dollar cost "give $K, get €1." This should make intuitive sense.

Tip 9.1.6. The textbook gives you the complex formula C$ (x0,K, T ) = x0KPf

µ1

x0,1

K,T

¶.

Do not memorize this formula or Equation 9.10. Memorizing complex formulasis often prone to errors. Just translate options into symbols. Then a simplesolution should emerge. See the next example.



Example 9.1.20. The current exchange rate is €0.9 per dollar. A Europeaneuro-denominated call on 1 dollar with a strike price €0.8 and 6 months toexpiration has a premium €0.0892. Calculate the price of a European dollar-denominated put option on 1 euro with a strike price $1.25.

Just translate the options into symbols. Then you’ll see a solution.

The current exchange rate is €0.9 per dollar. $1 =€0.9 or €1 = $1

0.9

Euro-denominated call on 1 dollar with strike price €0.8 has a premium €0.0892(€0.8→ $1)0 =€0.0892

Calculate the price of a dollar-denominated put on 1 euro with strike price $1.25(€1→ $1.25)0 = $?

(€1→ $1.25)0 = 1.25×µ€1

1.25→ $1

¶0

= 1.25× (€0.8→ $1)0

= 1.25×€0.0892 = 1.25× 0.08928× $ 10.9

= $0.124

9.1.5 Comparing options with respect to style, maturity,and strike

European vs American options

American options can be exercised at any time up to (and including) the matu-rity. In contrast, European options can be exercised only at the maturity. Sincewe can always convert an American option into a European option by exercisingthe American option only at the maturity date, American options are at leastas valuable as an otherwise identical European option.

CAmer (K,T ) ≥ CEur (K,T ) (9.11)

PAmer (K,T ) ≥ PEur (K,T ) (9.12)

Equation 9.11 and Equation 9.12 are not earth-shaking observations. Youshouldn’t have trouble memorizing them.

Maximum and minimum price of a call

1. The price of a call option is always non-negative. CAmer (K,T ) ≥CEur (K,T ) ≥ 0. Any option (American or European, call or put) is aprivilege with non-negative payoff. The price of a privilege can never benegative. The worst thing you can do is to throw away the privilege.

2. The price of a call option can’t exceed the current stock price.S0 ≥ CAmer (K,T ) ≥ CEur (K,T ). The best you can do with a call optionis to own a stock. So a call can’t be worth more than the current stock.



3. The price of a European call option must obey the put callparity. For a non-dividend paying stock, the parity is CEur (K,T ) =PEur (K,T ) + S0 − PV (K) ≥ S0 − PV (K)

Combining 1, 2, and 3, we have:For a non-dividend paying stock

S0 ≥ CAmer (K,T ) ≥ CEur (K,T ) ≥ max [0, S0 − PV (K)] (9.13)

For a discrete-dividend paying stock

S0 ≥ CAmer (K,T ) ≥ CEur (K,T ) ≥ max [0, S0 − PV (Div)− PV (K)] (9.14)

For a continuous-dividend paying stock

S0 ≥ CAmer (K,T ) ≥ CEur (K,T ) ≥ max£0, S0e

−δT − PV (K)¤

(9.15)

Tip 9.1.7. You don’t need to memorize Equation 9.13, 9.14, or 9.15. Just mem-orize basic ideas behind these formulas and derive the formulas from scratch.

Maximum and minimum price of a put

1. The price of a put option is always non-negative. PAmer (K,T ) ≥PEur (K,T ) ≥ 0.

2. The price of a European put option can’t exceed the presentvalue of the strike price. PEur (K,T ) ≤ PV (K). The best you cando with a European put option is to get the strike price K at T . So aEuropean put can’t be worth more than the present value of the strikeprice.

3. The price of an American put option can’t exceed the strikeprice. K ≥ CAmer (K,T ). The best you can do with an American putoption is to exercise it immediately after time zero and receive the strikeprice K. So an American put can’t be worth more than the strike price.

4. The price of a European put option must obey the put callparity. For a non-dividend paying stock, the parity is PEur (K,T ) =CEur (K,T ) + PV (K)− S0 ≥ PV (K)− S0

Combining 1, 2, and 3, we have:

K ≥ PAmer (K,T ) ≥ PEur (K,T ) ≥ 0 (9.16)

For a non-dividend paying stock



PV (K) ≥ PEur (K,T ) ≥ max [0, PV (K)− S0] (9.17)

For a discrete-dividend paying stock

PV (K) ≥ PEur (K,T ) ≥ max [0, PV (K) + PV (Div)− S0] (9.18)

For a continuous-dividend paying stock

PV (K) ≥ CEur (K,T ) ≥ max£0, PV (K)− S0e

−δT ¤ (9.19)

Tip 9.1.8. You don’t need to memorize Equation 9.16, 9.17, 9.18, or 9.19. Justmemorize the basic ideas behind these formulas and derive the formulas fromscratch.

Early exercise of American options

Suppose an American option is written at time zero. The option expires in dateT . Today’s date is t where 0 ≤ t < T . The stock price at the option expirationdate is ST . Today’s stock price is St. The continuously compounded risk-freeinterest rate is r per year.

Proposition 9.1.1. It’s never optimal to exercise an American call early on anon-dividend paying stock.

If at t you exercise an American call option , your payoff is St −K.If at t you sell the remaining call option, you’ll get at least CEur (St,K, T ) ≥

max [0, St − PV (K)] = max£0, St −Ke−r(T−t)

¤, the premium for a European

call option written in t and expiring in T .Clearly, CEur (St,K, T ) ≥ max [0, St − PV (K)] = max

£0, St −Ke−r(T−t)

¤>

St −K for r > 0It’s not optimal to exercise an American call option early if you can sell the

remaining call option.What if you can’t sell the remaining call option? If you can’t sell the call

option written in t and expiring in T ., you can short sell one stock at t, receivingSt and accumulating to Ster(T−t). Then at T , if ST > K, you exercise your call,paying K and receiving one stock. Next, you return the stock to the broker.Your total profit is Ster(T−t) −K > St −K for a positive r; if at T , ST < K,you let your call option expire worthless, purchase a stock in the market, andreturn it to the broker. Your profit is Ster(T−t)−ST > Ste

r(T−t)−K > St−Kfor a positive r.Intuition behind not exercising American call option early. If you exercise the

call option early at t, you pay the strike priceK at t and gain physical possessionof the stock at t. You lose the interest you could have earned during [t, T ] hadyou put K in a savings account, yet you gain nothing by physically owning a



stock during [t, T ] since the stock doesn’t pay any dividend. In addition, byexercising the call option t, you throw away the remaining call option during[t, T ].

Example 9.1.21. You purchase a 30-strike American call option expiring in 6months on a non-dividend paying stock. 2 months later the stock reaches a highprice of $90. You are 100% sure that the stock will drop to $10 in 4 months.You are attempted to exercise the call option right now (i.e. at t = 2/12 = 1/6year) and receive 90 − 30 = 60 profit. The continuously compounded risk-freeinterest rate is 6% per year. Explain why it’s not optimal to exercise the callearly.

Solution. The problem illustrates the pitfall in common thinking "If you I knowfor sure that the stock price is going to fall, shouldn’t I exercise the call now andreceive profit right away, rather than wait and let my option expire worthless?"

Suppose indeed the stock price will be $10 at the call expiration date T =6/12 = 0.5. If you exercise the call early at t = 2/12 = 1/6, you’ll gainSt − K = 90 − 30 = 60, which will accumulate to 60e0.06(4/12) = 61. 212 atT = 0.5.Instead of exercising the call early, you can short-sell the stock at t = 2/12 =

1/6. Then you’ll receive 90, which will accumulate to 90e0.06(4/12) = 91. 818 atT = 0.5. Then at T = 0.5, you purchase a stock from the market for 10 andreturn it to the brokerage firm where you borrow the stock for short sale. Yourprofit is 91. 818−10 = 81. 818, which is the greater than 61. 212 by 81. 818−61.212 = 20. 606.

Proposition 9.1.2. It might be optimal to exercise an American call optionearly for a dividend paying stock.

Suppose the stock pays dividend at tD.Time 0 ... ... tD ... ... T

Pro and con for exercising the call early at tD.

• +. If you exercise the call immediately before tD, you’ll receive dividendand earn interest during [tD, T ]

• −. You’ll pay the strike price K at tD, losing interest you could haveearned during [tD, T ]

• −. You throw away the remaining call option during [tD, T ]. Had youwaited, you would have the call option during [tD, T ]

However, if the accumulated value of the dividend is big enough, then it canoptimal to exercise the stock at tD.

Proposition 9.1.3. If it’s optimal to exercise an American call early, then thebest time to exercise the call is immediately before the dividend payment.



Here is the proof. Suppose the dividend is paid at tDTime 0 t1 ... ... tD t2 ... ... T

It’s never optimal to exercise an American call at t1. If you exercise the callat t1 instead of tD, you’ll

• lose interest that can be earned on K during [t1, tD]

• lose a call option during [t1, tD]

• gain nothing (there’s no dividend during [t1, tD])

It’s never optimal to exercise an American call at t2. If you exercise the callat t2 instead of tD, you’ll

• gain a tiny interest that can be earned during [tD, t2] but lose the dividendthat can be earned if the call is exercised at tD

So for a dividend paying stock, if it’s ever worthwhile to exercise an Americancall early, you should exercise the call immediately before the dividend payment,no sooner or later.Combining these two proposition, we have:

Proposition 9.1.4. It’s only optimal to exercise an American call option eitherat maturity or immediately before a dividend payment date. Any other time isnot optimal.

Proposition 9.1.5. It might be optimal to exercise an American put early.

Time 0 t1 ... ... t ... ... T

The pros and cons of exercising an American put at t instead of T

• +. You receive K at t and earn interest during [t, T ]

• −. You lose the remaining put option during [t, T ]. If you wait anddelay exercising the option, you’ll have a put option during [t, T ] and candecide whether to exercise it or discarding it. This is especially painful ifST > K. If ST > K and you exercise the put at t, you’ll get K at t, whichaccumulates to Ker(T−t) at T . If you wait and ST > K, you’ll let the putoption expire worthless and have ST at T . If ST > Ker(T−t), then youlose money by exercising the put at t. So one danger of exercising the putat t is that the stock might be worth more than K after t.

However, if the interest earned on K during [t, T ] is big enough, it can beoptimal to exercise the put at t instead of T .



Appendix 9.A: Parity bound for American options

Appendix 9.5A (and 9.5B) is not on the syllabus. However, I still recommendthat you study it.Appendix 9.5A has two important ideas

1. Why the put-call parity doesn’t hold for American options

2. How to calculate the non-arbitrage boundary price for American options

Why the put-call parity doesn’t hold for American options The put-call parity such as Equation 9.2 holds only for European options. It doesn’t holdfor American options. To understand why, let’s start from European options.For Equation 9.2 to hold, among other things, the call and the put must beexercised at the same time. Recall our proof of the put-call parity. At time twe have two portfolios. Portfolio #1 consists a European call option on a stockand PV (K), the present value of the strike price K. Portfolio #2 consists aEuropean put option on the stock and one share of the stock with current priceSt. Both the call and put have the same underlying stock, have the same strikeprice K, and the same expiration date T . We have found that Portfolio #1 andPortfolio #2 have an identical payoff of max (K,ST ) at the common exercisedate T . To avoid arbitrage, the two portfolios must cost us the same at timezero. Hence we have Equation 9.2.

Now suppose the call is exercised at T1 and the put is exercised at T2 whereT1 6= T2. Portfolio #1 consists of a European call option and PV (K) = Ke−rT1 ;Portfolio #2 consists an American put and one stock worth S0. Then atT1, Portfolio #1 has a payoff of max (K,ST1); Portfolio #2 has a payoff ofmax (K,ST2). Now the two payoffs differ in timing and the amount. As aresult, we don’t know whether the two portfolios have the same set-up cost.

Now you should understand why Equation 9.2 doesn’t hold for Americanoptions. American options can be exercised at any time up to (and including)the maturity. Even when an American call and an American put have the samematurity T , the American call can be exercised at T1 where 0 ≤ T1 ≤ T ; theAmerican put can be exercised at T1 where 0 ≤ T2 ≤ T . Hence an Americancall and an American put can be exercised at different times, they don’t followthe put-call parity.

Non-arbitrage boundary price for American options For a stock thatpays discrete dividend, the key formula is

S0 − PV (K) ≥ CAme (K,T )− PAme (K,T ) ≥ S0 − PV (Div)−K (9.20)

Many people find it hard to understand or memorize Equation 9.20. Here isan intuitive proof without using complex math.



First, we establish the boundary of American options:

CEur (K,T ) + PV (Div) ≥ CAme (K,T ) ≥ CEur (K,T ) (9.21)

PEur (K,T ) +K − PV (K) ≥ PAme (K,T ) ≥ PEur (K,T ) (9.22)

This is why Equation 9.21 holds. Clearly, CAme (K,T ) ≥ CEur (K,T ). Thisis because an American call option can always be converted to a European calloption.To understand why CEur (K,T ) + PV (Div) ≥ CAme (K,T ), consider an

American call option and an otherwise identical European call option. Bothcall options have the same underlying stock, the same strike price K, the sameexpiration date T . The European call option is currently selling for CEur (K,T ).How much more can the American call option sell for?The only advantage of an American option over an otherwise identical Eu-

ropean call option is that the American call option can be exercised early. Theonly good reason for exercising an American call early is to get the dividend.Consequently, the value of an American call option can exceed the value of anotherwise identical European call option by no more than the present value ofthe dividend. So CEur (K,T ) + PV (Div) ≥ CAme (K,T ). A rational personwill pay no more than CEur (K,T )+PV (Div) to buy the American call option.So Equation 9.21 holds.

Similarly, an American put is worth at least as much as an otherwise identicalEuropean put.In addition, the value of an American put exceeds the value of an otherwise

identical European put by no more than K − PV (K). The only advantage ofan American put over an otherwise identical European put is that the Americanput can be exercised early. The only good reason for exercising an Americanput early is to receive the strike price K immediately at time zero (as opposedto receiving K at T in a European put) and earn the interest on K from timezero to T . The maximum interest that can be earned on K during [0, T ] isK − PV (K) = K − Ke−rT . Consequently, PEur (K,T ) + K − PV (K) ≥PAme (K,T ). Equation 9.22 holds.

Next, we are ready to prove Equation 9.20. CAme (K,T ) − PAme (K,T )reaches its minimum value when CEur (K,T ) reaches it minimum value CEur (K,T )and PAme (K,T ) reaches its maximum value PEur (K,T ) +K − PV (K):

CAme (K,T )− PAme (K,T ) ≤ CEur (K,T )− [PEur (K,T ) +K − PV (K)]From the put-call parity, we have:CEur (K,T ) + PV (K) = PEur (K,T ) + S0 − PV (Div)→ CEur (K,T ) = PEur (K,T ) + S0 − PV (Div)− PV (K)→ CEur (K,T )− [PEur (K,T ) +K − PV (K)]= PEur (K,T ) + S0 − PV (Div)− PV (K)



− [PEur (K,T ) +K − PV (K)]= S0 − PV (Div)−K

→ CAme (K,T )− PAme (K,T ) ≤ S0 − PV (Div)−K

Similarly, CAme (K,T )−PAme (K,T ) reaches it maximum value when CAme (K,T )reaches its maximum value and PAme (K,T ) reaches it minimum value.

CAme (K,T )− PAme (K,T ) ≤ CEur (K,T ) + PV (Div)− PEur (K,T )CEur (K,T )+PV (Div)−PEur (K,T ) = [PEur (K,T ) + S0 − PV (Div)− PV (K)]

+PV (Div)− PEur (K,T )= S0 − PV (Div)

Hence S0 − PV (K) ≥ CAme (K,T ) − PAme (K,T ) ≥ S0 − PV (Div) −K.Equation 9.20 holds.

Tip 9.1.9. If you can’t memorize Equation 9.20, just memorize Equation 9.21,Equation 9.22, and Equation 9.3. Then can derive Equation 9.20 on the spot.

Example 9.1.22. If the interest rate is zero. Is it ever optimal to exercise anAmerican put on a stock?

Solution. According to Equation 9.22, if the risk-free interest rate is zero, thenK = PV (K) and PEur (K,T ) ≥ PAme (K,T ) ≥ PEur (K,T ). This gives usPAme (K,T ) = PEur (K,T ). So it’s never optimal to exercise an American putearly if the interest rate is zero.

Example 9.1.23. Is it ever optimal to exercise an American call on a non-dividend paying stock?

Solution. According to Equation 9.21, if Div = 0, then CEur (K,T ) ≥ CAme (K,T ) ≥CEur (K,T ) or CAme (K,T ) = CEur (K,T ). It’s never optimal to exercise theAmerican call option.

Example 9.1.24. An American call on a non-dividend paying stock withexercise price 20 and maturity in 5 months is worth 1.5. Suppose that the currentstock price is 1.9 and the risk-free continuously compounded interest rate is 10%per year. Calculate the non-arbitrage boundary price of the American put optionon the stock with strike price 20 and 5 months to maturity.

Solution.

If you can memorize Equation 9.20, thenS0 − PV (K) ≥ CAme (K,T )− PAme (K,T ) ≥ S0 − PV (Div)−K

19− 20e−0.1(5/12) ≥ 1.5− PAme (K,T ) ≥ 19− 0− 20−0.184 ≥ 1.5− PAme (K,T ) ≥ −11 ≥ PAme (K,T )− 1.5 ≥ 0.184



1 + 1.5 ≥ PAme (K,T ) ≥ 0.184 + 1.52. 5 ≥ PAme (K,T ) ≥ 1. 684If you can’t memorize Equation 9.20, this is how to solve the problem using

basic reasoning.

The American put is worth at least the otherwise identical European put.PAme (K,T ) ≥ PEur (K,T )Using the put-call parity: PEur (K,T ) = CEur (K,T ) + PV (K)− S0Since the stock doesn’t pay any dividend, CAme (K,T ) = CEur (K,T ) = 1.5PEur (K,T ) = CEur (K,T ) + PV (K)− S0

= 1.5 + 20e−0.1(5/12) − 19 = 1. 684

So PAme (K,T ) ≥ PEur (K,T ) = 1.684The value of an American put can exceed the value of an otherwise identical

European put by no more the early-exercise value. Since the only possible reasonto exercise an American put early is to receive K and earn interest K−PV (K)during [0, T ], so the maximum early-exercise value is

K − PV (K) = 20− 20e−0.1(5/12) = 0.816→ PAme (K,T ) ≤ PEur (K,T ) +K − PV (K) = 1.684 + 0.816 = 2. 5

Time to expiration

American option An American option (call or put) has more time to expi-ration is at least as valuable as an otherwise identical American option with lesstime to expiration. If options are on the same stock and T1 > T2, we have:

CAmer (K,T1) > CEur (K,T2) (9.23)

PAmer (K,T1) > PEur (K,T2) (9.24)

European option A European call on a non-dividend paying stock will be atleast as valuable as an otherwise identical European call option with a shortertime to expiration. This is because for a non-dividend paying stock, an Europeancall option is worth the same as an otherwise identical American call option.And an American call option with a longer time to expiration is more valuablethan an otherwise identical American option with a shorter time to expiration.If both options are on the same non-dividend paying stock and T1 > T2, we

have:CEur (K,T1) > CEur (K,T2) (9.25)

If the stock pays dividend, then a longer-lived European option may be lessvaluable than an otherwise identical but shorter-lived European option. Thetextbook gives two good examples.In Example #1, a stock is valuable only because of its dividend. The stock

pays a dividend at the end of Week 2. Once the dividend is paid, the stock dies



and is worth nothing. In this case, if T > 2 weeks, CEur (K,T ) = 0. If if T ≤ 2weeks, CEur (K,T ) might be worth something depending on how high the strikeprice K is.

In Example #2, a put is written on the asset of a bankrupt company. Theasset of the bankrupt company is a constant c. The put is worth PV (c) = ce−rT .If the risk-free interest rate r > 0, ce−rT will decrease if time to expiration Tincreases.

European options when the strike price grows over time Typically,a call or put has a fixed strike price K. However, there’s nothing to preventsomeone from inventing a European option whose strike price changes over time.Consider a European option whose strike price grows with the risk-free in-

terest rate. That is, KT = KerT . What can we say about the price of such aEuropean option?For a European option with strike priceKT = KerT , a longer-lived European

option is at least as valuable as an otherwise identical but shorter-lived Europeanoption.

CEur

¡KerT1 , T1

¢≥ CEur

¡KerT2 , T2

¢if T1 > T2 (9.26)

PEur¡KerT1 , T1

¢≥ PEur

¡KerT2 , T2

¢if T1 > T2 (9.27)

This is why Equation 9.26 holds. Suppose at time zero we buy two Europeancalls on the same stock. The first call expires at T1 and has a strike price KerT1 .The second call expires at T2 and has a strike price KerT2 , where T1 > T2.Let’s choose a common time T1 and compare the payoffs of these two calls

at T1. The payoff of the longer-lived call is max¡0, ST1 −KerT1

¢.

The payoff of the shorter-lived call is calculated as follows. First, we calculateits payoff at T2. Next, we accumulate this payoff from T2 to T1.The payoff of the shorter-lived call at T2 is max

¡0, ST2 −KerT2

¢. Next, we

accumulate this payoff max¡0, ST2 −KerT2

¢from T2 to T1.

As we’ll soon see, it’s much harder for a short-lived call to have a positivepayoff at T1.The longer-lived call will have a positive payoff at ST1 − KerT1 at T1 if

ST1 > KerT1 ; all else, the payoff is zero.On the other hand, the shorter-lived call will have a positive payoff ST1 −

KerT1 at T1 only if the following two conditions are met

• ST2 > KerT2

• ST1 > KerT1



If ST2 ≤ KerT2 , the shorter-lived call will expire worthless, leading to zeropayoff at T2, which accumulates to zero payoff at T1.If ST2 > KerT2 , we’ll receive a positive payoff ST2 − KerT2 at T2. If we

accumulate ST2 −KerT2 from T2 to T1, ST2 will accumulate to ST1 and KerT2

to KerT2er(T1−T2) = KerT1 , leading to a total amount ST1 −KerT1 at T1. Thetotal payoff amount ST1 −KerT1 is positive if ST1 > KerT1 .

In summary, both calls can reach the common positive payoff ST1−KerT1 atT1. The longer-lived call will reach this payoff if ST1 > KerT1 . The shorter-livedcall will reach this payoff if both ST2 > KerT2 and ST1 > KerT1 . Consequently,the long-lived call has a better payoff and should be at least as valuable as theshorter-lived call. Hence Equation 9.26 holds.If you still have trouble understanding why the longer-lived call has a richer

payoff, you can draw the following payoff table:The accumulated payoff of the shorter-lived call at T1

If ST1 ≤ KerT1 If ST1 > KerT1

If ST2 ≤ KerT2 Payoff = 0 Payoff = 0If ST2 > KerT2 Payoff = ST1 −KerT1 ≤ 0 Payoff = ST1 −KerT1 > 0

The payoff of the longer-lived call at T1If ST1 ≤ KerT1 If ST1 > KerT1

If ST2 ≤ KerT2 Payoff = 0 Payoff = ST1 −KerT1 > 0If ST2 > KerT2 Payoff = 0 Payoff = ST1 −KerT1 > 0

You can see that the longer-lived call has a slightly better payoff than theshorter-lived payoff. To avoid arbitrage, the longer-lived call can’t sell for lessthan the shorter-lived call. Hence Equation 9.26 holds. Similarly, you can proveEquation 9.27.

Different strike price

Proposition 9.1.6. A call (European or American) with a low strike priceis at least as valuable as an otherwise identical call with a higher strike price.However, the excess premium shouldn’t exceed the excess strike price.

0 ≤ C (K1, T )− C (K2, T ) ≤ K2 −K1 if K1 < K2 (9.28)

Equation 9.28 should make intuitive sense. The lower-strike call allows thecall holder to buy the underlying asset by the guaranteed lower strike price.Clearly, the payoff of a lower-strike call can never be less than the payoff ofan otherwise identical but higher-strike call. Consequently, 0 ≤ C (K1, T ) −C (K2, T ). And this is why C (K1, T )−C (K2, T ) ≤ K2−K1. The only advan-tage of a K1-strike call over the K2-strike call is that the guaranteed purchaseprice of the underlying asset is K2 −K1 lower in the K1-strike call; to buy theasset, the K1-strike call holder can pay K1 yet the K2-strike call holder will pay



K2. Consequently, no rational person will pay more than C (K2, T ) +K2 −K1

to buy the K1-strike call.

Proposition 9.1.7. A European call with a low strike price is at least as valu-able as an otherwise identical call with a higher strike price. However, the excesspremium shouldn’t exceed the present value of the excess strike price.

0 ≤ CEur (K1, T )− CEur (K2, T ) ≤ PV (K2 −K1) if K1 < K2 (9.29)

The only advantage of aK1-strike European call over theK2-strike Europeancall is that the guaranteed purchase price of the underlying asset is K2 − K1

lower in the K1-strike call at T . Consequently, no rational person will pay morethan CEur (K2, T ) + PV (K2 −K1) to buy the K1-strike call.Please note that CEur (K1, T )−CEur (K2, T ) ≤ PV (K2 −K1) doesn’t apply

to American call options because two American options can be exercised atdifferent dates.

Proposition 9.1.8. A put (European or American) with a higher strike priceis at least as valuable as an otherwise identical put with a lower strike price.However, the excess premium shouldn’t exceed the excess strike price.

0 ≤ P (K2, T )− P (K1, T ) ≤ K2 −K1 if K1 < K2 (9.30)

Clearly, a higher strike put is at least as valuable as an otherwise identical putwith a lower strike price. Since the only advantage of a K2-strike put over theK1-strike put is that the guaranteed sales price of the underlying asset isK2−K1

high in the K2-strike put, no rational person will pay C (K1, T ) +K2 −K1 tobuy the K2-strike put.

Proposition 9.1.9. A European put with a higher strike price is at least asvaluable as an otherwise identical put with a lower strike price. However, theexcess premium shouldn’t exceed the present value of the excess strike price.

0 ≤ PEur (K2, T )− PEur (K1, T ) ≤ PV (K2 −K1) if K1 < K2 (9.31)

Please note that PEur (K2, T )− PEur (K1, T ) ≤ PV (K2 −K1) won’t applyto two American put options because they can be exercised at two differentdates.

Proposition 9.1.10. A diversified option portfolio is at least as valuable as oneundiversified option. For K1 < K2 and 0 < λ < 1



C [λK1 + (1− λ)K2] ≤ λC (K1) + (1− λ)C (K2) (9.32)

P [λK1 + (1− λ)K2] ≤ λP (K1) + (1− λ)P (K2) (9.33)

Please note that Equation 9.32 and Equation 9.33 apply to both Europeanoptions and American options.A call portfolio consists of λ portion of K1-strike call and (1− λ) portion

of K2-strike call. The premium of this portfolio, λC (K1) + (1− λ)C (K2),can be no less than the premium of a single call with a strike price λK1 +(1− λ)K2. Similarly, a call portfolio consists of λ portion of K1-strike put and(1− λ) portion of K2-strike put. The premium of this portfolio, λP (K1) +(1− λ)P (K2), can be no less than the premium of a single call with a strikeprice λK1 + (1− λ)K2.Before proving Equation 9.32, let’s look at an example.

Example 9.1.25. (Textbook example 9.8 K2 and K3 switched) K1 = 50, K2 =65,K3 = 0.4 (50) + 0.6 (65) = 59. C (K1, T ) = 14, C (K2, T ) = 5. Explain whyC (59) ≤ 0.4C (50) + 0.6C (65).

Payoff ST < 50 50 ≤ ST < 59 59 ≤ ST < 65 ST ≥ 6559-strike call (a) 0 0 ST − 59 ST − 59

50-strike call (b) 0 ST − 50 ST − 50 ST − 5065-strike call (c) 0 0 0 ST − 650.4b+ 0.6c 0 0.4 (ST − 50) 0.4 (ST − 50) ST − 591

(0.4b+ 0.6c)− a 0 0.4 (ST − 50) ≥ 0 0.6 (65− ST ) > 02 0

The above table says the following:If we buy 0.4 unit of 50-strike call, buy 0.6 unit of 65-strike call, and sell

1 unit of 59-strike call, our initial cost is (0.4b+ 0.6c) − a and our payoff atT is always non-negative. To avoid arbitrage, the position of always having anon-negative payoff at expiration T surely has a non-negative cost at t = 0.Imagine what happens otherwise. For example, you always have a non-negativepayoff at T and it costs you −$10 (i.e. you receive $10) to set up this positionat t = 0. Then you’ll make at least $10 free money.

Hence we have (0.4b+ 0.6c)− a ≥ 0 no matter what ST is.Clearly, the call portfolio consisting of 40% 50-strike call and 60% 65-strike

call is at least as good as the 59-strike call. Consequently, the portfolio is atleast as valuable as the 59-strike call. C (59) ≤ 0.4C (50) + 0.6C (65) .

10.4 (ST − 50) + 0.6 (ST − 65) = ST − 5920.4 (ST − 50)− (ST − 59) = 0.6 (65− ST )



Example 9.1.26. Prove Equation 9.32.

Let K3 = λK1 + (1− λ)K2. Clearly, K1 < K3 < K2

Payoff ST < K1 K1 ≤ ST < K3 K3 ≤ ST < K2 ST ≥ K2

K3-strike call (a) 0 0 ST −K3 ST −K3

K1-strike call (b) 0 ST −K1 ST −K1 ST −K1

K2-strike call (c) 0 0 0 ST −K2

λb+ (1− λ) c 0 λ (ST −K1) λ (ST −K1) ST −K3

λb+ (1− λ) c− a 0 λ (ST −K1) ≥ 0 (1− λ) (K2 − ST ) > 0 0

Please noteλ (ST −K1) + (1− λ) (ST −K2) = ST − [λK1 + (1− λ)K2] = ST −K3

In addition,λ (ST −K1)− (ST −K3) = K3 − λK1 − (1− λ)ST= λK1 + (1− λ)K2 − λK1 − (1− λ)ST = (1− λ)K2 − (1− λ)ST= (1− λ) (K2 − ST )If we buy λ unit of K1-strike call, buy (1− λ) unit of K2 call, and sell 1 unit

of K3 = λK1 + (1− λ)K2 strike call, we’ll have a non-negative payoff at T . Toavoid arbitrage, the initial cost must be non-negative. Hence λb+(1− λ) c−a >0.Anyway, the call portfolio consisting of λ portion of K1-strike call and 1−λ

portion of K2-strike call is at least as good as the call with the strike priceK3 = λK1+(1− λ)K2. To avoid arbitrage, C [λK1 + (1− λ)K2] ≤ λC (K1)+(1− λ)C (K2).

Example 9.1.27. (Textbook example 9.9 K2 and K3 switched) K1 = 50, K2 =70,K3 = 0.75 (50) + 0.25 (70) = 55. P (K1, T ) = 4, P (K2, T ) = 16. Explainwhy P (55) ≤ 0.75P (50) + 0.25P (70).

Payoff ST < 50 50 ≤ ST < 55 55 ≤ ST < 70 ST ≥ 7055-strike put (a) 55− ST 55− ST 0 0

50-strike put (b) 50− ST 0 0 070-strike put (c) 70− ST 70− ST 70− ST 00.75b+ 0.25c 55− ST 0.25 (70− ST ) 0.25 (70− ST ) 0

(0.75b+ 0.25c)− a 0 0.75 (ST − 50) > 0 0.25 (70− ST ) > 03 0

Please note0.75 (50− ST ) + 0.25 (70− ST ) = 55− ST0.25 (70− ST )− (55− ST ) = 0.75 (ST − 50)30.4 (ST − 50)− (ST − 59) = 0.6 (65− ST )



(0.75b+ 0.25c) − a ≥ 0 no matter what ST is. Clearly, the put portfolioconsisting of 75% 50-strike put and 25% 70-strike put is at least as good asthe 55-strike put. Consequently, to avoid arbitrage, the portfolio is at least asvaluable as the 55-strike put. P (55) ≤ 0.75P (50) + 0.25P (70).

Example 9.1.28. Prove Equation 9.33.

Let K3 = λK1 + (1− λ)K2. Clearly, K1 < K3 < K2

Payoff ST < K1 K1 ≤ ST < K3 K3 ≤ ST < K2 ST ≥ K2

K3-strike put (a) K3 − ST K3 − ST 0 0

K1-strike put (b) K1 − ST 0 0 0K2-strike put (c) K2 − ST K2 − ST K2 − ST 0λb+ (1− λ) c K3 − ST (1− λ) (K2 − ST ) λ (K2 − ST ) ST −K3

λb+ (1− λ) c− a 0 λ (ST −K1) ≥ 0 λ (K2 − ST ) > 0 0

Please noteλ (K1 − ST ) + (1− λ) (K2 − ST ) = K3 − STIn addition,

(1− λ) (K2 − ST )− (K3 − ST )= (1− λ) (K2 − ST )− λK1 − (1− λ)K2 + ST = λ (ST −K1)λ (ST −K1)− (ST −K3) = K3 − λK1 − (1− λ)ST= λK1 + (1− λ)K2 − λK1 − (1− λ)ST = (1− λ)K2 − (1− λ)ST= (1− λ) (K2 − ST )The payoff is always non-negative. Consequently, the call portfolio consisting

of λ portion of K1-strike put and 1 − λ portion of K2-strike put is at least asgood as the put with the strike priceK3 = λK1+(1− λ)K2. To avoid arbitrage,P [λK1 + (1− λ)K2] ≤ λP (K1) + (1− λ)P (K2).

Exercise and moneyness

Proposition 9.1.11. If it’s optimal to exercise an option, it’s also optimal toexercise an otherwise identical option that’s more in-the-money.

This is just common sense. The textbook gives an example. Suppose it’soptimal to exercise a 50-strike American call on a dividend paying stock. Thecurrent stock price is 70. If it’s optimal to exercise the American a 50-strikeAmerican call, then it must also be optimal to exercise an otherwise identicalcall but with a strike price 40.




Chapter 10

Binomial option pricing: I

This chapter is one of the easiest chapters in Derivatives Markets. The textbookdid a good job explaining the mechanics of how to calculate the option price ina one-period binomial model. Besides learning the mechanics of option pricing,you should focus on understanding two basic ideas: the non-arbitrage pricingand the risk-neutral probabilities.

10.1 One-period binomial model: simple exam-ples

Example 10.1.1. Suppose we want to find the price of a 12-month Europeancall option on a stock with strike price $15. The stock currently sells for $20. In12 months, the stock can either go up to $30 or go down to $10. The continuouslycompounded risk-free interest rate per year is 10%.

3020

10Time 0 T = 1

15?

0Time 0 T = 1

Stock price binomial tree Call payoff binomial tree

It’s hard to directly calculate the price of the call option. So let’s buildsomething that behaves like a call, something that has the same payoff patternas the call. Suppose at time zero we create a portfolio by buying X stocks andputting Y dollars in a savings account. We want this portfolio to have the exactpayoff as the call.

30X Y e0.1 1520X + Y = ?

10X Y e0.1 0Time 0 T = 1 Time 0 T = 1 Time 0 T = 1

35


36 CHAPTER 10. BINOMIAL OPTION PRICING: I

In the above diagram, 2X stocks at time zero are worth either 30X or 10Xat T = 1. Putting Y dollars in a savings account at time zero will produce Y e0.1

at T = 1.We want our portfolio to behave like a call. So the payoff of our portfolio

should the same as the payoff of the call. We set up the following equation:½30X + Y e0.1 = 1510X + Y e0.1 = 0

X = 0.75 Y = −10 (0.75) e−0.1 = −6. 786

Y = −6. 786 means that we borrow 6. 786 at t = 0 and pays 6. 786e0.1 = 7.5 at T = 1If at t = 0 we buy 0.75 share of a stock and borrow $6. 786, then at T = 1,

this portfolio will have the same payoff as the call. To avoid arbitrage, theportfolio and the call should have the same cost at t = 0.

C = 20X + Y = 20 (0.75)− 6. 786 = 8. 214

Example 10.1.2. Find the price of a 12-month European put option on a stockwith strike price $15. The stock currently sells for $20. In 12 months, thestock can either go up to $30 or go down to $10. The continuously compoundedrisk-free interest rate per year is 10%.

Suppose at time zero we create a replicating portfolio by buying X stocksand investing Y dollars in a savings account. We want this portfolio to have theexact payoff as the put.

30X Y e0.1 020X + Y = ?

10X Y e0.1 5Time 0 T = 1 Time 0 T = 1 Time 0 T = 1½30X + Y e0.1 = 010X + Y e0.1 = 5

X = −0.25 Y = 6. 786

So the replicating portfolio is at t = 0 by short-selling 0.25 stock and invest-ing 6. 786 in a savings account.The price of the put at t = 0 is:P = 20X + Y = 20 (−0.25) + 6. 786 = 1. 786Let’s check whether the put-call parity holds.C + PV (K) = 8. 214 + 15e−0.1 = 21. 787P + S0 = 1. 786 + 20 = 21. 786Ignoring the rounding difference, we get: C + PV (K) = P + S0

10.2 General one-period binomial modelSuppose we have two points in time, t = 0 (today) and t = h (some point inthe future). The continuously compounded risk-free interest rate per year is r

( a positive constant). We have two assets: a stock that pays zero dividend and


10.2. GENERAL ONE-PERIOD BINOMIAL MODEL 37

a savings account. The savings account is the same as a zero-coupon bond. Attime h, the stock price is Sh; the bond price is Bh.The bond price is deterministic: B0 = 1 Bh = erh

The stock price is stochastic:

S0 = S Sh =

½SuSd

So at h the stock price either goes up to Su ("up state") or goes down to Sd("down state").

SuS

SdTime 0 Time h

We assume there’s no tax, no transaction cost, or margin requirements; oneis allowed to short sell any security and receive full proceeds. In addition, weassume that anyone can buy or sell any number of securities without affectingthe market price.We assume that the market is an invisible cop automatically enforcing Sd <

Serh < Su. For example, if Su > Sd > Serh, then the stock’s return is guar-anteed to be higher than the risk-free interest rate. If this is the case, then therisk-free interest rate was set too low and stock’s return is too good. Everyonewill jump on this opportunity, withdraw all his money from his savings account,and invest it in the stock. This will instantaneously bid up the price of the stockand the risk-free interest rate, forcing Sd < Serh < Su to hold.

What’s the non-arbitrage condition is the stock pays dividend continuouslyat the yield δ?Suppose you buy e−δh share of a stock at t = 0 (thus you pay Se−δh) , buy

reinvesting dividend in the stock, you’ll have exactly one e−δheδh = 1 stock attime h, which is worth either Su or Sd.

SuSe−δh

SdTime 0 Time h

Suppose you invest Se−δh in a savings account, then your wealth at h issimply

¡Se−δh

¢erh = Se(r−δ)h.Se(r−δ)h

Se−δh

Se(r−δ)h

Time 0 Time h

To avoid arbitrage, the following condition needs to be met:



• At good times (i.e. when the stock goes up), the return you earn from thestock should exceed the risk free interest rate. So Su > Se(r−δ)h

• At bad times (i.e. when the stock goes down), the return you earn fromthe stock should be less than the risk free interest rate. So Sd < Se(r−δ)h

If the condition is not met, we’ll end up in a weird situation where the stockis always better off than the savings account or the savings account is alwaysbetter off than the stock. Then everyone will invest his money in the betterperforming asset, instantly bidding up the price of the lower-performing assetand forcing the above condition to be met.To avoid arbitrage, Su, Sd, and r need to satisfy the following condition:

Su > Se(r−δ)h > Sd (10.1)

If Su = uS and Sd = dS, then Equation 10.1 becomes:

u > e(r−δ)h > d (10.2)

Equation 10.2 is the textbook Equation 10.4.

Let’s continue.Let C represent the option price at time zero. Let Cu and Cd represent the

payoff at time h of an option in the up state and down state respectively:

Cu

CCd

Time 0 Time h

Our task is to determine C by setting a portfolio that replicates the optionpayoff. We build the replicating portfolio by buying 4 shares of the stock andinvesting $B in a zero-coupon bond. So we set up the following equation:

4Su Berh Cu

4S + B = C4Sd Berh Cd

t = 0 t = h t = 0 t = h t = 0 t = h½4Su +Berh = Cu

4Sd +Berh = Cd

Solving these equations, we get:

4 =Cu − Cd

Su − Sd(10.3)



B = e−rhSuCd − SdCu

Su − Sd(10.4)

Rearranging Equation 10.3, we get 4Su − Cu = 4Sd − Cd. This equationis critical as we’ll soon see.The value of the portfolio at the up state is4Su +Berh = 4Su + e−rh (Cu −4Su) e

rh = Cu

The value of the portfolio at the down state is4Sd +Berh = 4Sd + e−rh (Cu −4Sd) e

rh = Cd

Using Equation 10.3 and Equation 10.4, we get (verify this for yourself):

C = 4S +B = e−rhµSerh − SdSu − Sd

Cu +Su − Serh

Su − SdCd

¶(10.5)

Define

πu = p∗ =Serh − SdSu − Sd

(10.6)

πd = q∗ =Su − Serh

Su − Sd(10.7)

Because market automatically enforces Sd < Serh < Su, we have

0 < πu < 1 (10.8)

0 < πp < 1 (10.9)

πu + πd = 1 (10.10)

According to Equation 10.8 , we can pretend that πu and πd are probabilities.Then Equation 10.5 becomes

C = e−rh (πuCu + πdCd) (10.11)

Doing some algebra, we also get (verify this for yourself):

S = e−rhµSerh − SdSu − Sd

Su +Su − Serh

Su − SdSd

¶= e−rh (πuSu + πdSd) (10.12)

According to Equation 10.11 and 10.12, once we set up faked probabilitiesπu and πd, the call price at t = 0 is simply the expected present value of the



call payoffs discounted at the risk-free interest rate; the stock price at t = 0 issimply the expected present value of the future stock prices discounted at therisk-free interest rate. Since the discounting rate is risk-free interest rate, πuand πd are called risk neutral probabilities.Please note that πu and πd are not real probabilities. They are artificially

created probabilities so that Equation 10.11 and 10.12 have simple and intuitiveexplanations.

Example 10.2.1. Using the risk-neutral probabilities, find the price of a 12-month European call option on a stock with strike price $15. The stock currentlysells for $20. In 12 months, the stock can either go up to $30 or go down to$10. The continuously compounded risk-free interest rate per year is 10%.

Solution.

Stock price tree Option terminal payoff

Time 0 TSu = 30

S = 20Sd = 10

Time 0 TCu = max (0, 30− 15) = 15

C =?Cd = max (0, 10− 15) = 0

πu =Serh − SdSu − Sd

=20e0.1(1) − 1030− 10 = 0.605

πd = 1− 0.605 = 0.395Cu = 30− 15 = 15 Cd = 0C = e−rh (πuCu + πdCd) = e−0.1(1) (0.605× 15 + 0.395× 0) = 8. 211

Example 10.2.2. Using the risk-neutral probabilities, find the price of a 12-month European put option on a stock with strike price $15. The stock currentlysells for $20. In 12 months, the stock can either go up to $30 or go down to$10. The continuously compounded risk-free interest rate per year is 10%.


Time 0 TSu = 30

S = 20Sd = 10

Time 0 TCu = max (0, 15− 30) = 0

C =?Cd = max (0, 15− 10) = 5

πu =Serh − SdSu − Sd

=20e0.1(1) − 1030− 10 = 0.605

πd = 1− 0.605 = 0.395

Cu = 0 Cd = 5C = e−rh (πuCu + πdCd) = e−0.1(1) (0.605× 0 + 0.395× 5) = 1. 787



Suppose the stock pays dividends at a continuously compounded rate δ peryear. At time zero, our replicating portfolio consists of 4 shares of stocks and$B in a bond (or a savings account). If we continuously reinvest dividends andbuy additional stocks during [0, h], our 4 shares at time zero will grow into4eδh shares at time h. Our 4eδh shares will be worth either 4eδhSu in the upstate or 4eδhSd in the down state. We want our replicating portfolio to havethe same payoff as the option at time h.½

4eδhSu +Berh = Cu

4eδhSd +Berh = Cd

Solving the equations, we get:

4 = e−δhCu − Cd

Su − Sd(10.13)


Su − Sd(10.14)

Notice whether the stock pays dividend or not, at time zero, we always need

to have e−rhSuCd − SdCu

Su − Sdin a savings account, which grows into

SuCd − SdCu

Su − Sd

dollars at t = h. If the stock doesn’t pay dividend, at t = 0 we holdCu − Cd

Su − Sd

shares of stock, which isCu − Cd

Su − Sdshares of stock at t = h ; if the stock pays

dividend at a continuously compounded rate δ, at t = 0 we hold e−δhCu − Cd

Su − Sd,

which growsCu − Cd

Su − Sdshares of stock at t = h.

So at time h we need to haveCu − Cd

Su − Sdunits of stocks and

SuCd − SdCu

Su − Sddollars in a savings account (or a bond), regardless of whether the stock paysdividend or not.

To see why, suppose our replicating portfolio at t = h (not t = 0) consistsof U shares of stocks and V dollars in a savings account. Then regardless ofwhether the stock pays dividend or not, we need to have:½

USu + V = Cu

USd + V = Cd

This gives us:

U =Cu − Cd

Su − SdV =

SuCd − SdCu

Su − Sd

To have V dollars in a savings account at t = h, we need to have V e−rh =

e−rhSuCd − SdCu

Su − Sdat t = 0. To have U =

Cu − Cd

Su − Sdshares of stocks at t = h,



if the stock pays dividend at a continuously compounded rate of δ, we need to

have Ue−δh = e−δhCu − Cd

Su − Sdshares of stocks at t = 0. If we use the dividends

received to buy additional stocks, then we’ll have U shares of stock at t = h.

Now let’s find the cost of the option on a stock that pays dividends at acontinuously compounded rate δ per year. The option cost at time zero is

4S +B = e−δhCu − Cd

Su − Sd× S + e−rh

SuCd − SdCu

Su − Sd

= e−rhµSe(r−δ)h − Sd

Su − SdCu +

Su − Se(r−δ)h

Su − SdCd

¶C = 4S +B = e−rh (πuCu + πdCd) (10.15)

where

πu =Se(r−δ)h − Sd

Su − Sd(10.16)

πd =Su − Se(r−δ)h

Su − Sd(10.17)

If Su = Su and Sd = Sd, then Equation 10.16 and 10.17 becomes:

πu =e(r−δ)h − d

u− d(10.18)

πd =u− e(r−δ)h

u− d(10.19)

Tip 10.2.1. If you don’t want to memorize Equation 10.12, 10.15, 10.16, 10.17,just set up the replication portfolio and calculate 4 and B from scratch.

Example 10.2.3. Find the price of a 12-month European call option on a stockwith strike price $15. The stock pays dividends at a continuously compoundedrate 6% per year. The stock currently sells for $20. In 12 months, the stock caneither go up to $30 or go down to $10. The continuously compounded risk-freeinterest rate per year is 10%.


Time 0 TSu = 30

S = 20Sd = 10

Time 0 TCu = max (0, 30− 15) = 15

C =?Cd = max (0, 10− 15) = 0



Replicating portfolioTime 0 T

(4u, Bu) = (0.75,−7. 50)(4, B) = (0.706 32,−6. 786 28)

(4d, Bd) = (0.75,−7. 50)Our replicating portfolio at t = 0 consists of 4 shares of stocks and $B in a

savings account.


Su − Sd= e−0.06(1)

15− 030− 10 = 0.706 32


Su − Sd= e−0.1(1)

30 (0)− 10 (15)30− 10 = −6. 786 28

C = 4S +B = 0.706 3 (20)− 6. 786 3 = 7. 34

The replicating portfolio at T is:4u = 4d = 4eδh = 0.706 32e0.06(1) = 0.75Bu = Bd = Berh = −6. 786 28e0.1(1) = −7. 5

Verify that the replicating portfolio and the option have the same value:The value of the replicating portfolio at the up state:4uSu +Bu = 0.75 (30)− 7. 50 = 15 = Cu

The value of the replicating portfolio at the down state:4dSd +Bd = 0.75 (10)− 7. 50 = 0 = Cd

The value of the portfolio at t = 0 :C = 4S +B = 0.706 32 (20)− 6. 78628 = 7. 34Alternatively,


Su − Sd=20e(0.1−0.06)1 − 10

30− 10 = 0.540 81

πd = 1− πu = 1− 0.540 81 = 0.459 19C = e−rh (πuCu + πdCd) = e−0.1(1) (0.540 81× 15 + 0.459 19× 0) = 7. 34

Example 10.2.4. Find the price of a 12-month European put option on a stockwith strike price $15. The stock pays dividends at a continuously compoundedrate 6% per year. The stock currently sells for $20. In 12 months, the stock caneither go up to $30 or go down to $10. The continuously compounded risk-freeinterest rate per year is 10%.

Solution.


Time 0 TSu = 30

S = 20Sd = 10

Time 0 TCu = max (0, 15− 30) = 0

C =?Cd = max (0, 15− 10) = 5




(4u, Bu) = (−0.25, 7. 50)(4, B) = (−0.235 4, 6. 786 3)

(4d, Bd) = (−0.25, 7. 50)Our replicating portfolio at t = 0 consists of 4 shares of stocks and $B in a

savings account.


Su − Sd= e−0.06(1)

0− 530− 10 = −0.25e

−0.06(1) = −0.235 4


Su − Sd= e−0.1(1)

30 (5)− 10 (0)30− 10 = 6. 786 3

C = 4S +B = −0.235 4 (20) + 6. 786 3 = 2. 078The replicating portfolio at T is:4u = 4d = 4eδh = −0.235 4e0.06(1) = −0.25Bu = Bd = Berh = 6. 786 3e0.1(1) = 7. 5

Verify that the replicating portfolio and the option have the same value:The value of the replicating portfolio at the up state:4uSu +Bu = −0.25 (30) + 7. 50 = 0 = Cu

The value of the replicating portfolio at the down state:4dSd +Bd = −0.25 (10) + 7. 50 = 5 = Cd

The value of the portfolio at t = 0 :C = 4S +B = −0.235 4 (20) + 6. 786 3 = 2. 078Alternatively,


Su − Sd=20e(0.1−0.06)1 − 10

30− 10 = 0.540 8

πd = 1− πu = 1− 0.540 8 = 0.459 2C = e−rh (πuCu + πdCd) = e−0.1(1) (0.540 8× 0 + 0.459 2× 5) = 2. 078

Arbitrage a mispriced option

If an option sells for more or less than the price indicated by Equation 10.15,we can make money by "buy low, sell high."

Example 10.2.5. A 12-month European call option on a stock has strike price$15. The stock pays dividends at a continuously compounded rate 6% per year.The stock currently sells for $20. In 12 months, the stock can either go up to$30 or go down to $10. The continuously compounded risk-free interest rate peryear is 10%. This call currently sells for $8. Design an arbitrage strategy.

Solution.




Time 0 TSu = 30

S = 20Sd = 10

Time 0 TCu = max (0, 30− 15) = 15

C =?Cd = max (0, 10− 15) = 0


(4u, Bu) = (0.75,−7. 50)(4, B) = (0.706 32,−6. 786 28)

(4d, Bd) = (0.75,−7. 50)

There are two calls. One is in the market selling for $8 at t = 0. The other isa synthetic call, which consists, at t = 0, of holding 0.706 3 stock and borrowing$6. 786 3 at risk-free interest rate. The synthetic call sells for $7. 34 at t = 0.These two calls have identical payoffs at t = 1.To make a riskless profit, we buy low and sell high. At t = 0, we sell a

call for $8 (sell high). Then at t = 1, if the stock price is $30, the call holderexercises the call and our payoff is 15 − 30 = −15; if the stock is $10, the callexpires worthless and our payoff is zero.

−158

0Time 0 Time t = 1Payoff of a written call

Simultaneously, at t = 0 we buy 0.706 3 stock and borrow 6. 786 3 dollarsat risk-free interest rate (buy low). This costs us 0.706 3 (20) − 6. 786 3 = 7.34 at t = 0. At t = 1, our initial 0.706 3 stock becomes 0.706 3e0.06(1) stockand our initial debt 6. 786 3 grows into 6. 786 3e0.1(1). Our portfolio is worth0.706 3e0.06(1)S1 − 6. 786 3e0.1(1).If S1 = 30, our portfolio is worth0.706 3e0.06(1) (30)− 6. 786 3e0.1(1) = 15If S1 = 10, our portfolio is worth0.706 3e0.06(1) (10)− 6. 786 3e0.1(1) = 0

15−7.34

0Time 0 Time t = 1

Payoff of a replicating portfolioSo at t = 0, we gain 8− 7. 34 = 0.66. At t = 1, the portfolio exactly offsets

our payoff in the call. We earn 0.66 sure profit at t = 0.



Example 10.2.6. A 12-month European call option on a stock has strike price$15. The stock pays dividends at a continuously compounded rate 6% per year.The stock currently sells for $20. In 12 months, the stock can either go up to$30 or go down to $10. The continuously compounded risk-free interest rate peryear is 10%. This call currently sells for $7. Design an arbitrage strategy.


Time 0 TSu = 30

S = 20Sd = 10

Time 0 TCu = max (0, 30− 15) = 15

C =?Cd = max (0, 10− 15) = 0


(4u, Bu) = (0.75,−7. 50)(4, B) = (0.706 32,−6. 786 28)

(4d, Bd) = (0.75,−7. 50)

There are two calls. One is in the market selling for $7 at t = 0. The other isa synthetic call, which consists, at t = 0, of holding 0.706 3 stock and borrowing6. 786 3 dollars at risk-free interest rate. The synthetic call sells for $7. 34 att = 0. These two calls have identical payoffs at t = 1.To make a riskless profit, we buy low and sell high. At t = 0, we buy a

call for $7 (buy low). Then at t = 1, if the stock price is $30, the call holderexercises the call and our payoff is 30 − 15 = 15; if the stock is $10, the callexpires worthless and our payoff is zero.

15−7

0Time 0 Time t = 1Payoff of a purchased call

Simultaneously, at t = 0 we sell the replicating portfolio. We short sell0.706 3 stock and lend 6. 786 3 dollars at risk-free interest rate (sell high). Wegain 0.706 3 (20)− 6. 786 3 = 7. 34 at t = 0.At t = 1, our initially borrowed 0.706 3 stock becomes 0.706 3e0.06(1) stock

and our lent principal 6. 786 3 grows into 6. 786 3e0.1(1). Our portfolio is worth6. 786 3e0.1(1) − 0.706 3e0.06(1)S1.

If S1 = 30, our portfolio is worth6. 786 3e0.1(1) − 0.706 3e0.06(1) (30) = −15

If S1 = 10, our portfolio is worth



6. 786 3e0.1(1) − 0.706 3e0.06(1) (10) = 0

−157.34

0Time 0 Time t = 1

Payoff of a replicating portfolioSo at t = 0, we gain7.34− 7 = 0.34. At t = 1, the call payoff exactly offsets

our liabilities in the replicating portfolio We earn 0.34 sure profit at t = 0.

Tip 10.2.2. An option and its replicating portfolio are exactly the same in termsof payoff and cost. If an option in the market sells for more than the fair priceindicated in Equation 10.15, we can make a sure profit by buying the replicatingportfolio and selling the option. If an option in the market sells for less thanthe fair price indicated in Equation 10.15, we can make a sure profit by sellingthe replicating portfolio and buying the option.

Risk neutral probability and forward price

If we use πu and πd to calculate the undiscounted stock price, we get:

πuSu + πdSd =Se(r−δ)h − Sd

Su − SdSu +

Su − Se(r−δ)h

Su − SdSd

= Se(r−δ)h = F0,h

πuSu + πdSd = Se(r−δ)h = F0,h (10.20)

F0,h is the delivery price at t = h of a forward contract signed at t = 0.The textbook uses the symbol Ft,t+h to indicate that the forward contract issigned at time t and the asset is to be delivered at t+h. If we treat the contractinitiation date t as time zero, then the asset deliver date is time h. So thenotation doesn’t matter. If you want to use the textbook notation, you’ll have

πuSu + πdSd = Se(r−δ)h = Ft,t+h (10.21)

If you want to use my notation, you’ll get Equation 10.20.Suppose you enter into a forward contract as a seller agreeing to deliver one

stock for a guaranteed price F0,h at t = h. To ensure you indeed can deliverone stock at time h, you’ll want to buy e−δh stock at time zero. If you reinvestdividend and buy additional stocks, your initial e−δh stock will grow into exactlyone stock at time h. Your cost of buying e−δh stock at time zero is Se−δh. Since

you’ll tie up your money Se−δh during [0, h], you’ll want the forward price toinclude the interest you could otherwise earn on Se−δh. So the forward price isjust the future value of Se−δh:

F0,h = Se−δherh = Se(r−δ)h



According to Equation 10.20, the undiscounted stock price equals the for-ward price under the risk neutral probability. If a problem gives you a forwardprice, you can use Equation 10.20 to calculate the risk-neutral probability.

Constructing a binomial tree

Suppose we are standing at time t. If we can be 100% certain about the stockprice at time t+ h, then investing in stocks doesn’t have any risk. Then stocksmust earn a risk-free interest rate. Since the stock already pays dividends atrate δ, to earn a risk free interest rate, the stock price just needs to grow atthe rate of r − δ. Hence the stock price at t + h is Se(r−δ)h, which is just theforward price Ft,t+h because Ft,t+h is also equal to Se(r−δ)h.However, the stock price is a random variable and we generally can’t be

100% certain about a stock’s future price. To incorporate uncertainty, we useFormula 11.16 in Derivatives Markets:

lnSt+hSt

= (r − δ)h± σ√h (Textbook 11.16)

In the above equation, lnSt+hSt

is the continuously compounded rate of return

during [t, t+ h]. This return consists of a known element (r − δ)h and a randomelement σ

√h, where σ is the annualized standard deviation of the continuously

compounded stock return. The variance of the stock return in one year is σ2.The variance during the interval [t, t+ h] (which is h year long) is σ2h and thisis why. [t, t+ h] can be broken down into h intervals, with each interval beingone year long. Assume stock return during each year is independent identicallydistributed. The total return during [t, t+ h] is the sum of the returns over hintervals. Then the total variance is just the sum of the variance over h intervals,σ2h.

So St+h = Ste(r−δ)h±σ

√h = Se(r−δ)h±σ

√h = Ft,t+he

±σ√h. In the binomial

model, the stock price either goes up to Se(r−δ)h+σ√h = Ft,t+he

σ√h or goes

down to Se(r−δ)h−σ√h = Ft,t+he

−σ√h. So we have

uSt = Ft,t+heσ√h (10.22)

dSt = Ft,t+he−σ√h (10.23)

If we set volatility σ to zero, then Equation 10.22 and 10.23 becomes uSt =dSt = Ft,t+h. This means that if the stock price is 100% certain, then the stockprice is just the forward price.Apply Equation 10.21 to Equation 10.22 and 10.23, we get:



u = e(r−δ)h+σ√h (10.24)

d = e(r−δ)h−σ√h (10.25)

10.2.1 Two or more binomial trees

Example 10.2.7. Let’s reproduce Derivatives Markets Figure 10.4. Here is therecap of the information on a European call option. The current stock price is 41.The strike price K = 40. The annualized standard deviation of the continuouslycompounded stock return is σ = 30%. The continuously compounded risk-freerate per year is r = 8%. The continuously compounded dividend rate per yearis δ = 0%.The option expiration date is T = 2 years. Use a 2-period binomialtree to calculate the option premium.

Each period is h =T

2= 1 year long.

Step 1 Draw a stock price tree.

u = e(r−δ)h+σ√h = e(0.08−0)1+0.3

√1 = 1. 462 28

d = e(r−δ)h−σ√h = e(0.08−0)1−0.3

√1 = 0.802 52

Stock pricePeriod 0 1 2

Su2

SuS Sud

SdSd2

=⇒

Period 0 1 2Suu = 87.6693

Su = 59.9537S = 41 Sud = 48.1139

Sd = 32.9033Sdd = 26.4055

Step 2 Draw the terminal payoffs at T = 2.47.6693 = max (0, 87.6693− 40)8.1139 = max (0, 48.1139− 40)

Option PayoffPeriod 0 1 2

Cuu = 47.6693Cu

C =? Cud = 8.1139Cd

Cdd = 0

Step 3 Work backward from right to left (called backwardization). Cal-culate the premium using the formula

C = e−rh (πuCu + πdCd)


u− d=

e(0.08−0)1 − 0.802 521. 462 28− 0.802 52 = 0.425 56



πd = 1− πu = 1− 0.42556 = 0.574 44

Option premiumPeriod 0 1 2

Cuu = 47.6693Cu = 23.0290

C = 10.7369 Cud = 8.1139Cd = 3.1875

Cdd = 0

Cu = e−rh (πuCuu + πdCud)= e−0.08(1) (0.425 56× 47.6693 + 0.574 44× 8.1139) = 23. 029 0

Cd = e−rh (πuCud + πdCud)= e−0.08(1) (0.425 56× 8.1139 + 0.574 44× 0) = 3. 187 5

The call premium isC = e−rh (πuCu + πdCd)

= e−0.08(1) (0.425 56× 23.0290 + 0.574 44× 3.1875) = 10.7369



Step 4 Calculate 4 and B using the formula


Su − SdB = e−rh

SuCd − SdCu

Su − SdTo avoid errors, put the stock price tree and the premium tree side by side:

Period 0 1 2Suu = 87.6693

Su = 59.9537S = 41 Sud = 48.1139

Sd = 32.9033Sdd = 26.4055

Period 0 1 2Cuu = 47.6693

Cu = 23.0290C = 10.7369 Cud = 8.1139

Cd = 3.1875Cdd = 0

Period 0 1

(4u, Bu) = (1.0,−36.9247)(4, B) = (0.7335,−19.3367)

(4d, Bd) = (0.3738,−9.1107)

4u = e−δhCuu − Cud

Suu − Sdd= e−0(1)

47.6693− 8.113987.6693− 48.1139 = 1.0

4d = e−δhCud − Cdd

Sud − Sdd= e−0(1)

8.1139− 048.1139− 26.4055 = 0.373 8


Su − Sd= e−0(1)

23.0290− 3.187559.9537− 32.9033 = 0.733 5

Bu = e−rhSuuCud − SudCuu

Suu − Sud

= e−0.08(1)87.6693 (8.1139)− 48.1139 (47.6693)

87.6693− 48.1139 = −36.9247

−36.9247 means that $36.9247 needs to be borrowed at a risk-free rate.

Bd = e−rhSduCdd − SddCdu

Sdu − Sdd

= e−0.08(1)48.1139 (0)− 26.4055 (8.1139)

48.1139− 26.4055 = −9. 110 7


Su − Sd

= e−0.08(1)59.9537 (3.1875)− 32.9033 (23.0290)

59.9537− 32.9033 = −19. 3367



Step 5 Verify that the portfolio replicates the premium tree. Here isrecap of the information:

Period 0 1 2Suu = 87.6693

Su = 59.9537S = 41 Sud = 48.1139

Sd = 32.9033Sdd = 26.4055

0 1 2Cuu = 47.6693

Cu = 23.0290C = 10.7369 Cud = 8.1139

Cd = 3.1875Cdd = 0

Period 0 1

(4u, Bu) = (1.0,−36.9247)(4, B) = (0.7335,−19.3367)

(4d, Bd) = (0.3738,−9.1107)

The value of (4, B) at time zero:4S +B = 0.7335 (41)− 19.3367 = 10. 736 8 = C

The value of(4u, Bu) at Node u (up state):4uSu +Bu = 1.0 (59.9537)− 36.9247 = 23. 029 = Cu

The value of(4d, Bd) at Node d (down state):4dSd +Bd = 0.3738 (32.9033)− 9.1107 = 3. 188 = Cd

Finally, let’s verify that the portfolio replicates the terminal payoff. First,we need to find the replicating portfolio at the expiration date.

Period 1 2(4uu, Buu) = (1.0,−40)

(4u, Bu) = (1.0,−36.9247)(4ud, Bud) = (1.0,−40) = (0.3738,−9. 869 5)

(4d, Bd) = (0.3738,−9.1107)(4dd, Bdd) = (0.3738,−9. 869 5)

If we reinvest dividends, 4ustocks grows into 4ueδh after h years; Bu grows

into Buerh after h years.

(4uu, Buu) =¡4ue

δh, Buerh¢=¡1.0e0(1),−36.9247e0.08(1)

¢= (1.0,−40)

4uuSuu +Buu = 1.0 (87.6693)− 40 = 47. 669 3 = Cuu

Similarly, (4dd, Bdd) =¡4de

δh, Bderh¢=¡0.3738e0(1),−9.1107e0.08(1)

¢=

(0.373 8,−9. 869 5)4ddSdd +Bdd = 0.3738 (26.4055)− 9. 869 5 = 0.00 = Cdd

Finally, (4ud, Bud) = (1.0,−40) = (0.3738,−9. 869 5)Portfolios (1.0,−40) and (0.3738,−9. 869 5) have equal values at Node ud.1.0 (48.1139)− 40 = 8. 113 9 = Cud

0.3738 (48.1139)− 9. 869 5 = 8. 115 = Cud (ignore rounding difference)



Tip 10.2.3. For a multi-binomial tree, using the risk neutral probability to findthe premium is faster than using the replicating portfolio. The risk neutral prob-abilities πu and πd are constant cross nodes. However, the replicating portfolio(4, B) varies by node.

Tip 10.2.4. For European options, you can calculate the premium using theterminal payoffs. This is how to quickly find the premium for this problem.

Node at T = 2 Payoff at T Risk neutral probability of reaching this node1

uu Cuu = 47.6693 π2u = 0.425 562 = 0.181 1

ud Cud = 8.1139 2πuπd = 2 (0.425 56) (0.574 44) = 0.488 92

dd Cdd = 0 π2d = 0.574 442 = 0.3300

Total (πu + πd)2 = 0.181 1 + 0.488 9 + 0.3300 = 1 3

The premium is the expected present value of the terminal payoff using therisk neutral probability.

C = e−rT¡π2uCuu + 2πuπdCud + π2dCdd

¢= e−0.08(2)

¡0.425 562 × 47.6693 + 0.488 9× 8.1139 + 0.33× 0

¢= 10. 736 9

Tip 10.2.5. If you purchased the textbook Derivatives Markets, you should seea CD attached to the back cover of the book. Install the CD in your computer.Run the spreadsheet titled "optall2’" or "optbasic2." These two spreadsheets cancalculate European and American option prices. When you solve a practice prob-lem, you can use either of these two spreadsheets to double check you answer.Please note that these two spreadsheets don’t calculate the replicating portfo-lio (4, B). So you can’t use them to verify your calculation of the replicatingportfolio.

Example 10.2.8. Let’s reproduce Derivatives Markets Figure 10.5. Here isthe recap of the information on a European call. The current stock price is 41.The strike price K = 40. The annualized standard deviation of the continuouslycompounded stock return is σ = 30%. The continuously compounded risk-freerate per year is r = 8%. The continuously compounded dividend rate per yearis δ = 0%.The option expiration date is T = 1 year. Use a 3-period binomialtree to calculate the option premium.

Solution.

Each period is h =1

3year long.

u = e(r−δ)h+σ√h = e(0.08−0)1/3+0.3

√1/3 = 1. 221 246

1The probabilities in this column are the 3 terms of (πu + πd)2 = π2d + 2πdπu + π2u

2There are two ways of reaching Node ud: up and dow or down and up.3The total probability is one. Otherwise, you made an error.



d = e(r−δ)h−σ√h = e(0.08−0)1/3−0.3

√1/3 = 0.863 693

Stock pricePeriod 0 1 2 3

Su3 = 74.6781Su2 = 61.1491

Su = 50.0711 Su2d = 52.8140S = 41 Sud = 43.2460

Sd = 35.4114 Sd2u = 37.3513Sd2 = 30.5846

Sd3 = 26.4157Calculate the premium by working backward from right to left.

Period 0 1 2 3Cu3 = max (0, 74.6781− 40) = 34. 678 1

Cu2 = 22. 201 6Cu = 12. 889 5 Cu2d = max (0, 52.8140− 40) = 12. 814

C = 7. 073 9 Cud = 5. 699 5Cd = 2. 535 1 Cd2u = max (0, 37.3513− 40) = 0

Cd2 = 0Cd3 = max (0, 26.4157− 40) = 0



u− d=

e(0.08−0)1/3 − 0.863 6931. 221 246 − 0.863 693 = 0.456 806

πd = 1− πu = 1− 0.456 806 = 0.543 194

Cu2 = e−rh (πuCu3 + πdCu2d)= e−0.08(1/3) (0.456 806× 34. 678 1 + 0.543 194× 12. 814) = 22. 201 7

Cud = e−rh (πuCu2d + πdCd2u)= e−0.08(1/3) (0.456 806× 12. 814 + 0.543 194× 0) = 5. 699 5

Cd2 = e−rh (πCd2u + πdCd3) = 0

Cu = e−rh (πuCu2 + πdCud)= e−0.08(1/3) (0.456 806× 22. 201 6 + 0.543 194× 5. 699 5) = 12. 889 5

Cd = e−rh (πuCud + πdCd2)= e−0.08(1/3) (0.456 806× 5. 699 5 + 0.543 194× 0) = 2. 535 1

C = e−rh (πuCu + πdCd)= e−0.08(1/3) (0.456 806× 12. 889 5 + 0.543 194× 2. 535 1) = 7. 073 9



Now I’m going to tell you a calculator shortcut I used when I was preparingfor the old Course 6. The above calculations are intense and prone to errors.To quickly and accurately calculate the premium at each node, use TI-30X IIScalculator because TI-30X IIS allows you to modify formulas easily.For example, to calculate Cu2 = e−rh (πuCu3 + πdCu2d), enter

eˆ(−0.08/3)(0.456 806× 34. 678 1 + 0.543 194× 12. 814Please note for TI-30 IIS, the above expression is the same aseˆ(−0.08/3)(0.456 806× 34. 678 1 + 0.543 194× 12. 814)

In other words, you can omit the ending parenthesis ")". I tell you thisbecause occasionally people emailed me saying they discovered a typo. This isnot a typo.

Now you have entered the formulaeˆ(−0.08/3)(0.456 806× 34. 678 1 + 0.543 194× 12. 814.Press "=" and you should get: 22.20164368

Next, to calculate Cud = e−rh (πudCu2d + πdCd2u), you don’t need to entera brand new formula. Just reuse the formula

eˆ(−0.08/3) (0.456 806× 34. 678 1 + 0.543 194× 12. 814)

Change 34. 678 1 to 12. 8140 (so 0.456 806×34. 678 1 becomes 0.456 806×12.8140).Change 12. 814 into 00.000 (so 0.543 194×12. 814 becomes 0.543 194×00.000).Now the modified formula iseˆ(−0.08/3) (0.456 806× 12. 8140 + 0.543 194× 00.000)Press "=" and you should get: 5.6994813To calculate Cu = e−rh (πuCu2 + πdCud), once again reuse a previous for-

mula. Change the formulaeˆ(−0.08/3) (0.456 806× 12. 8140 + 0.543 194× 00.000)

intoeˆ(−0.08/3) (0.456 806× 22. 201 6 + 0.543 194× 05. 699 5)Press "=" and you should get:12.8894166

Reusing formulas avoids the need to retype e−rh, πu, and πd (these threeterms are constant across all nodes) and increases your speed and accuracy. Byreusing formulas, you should quickly find C = 7. 073 9.



Next, we calculate the replicating portfolio.


Su − SdB = e−rh

SuCd − SdCu

Su − SdTo avoid errors, put the stock price tree and the premium tree side by side:

Period 0 1 2 3Su3 = 74.6781

Su2 = 61.1491Su = 50.0711 Su2d = 52.8140

S = 41 Sud = 43.2460Sd = 35.4114 Sd2u = 37.3513

Sd2 = 30.5846Sd3 = 26.4157

Period 0 1 2 3Cu3 = 34. 678 1

Cu2 = 22. 201 6Cu = 12. 889 5 Cu2d = 12. 814

C = 7. 073 9 Cud = 5. 699 5Cd = 2. 535 1 Cd2u = 0

Cd2 = 0Cd3 = 0

Period 0 1 2

(4, B)u2 = (1,−38. 947 4)(4, B)u = (0.921 8,−33. 263 6)

(4, B) = (0.706 3,−21. 885 2) (4, B)ud = (0.828 7,−30. 138 6)(4, B)d = (0.450 1,−13. 405 2)

(4, B)d2 = (0, 0)

4u2 = e−δhCu3 − Cu2d

Su3 − Su2d= e−0(1/3)

34. 678 1− 12. 81474.6781− 52.8140 = 1

Bu2 = e−rhSu3Cu2d − Su2dCu2d

Su3 − Su2d= e−0.08(1/3)

74.6781 (12. 814)− 52.8140 (34. 678 1)74.6781− 52.8140 =

−38. 947 4

4ud = e−δhCu2d − Cd2u

Su2d− Sd2u= e−0(1/3)

12. 814− 052.8140− 37.3513 = 0.828 7

Bud = e−rhSu2dCd2u − Sd2uCu2d

Su2d− Sd2u= e−0.08(1/3)

52.8140 (0)− 37.3513 (12. 814)52.8140− 37.3513 =

−30. 138 6

4d2 = e−δhCd2u − Cd3

Su2d− Sd3= 0



Bd2 = e−rhSu2dCd3 − Sd3Cd2u

Su2d− Sd3= 0

4u = e−δhCu2 − Cud

Su2 − Sud= e−0(1/3)

22. 201 6− 5. 699 561.1491− 43.2460 = 0.921 8

Bu = e−rhSu2Cud − SudCu2

Su2 − Sud= e−0.08(1/3)

61.1491 (5. 699 5)− 43.2460 (22. 201 6)61.1491− 43.2460 =

−33. 263 6

4d = e−δhCud − Cd2

Sud− Sd2= e−0(1/3)

5. 699 5− 043.2460− 30.5846 = 0.450 1

Bd = e−δhSudCd2 − Sd2Cud

Sud− Sd2= e−0.08(1/3)

43.2460 (0)− 30.5846 (5. 699 5)43.2460− 30.5846 =

−13. 405 2


Su− Sd= e−0(1/3)

12. 889 5− 2. 535 150.0711− 35.4114 = 0.706 3

B = e−rhCu − Cd

Su− Sd= e−0.08(1/3)

50.0711 (2. 535 1)− 35.4114 (12. 889 5)50.0711− 35.4114 =

−21. 885 2

If our goal is to calculate the premium without worrying about the replicatingportfolio, then we just need to know the risk neutral probability and the terminalpayoff.

Node Payoff Risk neutral prob of reaching this node4

u3 Cu3 = 34. 678 1 π3u = 0.456 8063

u2d Cu2d = 12. 814 3π2uπd = 3¡0.456 8062

¢0.543 194

ud2 Cd2u = 0 3πuπ2d = 3 (0.456 806) 0.543 194

2

d3 Cd3 = 0 π3d = 0.543 1943

The option premium is just the expected present value of the terminal payoffsusing the risk neutral probability.

C = e−rT¡Cu3π

3u + Cu2d × 3π2uπd + Cd2u × 3πuπ2d + Cd3π

3d

¢= e−0.08(1)

¡34. 678 1× 0.456 8063 + 12. 814× 3× 0.456 8062 × 0.543 194

¢= 7. 073 9

Example 10.2.9. Let’s reproduce Derivatives Markets Figure 10.6. Here isthe recap of the information on a European put. The current stock price is 41.The strike price K = 40. The annualized standard deviation of the continuouslycompounded stock return is σ = 30%. The continuously compounded risk-freerate per year is r = 8%. The continuously compounded dividend rate per yearis δ = 0%.The option expiration date is T = 1 year. Use a 3-period binomialtree to calculate the option premium.

4The probabilities of this column is just the four terms in (πu + πd)3 = π3u + 3πdπ

2u +

3π2dπu + π3d = 1.



Solution.

Each period is h =1

3year long.

u = e(r−δ)h+σ√h = e(0.08−0)1/3+0.3

√1/3 = 1. 221 246

d = e(r−δ)h−σ√h = e(0.08−0)1/3−0.3

√1/3 = 0.863 693


Su3 = 74.6781Su2 = 61.1491

Su = 50.0711 Su2d = 52.8140S = 41 Sud = 43.2460

Sd = 35.4114 Sd2u = 37.3513Sd2 = 30.5846

Sd3 = 26.4157

Calculate the premium by working backward from right to left.

Period 0 1 2 3Cu3 = max (0, 40− 74.6781) = 0

Cu2 = 0Cu = 0.740 9 Cu2d = max (0, 40− 52.8140) = 0

C = 2. 998 5 Cud = 1. 400 9Cd = 5. 046 2 Cd2u = max (0, 40− 37.3513) = 2. 648 7

Cd2 = 8. 362 9Cd3 = max (0, 40− 26.4157) = 13. 584 3



u− d=

e(0.08−0)1/3 − 0.863 6931. 221 246 − 0.863 693 = 0.456 806

πd = 1− πu = 1− 0.456 806 = 0.543 194

Cu2 = e−rh (πuCu3 + πdCu2d) = e−0.08(1/3) (0.456 806× 0 + 0.543 194× 0) =0

Cud = e−rh (πuCu2d + πdCd2u) = e−0.08(1/3) (0.456 806× 0 + 0.543 194× 2. 648 7) =1. 400 9

Cd2 = e−rh (πuCd2u + πdCd3) = e−0.08(1/3) (0.456 806× 2. 648 7 + 0.543 194× 13. 584 3) =8. 362 9

Cu = e−rh (πuCu2 + πdCud) = e−0.08(1/3) (0.456 806× 0 + 0.543 194× 1. 400 9) =0.740 9

Cd = e−rh (πuCud + πdCd2) = e−0.08(1/3) (0.456 806× 1. 400 9 + 0.543 194× 8. 362 9) =5. 046 2



C = e−rh (πuCu + πdCd) = e−0.08(1/3) (0.456 806× 0.740 9 + 0.543 194× 5. 046 2) =2. 998 5

Next, we calculate the replicating portfolio.


Su − SdB = e−rh

SuCd − SdCu

Su − SdPeriod 0 1 2

(4, B)u2 = (0, 0)(4, B)u = (−0.07 82, 4. 658 9)

(4, B) = (−0.293 7, 15. 039 5) (4, B)ud = (−0.171 3, 8. 808 8)(4, B)d = (−0.549 9, 24. 517 3)

(4, B)d2 = (1, 38. 947 4)

4u2 = e−δhCu3 − Cu2d

Su3 − Su2d= e−0(1/3)

0− 074.6781− 52.8140 = 0

Bu2 = e−rhSu3Cu2d − Su2dCu2d

Su3 − Su2d= e−0.08(1/3)

74.6781 (0)− 52.8140 (0)74.6781− 52.8140 = 0

4ud = e−δhCu2d − Cd2u

Su2d− Sd2u= e−0(1/3)

0− 2. 648 752.8140− 37.3513 = −0.171 3

Bud = e−rhSu2dCd2u − Sd2uCu2d

Su2d− Sd2u= e−0.08(1/3)

52.8140 (2. 648 7)− 37.3513 (0)52.8140− 37.3513 =

8. 808 8

4d2 = e−δhCd2u − Cd3

Su2d− Sd3= e−0(1/3)

2. 648 7− 13. 584 337.3513− 26.4157 = −1.0

Bd2 = e−rhSu2dCd3 − Sd3Cd2u

Su2d− Sd3= e−0.08(1/3)

37.3513 (13. 584 3)− 26.4157 (2. 648 7)37.3513− 26.4157 =

38. 947 4

4u = e−δhCu2 − Cud

Su2 − Sud= e−0(1/3)

0− 1. 400 961.1491− 43.2460 = −0.07 82

Bu = e−rhSu2Cud − SudCu2

Su2 − Sud= e−0.08(1/3)

61.1491 (1. 400 9)− 43.2460 (0)61.1491− 43.2460 =

4. 659

4d = e−δhCud − Cd2

Sud− Sd2= e−0(1/3)

1. 400 9− 8. 362 943.2460− 30.5846 = −0.549 9

Bd = e−δhSudCd2 − Sd2Cud

Sud− Sd2= e−0.08(1/3)

43.2460 (8. 362 9)− 30.5846 (1. 400 9)43.2460− 30.5846 =

24. 517 3


Su− Sd= e−0(1/3)

0.740 9− 5. 046 250.0711− 35.4114 = −0.293 7

B = e−rhCu − Cd

Su− Sd= e−0.08(1/3)

50.0711 (5. 046 2)− 35.4114 (0.740 9)50.0711− 35.4114 = 15.

039 5



If our goal is to calculate the premium without worrying about the replicatingportfolio, then we just need to know the risk neutral probability and the terminalpayoff.

Node Payoff Risk neutral prob of reaching this node5

u3 Cu3 = 0 π3u = 0.456 8063

u2d Cu2d = 0 3π2uπd = 3¡0.456 8062

¢0.543 194

ud2 Cd2u = 2. 648 7 3πuπ2d = 3 (0.456 806) 0.543 194

2

d3 Cd3 = 13. 584 3 π3d = 0.543 1943

The option premium is just the expected present value of the terminal payoffsusing the risk neutral probability.

C = e−rT¡Cd2u × 3πuπ2d + Cd3π

3d

¢= e−0.08(1)

¡2. 648 7× 3× 0.456 806× 0.543 1942 + 13. 584 3× 0.543 1943

¢=

2. 998 5

Example 10.2.10. Let’s reproduce Derivatives Markets Figure 10.7. Here isthe recap of the information on an American put. The current stock price is 41.The strike price K = 40. The annualized standard deviation of the continuouslycompounded stock return is σ = 30%. The continuously compounded risk-freerate per year is r = 8%. The continuously compounded dividend rate per yearis δ = 0%.The option expiration date is T = 1 year. Use a 3-period binomialtree to calculate the option premium.

Solution.

Each period is h =1

3year long.

u = e(r−δ)h+σ√h = e(0.08−0)1/3+0.3

√1/3 = 1. 221 246

d = e(r−δ)h−σ√h = e(0.08−0)1/3−0.3

√1/3 = 0.863 693

Step 1 Find the stock price treePeriod 0 1 2 3

Su3 = 74.6781Su2 = 61.1491

Su = 50.0711 Su2d = 52.8140S = 41 Sud = 43.2460

Sd = 35.4114 Sd2u = 37.3513Sd2 = 30.5846

Sd3 = 26.4157

5The probabilities of this column is just the four terms in (πu + πd)3 = π3u + 3πdπ

2u +

3π2dπu + π3d = 1.



Step 2 Calculate the payoff at expiration (terminal payoff)

Period 0 1 2 3Vu3 = max (0, 40− 74.6781) = 0

Vu2

Vu Vu2d = max (0, 40− 52.8140) = 0V Vud

Vd Vd2u = max (0, 40− 37.3513) = 2. 648 7Vd2

Vd3 = max (0, 40− 26.4157) = 13. 584 3

Step 3 Calculate the value of the American put one step left to theexpiration.An American put can be exercised immediately. The value of an American

option if exercised immediately is called the exercise value (EV ) or intrinsicvalue. At Period 2, we compare the value calculated using backwardization andthe exercise value. We take the greater of the two as the value of the Americanput.

The backwardized values at Period 2 are:Cu2 = e−rh (πuCu3 + πdCu2d) = e−0.08(1/3) (0.456 806× 0 + 0.543 194× 0) =

0Cud = e−rh (πuCu2d + πdCd2u) = e−0.08(1/3) (0.456 806× 0 + 0.543 194× 2. 648 7) =

1. 400 9Cd2 = e−rh (πuCd2u + πdCd3) = e−0.08(1/3) (0.456 806× 2. 648 7 + 0.543 194× 13. 584 3) =

8. 362 9

The exercise values at Period 2 are:

EVu2 = max¡0,K − Su2

¢= max (0, 40− 61.1491) = 0

EVud = max (0,K − Sud) = max (0, 40− 43.2460) = 0EVd2 = max

¡0,K − Sd2

¢= max (0, 40− 30.5846) = 9. 415 4

We take the greater of the two as the value of the American put.Vu2 = max (Cu2 , EVu2) = max (0, 0) = 0Vud = max (Cud, EVud) = max (1. 400 9, 0) = 1. 400 9Vd2 = max (Cd2 , EVd2) = max (8. 362 9, 9. 415 4) = 9. 415 4

Now we have:Period 0 1 2 3

Vu3 = 0Vu2 = 0

Vu Vu2d = 0V Vud = 1. 400 9

Vd Vd2u = 2. 648 7Vd2 = 9. 415 4

Vd3 = 13. 584 3



Step 3 Move one step to the left. Repeat Step 2. Compare the back-wardized value and the exercise value. Choose the greater.The backwardized values are:Cu = e−0.08(1/3) (0.456 806× 0 + 0.543 194× 1. 400 9) = 0.740 9

Cu = e−0.08(1/3) (0.456 806× 1. 400 9 + 0.543 194× 9. 415 4) = 5. 602 9The exercise value at Period 1 is:

EVu = max (0,K − Su) = max (0, 40− 50.0711) = 0EVd = max (0,K − Sd) = max (0, 40− 35.4114) = 4. 588 6

Vu = max (Cu, EVu) = max (0.740 9, 0) = 0.740 9Vud = max (Cd, EVd) = max (5. 602 9, 4. 588 6) = 5. 602 9Hence we have:

Period 0 1 2 3Vu3 = 0

Vu2 = 0Vu = 0.740 9 Vu2d = 0

V Vud = 1. 400 9Vd = 5. 602 9 Vd2u = 2. 648 7

Vd2 = 9. 415 4Vd3 = 13. 584 3

Step 4 Repeat Step 3. Move one step left. Compare the backwardizedvalue and the exercise value. Choose the greater value.The backwardized value at t = 0 is:C = e−0.08(1/3) (0.456 806× 0.740 9 + 0.543 194× 5. 602 9) = 3. 292 9The exercise value is:EV = max (0,K − S) = max (0, 40− 41) = 0Hence the premium for the American put is:V = max (C,EV )= max (3. 292 9, 0) = 3. 292 9

Now we have:

Period 0 1 2 3Vu3 = 0

Vu2 = 0Vu = 0.740 9 Vu2d = 0

V = 3. 292 9 Vud = 1. 400 9Vd = 5. 602 9 Vd2u = 2. 648 7

Vd2 = 9. 415 4Vd3 = 13. 584 3

Step 5 Calculate the replicating portfolio at each node.




Su − SdB = e−rh

SuCd − SdCu


(4, B)u2 = (0, 0)(4, B)u = (−0.07 82, 4. 658 9)

(4, B) = (−0.331 66, 16. 890 9) (4, B)ud = (−0.171 3, 8. 808 8)(4, B)d = (−0.632 99, 28. 017 8)

(4, B)d2 = (1, 38. 947 4)

4u2 = e−δhVu3 − Vu2dSu3 − Su2d

= e−0(1/3)0− 0

74.6781− 52.8140 = 0

Bu2 = e−rhSu3Vu2d − Su2dVu2d

Su3 − Su2d= e−0.08(1/3)

74.6781 (0)− 52.8140 (0)74.6781− 52.8140 = 0

4ud = e−δhVu2d − Vd2uSu2d− Sd2u

= e−0(1/3)0− 2. 648 7

52.8140− 37.3513 = −0.171 3

Bud = e−rhSu2dVd2u − Sd2Vu2d

Su2d− Sd2u= e−0.08(1/3)

52.8140 (2. 648 7)− 37.3513 (0)52.8140− 37.3513 =

8. 808 8

4d2 = e−δhVd2u − Vd3

Su2d− Sd3= e−0(1/3)

2. 648 7− 13. 584 337.3513− 26.4157 = −1.0

Bd2 = e−rhSu2dVd3 − Sd3Vd2u

Su2d− Sd3= e−0.08(1/3)

37.3513 (13. 584 3)− 26.4157 (2. 648 7)37.3513− 26.4157 =

38. 947 4

4u = e−δhVu2 − VudSu2 − Sud

= e−0(1/3)0− 1. 400 9

61.1491− 43.2460 = −0.07 82

Bu = e−rhSu2Vud − SudVu2

Su2 − Sud= e−0.08(1/3)

61.1491 (1. 400 9)− 43.2460 (0)61.1491− 43.2460 =

4. 659

4d = e−δhVud − Vd2

Sud− Sd2= e−0(1/3)

1. 400 9− 9. 415 443.2460− 30.5846 = −0.632 99

Bd = e−δhSudVd2 − Sd2Vud

Sud− Sd2= e−0.08(1/3)

43.2460 (9. 415 4)− 30.5846 (1. 400 9)43.2460− 30.5846 =

28. 017 8

4 = e−δhVu − VdSu− Sd

= e−0(1/3)0.740 9− 5. 602 950.0711− 35.4114 = −0.331 66

B = e−rhVu − VdSu− Sd

= e−0.08(1/3)50.0711 (5. 602 9)− 35.4114 (0.740 9)

50.0711− 35.4114 = 16.

890 9

Please note that for an American option, you can’t use the following ap-proach to find the option price:



Node Payoff Risk neutral prob of reaching this nodeu3 Vu3 = 0 π3u = 0.456 806

3

u2d Vu2d = 0 3π2uπd = 3¡0.456 8062

¢0.543 194

ud2 Vd2u = 2. 648 7 3πuπ2d = 3 (0.456 806) 0.543 194

2

d3 Vd3 = 13. 584 3 π3d = 0.543 1943

V = e−rT¡Cd2u × 3πuπ2d + Cd3π

3d

¢= e−0.08(1)

¡2. 648 7× 3× 0.456 806× 0.543 1942 + 13. 584 3× 0.543 1943

¢=

2. 998 5

This approach is wrong because it ignores the possibility that the Americanoption can be exercised early.

10.2.2 Options on stock index

The price of an option on stock index can be calculated the same way as theprice of an option on a stock is calculated.

Example 10.2.11. Let’s reproduce Derivatives Markets Figure 10.8. Here isthe recap of the information on an American call. The current stock indexis 110. The strike price K = 100. The annualized standard deviation of thecontinuously compounded stock index return is σ = 30%. The continuouslycompounded risk-free rate per year is r = 5%. The continuously compoundeddividend rate per year is δ = 3.5%.The option expiration date is T = 1 year.Use a 3-period binomial tree to calculate the option premium.

Solution.

Each period is h =1

3year long.

u = e(r−δ)h+σ√h = e(0.05−0.035)1/3+0.3

√1/3 = 1. 195 07

d = e(r−δ)h−σ√h = e(0.05−0.035)1/3−0.3

√1/3 = 0.845 18


u− d=

e(0.05−0.035)1/3 − 0.845 181. 195 07 − 0.845 18 = 0.456 807

πd = 1− πu = 1− 0.456 807 = 0.543 193Step 1 Find the stock price treePeriod 0 1 2 3

Su3 = 187.7471Su2 = 157.1013

Su = 131.4577 Su2d = 132.7789S = 110 Sud = 111.1055

Sd = 92.9699 Sd2u = 93.9042Sd2 = 78.5763

Sd3 = 66.4112




Period 0 1 2 3Vu3 = max (0, 187.7471− 100) = 87. 747 1

Vu2

Vu Vu2d = max (0, 132.7789− 100) = 32. 778 9V Vud

Vd Vd2u = max (0, 93.9042− 100) = 0Vd2

Vd3 = max (0, 66.4112− 100) = 0

Step 3 Calculate the value of the American call at Period 2 by takingthe greater of the backwardized value and the exercise value at each node.

The backwardized values at Period 2 are:Cu2 = e−rh (πuVu3 + πdVu2d) = e−0.05(1/3) (0.456 807× 87. 747 1 + 0.543 193× 32. 778 9) =

56. 931 9Cud = e−rh (πuVu2d + πdVd2u) = e−0.05(1/3) (0.456 807× 32. 778 9 + 0.543 193× 0) =

14. 726 1Cd2 = e−rh (πudVd2u + πdVd3) = e−0.05(1/3) (0.456 807× 0 + 0.543 193× 0) =

0


EVu2 = max¡0, Su2 −K

¢= max (0, 157.1013− 100) = 57. 101 3

EVud = max (0, Sud−K) = max (0, 111.1055 − 100) = 11. 105 5EVd2 = max

¡0, Sd2 −K

¢= max (0, 78.5763 − 100) = 0

We take the greater of the two as the values.Vu2 = max (Cu2 , EVu2) = max (56. 931 9, 57. 101 3) = 57. 101 3Vud = max (Cud, EVud) = max (14. 726 1, 11. 105 5) = 14. 726 1Vd2 = max (Cd2 , EVd2) = max (0, 0) = 0


Vu3 = 87. 747 1Vu2 = 57. 101 3

Vu Vu2d = 32. 778 9V Vud = 14. 726 1

Vd Vd2u = 0Vd2 = 0

Vd3 = 0


Cu = e−rh (πuVu2 + πdVud) = e−0.05(1/3) (0.456 807× 57. 101 3 + 0.543 193× 14. 726 1) =33. 520 02



Cd = e−rh (πuVud + πdVd2) = e−0.05(1/3) (0.456 807× 14. 726 1 + 0.543 193× 0) =6. 615 8

EVu = max (0, Su −K) = max (0, 131.4577 − 100) = 31. 457 7EVd = max (0, Sd −K) = max (0, 92.9699− 100) = 0Vu = max (Cu, EVu) = max (33. 520 02, 31. 457 7) = 33. 520 02Vd = max (Cd, EVd) = max (6. 615 8, 0) = 6. 615 8


Vu3 = 87. 747 1Vu2 = 57. 101 3

Vu = 33. 520 02 Vu2d = 32. 778 9V Vud = 14. 726 1

Vd = 6. 615 8 Vd2u = 0Vd2 = 0

Vd3 = 0


C = e−rh (πuVu + πdVd) = e−0.05(1/3) (0.456 807× 33. 520 02 + 0.543 193× 6. 615 8) =18. 593 35

EV = max (0, S −K) = max (0, 110 − 100) = 10V = max (C,EV ) = max (18. 593 35, 10) = 18. 593 35

Step 6 Calculate the replicating portfolioOur goal is to replicate the following values:

Period 0 1 2 3Vu3 = 87. 747 1

Vu2 = 57. 101 3Vu = 33. 520 02 Vu2d = 32. 778 9

V = 18. 593 35 Vud = 14. 726 1Vd = 6. 615 8 Vd2u = 0

Vd2 = 0Vd3 = 0


Su − SdB = e−rh

SuCd − SdCu


4u2 = 0.988 4014u = 0.910 598

4 = 0.690 923 4ud = 0.988 4014d = 0.447 45

4d2 = 0



Period 0 1 2

Bu2 = −98. 347 1Bu = −86. 185 1

B = −57. 408 1 Bud = −77. 870 8Bd = −34. 983 9

Bd2 = 0

4u2 = e−δhVu3 − Vu2dSu3 − Su2d

= e−0.035(1/3)87. 747 1− 32. 778 9187.7471− 132.7789 = 0.988 401

Bu2 = e−rhSu3Vu2d − Su2dVu3

Su3 − Su2d= e−0.05(1/3)

187.7471 (32. 778 9)− 132.7789 (87. 747 1)187.7471− 132.7789 =

−98. 347 1

4ud = e−δhVu2d − Vud2

Su2d− Sud2= e−0.035(1/3)

32. 778 9− 0187.7471− 132.7789 = 0.988 401

Bud = e−rhSu3dVu2d − Su2dVu3

Su3 − Su2d= e−0.05(1/3)

187.7471 (0)− 132.7789 (32. 778 9)187.7471− 132.7789 =

−77. 870 8

4d2 = e−δhVd2u − Vd3

Sd2u− Sd3= 0

Bd2 = e−rhSd2uVd3 − Sd3Vd2u

Sd2u− Sd3= 0

4u = e−δhVu2 − VudSu2 − Sud

= e−0.035(1/3)57. 101 3− 14. 726 1157.1013− 111.1055 = 0.910 598

Bu = e−rhSu2Vud − SudVu2

Su2 − Sud= e−0.05(1/3)

157.1013 (14. 726 1)− 111.1055 (57. 101 3)157.1013− 111.1055 =

−86. 185 1

4d = e−δhVud − Vd2

Sud− Sd2= e−0.035(1/3)

14. 726 1− 0111.1055− 78.5763 = 0.447 45

Bd = e−rhSu2Vud − SudVu2

Su2 − Sud= e−0.05(1/3)

111.1055 (0)− 78.5763 (14. 726 1)111.1055− 78.5763 =

−34. 983 9

4 = e−δhVu − VdSu− Sd

= e−0.035(1/3)33. 520 02− 6. 615 8131.4577− 92.9699 = 0.690 923

B = e−rhSuVd − SdVuSu− Sd

= e−0.05(1/3)131.4577 (6. 615 8)− 92.9699 (33. 520 02)

131.4577− 92.9699 =

−57. 408 1

10.2.3 Options on currency

Now let’s find the price of a European call option on €1. The underlying asset is€1. The option expires in h years. The current dollar value of the underlying is



S (so €1 = $S at t = 0). In h years the dollar value of the underlying asset €1can go up to $Su or go down to $Sd. The strike price is $K. The continuouslycompounded risk-free interest rate on dollars is r per year (so dollars earn areturn r). The continuously compounded interest rate on the underlying assetof €1 is δ per year (so euros earn a return δ).Let’s calculate the call price using one-period binomial tree. The asset price

tree and the option payoff tree are as follows:Asset price tree (in dollars) Option payoff (in dollars)

time 0 hSu

SSd

time 0 hCu = max (0, Su −K)

CCd = max (0, Sd −K)

To replicate the payoff, at t = 0 we’ll buy ∆ units of the underlying (i.e.buy ∆ euros) and simultaneously invest B dollars in a savings account. Sincethe underlying asset €∆ earns interest at a continuous interest δ, it will growinto €∆eδh at T , which is worth $∆eδh Su in the up state and $∆eδh Sd in thedown state.

We want our replicating portfolio to match the option payoff. So we have:∆eδh Su Berh Cu

4S + B = C∆eδh Sd Berh Cd

t = 0 t = h t = 0 t = h t = 0 t = h½∆eδh Su +Berh = Cu

∆eδh Sd +Berh = Cd



Su − Sd(10.26)


Su − Sd(10.27)

Equation 10.26 and 10.27 are exactly the same as Equation 10.13 and 10.14.This tells us that if we treat the currency as a stock and treat the euro returnδ as the stock’s dividend rate, we can find the currency option’s price and thereplicating portfolios using all the formulas available for a stock option.

Example 10.2.12. Reproduce Derivatives Markets Figure 10.9. Here is the re-cap of the information on an American put option on €1. The current exchangerate is S = $1.05/€. The strike price is K = $1.10. The annualized standarddeviation of the continuously compounded return on dollars is σ = 10%. Thecontinuously compounded risk-free rate on dollars is r = 5% per year. Thecontinuously compounded return on euros is δ = 3.1% per year. The option



expiration date is T = 0.5 year. Use a 3-period binomial tree to calculate theoption premium.

Solution.

The length period is h =T

3=1

6

u = e(r−δ)h+σ√h = e(0.055−0.031)1/6+0.1

√1/6 = 1. 045 845

d = e(r−δ)h−σ√h = e(0.055−0.031)1/6−0.1

√1/6 = 0.963 845


u− d=

e(0.055−0.031)1/6 − 0.963 8451. 045 845 − 0.963 845 = 0.489 795

πd = 1− πu = 1− 0.489 795 = 0.510 205

Step 1 Find the underlying asset price treePeriod 0 1 2 3

Su3 = 1.2011Su2 = 1.1485

Su = 1.0981 Su2d = 1.1070S = 1.05 Sud = 1.0584

Sd = 1.0120 Sd2u = 1.0202Sd2 = 0.9754

Sd3 = 0.9402


Period 0 1 2 3Vu3 = max (0, 1.10− 1.2011) = 0

Vu2

Vu Vu2d = max (0, 1.10− 1.1070) = 0V Vud

Vd Vd2u = max (0, 1.10− 1.0202) = 0.079 8Vd2

Vd3 = max (0, 1.10− 0.9402 ) = 0.159 8

Step 3 Calculate the value of the American put at Period 2 by takingthe greater of the backwardized value and the exercise value at each node.

The backwardized values at Period 2 are:Cu2 = e−rh (πuVu3 + πdVu2d) = e−0.055(1/6) (0.489 795× 0 + 0.510 205 × 0) =

0Cud = e−rh (πuVu2d + πdVd2u) = e−0.055(1/6) (0.489 795× 0 + 0.510 205 × 0.079 8) =

0.04 034Cd2 = e−rh (πudVd2u + πdVd3) = e−0.055(1/6) (0.489 795× 0.079 8 + 0.510 205 × 0.159 8) =

0.119 5




EVu2 = max¡0,K − Su2

¢= max (0, 1.10− 1.1485) = 0

EVud = max (0,K − Sud) = max (0, 1.10− 1.0584) = 0.041 6EVd2 = max

¡0,K − Sd2

¢= max (0, 1.10− 0.9754) = 0.124 6

We take the greater of the two as the values.Vu2 = max (Cu2 , EVu2) = max (0, 0) = 0Vud = max (Cud, EVud) = max (0.04 034 , 0.041 6) = 0.041 6Vd2 = max (Cd2 , EVd2) = max (0.119 5, 0.124 6) = 0.124 6


Vu3 = 0Vu2 = 0

Vu Vu2d = 0V Vud = 0.041 6

Vd Vd2u = 0.079 8Vd2 = 0.124 6

Vd3 = 0.159 8


Cu = e−rh (πuVu2 + πdVud) = e−0.055(1/6) (0.489 795× 0 + 0.510 205 × 0.041 6) =0.02 10

Cd = e−rh (πuVud + πdVd2) = e−0.055(1/6) (0.489 795× 0.041 6 + 0.510 205 × 0.124 6) =0.08 31

EVu = max (0,K − Su) = max (0, 1.10− 1.0981) = 0.001 9EVd = max (0,K − Sd) = max (0, 1.10− 1.0120 ) = 0.088Vu = max (Cu, EVu) = max (0.02 10 , 0.001 9) = 0.021Vd = max (Cd, EVd) = max (0.088 , 0.021) = 0.088


Vu3 = 0Vu2 = 0

Vu = 0.021 Vu2d = 0V Vud = 0.041 6

Vd = 0.088 Vd2u = 0.079 8Vd2 = 0.124 6

Vd3 = 0.159 8




C = e−rh (πuVu + πdVd) = e−0.055(1/6) (0.489 795× 0.021 + 0.510 205 × 0.088) =0.05 47

EV = max (0,K − S ) = max (0, 1.10 − 1.05) = 0.05V = max (C,EV ) = max (0.05 47, 0.05) = 0.054 7


Vu3 = 0Vu2 = 0

Vu = 0.021 Vu2d = 0V = 0.054 7 Vud = 0.041 6

Vd = 0.088 Vd2u = 0.079 8Vd2 = 0.124 6

Vd3 = 0.159 8

You can verify that the replicating portfolios are as follows:Period 0 1 2

4u2 = 0.00004u = −0.4592

4 = −0.7736 4ud = −0.91514d = −0.9948

4d2 = −0.9948

Period 0 1 2

Bu2 = $0.0000Bu = $0.5253

B = $0.8669 Bud = $1.0089Bd = $1.0900

Bd2 = $1.0900

10.2.4 Options on futures contracts

Suppose we want to find the price of a European call option on a stock futurescontract. The underlying asset is futures. The option expires in h years. Thecurrent price of the underlying asset is F0,h = F , where F0,h is the price of afutures contract signed at t = 0 and expiring on date h.In h years the futuresprice can go up to Fh,h = Fu = Fu or go down to Fh,h = Fd = Fd, where Fh,his the price of a futures contract signed at t = h and expiring on date h years(i.e. expiring immediately). Fh,h is equal to the spot price Sh. The fact thatFh,h can be either Fu or Fd is the same as fact that the stock price at t = h is



either Fu or Fd. The continuously compounded risk-free interest rate is r peryear. Stocks pay dividend at a continuously compounded rate of δ per year.The strike price is K.The find the European call price on the futures, we draw the price tree of

the underlying asset and the payoff tree.Asset price tree Option payoff

time 0 hFu = Fu

FFd = Fd

time 0 hCu = max (0, Fu −K)

CCd = max (0, Fd −K)

We form a replicating portfolio at t = 0 by entering ∆ futures contracts asa buyer and simultaneously putting $B in the savings account. Assume thatno margin account is needed before one enters a futures contract. Then thecost of entering a futures contracts is zero. At the contract expiration date h,the ∆ futures contracts are settled in cash. If the futures price at expiration isFu > F , then the seller in the futures pays ∆ (Fu − F ) to us, the buyer.If on the other hand, the futures price at h is Fd < F , then we pay the seller

∆ (F − Fd). Paying ∆ (F − Fd) is the same as receiving ∆ (Fd − F ).We assume that Fd < F < Fu holds so there’s no arbitrage.

So the cash flow of the underlying asset (futures) is:time 0 h

$∆ (Fu − F )$0

$∆ (Fd − F )

We want the replicating portfolio and the option have the same payoff.∆ (Fu − F ) Berh Cu

0 + B = C∆ (Fd − F ) Berh Cd

t = 0 t = h t = 0 t = h t = 0 t = h½∆ (Fu − F ) +Berh = Cu

∆ (Fd − F ) +Berh = Cd


4 =Cu − Cd

Fu − Fd=

Cu − Cd

F (u− d)(10.28)

B = e−rhµCu1− d

u− d+ Cd

u− 1u− d

¶(10.29)

V = 4× 0 +B = B = e−rhµCu1− d

u− d+ Cd

u− 1u− d

¶(10.30)



Define

πu =F − FdFu − Fd

=1− d

u− d(10.31)

πd =Fu − 1Fu − Fd

=u− 1u− d

(10.32)

ThenV = e−rh (Cuπu + Cdπd) (10.33)

Equation 10.30 is the same as Equation 10.16 if we set r− δ = 0 and S = F .Consequently, we can find the price of a futures option by using Equation 10.16

if we set δ = r. We just need to remember that 4 =Cu − Cd

Fu − Fdinstead of

4 = e−rδCu − Cd

Fu − Fdand that B = V = e−rh

µCu1− d

u− d+ Cd

u− 1u− d

¶instead of


Su − Sd.

How we can specify u or d?

u =up price of Fh,h

F0,hu =

down price of Fh,hF0,h

We know that Fh,h = Sh =

(S0e

(r−δ)h+σ√h

S0e(r−δ)h−σ

√h. In addition, F0,h = S0e

(r−δ)h.

u =up price of Fh,h

F0,h=

S0e(r−δ)h+σ

√h

S0e(r−δ)h= eσ

√h

d =down price of Fh,h

F0,h=

S0e(r−δ)h−σ

√h

S0e(r−δ)h= e−σ

√h

u = eσ√h (10.34)

d = e−σ√h (10.35)

Equation 10.34 and 10.35 are the same as Equation 10.24 and 10.25 if weset δ = r. We can use the stock option’s formula on u and d for futures options.

Tip 10.2.6. To find the price of a futures option, just use the price formula fora stock option and set δ = r. However, remember that for a futures option, 4 =Cu − Cd

Fu − Fdinstead of4 = e−rh

Cu − Cd

Fu − Fdand B = V = e−rh

µCu1− d

u− d+ Cd

u− 1u− d

¶instead of B = e−rh

SuCd − SdCu

Su − Sd.



Example 10.2.13. Let’s reproduce Derivatives Markets Figure 10.10. Here isthe recap of the information on an American call on a futures contract. Thecurrent futures price is S = 300. The strike price K = 300. The annualizedstandard deviation of the continuously compounded stock index return is σ =10%. The continuously compounded risk-free rate per year is r = 5%.The optionexpiration date is T = 1 year. Use a 3-period binomial tree to calculate theoption premium.

Solution.

We’ll reuse the stock option formula and set δ = r.

Each period is h =1

3year long.

u = e(r−δ)h+σ√h = eσ

√h = e0.1

√1/3 = 1. 059 434

d = e(r−δ)h−σ√h = e−σ

√h = e−0.1

√1/3 = 0.943 900


u− d=1− d

u− d=

1− 0.943 9001. 059 434 − 0.943 900 = 0.485 57

πd = 1− πu = 1− 0.485 57 = 0.514 43

Step 1 Find the underlying asset price treePeriod 0 1 2 3

Su3 = 356.7330Su2 = 336.7203

Su = 317.8303 Su2d = 317.8303F = 300 Sud = 300.0000

Sd = 283.1700 Sd2u = 283.1700Sd2 = 267.2842

Sd3 = 252.2895


Period 0 1 2 3Vu3 = max (0, 356.7330− 300) = 56. 733

Vu2

Vu Vu2d = max (0, 317.8303− 300) = 17. 830 3V Vud

Vd Vd2u = max (0, 283.1700− 300) = 0Vd2

Vd3 = max (0, 252.2895− 300) = 0


The backwardized values at Period 2 are:



Cu2 = e−rh (πuVu3 + πdVu2d) = e−0.05(1/3) (0.485 57× 56. 733 + 0.514 43× 17. 830 3) =36. 113 3

Cud = e−rh (πuVu2d + πdVd2u) = e−0.05(1/3) (0.485 57× 17. 830 3 + 0.514 43× 0) =8. 514 7

Cd2 = e−rh (πudVd2u + πdVd3) = e−0.05(1/3) (0.485 57× 0 + 0.514 43× 0) =0


EVu2 = max¡0, Su2 −K

¢= max (0, 336.7203− 300) = 36. 720 3

EVud = max (0, Sud−K) = max (0, 300.0000 − 300) = 0EVd2 = max

¡0, Sd2 −K

¢= max (0, 267.2842 − 300) = 0

We take the greater of the two as the values.Vu2 = max (Cu2 , EVu2) = max (36. 113 3, 36. 720 3) = 36. 720 3

Vud = max (Cud, EVud) = max (8. 514 7, 0) = 8. 514 7

Vd2 = max (Cd2 , EVd2) = max (0, 0) = 0


Vu3 = 56. 733Vu2 = 36. 720 3

Vu Vu2d = 17. 830 3V Vud = 8. 514 7

Vd Vd2u = 0Vd2 = 0

Vd3 = 0


Cu = e−rh (πuVu2 + πdVud) = e−0.05(1/3) (0.485 57× 36. 720 3 + 0.514 43× 8. 514 7) =21. 843 4

Cd = e−rh (πuVud + πdVd2) = e−0.05(1/3) (0.485 57× 8. 514 7 + 0.514 43× 0) =4. 066 2

EVu = max (0, Su −K) = max (0, 317.8303 − 300) = 17. 830 3EVd = max (0, Sd −K) = max (0, 283.1700− 300) = 0Vu = max (Cu, EVu) = max (21. 843 4, 17. 830 3) = 21. 843 4

Vd = max (Cd, EVd) = max (4. 066 2, 0) = 4. 066 2

Now we have:



Period 0 1 2 3Vu3 = 56. 733

Vu2 = 36. 720 3Vu = 21. 843 4 Vu2d = 17. 830 3

V Vud = 8. 514 7Vd = 4. 066 2 Vd2u = 0

Vd2 = 0Vd3 = 0


C = e−rh (πuVu + πdVd) = e−0.05(1/3) (0.485 57× 21. 843 4 + 0.514 43× 4. 066 2) =12. 488 4

EV = max (0, S −K) = max (0, 300 − 300) = 0V = max (C,EV ) = max (12. 488 4, 0) = 12. 488 4Now we have:

Period 0 1 2 3Vu3 = 56. 733

Vu2 = 36. 720 3Vu = 21. 843 4 Vu2d = 17. 830 3

V = 12. 488 4 Vud = 8. 514 7Vd = 4. 066 2 Vd2u = 0

Vd2 = 0Vd3 = 0

Next, we need to find the replicating portfolio. Our goal is to replicate thefollowing values:

Period 0 1 2 3Vu3 = 56. 733

Vu2 = 36. 720 3Vu = 21. 843 4 Vu2d = 17. 830 3

V = 12. 488 4 Vud = 8. 514 7Vd = 4. 066 2 Vd2u = 0

Vd2 = 0Vd3 = 0

Using Equation 10.29 and 10.29, you should get:Period 0 1 2

4u2 = 14u = 0.7681

4 = 0.5129 4ud = 0.51444d = 0.2603

4d2 = 0



Period 0 1 2

Bu2 = $36. 720 3Bu = $21. 843 4

B = $12. 488 4 Bud = $8. 514 7Bd = $4. 066 2

Bd2 = $0

For example,

4u2 =Vu3 − Vu2dSu3 − Su2d

=56. 733− 17. 830 3356.7330− 317.8303 = 1.0

Bu2 = Vu2 = $36. 720 3

4 =Vu − VdSu − Sd

=21. 843 4− 4. 066 2317.8303− 283.1700 = 0.512 9

B = V = $12. 488 4I recommend that you reproduce my replicating portfolio in each node.

Options on commodities

The textbook is brief on this topic. So you don’t need to spend lot of time on it.This is the main idea: we can price commodity options using the same frameworkfor pricing stock options if we can build a replicating portfolio using commoditiesand bonds with zero transaction cost. In reality, it’s hard to build a replicatingportfolio using commodities. Unlike stocks, commodities such as corn may incurstorage cost or other cost. It may be impossible to short sell commodities.As such, our ability to build a replicating portfolio is limited. However, ifwe can build any replicating portfolios using commodities instantaneously andeffortlessly, commodity options and stock options are conceptually the same.We can use the same framework to calculate the price of a commodity optionand the price of a stock option.

Example 10.2.14. Here is the information on an American call on a com-modity. The current commodity price is 110. The strike price is K = 100.The annualized standard deviation of the continuously compounded return onthe commodity is σ = 30%. The continuously compounded risk-free rate peryear is r = 5%. The continuously compounded lease rate of the commodity isδ = 3.5% per year. The option expiration date is T = 1 year. Use a 3-periodbinomial tree to calculate the option premium. Assume that you can effortlesslyand instantaneously build any replicating portfolio using the commodity and thebond.

We just treat the commodity as a stock. The lease rate δ = 3.5% is the sameas the stock dividend rate. We can use the framework for pricing stock optionsto price this commodity option. The solution to this problem is in the textbookFigure 10.8.



Options on bonds

The textbook points out two major differences between bonds and stocks:

1. A bond’s volatility decreases over time as the bond approaches its matu-rity. A stock’s volatility doesn’t have this pattern.

2. When pricing a stock option, we assume that the interest rate is constantover time. The random variable is the stock price under a fixed interest.However, when pricing a bond, we can’t assume that the interest rate isconstant any more. If the interest rate is constant, then the bond’s priceis known with 100% certainty. If the bond price doesn’t change randomly,an option on the bond has zero value. Who wants to buy a call or put onan asset whose price is known with 100% certainty?

Because of these differences, options on bonds should be treated differentlyfrom options on stocks. This is all you need to know about bond options rightnow. Chapter 24 will cover more on options on bonds.


Chapter 11

Binomial option pricing: II

11.1 Understanding early exercise

Pros and cons of exercising an American call option earlyPro:

• Receive the stock and future dividends

Cons

• Pay the strike price early and lost interest that could have earned on thestrike price.

• Lose the insurance implicit in the call. If you hold the option, the stockprice might be below the strike price at expiration, in which case youwould not exercise the option. However, if you exercise the American callearly, you lose the privilege of not exercising it. To understand this point,suppose you go to a garage sale and find a book you like that sells foronly $1. You think "How cheap the book is. I must buy it." You pay $1and buy the book. You think you get a good deal. However, if you resistthe temptation to buy the book immediately and wait till the end of thegarage sale, the book’s price may drop to $0.25. Better yet, you may evenget the book for free. That same thing may happen when you exercisean American call early. At the moment, the stock price is high and youmight be attempted to exercise the call. However, if we wait for a while,the stock price may drop below the strike price.

Next, the textbook gives us a rule to determine when it’s optimal to exercisea perpetual 1American call early. For a perpetual American call with zerovolatility, it’s optimal to exercise a perpetual American call early if the dividendto be received exceeds the interest savings on the strike price:

1The textbook errata say the formulas work for an infinitely-lived American call option.

79


80 CHAPTER 11. BINOMIAL OPTION PRICING: II

δST > rK

The annual dividend you gain if you exercise the call early is ST¡eδ − 1

¢=

ST

µ1 + δ +

1

2δ2 + ...− 1

¶≈ δST for a small δ. The annual interest earned on

K is K (er − 1) = K

µ1 + r +

1

2r2 + ...− 1

¶≈ rK for a small r. If you early

exercise, you’ll pay K and receive ST ; during a year you’ll receive δST dividendbut you will lose rK interest. Hence early exercise is optimal if the annualdividend exceeds the annual interest, δST > rK.And it’s optimal to defer exercising a perpetual American call if the interest

savings on the strike price exceeds dividends lost:δST < rK

Example 11.1.1. A perpetual American call option on a stock has a strikeprice $50. The stock pays dividend at a continuously compounded rate of 8%per year. The continuously compounded risk-free interest rate is 6% per year.The volatility of the stock price is zero. When is it optimal to exercise thisAmerican call option early? When is it optimal to defer exercise?

Sδ > rK

S >rK

δ=0.06 (50)

0.08= 37. 5

Once the stock price becomes greater than 37. 5, then it’s optimal to exercisethis perpetual American call option early.If the stock price is less than 37. 5, then it’s optimal to defer exercising this

perpetual American call option early.Please note that zero volatility doesn’t mean that the stock price is a con-

stant. It means that the stock price is known in advance with 100% certainty.

Next, the textbook says that the decision to exercise a perpetual Americancall option early is complex if the volatility of the stock price is greater thanzero. In this case, the insurance in the call option is greater than zero. Foreach non-zero volatility, there’s a lowest stock price at which the early exerciseis optimal.

11.2 Understanding risk-neutral probability

Risk neutral probability is explained in the previous chapter. There’s not muchnew information about the risk neutral probability in this chapter. The keypoint to remember is that risk neutral probability is a shortcut or a math trickthat enables us to quickly find the price of an option. We can use risk neutralprobability to find the correct price of an option whether consumers are reallyrisk neutral or not.The risk neutral probability is similar to the moment generating function in

Exam P. The moment generating function (MGF ) is merely a math trick that


11.2. UNDERSTANDING RISK-NEUTRAL PROBABILITY 81

allows us to quickly find the mean and variance of a random variable (hencethe name "moment generating"). If we don’t use GMF , we can still find themean and variance, but we have to work a lot harder. With the help of GMF ,we can quickly find the mean and variance. Similarly, if we don’t use the riskneutral probability, we can still find the option price, but we have to work a lotharder. Once we use risk-neutral probability, we can quickly find an option’sprice. Risk neutral probability is merely a math risk.By the way, one investment consultant told me that risk neutral probability

is often hard to non-technical clients to understand. If you tell a non-technicalclient that an option price is calculated using risk-neutral probability and thatthe risk neutral probability not real, the client often immediately ask "So theprice you calculated is wrong then?" It may take the consultant a while toexplain why the risk neutral probability is not real yet the price is still correct.

11.2.1 Pricing an option using real probabilities

Next, the textbook answers a frequently asked question: Can we calculate theoption price as the expected payoff using real probabilities of the stock price?The answer is "Yes if you know the discount rate."Let p and q = 1−p represent the real probability of stock going up and down

respectively. Let γ represent the real discount rate (instead of the risk-free rate,which is used to discount payoff in the risk neutral world).

uS with real probability pS

dS with real probability qt = 0 t = h

Cu with real probability pC =?

Cd with real probability qt = 0 t = h

Then the price of a European option expiring in h years is:C = e−γh (pCu + qCd)

How can we find the real probability p and the real discount rate γ? Supposewe know that the expected return on the stock during [0, h] is α. Assume thatthe continuously compounded dividend rate is δ per year. If we have one stockat t = 0, then at t = h we’ll have eδh stocks. The value at t = 0 is the expectedvalue at t = h discounted at rate α.

uSeδh with real probability pS

dSeδh with real probability qValue t = 0 Value t = h

S = e−αh¡puSeδh + qdSeδh

¢→ p =

e(α−δ)h − d

u− d

p =e(α−δ)h − d

u− d(11.1)



q =u− e(α−δ)h

u− d(11.2)

Tip 11.2.1. If we set r = α Equation 10.16 and 10.17 become Equation 11.1and 11.2. So you just need to memorize Equation 10.16 and 10.17. To get theformulas for the real probability, just set r = α.

We can use the replicating portfolio to find γ. Suppose the replicating port-folio at t = 0 consists of ∆ shares of stock and putting $B in a savings account.We already know that we can calculate ∆ and B using Equation 10.13 and10.13.At t = 0, the replicating portfolio is worth ∆S+B. At t = h, the replicating

portfolio consists of ∆eδh shares of stock and $Berh in a savings account, whichis worth ∆eδhdS +Berh in the up state and ∆eδhdS −Berh in the down state.

uS∆eδh +Berh with real probability p∆S +B

dS∆eδh +Berh with real probability qValue t = 0 Value t = h

The value at t = 0 is the expected value at t = h discounted at rate γ:∆S +B =

£p¡uS∆eδh +Berh

¢+ q

¡dS∆eδh +Berh

¢¤e−γh

=£∆¡puSeδh + qdSeδh

¢+Berh (p+ q)

¤e−γh

=¡∆Seαh +Berh

¢e−γh

e−γh =∆S +B

∆Seαh +Berh(11.3)

C = e−γh (pCu + qCd) =∆S +B

∆Seαh +Berh(pCu + qCd)

Since C = ∆S +B, we just need to prove that pCu+ qCd = ∆Seαh+Berh.

pCu + qCd

=e(α−δ)h − d

u− dCu +

u− e(α−δ)h

u− dC

=

∙e(r−δ)h − d

u− dCu +

e(α−δ)h − e(r−δ)h

u− dCu

¸+

∙u− e(r−δ)h

u− dCd +

e(r−δ)h − e(α−δ)h

u− dCd

¸=

∙e(r−δ)h − d

u− dCu +

u− e(r−δ)h

u− dCd

¸+

e(α−δ)h − e(r−δ)h

u− d(Cu − Cd)

According to Equation 10.15:e(r−δ)h − d

u− dCu +

u− e(r−δ)h

u− dCd = erh (∆S +B)



According to Equation 10.13,Cu − Cd

u− d= ∆eδhS

pCu + qCd

= erh (∆S +B) +∆eδhS£e(α−δ)h − e(r−δ)h

¤= erh (∆S +B) +∆S

¡eαh − erh

¢= eαh∆S +Berh

→ C =∆S +B

∆Seαh +Berh(pCu + qCd) = ∆S +B

The above derivation tell us that

• Real probabilities lead to the same answer as the risk neutral probability

• Any consistent pair of (α, γ) will produce the correct answer. The abovederivation doesn’t require that α has to be reasonable or precise. Any αwill generate a corresponding γ. Together, α and γ will produce the optionprice correctly.

• The simplest calculation is to set α = γ = r. Setting α = γ = r meansusing risk neutral probabilities.

Example 11.2.1. Reproduce the textbook Figure 11.3 (which is the same as thetextbook Figure 10.5). A European call option has strike price K = 40. Thecurrent price is S = 41. The annualized standard deviation of the continuouslycompounded stock return is σ = 30%. The continuously compounded risk-freerate per year is r = 8%. The continuously compounded dividend rate per yearis δ = 0%. The continuously compounded expected return on the stock per yearis α = 15%. The option expiration date is T = 1 year. Use a 1-period binomialtree and real probabilities to calculate the option premium.

Solution.

u = e(r−δ)h+σ√h = e(0.08−0)1+0.3

√1 = 1. 462 3

u = e(r−δ)h−σ√h = e(0.08−0)1−0.3

√1 = 0.802 5

1. 462 3 (41) = 59. 954 with real probability p41

0.802 5 (41) = 32. 9023 with real probability qt = 0 t = h = 1

Cu = max (0, 59. 954− 40) = 19. 954 with real probability pC =?

Cd = max (0, 32. 9023− 40) = 0 with real probability qt = 0 t = h = 1

Calculate the real probabilities:41e0.15(1) = 59. 954p+ 32. 9023 (1− p)



p = 0.544 6 q = 1− p = 1− 0.544 6 = 0.455 4

Calculate the replicating portfolio:


Su − Sd= e−0(1)

19. 954− 059. 954− 32. 9023 = 0.737 6


Su − Sd= e−0.08(1)

59. 954 (0)− 32. 9023 (19. 954)59. 954− 32. 9023 = −22.

403 6

Calculate the discounting rate:∆S +B =

¡∆Seαh +Berh

¢e−γh

(∆S +B) eγh = ∆Seαh +Berh

(0.737 6× 41− 22. 403 6) eγ(1) = 0.737 6× 41e0.15(1) − 22. 403 6e0.08(1)eγ(1) = 1. 386 γ = 0.326 4

Calculate the option price:C = e−γh (pCu + qCd) = e−0.326 4(1) (0.544 6× 19. 954 + 0.455 4× 0) = 7. 84This option price is the same as the price in the textbook Figure 10.3.

Example 11.2.2. Reproduce the textbook Figure 11.4 (the risk neutral solutionto an otherwise identical European call is in the textbook Figure 10.5). Here isthe recap of the information on an American call. The current stock price is 41.The strike price K = 40. The annualized standard deviation of the continuouslycompounded stock return is σ = 30%. The continuously compounded risk-freerate per year is r = 8%. The continuously compounded expected return on thestock per year is α = 15%.The continuously compounded dividend rate per yearis δ = 0%.The option expiration date is T = 1 year. Use a 3-period binomialtree and real probabilities to calculate the option premium.

Solution.

Each period is h =1

3year long.

u = e(r−δ)h+σ√h = e(0.08−0)1/3+0.3

√1/3 = 1. 221 246

d = e(r−δ)h−σ√h = e(0.08−0)1/3−0.3

√1/3 = 0.863 693


Su3 = 74.6781Su2 = 61.1491

Su = 50.0711 Su2d = 52.8140S = 41 Sud = 43.2460

Sd = 35.4114 Sd2u = 37.3513Sd2 = 30.5846

Sd3 = 26.4157



The replicating portfolios are copied over from the textbook Figure 10.5.

Period 0 1 2

(4, B)u2 = (1,−38. 947 4)(4, B)u = (0.921 8,−33. 263 6)

(4, B) = (0.706 3,−21. 885 2) (4, B)ud = (0.828 7,−30. 138 6)(4, B)d = (0.450 1,−13. 405 2)

(4, B)d2 = (0, 0)

In addition, we need to calculate the discount rate for each node. We putthe stock price table and the replicating portfolio table side by side:Stock price (4, B)Period 0 1 2 3

74.678161.1491

50.0711 52.814041 43.2460

35.4114 37.351330.5846

26.4157

Period 0 1 2

(1,−38. 947 4)(0.921 8,−33. 263 6)

(0.706 3,−21. 885 2) (0.828 7,−30. 138 6)(0.450 1,−13. 405 2)

(0, 0)

Calculate the common discounting factor Node u→Node 0 and Node d→Node0 :

e−γ(1/3) =∆S +B

∆Seαh +Berh=

0.706 3 (41)− 21. 885 20.706 3 (41) e0.15(1/3) − 21. 885 2e0.08(1/3) =

0.887 87e−γ(1/3) = 0.887 87 γ = 0.356 8

Calculate the common discounting factor Node u2 →Node u and Nodeud→Node u :

e−γ(1/3) =∆S +B

∆Seαh +Berh=

0.921 8 1 (50.0711)− 33. 263 60.921 8 (50.0711) e0.15(1/3) − 33. 263 6 2e0.08(1/3) =

0.897 84e−γ(1/3) = 0.897 84 γ = 0.323 3

Calculate the common discounting factor Node ud →Node d and Noded2 →Node d :

e−γ(1/3) =∆S +B

∆Seαh +Berh=

0.450 1 (35.4114)− 13. 405 20.450 1 (35.4114) e0.15(1/3) − 13. 405 2 64e0.08(1/3) =

0.847 79e−γ(1/3) = 0.847 79 γ = 0.495 4



Calculate the common discounting factor Node u3 →Node u2 and Nodeu2d→Node u2 :

e−γ(1/3) =∆S +B

∆Seαh +Berh=

1 (61.1491)− 38. 947 41 (61.1491) e0.15(1/3) − 38. 947 4e0.08(1/3) =

0.914 24e−γ(1/3) = 0.914 24 γ = 0.2690

Calculate the common discounting factor Node u2d →Node ud and Nodeud2 →Node ud :

e−γ(1/3) =∆S +B

∆Seαh +Berh=

0.828 7 (43.2460)− 30. 138 60.828 7 (43.2460) e0.15(1/3) − 30. 138 6e0.08(1/3) =

0.847 83e−γ(1/3) = 0.847 83 γ = 0.495 2

Calculate the common discounting factor Node ud2 →Node d2 and Noded3 →Node d2 :

e−γ(1/3) =∆S +B

∆Seαh +Berh=

0 (30.5846)− 00 (30.5846) e0.15(1/3) − 0 6e0.08(1/3) = N/A

e−γ(1/3) = N/A γ = N/A

We calculate the premium by working backward from right to left. At eachnode, we take the greater of the backwardized value and the exercise value.Calculate the common real probability of stock going up and down at each

node.41e0.15(1/3) = 50.0711p+ 35.4114 (1− p)p = 0.524 6 q = 1− 0.524 6 = 0.475 4

Period 0 1 2 3Cu3 = max (0, 74.6781− 40) = 34. 678 1

γ = 0.2690γ = 0.323 3 Cu2d = max (0, 52.8140− 40) = 12. 814

γ = 0.356 8 γ = 0.495 2γ = 0.495 4 Cd2u = max (0, 37.3513− 40) = 0

γ = N/ACd3 = max (0, 26.4157− 40) = 0

Period 0 1 2 3Su3 = 74.6781

Su2 = 61.1491Su = 50.0711 Su2d = 52.8140

S = 41 Sud = 43.2460Sd = 35.4114 Sd2u = 37.3513

Sd2 = 30.5846Sd3 = 26.4157



Cu2 = (34. 678 1× 0.524 6 + 12. 814× 0.475 4) e−0.2690(1/3) = 22. 201 2EVu2 = max (0, 61.1491− 40) = 21. 149 1

Vu2 = max (22. 201 2, 21. 149 1) = 22. 201 2

Cud = (12. 814× 0.524 6 + 0× 0.475 4) e−0.495 2(1/3) = 5. 699 4EVud = max (0, 43.2460− 40) = 3. 246

Vud = max (5. 699 4, 3. 246) = 5. 699 4Cd2 = 0EVd2 = max (0, 30.5846− 40) = 0

Vd2 = max (0, 0) = 0Now we have:Period 0 1 2 3

Vu3 = 34. 678 1Vu2 = 22. 201 2

γ = 0.323 3 Vu2d = 12. 814γ = 0.356 8 Vud = 5. 699 4

γ = 0.495 4 Vud2 = 0Vd2 = 0

Vd3 = 0

Similarly,Cu = (22. 201 2× 0.524 6 + 5. 699 4× 0.475 4) e−0.323 3(1/3) = 12. 889 6EVu = max (0, 50.0711− 40) = 10. 071 1

Vu = max (12. 889 6, 10. 071 1) = 12. 889 6

Cd = (5. 699 4× 0.524 6 + 0× 0.475 4) e−0.495 4(1/3) = 2. 534 8EVd = max (0, 35.4114− 40) = 0

Vd = max (2. 534 8, 0) = 2. 534 8Now we have:Period 0 1 2 3

Vu3 = 34. 678 1Vu2 = 22. 201 2

Vu = 12. 889 6 Vu2d = 12. 814γ = 0.356 8 Vud = 5. 699 4

Vd = 2. 534 8 Vud2 = 0Vd2 = 0

Vd3 = 0Finally,C = (12. 889 6× 0.524 6 + 2. 534 8× 0.475 4) e−0.356 8(1/3) = 7. 0734EV = max (0, 41− 40) = 1



V = max (7. 0734, 1) = 7. 073 4So we have:Period 0 1 2 3

Vu3 = 34. 678 1Vu2 = 22. 201 2

Vu = 12. 889 6 Vu2d = 12. 814V = 7. 073 4 Vud = 5. 699 4

Vd = 2. 534 8 Vud2 = 0Vd2 = 0

Vd3 = 0

Tip 11.2.2. Real probability pricing requires intensive calculation. Not onlydo we need to find the real probability of up and down, we also need to findthe replicating portfolio at each node. In contrast, in risk neutral pricing, weeither use risk neutral probabilities or use replicating portfolio but not both. Incomparison, risk neutral pricing is more efficient than real probability pricing.

11.2.2 Binomial tree and lognormality

Random Walk model

Here’s a brief review of the random walk model. There are 3 schools of thoughts:the chartist approach (or technical analysis), fundamental analysis, and therandom walk model. Those who use the chartist approach draw charts to predictstock future prices. They believe that history repeats itself and that past stockprices help predict future stock prices. Fundamental analysis believes that atany point the stock has an intrinsic value that depends on the earning potentialof the stock. Random Walk model, on the other hand, believes that the price ofa stock is purely random and that past price can’t help predict the stock pricein the future.Is the random walk model true? Is stock price purely random? Some scholars

challenged the random walk theory. If interested, you can look into the bookA Non-Random Walk Down Wall Street at Amazon.com http://www.amazon.com/Now let’s look at the random walk math model. Imagine that an object zero

starts from point zero and travels along a straight line (such as Y axis). At eachstep, the object either move up by 1 unit with probability p or move down by 1unit with probability 1− p. An equivalent description of the movement is this:at each move a coin is tossed. If we get a head, the object moves up by 1 unit;if we get a tail, the object moves down by 1 unit.Let Yi represent the movement in the i-th step. Then

Yi =

½1 with probability p−1 with probability 1− p

Let Zn represent the position of the article after n steps. ThenZn = Y1 + Y2 + ...+ YnY1, Y2, ...Yn are independent identically distributed.



To apply the random walk model to stock prices, we can use Yi to representthe price movement during an interval of time. We can use Zn to represent theending price of a stock after n equal intervals.

To get a good feel of the random variable Zn, check out the simulation ofthe random walk model athttp://math.furman.edu/~dcs/java/rw.htmll

Just type in the url in your web browser.

Modeling stock as a random walk

There are 3 problems if we use Zn = Y1 + Y2 + ...+ Yn to model the price of astock:

1. Zn can be negative yet the price of a stock can’t be negative.

2. The incremental change of any stock price Yi is either 1 or −1. Though$1 change might be OK for modeling the change of a low priced stock, itmay be inappropriate to model the change of a high priced stock.

3. The expected return on stock whose price following a random walk is zero,that is E (Zn) = nE (Y ) = n× 0 = 0. However, stock on average shouldhave a positive return.

Continuously compounded returns

Let rt,t+h represent the continuously compounded return earned during the timeinterval [t, t+ h]. Let St and St+h represent the stock price at time t and t+ hrespectively. Then

St ert,t+h = St+h ert,t+h =

St+hSt

rt,t+h = lnSt+hSt

Continuously compounded returns are additive. Consider the time interval[t, t+ nh]. We have:

ert,t+nh =St+nhSt

=St+hSt

× St+2hSt+h

× St+3hSt+2h

× ...× St+nhSt+(n−1)h

= ert,t+h ert+h,t+2h ert+2h,t+3h ...ert+(n−1)h,t+nh

⇒ rt,t+nh = rt,t+h +rt+h,t+2h +rt+2h,t+3h + ...+ rt+(n−1)h,t+nh

Continuously compounded returns can be negative. Even if r < 0, we stillhave er > 0. Thus if lnS follows a random walk, S can’t be negative.

Standard deviation of returns

The annual return is the sum of the returns in each of the 12 months:rannual = rJan + rFeb + ...+ rDec

Assume each monthly return is independent identical distributed with com-mon variance σ2Monthly. Let σ

2 represent the variance of the return over 1 yearperiod.



V ar (rannual) = V ar (rJan + rFeb + ...+ rDec) = 12σ2Monthly

σ2 = 12σ2Monthly σMonthly =σ√12

Suppose we split one year into n intervals with each interval being h long (sonh = 1). Let σh represent the standard deviation of the return over an intervalof h long, then

rannual = r0,h + rh,2h + r2h,3h + ...+ r(n−1)h,nhσ2 = V ar (rannual) = V ar

£r0,h + rh,2h + r2h,3h + ...+ r(n−1)h,nh

¤= n σ2h

σh =σ√n

However, nh = 1. So we have: σh =σ√n= σ√h

Binomial model

Previously, we set u = e(r−δ)h+σ√h and d = e(r−δ)h−σ

√h. Now let’s see why do-

ing so is reasonable. Setting u = e(r−δ)h+σ√h and d = e(r−δ)h−σ

√h is equivalent

to settingSt+h = Ste

(r−δ)h±σ√h, which is equivalent to

rt,t+h = (r − δ)h± σ√h (11.4)

Let’s see why Equation 11.4 solves the three problems in the random walkmodel:

1. Even when rt,t+h is negative, the stock price St+h is always positive.

2. The change in stock price is proportional to the stock price. ∆S = St+h−St = St

he(r−δ)h±σ

√h − 1

i.

3. The expected return during [t, t+ h] is largely driven by the constant term(r − δ)h. Hence the expected return is no long always zero.

Alternative binomial tree

The Cox-Ross-Rubinstein binomial tree

u = eσ√h (11.5)

d = e−σ√h (11.6)

The lognormal tree (also called the Jarrow-Rudd binomial model)

u = e(r−δ−0.5σ2)h+σ

√h (11.7)

d = e(r−δ−0.5σ2)h−σ

√h (11.8)



11.2.3 Estimate stock volatility

Formulasn+ 1 number of stock prices observed: S0, S1,...,Sn−1,and Snn number of stock returns observed. Continuously compounded

return per period is estimated as

r1 = lnS1S0

r2 = lnS2S1

r3 = lnS3S2

...rn = lnSnSn−1

Since our focus is stock returns not stock price, the number of observationsis n (i.e. the number of stock prices observed minus one). Remember this point.The expected return is:

∧r =

r1 + r2 + ...+ rnn

The estimated standard deviation is:

∧σ =

vuut³r1 −

∧r´2+³r2 −

∧r´2+ ...+

³rn −

∧r´2

n− 1

Example 11.2.3. Reproduce the textbook Figure 11.1 and estimate the standarddeviation of the continuously compounded return per year earned by S&P 50index.

Week S&P 500 price rt=lnStSt−1

³rt −

∧r´2

0 829.851 804.19 −0.0314 0.0018462 874.02 0.0833 0.0051433 869.95 −0.0047 0.0002634 880.9 0.0125 0.0000015 865.99 −0.0171 0.0008196 879.91 0.0159 0.0000197 919.02 0.0435 0.0010208 916.92 −0.0023 0.0001919 929.62 0.0138 0.00000510 939.28 0.0103 0.00000111 923.42 −0.0170 0.00081712 953.22 0.0318 0.000409

Total 0.1386 0.010534

The expected continuously compounded return per week is estimated as:∧r =

r1 + r2 + ...+ r12n

=0.1386

12= 0.011 55

The standard deviation of the continuously compounded return per week isestimated as:



∧σ =

r0.010534

12− 1 = 0.03 095

1 Year = 52 Weeks

Let Y represent the continuously compounded return per year and Xi rep-resent the continuously compounded return earned in the i-th week.Then Y = X1 +X2 + ...+X52

WhereX1,X2, ...,X52 are assumed to be independent identically distributed.V ar (Y ) = V ar (X1 +X2 + ...+X52) = 52V ar (X)σY =

√52σY =

√52 (0.03 095) = 0.223 18

Please note my calculation was done using Excel. If you can’t perfectlyreproduce my numbers, that’s fine.By the way, in Excel, the formula for the mean is AVERAGE; the formula

for the sample variance is VAR

This is how to use Excel to calculate the continuously compounded returnper year earned by S&P 500 index. Suppose the stock prices are entered inColumn C (from C3 to C15) and the weekly returns are calculated in ColumnD (from D4 to D15).

1 Column B Column C Column D

2 Week S&P 500 price rt=lnStSt−1

3 0 829.854 1 804.19 −0.03145 2 874.02 0.08336 3 869.95 −0.00477 4 880.9 0.01258 5 865.99 −0.01719 6 879.91 0.015910 7 919.02 0.043511 8 916.92 −0.002312 9 929.62 0.013813 10 939.28 0.010314 11 923.42 −0.017015 12 953.22 0.0318The expected continuously compounded return per week is:

∧r =AVERAGE(D4:D15)= 0.01155006

The sample variance of the continuously compounded return per week isestimated as:∧σ2=VAR(D4:D15)= 0.000957682

∧σ =√0.000957682 = 0.030946434

Make sure you don’t use the population variance formula:VARP(D4:D15)= 0.000877875



In Excel, VARP is:

V ARP =

³r1 −

∧r´2+³r2 −

∧r´2+ ...+

³rn −

∧r´2

n

While VAR is calculated as

V AR =

³r1 −

∧r´2+³r2 −

∧r´2+ ...+

³rn −

∧r´2

n− 1

So V AR = V ARP × n

n− 1

Finally, I’ll give you a BA II Plus (or BA II Plus Professional) calculatorshortcut for quickly finding the sample mean and the sample standard deviation.BA II Plus and BA II Plus Professional have a 1-V Statistics Worksheet. Thisworksheet can calculate the sample mean, the sample standard deviation, thepopulation mean (which is the same as the sample mean), and the populationstandard deviation.In 1-V Statistics Worksheet, enter:1 Column B Column C Column D

2 Week S&P 500 price rt=lnStSt−1

3 0 829.854 1 804.19 −0.03145 2 874.02 0.08336 3 869.95 −0.00477 4 880.9 0.01258 5 865.99 −0.01719 6 879.91 0.015910 7 919.02 0.043511 8 916.92 −0.002312 9 929.62 0.013813 10 939.28 0.010314 11 923.42 −0.017015 12 953.22 0.0318

X01 = ln804.19

829.85= −0.03 140 940

Y 01 = 1 (we observed X01 only once)

X02 = ln874.02

804.19= 0.08 326 770


So on and so forth. The final entry is:

X12 = ln953.22

923.42= 0.03 176 156




Press 2nd STAT of the calculator key, you should get:n = 12X = 0.01155006This is the sample mean (or population mean) of the continuously com-

pounded return per week.SX = 0.03094643This is the estimated standard deviation (or sample standard deviation) of

the continuously compounded return per week.σX = 0.02962895This is the estimated standard deviation (or population standard deviation)

of the continuously compounded return per week.

You should use SX and discard σX when estimating the stock volatility.

Example 11.2.4. Reproduce the textbook Figure 11.1 and estimate the standarddeviation of the continuously compounded return per year earned by IBM.

Week S&P 500 price rt=lnStSt−1

0 77.731 75.18 −0.03342 82 0.08683 81.55 −0.00554 81.46 −0.00115 78.71 −0.03436 82.88 0.05167 85.75 0.03408 84.9 −0.01009 86.68 0.020710 88.7 0.023011 86.18 −0.028812 87.57 0.0160

In BA II Plus (or BA II Plus Professional) 1-V Statistics Worksheet, enter:X01 = −0.0334 Y 01 = 1X02 = 0.0868 Y 02 = 1X03 = −0.0055 Y 03 = 1......X13 = 0.0160 Y 12 = 1

You should get: SX = 0.0365So the estimated standard deviation (or sample standard deviation) of the

continuously compounded return per week is:σX = 0.0365

The standard deviation of the continuously compounded return per year is:


11.3. STOCKS PAYING DISCRETE DIVIDENDS 95

√52 (0.0365) = 0.263 2

11.3 Stocks paying discrete dividends

Previously, we assume that dividends are paid continuously at rate δ. However,in reality, dividends are paid discretely (such as quarterly or annually). Nowlet’s build a binomial tree to calculate the price of a stock that pays discretedividends.Suppose we have a European option on a stock that pays a discrete dividend.

The option is written at t (today) and expires in t+h. The stock pays a dividendor several dividends during [t, t+ h]. The future value of the dividends at t+ his D. The continuously compounded risk-free interest rate per year is r ( a

positive constant). The stock price today is St. At t+ h, the stock price eithergoes up to Sut = uSt or goes down to Sdt = dSt. The standard deviation of thecontinuously compounded return earned by the stock per year is σ. We wantto calculate the option price.

Using Equation 10.22 and 10.23, we have:Sut = uSt = Ft,t+he

σ√h Sdt = dSt = Ft,t+he

−σ√h

However, Ft,t+h = Sterh −D

This gives us:

Sut =¡Ste

rh −D¢eσ√h (11.9)

Sdt =¡Ste

rh −D¢e−σ√h (11.10)

To find the price of the European option, we calculate the cost of the repli-cating portfolio. We have two assets: the stock and a savings account. Thesavings account is the same as a zero-coupon bond. At time t + h, the stockprice is Sh; the bond price is Bt+h. The bond price is deterministic:

Bt = 1 Bt+h = erh

The stock price at t+ h is stochastic:

St+h =

(Sut =

¡Ste

rh −D¢eσ√h

Sdt =¡Ste

rh −D¢e−σ√h

So at t + h the stock price either goes up to Sut ("up state") or goes downto Sdt ("down state").



SutSt

SdtTime t Time t+ h

Our task is to determine C by setting a portfolio that replicates the optionpayoff of Cu in the up state and Cd in the down state. We build the replicatingportfolio by buying 4 stocks and investing $B in a zero-coupon bond.

If we own one stock at t, then at t + h our total wealth is St+h + D. Wenot only can sell the stock in the market for St+h, we’ll also have D, the futurevalue of the dividend earned during [t, t+ h].

So we need to set up the following equation:

4 (Sut +D) Berh Cu

4St + B = C4¡Sdt +D

¢Berh Cd

t t+ h t t+ h t t+ h½4 (Sut +D) +Berh = Cu

4¡Sdt +D

¢+Berh = Cd


4 =Cu − Cd

Sut − Sdt(11.11)

B = e−rhµSut Cd − Sdt Cu

Sut − Sdt−4D

¶(11.12)

11.3.1 Problems with discrete dividend tree

One major problem with the stock price tree using Equation 11.9 and 11.10 isthat the tree doesn’t complete recombine after the discrete dividend.

Example 11.3.1. Reproduce the textbook Figure 11.1 but just for the 2 periods.Show that the stock price tree doesn’t recombine at Period 2. This is the recapof the information. The current stock price is 41. The stock pays a dividendduring Period 1 and Period 2. The future value of the dividend accumulatedat the risk-free interest rate r from Period 1 to Period 2 is 5. Other data are:r = 0.08, σ = 0.3, t = 1,and h = 1/3.



Period 0 1 2 3Suuut = 67. 417 15

Suut = 55. 203Sut = 50. 071 Suudt = Sudut = 47. 678 91

Sudt = 39. 041Suddt = 33. 719 59

St = 41Sduut = 45. 553 05

Sdut = 37. 300Sdt = 35. 411 Sdudt = Sddut = 32. 216 14

Sddt = 26. 380Sdddt = 22. 783 97

No dividend is paid during Period 0 and Period 1.

Sut =¡Ste

rh −D¢eσ√h =

¡41e0.08×1/3 − 0

¢e0.3√1/3 = 50. 071 09

Sdt =¡Ste

rh −D¢e−σ√h =

¡41e0.08×1/3 − 0

¢e−0.3

√1/3 = 35. 411 39

Dividend is paid during Period 1 to Period 2, whose value at Period 2 is 5.

Suut =¡Sut e

rh −D¢eσ√h =

¡50. 071 09e0.08×1/3 − 5

¢e0.3√1/3 = 55. 203 57

Sudt =¡Sut e


¡50. 071 09e0.08×1/3 − 5

¢e−0.3

√1/3 = 39. 041 20

Sdut =¡Sdt e

rh −D¢eσ√h =

¡35. 411 39e0.08×1/3 − 5

¢e0.3√1/3 = 37. 300 47

Sddt =¡Sdt e


¡35. 411 39e0.08×1/3 − 5

¢e−0.3

√1/3 = 26. 379 73

Now you see that Sudt 6= Sdut , . The tree doesn’t recombine.No dividend is paid during Period 2 and Period 3.

Suuut =¡Suut erh −D

¢eσ√h =

¡55. 203 57e0.08×1/3 − 0

¢e0.3√1/3 = 67. 417 15

Suudt =¡Suut erh −D

¢e−σ√h =

¡55. 203 57e0.08×1/3 − 0

¢e−0.3

√1/3 = 47.

678 91Sudut =

¡Sudt erh −D

¢eσ√h =

¡39. 041 20e0.08×1/3 − 0

¢e0.3√1/3 = 47. 678 91

Suddt =¡Sudt erh −D

¢e−σ√h =

¡39. 041 20e0.08×1/3 − 0

¢e−0.3

√1/3 = 33.

719 59

Sduut =¡Sdut erh −D

¢eσ√h =

¡37. 300 47e0.08×1/3 − 0

¢e0.3√1/3 = 45. 553 05

Sdudt =¡Sdut erh −D

¢e−σ√h =

¡37. 300 47e0.08×1/3 − 0

¢e−0.3

√1/3 = 32.

216 14Sddut =

¡Sddt erh −D

¢eσ√h =

¡26. 379 73e0.08×1/3 − 0

¢e0.3√1/3 = 32. 216 14

Sdddt =¡Sddt erh −D

¢e−σ√h =

¡26. 379 73e0.08×1/3 − 0

¢e−0.3

√1/3 = 22.

783 97

Please note that in this example Suudt = Sudut and Sdudt = Sddut . This is nota coincidence.



Let’s verify that Suudt = Sudut .

Sut =¡Ste

rh − 0¢eσ√h = Ste

rheσ√h

Suut =¡Sut e

rh − 5¢eσ√h =

³Ste

rheσ√h − 5

éσ√h

Suudt =¡Suut erh − 0

¢e−σ√h =

³³Ste

rheσ√h − 5

éσ√herh

é−σ√h

=³Ste

rheσ√h − 5

érh

Sudt =¡Sut e

rh − 5¢e−σ√h =

³Ste

rheσ√h − 5

é−σ√h

Sudut =¡Sudt erh − 0

¢eσ√h =

³³Ste

rheσ√h − 5

é−σ√herh − 0

éσ√h

=³Ste

rheσ√h − 5

érh

Clearly, Suudt = Sudut . Similarly, you can verify for yourself that Sdudt =Sddut .

In this problem, Period 2 had 4 prices. If the stock pays continuous dividend,Period 2 will have only 3 prices.Similarly, in this problem, Period 3 has 5 distinct prices. In contrast, if the

stock pays continuous dividend, Period 3 will have only 4 prices.In addition to non-combining stock prices, the above method may produce

negative stock prices.

11.3.2 Binomial tree using prepaid forward

Schroder presents a method that overcomes the two shortcomings of the abovemethod.

This is the idea behind Schroder’s method. Instead of directly building astock price tree (which proves to be non-combining after the discrete dividend ispaid), we’ll build a tree of a series of prepaid forward prices on the same stock.Hopefully, the prepaid forward price tree is recombining and looks like this:³

FPt+2h,T

úu³FPt+h,T

úFPt,T

³FPt+2h,T

úd³FPt+h,T

´d ³FPt+2h,T

´ddIn the above table,

• FPt,T is the prepaid price of a forward contract signed at t expiring on date

T



• FPt+h,T is the prepaid price of a forward contract signed at t+ h expiringon date T

• FPt+2h,T is the prepaid price of a forward contract signed at t+2h expiringon date T

Since there’s a one-to-one relationship between a prepaid forward price andthe stock price, we can change the (recombining) prepaid forward price tree intoa (recombining) stock price tree. Once we have a recombining stock price tree,we’ll can easily find the price of the option using either risk-neutral probabilityor the replicating portfolio.First, let’s find the relationship between the stock price and the prepaid

forward price on the stock. Suppose today is time t . At t we enter into aforward contract agreeing to buy a stock on date T . The stock will pay dividendD on date TD where TD < T .

Suppose we want to prepay the seller at t. The price of this prepaid forwardcontract is the current stock price St minus the present value of the dividend:

FPt,T = St − PVt (D) = St −

½De−r(TD−t) if TD ≥ t

0 if TD < t

→ St = FPt,T +

½De−r(TD−t) if TD ≥ t

0 if TD < t

Similarly,

FPt+h,T = St+h − PVt+h (D) = St+h −

½De−r(TD−t−h) if TD ≥ t+ h

0 if TD < t+ h

→ St+h = FPt+h,T +

½De−r(TD−t−h) if TD ≥ t+ h

0 if TD < t+ h

So there’s a one-to-one mapping between the prepaid forward price and thestock price.Next, let’s find out how to build a prepaid forward price tree. We need to

know how the prepaid forward price changes over time. Suppose today is timet . At t+ h we enter into a forward contract agreeing to buy a stock at date Twhere T > t+ h. The stock will pay dividend D in date TD where t < TD < T .If the stock volatility is zero (meaning that the future stock price is known todaywith 100% certainty), then the price of the prepaid forward contact at t+ h is

FPt+h,T = FP

t,T erh (11.13)

FPt+h,T

FPt,T

= erh (11.14)

This is why Equation 11.13 holds. If we pay FPt,T at t, we’ll receive one stock

at T . This gives us FPt,T = ST e

−r(T−t). Similarly, if we pay FPt+h,T at t+h, we’ll



also receive a stock at T . This gives us FPt+h,T = ST e

−r(T−t−h). Then Equation11.13 holds.Now suppose that the forward price has a volatility of σF per year. Then

it’s reasonable to assume that

FPt+h,T =

(FPt,T e

rheσ√h = FP

t,Tu in the up stateFPt,T e

rhe−σ√h = FP

t,T d in the down state

where u = erh+σF√h and d = erh−σF

√h

Similarly, FPt+2h,T =

½FPt+h,Tu in the up state

FPt+h,Td in the down state

Typically, we know the volatility of the stock, σS . We need to calculate σFusing the following approximation:

σF = σS ×StFPt,T

Now the prepaid forward price tree is:³FPt+2h,T

úu= FP

t,Tu2³

FPt+h,T

ú= FP

t,Tu

FPt,T

³FPt+2h,T

úd=³FPt+2h,T

´du= FP

t,Tud³FPt+h,T

´d= FP

t,Td ³FPt+2h,T

´dd= FP

t,Td2

Once we have the prepaid forward price tree, we’ll transform it into the stockprice tree:

Suut+2h =³FPt+2h,T

úu+ PVt+2h (D)

Sut+h =³FPt+h,T

ú+ PVt+h (D)

St = FPt,T + PVt (D) Sudt+2h =

³FPt+2h,T

úd+ PVt+2h (D)

Sdt+h =³FPt+h,T

´d+ PVt+h (D)

Sddt+2h =³FPt+2h,T

´dd+ PVt+2h (D)

Example 11.3.2. Let’ s reproduce the textbook Figure 11.11. Here is the recapof the information on an American call option. The stock pays a dividend of $5in 8 months. Current stock price is $41. The strike price K = $40. The stockvolatility is σS = 0.3. The continuously compounded risk-free rate is r = 0.08.The option expires in T = 1 year. Use a 3-period binomial tree to calculate theoption price.



First, we’ll build a prepaid forward price tree.

σF = σS ×StFPt,T

FPt,T = St − PVt (D) = 41− 5e−0.08(8/12) = 36. 26

σF = σS ×StFPt,T

= 0.3× 41

36. 26= 0.339 2

u = erh+σF√h = e0.08(1/3)+0.3392

√1/3 = 1. 249 20

d = erh−σF√h = e0.08(1/3)−0.3392

√1/3 = 0.844 36

Prepaid forward price tree:

³FPt+3h,T

úuu³FPt+2h,T

úu³FPt+h,T

ú ³FPt+3h,T

úudFPt,T

³FPt+2h,T

úd³FPt+h,T

´d ³FPt+3h,T

údd³FPt+2h,T

´dd ³FPt+3h,T

´ddd³FPt+h,T

ú= FP

t,Tu = 36. 26× 1. 249 20 = 45. 296³FPt+h,T

´d= FP

t,T d = 36. 26× 0.844 36 = 30. 616³FPt+2h,T

úu= FP

t,Tu2 =

³FPt+h,T

úu = 45. 296× 1. 249 20 = 56. 584³

FPt+2h,T

úd=³FPt+h,T

úd = 45. 296× 0.844 36 = 38. 246³

FPt+3h,T

´dd= FP

t,Td2 =

³FPt+h,T

´dd = 30. 616 × 0.844 36 = 25. 851³

FPt+3h,T

úuu=³FPt+2h,T

úuu = 56. 584× 1. 249 20 = 70. 685³

FPt+3h,T

úud=³FPt+2h,T

úud = 56. 584× 0.844 36 = 47. 777³

FPt+3h,T

údd=³FPt+2h,T

údd = 38. 246 × 0.844 36 = 32. 293³

FPt+3h,T

´ddd=³FPt+2h,T

´ddd = 25. 851× 0.844 36 = 21. 828




70. 68556. 584

45. 296 47. 77736. 26 38. 246

30. 616 32. 29325. 851

21. 828

Next, we change the prepaid forward price tree into a stock price tree. Theone-to-one mapping between the prepaid forward price and the stock price is

St+∆t = FPt+∆t,T +

½De−r(TD−t−∆t) if TD ≥ t+∆t

0 if TD < t+∆twhere 0 ≤ ∆t ≤ T − t ¡

St+3/3¢uuu¡

St+2/3¢uu¡

St+1/3¢u ¡

St+3/3¢uud

St¡St+2/3

¢ud¡St+1/3

¢d ¡St+3/3

¢udd¡St+2/3

¢dd ¡St+3/3

¢dddIn this problem, TD =

8

12+ t, D = 5,T = t+ 1

St = FPt,T +De−r(TD−t) = 36. 26 + 5e−0.08(8/12) = 41. 000

¡St+1/3

¢u=³FPt+1/3,T

ú+De−r(TD−t−1/3) = 45. 296 + 5e−0.08(8/12−1/3) =

50. 164¡St+1/3

¢d=³FPt+1/3,T

´d+De−r(TD−t−1/3) = 30. 616 + 5e−0.08(8/12−1/3) =

35. 484¡St+2/3

¢uu=³FPt+2/3,T

úu+De−r(TD−t−2/3) = 56. 584+5e−0.08(8/12−2/3) =

61. 584¡St+2/3

¢ud=³FPt+2/3,T

úd+De−r(TD−t−2/3) = 38. 246+5e−0.08(8/12−2/3) =

43. 246¡St+2/3

¢dd=³FPt+2/3,T

´dd+De−r(TD−t−2/3) = 25. 851+5e−0.08(8/12−2/3) =

30. 851¡St+3/3

¢uuu=³FPt+3/3,T

úuu= 70. 685 (because TD < t+ 3/3)

Similarly,¡St+3/3

¢uud=³FPt+3/3,T

úud= 47. 777¡

St+3/3¢udd

=³FPt+3/3,T

údd= 32. 293¡

St+3/3¢ddd

=³FPt+3/3,T

´ddd= 21. 828



So the stock price tree is:70. 685

61. 58450. 164 47. 777

41. 000 43. 24635. 484 32. 293

30. 85121. 828

After getting the stock price tree, we calculate the price of the Americancall option as usual. We work backward from right to left. At each node, wecompare the backwardized value with the exercise value, taking the maximumof the two.

The risk neutral probabilities are:

πu =erh − d

u− d=

e0.08(1/3) − 0.844 361. 249 20− 0.844 36 = 0.451 2

πd = 1− 0.451 2 = 0.548 8

Payoff tree:Vuuu = 30. 685

Vuu = 21. 584Vu = 11. 308 Vuud = 7. 777

V = 5.770 Vud = 3. 417Vd = 1. 501 Vudd = 0

Vdd = 0Vddd = 0

Vuuu = Cuuu = max (0, 70. 685− 40) = 30. 685Vuud = Cuud = max (0, 47. 777− 40) = 7. 777Vudd = Cudd = max (0, 32. 293− 40 ) = 0Vddd = Cddd = max (0, 21. 828− 40) = 0Cuu = (30. 685× 0.451 2 + 7. 777× 0.548 8) e−0.08(1/3) = 17. 636EVuu = max (0, 61. 584− 40) = 21. 584Vuu = max (17. 636 , 21. 584) = 21. 584

Cud = (7. 777× 0.451 2 + 0× 0.548 8) e−0.08(1/3) = 3. 417EVud = max (0, 43. 246− 40) = 3. 246Vud = max (3. 417 , 3. 246) = 3. 417

Cdd = 0EVdd = max (0, 30. 851− 40) = 0Vud = 0

Cu = (21. 584× 0.451 2 + 3. 417× 0.548 8) e−0.08(1/3) = 11. 308EVu = max (0, 50. 164− 40) = 10. 164



Vu = max (11. 308, 10. 164) = 11. 308

Cd = (3. 417× 0.451 2 + 0× 0.548 8) e−0.08(1/3) = 1. 501EVd = max (0, 35. 484− 40) = 0Vd = max (1. 501 , 0) = 1. 501

C = (11. 308× 0.451 2 + 1. 501× 0.548 8) e−0.08(1/3) = 5. 770EV = max (0, 41. 000− 40) = 1.0V = max (5. 770, 1.0) = 5. 77


Chapter 12

Black-Scholes

Except the option Greeks and the barrier option price formula, this chapter isan easy read.

12.1 Introduction to the Black-Scholes formula

12.1.1 Call and put option price

The price of a European call option is:

C (S,K, σ, r, T, δ) = Se−δTN (d1)−Ke−rTN (d2) (12.1)

The price of a European put option is:

P (S,K, σ, r, T, δ) = −Se−δTN (−d1) +Ke−rTN (−d2) (12.2)

d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

(12.3)

d2 = d1 − σ√T (12.4)

Notations used in Equation 12.1, 12.3, and 12.4:

• S, the current stock price (i.e. the stock price when the call option iswritten)

• K, the strike price

• r, the continuously compounded risk-free interest rate per year

• δ, the continuously compounded dividend rate per year

105


106 CHAPTER 12. BLACK-SCHOLES

• σ, the annualized standard deviation of the continuously compoundedstock return (i.e. stock volatility)

• T , option expiration time

• N (d) = P (z ≤ d) where z is a standard normal random variable

• C (S,K, σ, r, T, δ), the price of a European call option with parameters(S,K, σ, r, T, δ)

• P (S,K, σ, r, T, δ), the price of a European put option with parameters(S,K, σ, r, T, δ)

Tip 12.1.1. To help memorize Equation 12.2, we can rewrite Equation 12.2similar to Equation 12.1 as P (S,K, σ, r, T, δ) = (−S) e−δTN (−d1)+(−K) e−rTN (−d2).In other words, change S,K,d1,and d2 in Equation 12.1 and you’ll get Equation12.2.

Example 12.1.1. Reproduce the textbook example 12.1. This is the recap ofthe information. S = 41, K = 40,r = 0.08, σ = 0.3, T = 0.25 (i.e. 3 months),and δ = 0. Calculate the price of the price of a European call option.

d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

=

ln41

40+

µ0.08− 0 + 1

2× 0.32

¶0.25

0.3√0.25

= 0.3730

d2 = d1 − σ√T = 0.3730− 0.3

√0.25 = 0.2230

N (d1) = 0.645 4 N (d2) = 0.588 2

C = 41e−0(0.25)0.645 4− 40e−0.08(0.25)0.588 2 = 3. 399

Example 12.1.2. Reproduce the textbook example 12.2. This is the recap ofthe information. S = 41, K = 40,r = 0.08, σ = 0.3, T = 0.25 (i.e. 3 months),and δ = 0. Calculate the price of the price of a European put option.

N (−d1) = 1−N (d1) = 1− 0.645 4 = 0.354 6

N (−d2) = 1−N (d2) = 1− 0.588 2 = 0.411 8P = −41e−0(0.25)0.354 6 + 40e−0.08(0.25)0.411 8 = 1. 607


12.2. APPLYING THE FORMULA TO OTHER ASSETS 107

12.1.2 When is the Black-Scholes formula valid?

Assumptions under the Black-Scholes formula:

Assumptions about the distribution of stock price:

1. Continuously compounded returns on the stock are normally distributed(i.e. stock price is lognormally distributed) and independent over time

2. The volatility of the continuously compounded returns is known and con-stant

3. Future dividends are known, either as a dollar amount (i.e. D and TD areknown in advance) or as a fixed dividend yield (i.e. δ is a known constant)

Assumptions about the economic environment

1. The risk-free rate is known and fixed (i.e. r is a known constant)

2. There are no transaction costs or taxes

3. It’s possible to short-sell costlessly and to borrow at the risk-free rate

12.2 Applying the formula to other assets

12.2.1 Black-Scholes formula in terms of prepaid forwardprice

The prepaid forward price for the stock is: FP0,T (S) = Se−δT

The prepaid forward price for the strike asset is: FP0,T (K) = PV (K) =

Ke−rT

Define V (T ) = σ√T

The price of a European call option in terms of repaid forward is:

C¡FP0,T (S) , F

P0,T (K) , V (T )

¢= FP

0,T (S)N (d1)− FP0,T (K)N (d2) (12.5)

The price of a European put option in terms of prepaid forward is:

P¡FP0,T (S) , F

P0,T (K) , V (T )

¢= −FP

0,T (S)N (−d1) +FP0,T (K)N (−d2) (12.6)

d1 =

lnFP0,T (S)

FP0,T (K)

+1

2V 2 (T )

V (T )(12.7)

d2 = d1 − V (T ) (12.8)



12.2.2 Options on stocks with discrete dividends

When the stock pays discrete dividends, the prepaid forward price is:FP0,T (S) = S − PV0,T (Div)

Apply FP0,T (S) in Equation 12.5 and 12.6, you should get the price of the

European call and put where the stock pays discrete dividends.

Example 12.2.1. Reproduce the textbook example 12.3. Here is the recap ofthe information. S = 41, K = 40, σ = 0.3, r = 0.08, and T = 0.25 (i.e. 3months). The stock pays dividend of 3 in 1 month, but makes no other payoutsover the life of the option (so δ = 0). Calculate the price of the European calland put.

FP0,T (S) = S − PV0,T (Div) = 41− 3e−(0.08)1/12 = 38. 020

FP0,T (K) = PV (K) = Ke−rT = 40e−0.08(0.25) = 39. 208

V (T ) = σ√T = 0.3

√0.25 = 0.15

d1 =

lnFP0,T (S)

FP0,T (K)

+1

2V 2 (T )

V (T )=ln38. 020

39. 208+1

20.152

0.15= −0.130 1

d2 = d1 − V (T ) = −0.130 1− 0.15 = −0.280 1N (d1) = 0.448 2 N (d2) = 0.389 7N (−d1) = 1− 0.448 2 = 0.551 8N (−d2) = 1− 0.389 7 = 0.610 3C = FP

0,T (S)N (d1)− FP0,T (K)N (d2)

= 38. 020 (0.448 2)− 39. 208 (0.389 7) = 1. 76

P = −FP0,T (S)N (−d1) + FP

0,T (K)N (−d2)= −38. 020 (0.551 8) + 39. 208 (0.610 3) = 2. 95

12.2.3 Options on currencies

Notation

• x, the current dollar value of €1

• K, the strike price in dollars of €1

• r, the continuously compounded risk-free rate earned by $1

• rf , the continuously compounded risk-free rate earned by €1

• σ, the annualized standard deviation of the continuously compoundedreturn on dollars

• T , the option expiration date

• C (x0,K, σ, r, T, rf ), the price of a European call option with parameters(x0,K, σ, r, T, rf )


12.2. APPLYING THE FORMULA TO OTHER ASSETS 109

• P (x0,K, σ, r, T, rf ), the price of a European put option with parameters(x0,K, σ, r, T, rf )

The price of a European call option is:

C (x,K, σ, r, T, rf ) = xe−rfTN (d1)−Ke−rTN (d2) (12.9)


P (x,K, σ, r, T, rf ) = −xe−rfTN (−d1) +Ke−rTN (−d2) (12.10)

d1 =

lnx

K+

µr − rf +

1

2σ2¶T

σ√T

(12.11)

d2 = d1 − σ√T (12.12)

Tip 12.2.1. For currency options, just set S = x and δ = rf and apply theBlack-Scholes formulas on European call and put. The same thing happened inEquation 10.26 and 10.27.

Example 12.2.2. Reproduce the textbook example 12.4. Here is the recap of theinformation. The current dollar price of €1 is $0.92. The strike dollar price of€1 is $0.9. The annualized standard deviation of the continuously compoundedreturn on dollars is σ = 0.1. The continuously compounded risk-free rate earnedby dollars is r = 6%. The the continuously compounded risk-free rate earnedby €1 is rf = 3.2%. The option expires in 1 year. Calculate the price of theEuropean call and put on €1.

d1 =

lnx

K+

µr − rf +

1

2σ2¶T

σ√T

=

ln0.92

0.9+

µ0.06− 0.032 + 1

2× 0.12

¶1

0.1√1

= 0.549 8

d2 = d1 − σ√T = 0.549 8− 0.1

√1 = 0.449 8

N (d1) = 0.708 8N (−d1) = 1−N (d1) = 1− 0.708 8 = 0.291 2N (d2) = 0.673 6N (−d2) = 1−N (d2) = 1− 0.673 6 = 0.326 4C = xe−rfTN (d1)−Ke−rTN (d2)= 0.92e−0.032(1)0.708 8− 0.9e−0.06(1)0.673 6 = 0.06 06

P = −xe−rfTN (−d1) +Ke−rTN (−d2)= −0.92e−0.032(1)0.291 2 + 0.9e−0.06(1)0.326 4 = 0.017 2



12.2.4 Options on futures

For a futures contract, the prepaid price is just the present value of the futuresprice. Set FP

0,T (F ) = Fe−rT and FP0,T (K) = Ke−rT , we get:

d1 =

lnFP0,T (F )

FP0,T (K)

+1

2σ2T

σ√T

(12.13)

d2 = d1 − σ√T (12.14)

C (F,K, σ, r, T ) = Fe−rTN (d1)−Ke−rTN (d2) (12.15)

P (F,K, σ, r, T ) = −Fe−rTN (−d1) +Ke−rTN (−d2) (12.16)

Example 12.2.3. Reproduce the textbook example 12.5. Here is the recap ofthe information about the European option on a 1-year futures contract. Thecurrent futures price for natural gas is $2.10. The strike price is K = 2.10. Thevolatility is σ = 0.25. r = 0.055, T = 1. Calculate the price of the Europeancall and put.

d1 =ln

F

K+1

2σ2T

σ√T

=ln2.10

2.10+1

2× 0.252 (1)

0.25¡√1¢ = 0.125

d2 = d1 − σ√T = 0.125− 0.25

¡√1¢= −0.125

N (d1) = 0.549 7 N (d2) = 0.450 3

N (−d1) = 1−N (d1) = 1− 0.549 7 = 0.450 3N (−d2) = 1−N (d2) = 1− 0.450 3 = 0.549 7C = Fe−rTN (d1)−Ke−rTN (d2)

= 2.10e−0.055(1)0.549 7− 2.10e−0.055(1)0.450 3 = 0.197 6P = −Fe−rTN (−d1) +Ke−rTN (dd2)

= −2.10e−0.055(1)0.450 3 + 2.10e−0.055(1)0.549 7 = 0.197 6

12.3 Option the Greeks

The learning objectives in the SOA’s syllabus is to Interpret the option Greeks.The learning objective in CAS Exam 3 Financial Economics is to Interpretthe option Greeks and elasticity measures. How to derive the option Greeksis explained in the textbook Appendix 12.B, but Appendix 21.B is excludedfrom the SOA MFE and CAS FE. So you might want to focus on the learningobjective of the exam.Of all the Greeks, delta and gamma are the most important.


12.3. OPTION THE GREEKS 111

12.3.1 Delta

Delta ∆. You already see ∆ =Cu − Cd

Su − Sdwhen we try to find the replicating

portfolio of a European call or put. It’s the number of stocks you need toown at time zero to replicate the discrete payoff of a European call or put at

expiration date T . If the payoff is continuous, then ∆ =∂C

∂S. Here ∆ =

∂C

∂Sis the number of stocks you need to have now to replicate the payoff of thenext instant (i.e. the payoff one moment later). The European call price is

C = Se−δTN (d1) − Ke−rTN (d2) and the delta for a call is ∆call =∂C

∂S=

e−δTN (d1).One less visible thing to know is that d1is also a function of S. So it’ll tame

some work to derive ∆call =∂C

∂S= e−δTN (d1). One naive approach is treat

N (d1) as a constant and get ∆call =∂C

∂S= e−δTN (d1). Interestingly, this gives

the correct answer!Since deriving delta is not on the syllabus, you don’t need to go through the

messy math and prove ∆call =∂C

∂S= e−δTN (d1). Just memorize that ∆call =

e−δTN (d1) for a European call and∆put = −e−δTN (−d1) = −e−δT [1−N (d1)] =∆call − e−δT for a European put.Other results you might want to memorize:0 ≤ ∆call ≤ 1−1 ≤ ∆put ≤ 0

Example 12.3.1. Calculate the delta of the following European call and put.The information is: S = 25,K = 20, σ = 0.15, r = 6%, δ = 2%, and T = 1year.

d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

=

ln25

20+

µ0.06− 0.02 + 1

2× 0.152

¶1

0.15√1

=

1. 829 3N (d1) = 0.966 3N (−d1) = 1− 0.966 3 = 0.033 7∆call = e−δTN (d1) = e−0.02(1)0.966 3 = 0.947 2∆put = −e−δTN (−d1) = −e−0.02(1)0.033 7 = −0.03 30

12.3.2 Gamma

Gamma is a measure of the change in delta regarding change in the underlyingstock price.

Γ =∂∆

∂S=

∂2C

∂S2If gamma is too large a small change in stock price will cause a big change

in ∆. The bigger Γ, the more often you need to adjust your holding of theunderlying stocks.



Since ∆put = ∆call − e−δT , then∂∆put

∂S=

∂∆call

∂Sor Γcall = Γput.

12.3.3 Vega

Vega is the change of option price for 1% change of stock volatility (you canthink that the letter V stands for volatility).

V ega =∂C

100∂σ

12.3.4 Theta

Theta is the change of option price regarding change in time when the optionis written (you can think that the letter T represents time). Let t represent thetime when the option is written and T the expiration date. Then

θ =∂C (T − t)

∂t

12.3.5 Rho

Rho is a measure of the change in option value regarding a 1% change in therisk free interest rate (you can think the letter R represent r)

ρ =∂C

100∂r

12.3.6 Psi

Psi is a measure of the change in option value regarding a 1% change in thedividend yield.

Ψ =∂C

100∂δ

12.3.7 Greek measures for a portfolio

The portfolio’s Greek is just the sum of the individual Greek.

Example 12.3.2. A portfolio consists of 10 European calls and 50 otherwiseidentical puts. The information about the European call and put is as follows:S = 60,K = 65, σ = 0.25, r = 6%, δ = 4%, and T = 0.75 year. Calculate theportfolio delta.

d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

=

ln60

65+

µ0.06− 0.04 + 1

2× 0.252

¶0.75

0.25√0.75

=

−0.192 2N (d1) = N (−0.192 2) = 1−N (0.192 2)N (0.192 2) = 0.576 2N (d1) = 1− 0.576 2 = 0.423 8N (−d1) = N (0.192 2) = 0.576 2


12.3. OPTION THE GREEKS 113

∆call = e−δTN (d1) = e−0.04(0.75)0.423 8 = 0.411 3∆put = −e−δTN (−d1) = −e−0.04(0.75)0.576 2 = −0.559 2∆portfolio = 10 (0.411 3) + 50 (−0.559 2) = −23. 847

Example 12.3.3. You buy 20 European calls and simultaneously write 35 Eu-ropean puts on the same stock. The call expires in 3 months. The put bothexpires in 9 months. The current stock price is 40. The call strike price is 35.The put strike price is 45. The volatility is 20%. r = 8%, δ = 3%. Calculatethe delta of your portfolio.

Calculate ∆call = e−δTN (d1)

d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

=

ln40

35+

µ0.08− 0.03 + 1

2× 0.22

¶0.25

0.2√0.25

=

1. 510 3N (d1) = 0.934 5∆call = e−δTN (d1) = e−0.03(0.25)0.934 5 = 0.927 5

Calculate ∆put = −e−δTN (−d1)

d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

=

ln40

45+

µ0.08− 0.03 + 1

2× 0.22

¶0.75

0.2√0.75

=

−0.376 9N (−d1) = 0.646 9∆put = −e−0.03(0.75)0.646 9 = −0.632 5∆portfolio = 20 (0.927 5)− 35 (−0.632 5) = 40. 687 5Since your write 35 puts, the delta of 35 puts is −35 (−0.632 5).

12.3.8 Option elasticity and volatility

The elasticity is

Ω =% change in option price% change in stock price

Suppose the stock price increase by where can be positive or negative.Then the option price will change by ∆.% change in stock price=

S

% change in option price=∆

C

Ω =

∆

C

S

=∆S

C(12.17)

The option volatility is

σoption = σstock × |Ω| (12.18)



Example 12.3.4. Calculate the elasticity and the volatility of a European calloption and an otherwise identical European put option. The information is:S = 80,K = 70, σ = 0.35, r = 5%, δ = 3%, and T = 0.5.

d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

=

ln80

70+

µ0.05− 0.03 + 1

2× 0.352

¶0.5

0.35√0.5

=

0.703 7N (d1) = 0.759 2 N (−d1) = 1− 0.759 2 = 0.240 8d2 = d1 − σ

√T = 0.703 7− 0.35

√0.5 = 0.456 2

N (d2) = 0.675 9 N (−d2) = 1− 0.675 9 = 0.324 1C = Se−δTN (d1)−Ke−rTN (d2) = 80e

−0.03(0.5)0.759 2−70e−0.05(0.5)0.675 9 =13. 687

P = −Se−δTN (−d1)+Ke−rTN (−d2) = −80e−0.03(0.5)0.240 8+70e−0.05(0.5)0.324 1 = 3. 150∆call = e−δTN (d1) = e−0.03(0.5)0.759 2 = 0.747 9∆put = −e−δTN (−d1) = −e−0.03(0.5)0.240 8 = −0.237 2Ωcall =

∆callS

C=0.747 9× 8013. 687

= 4. 371

Ωput =∆putS

P=−0.237 2× 80

3. 150= −6. 024

σcall = σstock × |Ωcall| = 0.35 (4. 371) = 1. 530σput = σstock × |Ωput| = 0.35 (6. 024) = 2. 108

12.3.9 Option risk premium and Sharp ratio

At t = 0 the option is worth C = ∆S + B. Suppose the option expires in hyears. Let α represent the expected annual return on the stock, r the contin-uously compounded risk-free rate per year, and γ the expected continuouslycompounded return earned on the option per year. According to Equation 11.3,we have

e−γh =∆S +B

∆Seαh +BerhHowever, ∆S +B = C

eγh =∆Seαh +Berh

C=∆S

Ceαh +

µ1− ∆S

C

¶erh

eγh = Ωeαh + (1− Ω) erh (12.19)

Equation 12.19 holds for any h. Using the Taylor’s expansion, we have:

1 + γh+1

2(γh)2 + ...

= Ω

∙1 + ah+

1

2(ah)2 + ...

¸+ (1− Ω)

∙1 + rh+

1

2(rh)2 + ...

¸


12.4. PROFIT DIAGRAMS BEFORE MATURITY 115

For the above equation to hold for any h, it seems reasonable to assume that1 = Ω+ (1− Ω)

γh = Ωah+ (1− Ω) rh

1

2(γh)

2+ ... = Ω

∙1

2(ah)

2+ ...

¸+ (1− Ω)

∙1

2(rh)

2+ ...

¸So we have γh = Ωah+ (1− Ω) rh or

γ − r = Ω (α− r) (12.20)

The Sharp ratio of an asset is the asset’s risk premium divided by the asset’svolatility:

Sharp Ratio =α− r

σ(12.21)

The Sharp ratio of an option is

Sharp Ratiooption =Ω (α− r)

Ωσstock=

α− r

σstock= Sharp Ratiostock (12.22)

So the Sharp ratio of an option equals the Sharp ratio of the underlyingstock.

12.3.10 Elasticity and risk premium of a portfolio

The elasticity of a portfolio is the weighted average of the elasticities of theportfolio components. In contrast, the Greek of a portfolio is just the sum ofthe Greeks of the portfolio components.The risk premium of a stock portfolio is just the portfolio’s elasticity times

the stock’s risk premium:

(γ − r)portfolio = Ωportfolio (α− r) (12.23)

12.4 Profit diagrams before maturity

12.4.1 Holding period profit

Example 12.4.1.

You buy a European call option that expires in 1 year and hold it for oneday. Calculate your holding profit. Information is:

• The stock price is 40 when you buy the option.



• The stock price is still 40 one day later.

• K = 40

• r = 0.08

• δ = 0

• σ = 30%

Solution.

At time zero, you buy a 1-year European option. Your purchase price is thecall price.

d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

=

ln40

40+

µ0.08− 0 + 1

2× 0.32

¶1

0.3√1

= 0.416 7

d2 = d1 − σ√T = 0.416 7− 0.3

√1 = 0.116 7

N (d1) = 0.661 6 N (d2) = 0.546 5C = 40e−0(1)0.661 6− 40e−0.08(1)0.546 5 = 6. 285

One day later, your option is worth:

d1 =

ln40

40+

µ0.08− 0 + 1

2× 0.32

¶364

365

0.3

r364

365

= 0.416 1

d2 = d1 − σ√T = 0.416 1− 0.3

r364

365= 0.116 5

N (d1) = 0.661 33N (d2) = 0.546 37C = 40e−0(364/365)0.661 33− 40e−0.08(364/365)0.546 37 = 6. 274

Suppose you buy the option at t = 0 by paying 6. 285 and sell the optionone day later for 6. 274 . Your holding period profit is:6. 274− 6. 285e0.08(1/365) = −0.01 2You buy a European call option that expires in 1 year and hold it for 6

months. Calculate your holding profit. Information is:


• The stock price is 40 after 6 months.

• K = 40

• r = 0.08


12.4. PROFIT DIAGRAMS BEFORE MATURITY 117

• δ = 0

• σ = 30%

At time zero, you buy a 1-year European option. Your purchase price is thecall price. As calculated before, the call price is 6. 285

6 months later, the call is worth:

d1 =

ln40

40+

µ0.08− 0 + 1

2× 0.32

¶0.5

0.3√0.5

= 0.294 6

d2 = d1 − σ√T = 0.294 6− 0.3

√0.5 = 0.082 5

N (d1) = 0.615 9 N (d2) = 0.532 9C = 40e−0(0.5)0.615 9− 40e−0.08(0.5)0.532 9 = 4. 156Your holding profit is:4. 156− 6. 285e0.08(0.5) = −2. 385You buy a European put option that expires in 1 year and hold it for 6

months. Calculate your holding profit. Information is:


• The stock price is 42 after 6 months.

• K = 40

• r = 0.08

• δ = 0.02

• σ = 30%

At time zero, you buy a 1-year European option. Your purchase price is thecall price.

d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

=

ln40

40+

µ0.08− 0.02 + 1

2× 0.32

¶1

0.3√1

= 0.35

d2 = d1 − σ√T = 0.35− 0.3

√1 = 0.05

N (−d1) = 0.363 2 N (−d2) = 0.480 1P = −40e−0.02(1)0.363 2 + 40e−0.08(1)0.480 1 = 3. 487

One day later, your option is worth:

d1 = =

ln42

40+

µ0.08− 0.02 + 1

2× 0.32

¶0.5

0.3√0.5

= 0.477 5

d2 = d1 − σ√T = 0.477 5− 0.3

√0.5 = 0.265 4



N (−d1) = 0.316 5 N (−d2) = 0.395 4P = −42e−0.02(0.5)0.316 5 + 40e−0.08(0.5)0.395 4 = 2. 035

Suppose you buy the option at t = 0 by paying 3. 487 and sell the option 6months later for 2. 035. Your holding period profit is:2. 035− 3. 487e0.08(0.5) = −1. 59

12.4.2 Calendar spread

The textbook has an intimidating diagram (Figure 12.14). Don’t worry aboutthis diagram. Just focus on understanding what a calendar spread is.A calendar spread (also called time spread or horizontal spread) is an option

strategy that takes advantage of the deteriorating time value of options. Acalendar spread involves selling one option that has a shorter expiring date andsimultaneously buying another option that has a longer expiration date, withboth options on the same stock and having the same strike price.

Suppose that Microsoft is trading for $40 per share. To have a calendarspread, you can sell a $40-strike call on a Microsoft stock with option expiringin 2 month. Simultaneously, you buy a $40-strike call on a Microsoft stockwith option expiring in 3 months. Suppose the price of a $40-strike 2-month toexpiration call is $2; the price of a $40-strike 3-month to expiration call is $5.So your net cost of having a calendar spread at time zero is $3.Then as time goes by, suppose the stock price doesn’t move much and is still

around $40, then the value of your sold call and purchased call both deterioratebut at a different deteriorating speed. The value of the $40-strike 2-month toexpiration call deteriorates much faster. With each day passing, this option hasless and less value left. If there are only several days left before expiration, thevalue of the sold call will be close to zero.With each day passing, the value of the $40-strike 3-month to expiration call

also decreases but at a slower speed.

For example, one month later, the sold call has 1 month to expiration andis worth only $1. The purchased call has 2-month to expiration and is worth$4.5. Now the calendar spread is worth $3.5. You can close out your position bybuying a $40-strike 1-month to expiration call (price: $1) and sell a $40-strike2-month to expiration call (price: $4.5). If you close out your position, you’llget $3.5.At time zero, you invest $3 to set up a calendar spread. One month later,

you close out your position and get $3.5. Your profit (assuming no transactioncost) is $0.5.Time zero: your cost is $3

$40-strike call $40-strike callValue $2 $5Time to expiration 2 months 3 months


12.5. IMPLIED VOLATILITY 119

One month later: your total value is $3.5$40-strike call $40-strike call

Value $1 $4.5Time to expiration 1 month 2 months

A calendar spread can create value because as time passes the sold option(which is your liability) can quickly become worthless yet the purchased option(your asset) is still worth something.For more examples, please refer to

• http://www.optionsxpress.com/educate/strategies/calendarspread.aspx

• http://www.highyieldstrategy.com/artclndrsprds.htm.

12.5 Implied volatility

12.5.1 Calculate the implied volatility

Volatility cannot be observed. One approach to estimating volatility is use pastreturns to calculate the historical volatility. However, past volatility may be apoor estimate of future volatility because the market condition may change.Another approach to estimating volatility is to calculate the implied volatil-

ity. The call and put values depend on (S,K, T, r, δ, σ). Given (S,K, T, r, δ)and the option price, we can calculate σ. This is called the implied volatility.Calculate the implied volatility given the following information about a Eu-

ropean call.

• CEuropean = 7.25

• S = 60

• K = 55

• T = 0.75 (i.e. 9 months)

• r = 0.06

• δ = 0.02

Solution.

This is a difficult problem to solve manually. However, the calculation pro-cedure is conceptually simple.Implied volatility is solved by trial and error. You use a trial σ and see

whether the computed option price under the trial σ reproduces the actualoption price. If the computed option price is lower than the observed optionprice, use a higher trial σ and try again; if the computed option price is higher



than the observed option price, use a lower trial σ and try again. Keep doingthis until you find a σ such the computed option price equals the observed optionprice.First, let’s try σ = 10%

d1 =

ln60

55+

µ0.06− 0.02 + 1

20.12

¶0.75

0.1√0.75

= 1. 3944

d2 = d1 − σ√T = 1. 3944− 0.1

√0.75 = 1. 307 8

N (d1) = 0.918 4 N (d2) = 0.904 5

C = 60e−0.02(0.75)0.918 4− 55e−0.06(0.75)0.904 5 = 6. 7256. 725 < CEuropean = 7.25. So increase σ and try again.Try σ = 20%

d1 =

ln60

55+

µ0.06− 0.02 + 1

20.22

¶0.75

0.2√0.75

= 0.762 2

d2 = d1 − σ√T = 0.762 2− 0.2

√0.75 = 0.5890

N (d1) = 0.777 0 N (d2) = 0.722 1

C = 60e−0.02(0.75)0.777 0− 55e−0.06(0.75)0.722 1 = 7. 9587. 958 > CEuropean = 7.25. So decrease σ and try again

Try σ = 15%

d1 =

ln60

55+

µ0.06− 0.02 + 1

20.152

¶0.75

0.15√0.75

= 0.965 7

d2 = d1 − σ√T = 0.965 7− 0.15

√0.75 = 0.835 8

N (d1) = 0.832 9 N (d2) = 0.798 4

C = 60e−0.02(0.75)0.832 9− 55e−0.06(0.75)0.798 4 = 7. 257. 25 = CEuropean = 7.25.So the implied σ is 15%.

12.5.2 Volatility skew

Volatility skew refers to

• Options on the same stock with different strike price and expiration dateshould have the same implied volatility. However, in reality, options onthe same stock with different strike price and expiration date don’t havethe same implied volatility.

• In addition, the different implied volatilities often form a pattern such as"smiles," "frowns," and "smirks."

Volatility is explained more in Derivatives Markets Chapter 23.


12.6. PERPETUAL AMERICAN OPTIONS 121

12.5.3 Using implied volatility

Implied volatility is important because it helps us

• We can generate option price that’s consistent with the price of othersimilar options

• We can quote the option in terms of volatility rather than a dollar price

• Volatility skew helps us see how well an option pricing formula works.Volatility skew shows that the Black-Scholes formula and assumptions arenot perfect.

This is all you need to know about how to use the implied volatility.

12.6 Perpetual American options

Perpetual American options are excluded from the exam syllabus. Please ignorethis chapter.I included this section for completeness, but you don’t need to read it.

12.6.1 Perpetual calls and puts

Our task here is to derive the price formula for a perpetual American call withstrike price K.A perpetual American option never expires and the option holder can exer-

cise the option at any time. It’s a typical American option where the expirationdate T = +∞.The theoretical framework behind the perpetual option formula is the Black-

Scholes partial differential equation (called the Black-Scholes PDE). The Black-Scholes PDE is Derivatives Market Equation 21.11 (page 682):

Vt +1

2σ2S2VSS + (r − δ)SVS − rV = 0 (Textbook 20.11)

The above equation is also presented in Derivatives Market Equation 13.10(page 430):

θ +1

2σ2S2t Γt + rSt∆t − rV (St) = 0 (Textbook 13.10)

Please note that Equation 13.10 assumes δ = 0. In addition,it uses C (in-stead of V ) to represent the option price.The Black-Scholes PDE is commonly written as:

∂V (t, St)

∂t+1

2σ2S2t

∂2V (t, St)

∂S2t+ (r − δ)St

∂V (t, St)

∂St− rV (t, St) = 0 (12.24)



In the above equation, V (t, St) is the option price at time t where the stockprice is St.If you are interested in learning how to derive the Black-Scholes PDE, refer

to the textbook. For now let’s accept Equation 12.24.

For a perpetual option, its value doesn’t depends on time. Hence∂V

∂t= 0.

The Black-Scholes PDE becomes an ordinary differential equation:

1

2σ2S2t

d2V (t, St)

dS2t+ (r − δ)St

dV (t, St)

dSt− rV (t, St) = 0 (12.25)

To find the solution to Equation 12.25, let’s simplify the equation as

S2td2V (t, St)

dS2t+ St

dV (t, St)

dSt− V (t, St) = 0

We can guess the solution is in the form of V (t, St) = Sht . ThendV (t, St)

dSt= hSh−1t

d2V (t, St)

dS2t= h (h− 1)Sh−2t

S2td2V (t, St)

dS2t+ St

dV (t, St)

dSt− V (t, St)

= S2t h (h− 1)Sh−2t + SthSh−1t − Sht

= Sht [h (h− 1) + h− 1] = Sht¡h2 − 1

¢So as long as

¡h2 − 1

¢= 0, or h = ±1, Equation S2t

d2V (t, St)

dS2t+St

dV (t, St)

dSt−

V (t, St) = 0 has a solution.Of course, if Sht is a solution, aS

ht must also be a solution.

Similarly, we can guess that the solution to Equation 12.25 is in the form ofV (t, St) = Sht . Some brilliant thinker guessed the following solution:

V (t, St) = (H∗ −K)

µStH∗

¶h=

H∗ −K

(H∗)hSht = aSht

Here H∗ the stock price where exercise is optimal (H∗ is a constant). H∗−K

is the terminal payoff at exercise time.µStH∗

¶his an indicator telling us how

close the stock price approaches H∗.dV (t, St)

dSt=

d

dS

¡aSht

¢= ahSh−1t

d2V (t, St)

dS2t= ah (h− 1)Sh−2t

Equation 12.25 becomes:1

2σ2S2t ah (h− 1)Sh−2t + (r − δ)StahS

h−1t − raSht = 0

1

2σ2h (h− 1) + (r − δ)h− r = 0

1

2σ2h2 +

µr − δ − 1

2σ2¶h− r = 0

h =

1

2σ2 − (r − δ)±

sµr − δ − 1

2σ2¶2+ 2σ2r

σ2



=1

2− r − δ

σ2±

sµr − δ

σ2− 12

¶2+2r

σ2

=1

2− r − δ

σ2±

sµ1

2− r − δ

σ2

¶2+2r

σ2

But how can we find H∗? Since the perpetual American option can be ex-ercised at any time, the option holder will choose H∗ such that V (t, St) =

(H∗ −K)

µStH∗

¶hreaches its maximum value. This requires setting

dV (t, St)

dH∗=

0.dV (t, St)

dH∗=

d

dH∗

"(H∗ −K)

µStH∗

¶h#= Sht

d

dH∗

h(H∗)1−h −K (H∗)−h

i= Sht

h(1− h) (H∗)−h −K (−h) (H∗)−h−1

i= 0

(1− h) (H∗)−h −K (−h) (H∗)−h−1 = 0(1− h)H∗ −K (−h) = 0H∗ =

K (−h)1− h

=h

h− 1K

H∗ −K =h

h− 1K −K =1

h− 1K

So V (t, St) = (H∗ −K)

µStH∗

¶h=

K

h− 1

µh− 1h

StK

¶hSet t = 0. Let S represent the stock price at time zero (i.e. S = S0). Then

the option value at time zero is

V (0, S) = (H∗ −K)

µS

H∗

¶h=

K

h− 1

µh− 1h

S

K

¶hBut how do we choose h since there are two possible value of h? H∗ −K =1

h− 1K. To avoid h− 1 becoming negative, we choose the bigger h:

hcall =1

2− r − δ

σ2+

sµ1

2− r − δ

σ2

¶2+2r

σ2

A similar logic can be applied to a perpetual American put. We can guessthe put value is:

V (t, St) = (K −H∗)

µStH∗

¶h=

K −H∗

(H∗)hSht = bSht

Once again, we get the following equation:



1

2σ2h (h− 1) + (r − δ)h− r = 0

1

2σ2h2 +

µr − δ − 1

2σ2¶h− r = 0

h =1

2− r − δ

σ2±

sµ1

2− r − δ

σ2

¶2+2r

σ2

SetdV (t, St)

dH∗= 0:

dV (t, St)

dH∗=

d

dH∗

"(K −H∗)

µStH∗

¶h#= (St)

h d

dH∗

h(K −H∗) (H∗)−h

i= (St)

h d

dH∗

hK (H∗)−h − (H∗)1−h

i= (St)

hhK (−h) (H∗)−h−1 − (1− h) (H∗)−h

i= 0

K (−h) (H∗)−h−1 − (1− h) (H∗)−h = 0

K (−h)− (1− h)H∗ = 0 → H∗ = Kh

h− 1→ K −H∗ =

µ1− h

h− 1

¶K =

1

1− hK

To avoid 1− h becoming negative, we choose the smaller h:

hput =1

2− r − δ

σ2−

sµ1

2− r − δ

σ2

¶2+2r

σ2

Summary of the formulas for perpetual American calls and perpetual Amer-ican puts:

Cperpetual = (H∗Call −K)

µS

H∗Call

¶hcall=

K

hcall − 1

µhcall − 1hcall

S

K

¶hcall(12.26)

H∗Call =hcall

hcall − 1K (12.27)

hcall =

µ1

2− r − δ

σ2

¶+

sµ1

2− r − δ

σ2

¶2+2r

σ2(12.28)

Pperpetual =¡K −H∗put

¢µ S

H∗put

¶hput=

K

1− hput

µhput − 1hput

S

K

¶hput(12.29)

H∗put =hput

hput − 1K (12.30)



hput =

µ1

2− r − δ

σ2

¶−

sµ1

2− r − δ

σ2

¶2+2r

σ2(12.31)

hcall and hput satisfies:1

2σ2h (h− 1) + (r − δ)h− r = 0 (12.32)

Example 12.6.1. Calculate the price of a perpetual American call and the priceof an otherwise identical perpetual American put. The information is as follows.The current stock price is S = 50. The strike price is K = 45. The continuouslycompounded risk-free rate is r = 6%. The continuously compounded dividendyield is 2%. The stock volatility is σ = 25%.

Solution.

Solve for h.1

2σ2h (h− 1) + (r − δ)h− r = 0

1

2× 0.252h (h− 1) + (0.06− 0.02)h− 0.06 = 0

h1 = 1. 252 7 h2 = −1. 532 7Use the bigger h for call and the smaller h for put.

Next, calculate the stock price where exercising the option is optimal.

H∗Call =hcall

hcall − 1K =

1. 252 7

1. 252 7− 1 × 45 = 223. 08

H∗put =hput

hput − 1K =

−1. 532 7−1. 532 7− 1 × 45 = 27. 23


µS

H∗Call

¶hcall= (223. 08− 45)

µ50

223. 08

¶1. 252 7=

27. 35

Pperpetual =¡K −H∗put

¢µ S

H∗put

¶hput= (45− 27. 23 )

µ50

27. 23

¶−1. 532 7= 7.

00

Tip 12.6.1. The CD attached to the textbook Derivatives Markets has a spread-sheet that calculates the price of a perpetual American call and a perpetual Amer-ican put. The spreadsheet is titled "optbasic2." You can use this spreadsheet todouble check your solution.

12.6.2 Barrier present values

Consider the following barrier option. If the stock price first reaches a presetprice H from below, then the payoff of $1 is received. This is a special case of aperpetual American call option by setting the terminal payoff H∗Call −K as $1and by setting H∗Call = H.



The value at time zero of $1 received when the stock price first reaches Hfrom below (i.e. the stock first rises to H) isµ

S

H

¶h1(12.33)

h1 =

µ1

2− r − δ

σ2

¶+

sµ1

2− r − δ

σ2

¶2+2r

σ2(12.34)

Similarly, the value at time zero of $1 received when the stock price firstreaches H from above (i.e. the stock first falls to H) isµ

S

H

¶h2(12.35)

h2 =

µ1

2− r − δ

σ2

¶−

sµ1

2− r − δ

σ2

¶2+2r

σ2(12.36)

h1 and h2 satisfies:1

2σ2h (h− 1) + (r − δ)h− r = 0 (12.37)

Example 12.6.2.

Calculate the value of a $1 paid if the stock price first reaches $100 and $60respectively. The information is:

• S = 80

• r = 6%

• δ = 2%

• σ = 30%

Solution.

• calculate the value of a $1 paid if the stock price first reaches $100

H = 100 > S. So we need to calculate the price of $1 payoff when the stockprice first rises to H from below

1

2σ2h (h− 1) + (r − δ)h− r = 0

1

20.32h (h− 1) + (0.06− 0.02)h− 0.06 = 0

h1 = 1. 211 6 h2 = −1. 100 5Use the bigger hµS

H

¶h1=

µ80

100

¶1. 211 6= 0.763



• calculate the value of a $1 paid if the stock price first reaches $90

H = 60 < S.So we need to calculate the price of $1 payoff when the stockprice first falls to H from above.

h1 = 1. 211 6 h2 = −1. 100 5Use the smaller hµS

H

¶h2=

µ80

60

¶−1. 100 5= 0.729

Tip 12.6.2. The CD attached to the textbook Derivatives Markets has a spread-sheet that calculates the price of a barrier option. The spreadsheet is titled "opt-basic2." You can use this spreadsheet to double check your solution.




Chapter 13

Market-making anddelta-hedging

13.1 Delta hedging

Here is the main idea behind delta hedging. Delta of an option is ∆ =∂V

∂S.

Hence for a small change in stock price, the change of the option value is ap-proximately V1−V0 ≈ ∆ (S1 − S0). Suppose you sell one European call option.If you hold ∆ shares of stock, you are immunized against a small change of thestock price.If the stock price goes up from S0 to S1, then the call will be more valuable

to the buyer and your are exposed to more risk. Suppose the value of the callgoes up from V0 to V1 as the stock price goes up from S0 to S1, then yourliability will increase by V1 − V0. At the same time, the value of your ∆ sharesof stock will go up by ∆ (S1 − S0). Because V1−V0 ≈ ∆ (S1 − S0), the increaseof your liability will be roughly offset by the increase of your asset.However, under delta hedging, you are immunized against only a small

change of the stock price (just like immunization by duration matching assetsand liabilities is only good for a small change of interest rate). If a big changeof stock price knocks off your hedging, you’ll need to rebalance your hedging.However, in the real world, continuously rebalancing the hedging portfolio

is impossible. Traders can only do discrete rebalancing.

13.2 Examples of Delta hedging

Make sure you can reproduce the textbook calculation of delta-hedging for 2days. In addition, make sure you can reproduce the textbook table 13.2 and13.3Exam problems may ask you to outline hedging transactions or calculate the

hedging profit.

129


130 CHAPTER 13. MARKET-MAKING AND DELTA-HEDGING

The major difficulty many candidates face is not knowing how to hedge.They wonder "Should the market-maker buy stocks? Should he sell stocks?"To determine how to hedge a risk, use the following ideas:

• The goal of hedging is to break even. If a trader makes moneyon option, he must lose money on stock; if he loses money onoption, he must make money on stock.

• To determine whether a trader should buy stocks or sell stocks,ask "If the stock price go up (or down), will the trader makemoney or lose money on the option?"

• If a trader loses money on the option as the stock price goes up,then the trader needs to initially own (i.e. buy) stocks. Thisway, the value of the trader’s stocks will go up and the traderwill make money on his stocks. He can use this profit to offsethis loss in the option.

• If a trader makes money on the option as the stock price goesup, then the trader needs to initially short sell stocks. This way,as the stock price goes up, the trader will lose money on theshort sale (because he needs to buy back the stocks at a higherprice). His loss in short sale can offset his profit in option.

Example 13.2.1. The trader sells a call. How can he hedge his risk,?

If the stock price goes up, the call payoff is higher and the trader will losemoney. To hedge this risk, the trader should buy stocks. This way, if the stockprice goes up, the trader makes money in the stocks. This profit can be used tooffset the trader’s loss in the written call.You can also ask the question "If the stock goes down, will the trader make

money or lose money on the option?" If the stock goes down, the call payoff islower and the trader will make money. To eat up his profit in the option, thetrader needs to buy stocks. This way, as the stock price goes down, the valueof the trader’s stocks will go down too and the trader will lose money in hisstocks. This loss will offset the trader’s profit in the written call.

Example 13.2.2. The trader sells a put. How can he hedge his risk?

If the stock price goes down, the put payoff is higher and the trader willmake money. To hedge his risk, the trader should short sell stocks. This way,if the stock price goes down, the trader can buy back stocks at lower price,making a profit on stocks. This profit can be used to offset the trader’s loss inthe written put.You can also ask the question "If the stock goes up, will the trader make

money or lose money on the option?" If the stock goes up, the put payoff islower and the trader will make money on the written put. To eat up his profitin the option, the trader needs to short sell stocks. This way, as the stock price


13.2. EXAMPLES OF DELTA HEDGING 131

goes up, the trader needs to buy back stocks at a higher price. The trader willlose money in his stocks. This loss will offset the trader’s profit in the writtenput.

Example 13.2.3. The trader buys a call. How can he hedge his risk?

If a trader buys a call, the most he can lose is his premium and there’s noneed to hedge. This is different from selling a call, where the call seller hasunlimited loss potential.However, if a trader really wants to hedge his limited risk, he can short sell

stocks.

Example 13.2.4. The trader buys a put. How can he hedge his risk?

If a trader buys a put, the most he can lose is his premium and there’s noneed to hedge. This is different from selling a put, where the call seller has abig loss potential.However, if a trader really wants to hedge his limited risk, he can buy stocks.

Example 13.2.5.

Reproduce the textbook example of delta-hedging for 2 days. Here is therecap of the information. At time zero the market maker sells 100 Europeancall options on a stock. The option expires in 91 days.

• K = $40

• r = 0.08

• δ = 0

• σ = 0.3

• The stock price at t = 0 is $40

• The stock price at t = 1/365 (one day later) is $40.50

• The stock price at t = 2/365 (two days later) is $39.25

The market-maker delta hedges its position daily. Calculate the market-maker’s daily mark-to-market profit

Solution.

First, let’s calculate the call premium and delta at Day 0, Day 1, and Day 2.I used my Excel spreadsheet to do the following calculation. If you can’t fullymatch my numbers, it’s OK.



Time t Day 0 (t = 0) Day 1 Day 2Expiry T (Yrs) 91/365 90/365 89/365

St 40 40.50 39.25

d1 =

lnSt40+

µ0.08 +

1

2× 0.32

¶T

0.3√T

0.208048 0.290291 0.077977

N (d1) 0.582404 0.614203 0.531077

d2 = d1 − 0.3√T 0.058253 0.141322 −0.070162

N (d2) 0.523227 0.556192 0.472032C = StN (d1)− 40e−0.08TN (d2) 2.7804 3.0621 2.3282

∆ = e−δ(T−t)N (d1) 0.58240 0.61420 0.53108

The above table can be simplified as follows:Time t Day 0 t = 0 Day 1 t = 1/365 Day 2 t = 2/365

Expiry T T0 = 91/365 T1 = 90/365 T2 = 89/365St S0 = 40 S1 = 40.50 S2 = 39.25Ct C0 = 100× 2.7804 = 278. 04 C1 = 306.21 C2 = 232.82∆t ∆0 = 100× 0.58240 = 58. 24 ∆1 = 61.420 ∆2 = 53.108

Beginning of Day 0Trader #0 goes to workDay 0 t = 0T0 = 91/365

S0 = 40C0 = 278.04∆0 = 58.240

Trader #0 goes to work. The brokerage firm (i.e. the employer of Trader#0) gives Trader #0 C0 = $278. 04. This is what the call is worth today.Trader #0 needs to hedge the risk of the written call throughout Day 0.To hedge the risk, Trader #0 buys ∆0 stocks, costing ∆0S0 = 58. 24×40 = $

2329. 6. Since Trader #0 gets $278. 04 from the brokerage firm, he needs toborrow:∆0S0 − C0 = 2329. 6− 278. 04 = $2051. 56.The trader can borrow $2051. 56 from a bank or use this own money. Either

way, this amount is borrowed. The borrowed amount earns a risk free interestrate.Now Trader #0’s portfolio is:Component Value∆0 = 58.24 stocks 2487. 51call liability −278. 04borrowed amount 2051. 56Net position 0

End of Day 0 (or Beginning of Day 1)



Mark to market without rebalancing the portfolioDay 1 t = 1/365

T1 = 90/365S1 = 40.50

C1 = 306.21

Method 1To cancel out his position, Trader #0 can at t = 0

• buy a call (we call this the 2nd call) from the market paying C1 = $306.21. At expiration, the payoff of this 2nd call will exactly offset the payoffof the 1st call. The 1st call is the call sold by the brokerage firm at t = 0to the customer who bought the call. For example, if at expiration thestock price is ST = 100, then both calls are exercised. The trader getsST −K = 100− 40 = 60 from the 2nd call. The liability of the first callis also $60. These two calls cancel each other out.

• sell out ∆0 = 58. 24 stocks for ∆0S1 = 58. 24× 40.5 = $2358. 72

• pay off the loan. The payment is (∆0S0 − C0) erh = 2051. 56e0.08×1/365 =

2052. 01

At the end of Day 0, the trader’s profit is −C1+∆0S1− (∆0S0 − C0) erh =

−306. 21 + 2358. 72− 2052. 01 = 0.5Trader #0 hands in $0.5 profit to his employer and goes home.

Method 2 We consider the change between Day 0 and Day 1.

• In the beginning of Day 0, the trader’s stock is worth∆0S0 = 58. 24 (40) =2329. 6. In the end of Day 0 (or the beginning of Day 1), the trader’s stockis worth ∆0S1 = 58. 24 (40.5) = 2358. 72. The value of the trader’s 58. 24stocks goes up by 58. 24 (40.5− 40) = 29. 12. This is good for the trader.

• In the beginning of Day 0, the call is worth C0 = $278. 04. In the end ofDay 0 (or the beginning of Day 1), the call is worth C1 = 306.21. The callvalue is the trader’s liability. Now the trader’s liability increases by 306.21−278. 04 = 28. 17. So the trader has a loss 28. 17 (or a gain of −28.75).Recall under Method 1, the trader has to buy a call for 306. 21 to cancelout the call he sold. So call value increase is bad for the trader.

• In the beginning of Day 0, the borrowed amount is ∆0S0−C0 = 2051. 56.In the end of Day 0 (or the beginning of Day 1), this borrowed amountgrows to (∆0S0 − C0) e

rh = 2051. 56e0.08×1/365 = 2052. 01. The increase(∆0S0 − C0)

¡erh − 1

¢= 2051. 56e0.08×1/365 − 2051. 56 = 0.45. This is

the interest paid on the amount borrowed. No matter the trader borrowsmoney from a bank or uses his own money, the borrowed money needs toearn a risk free interest rate.



At the end of Day 0, the trader’s profit is:∆0S1−∆0S0− (C1 − C0)− (∆0S0 − C0)

¡erh − 1

¢= 29. 12−28. 17−0.45 =

0.5You can verify that∆0S1−∆0S0−(C1 − C0)−(∆0S0 − C0)

¡erh − 1

¢= −C1+∆0S1−(∆0S0 − C0) e

rh

Method 3

On Day 0, the trader owns ∆0 = 58.240 stocks to hedge the call liabilityC0 = $278. 04. The trader’s net asset is

MV (0) = ∆0S0 − C0 = 58. 24× 40− 278. 04 = $2051. 56

In the end of Day 0 (or the beginning of Day 1) before the trader rebalanceshis portfolio, the trader’s asset is:

MV BR (1) = ∆0S1 − C1 = 58. 24× 40.5− 100× 3.0621 = 2052. 51BR stands for before rebalancing.The trader’s profit at the end of Day 0 is:MV BR (1) −MV (0) e0.08×1/365 = (∆0S1 − C1) − (∆0S0 − C0) e

rh = 2052.51− 2051. 56e0.08×1/365 = 0.50

Please note that Trader #0 doesn’t need to rebalance the portfolio. Theportfolio is rebalanced by the next trader.

Beginning of Day 1Trader #1 goes to workDay 1 t = 1/365

T1 = 90/365S1 = 40.50

C1 = 306.21∆1 = 61.420

Trader #1 goes to work. He starts from a clean slate. The portfolio isautomatically rebalanced since Trader #1 starts from scratch.

The brokerage firm gives Trader #1 C1 = $306.21.Trader #1 needs to hedge the risk of the written call throughout Day 1.To hedge the risk, Trader #1 buys ∆1 = 61.420 stocks, costing ∆1S1 =

61.42×40.5 = $2487. 51. Since Trader #1 gets $306.21 from the brokerage firm,he needs to borrow:∆1S1 − C1 = 2487. 51− 306.21 = 2181. 3The trader can borrow $2181. 3 from a bank or use this own money. Either

way, this amount is borrowed. The borrowed amount earns a risk free interestrate.Now Trader #1’s portfolio is:



component value∆1 = 61.42 stocks 2487. 51call liability −306.21borrowed amount 2181. 3Net position 0

One question arises, "What if Trader #1 doesn’t start from a clean slate?"Next, we’ll answer this question.Instead of starting from scratch, Trader #1 can start off with Trader #0’s

portfolio. At the end of Day 0, Trader #0 has

• ∆0 = 58.24 stocks

• a borrowed amount (∆0S0 − C0) erh = 2051. 56e0.08×1/365 = 2052. 01

• 0.5 profit

At the end of Day 0 or the beginning of Day 1, ∆1 = 61.420. So Trader #1buys additional shares:∆1 −∆0 = 61.42− 58.24 = 3. 18The cost of these additional shares is (∆1 −∆0)S1 = 3. 18× 40.5 = 128. 79Since these additional shares are bought at the current market price, Trader

#1 can sell these shares at the same price he bought them. This doesn’t affectthe mark-to-market profit.Trader #1 can borrow 128. 79 to pay for the purchase of 3. 18 stocks.Now Trader #1 has a total of∆1 = 61.42 shares worth∆1S1 = 61.42×40.5 =

2487. 51His liability is now C1 = 306.21The borrowed amount is now 2052. 01+128. 79+0.5 = 2181. 3. The 0.5 (the

profit made by Trader #0) is the amount of money Trader #1 borrows fromTrader #0.

Now Trader #1’s portfolio is:component value∆1 = 61.42 stocks $2487. 51call liability −306.21borrowed amount 2181. 3Net position 0The portfolio is the same as the portfolio if Trader #1 starts from scratch.

Calculation is simpler and cleaner if we start from scratch.

End of Day 1 (or Beginning of Day 2)Mark to market without rebalancing the portfolioDay 2 t = 2/365

T2 = 89/365S2 = 39.25

C2 = 232.82



Method 1To cancel out his position, the trader can

• buy a call from the market paying C2 = 232.82. The payoff of this callwill exactly offset the payoff of the call sold by the brokerage firm to thecustomer. These two call have the common expiration date T1 = 89/365and the same payoff. They will cancel each other out.

• sell out ∆1 = 61.420 stocks for ∆1S2 = 61.420× 39.25 = 2410. 735

• pay off the loan. The payment is (∆1S1 − C1) erh = 2181. 3e0.08×1/365 =

2181. 778

The trader’s profit at the end of Day 1 is−C2 +∆1S2 − (∆1S1 − C1) e

rh = 232.82 + 2410. 735− 2181. 778 = −3. 863Trader #1 hands in −3. 863 profit to his employer and goes home.

Method 2 We consider the change between the beginning of Day 1 andthe end of Day 1 (or the beginning of Day 2).

• In the beginning of Day 1, the trader’s stock is worth ∆1S1; In the end ofDay 1 (or the beginning of Day 2), the trader’s stock is worth ∆1S2. Thevalue of the trader’s stocks goes up by ∆1S2 −∆1S1

• In the beginning of Day 1, the call is worth C1; In the end of Day 1 (or thebeginning of Day 2), the call is worth to C2. The call value is the trader’sliability. Now the trader’s liability increases by C2 − C1

• In the beginning of Day 1, the borrowed amount is ∆1S1 − C1. In theend of Day 1 (or the beginning of Day 2), this borrowed amount growsto (∆1S1 − C1) e

rh. The increase is (∆1S1 − C1)¡erh − 1

¢. This is the

interest paid on the amount borrowed. No matter the trader borrowsmoney from a bank or uses his own money, the borrowed money needs toearn a risk free interest rate.

The trader’s profit at the end of Day 1:(∆1S2 −∆1S1)− (C2 − C1)− (∆1S1 − C1)

¡erh − 1

¢= −C2 +∆1S2 − (∆1S1 − C1) e

rh = −3. 863

Method 3

In the beginning of Day 1, the trader’s net asset isMV (1) = ∆1S1 − C1 = 61.420× 40.5− 306.21 = 2181. 3In the end of Day 1 before the trader rebalances his portfolio, the trader’s

asset is:MV BR (2) = ∆1S2 − C2 = 61.420× 39.25− 232.82 = 2177. 915The trader’s profit at the end of Day 1 is:MV BR (2)−MV (1) erh = 2177. 915− 2181. 3e0.08×1/365 = −3. 863



You can verify thatMV BR (2)−MV (1) erh

= (∆1S2 −∆1S1)− (C2 − C1)− (∆1S1 − C1)¡erh − 1

¢= −C2 +∆1S2 − (∆1S1 − C1) e

rh

Method 3 is often faster. Under this method,Profit during a day=Asset at the end of Day before rebalancing the portfolio

- Future value of the asset at the beginning of the day

Asset at the end of Day before rebalancing the portfolio = delta at thebeginning of the day × stock price at the end of the dayProfit at the end of Day t =MV BR (t+ 1)−MV (t) erh = (∆tSt+1 − Ct+1)−

(∆tSt − Ct) erh

Now this completes the textbook example on page 417 to 418.Next, let’s do additional calculations and calculate the profit at the end of

Day 2, 3, and 4 (or the profit at the beginning of Day 3,4,5)

Day Day 2 Day 3 Day 4 Day 5Time t t = 2/365 t = 3/365 t = 4/365 t = 5/365

Expiry T T2 = 89/365 T3 = 88/365 T4 = 87/365 T5 = 86/365St S2 = 39.25 S3 = 38.75 S4 = 40 S5 = 40Ct C2 = 232.82 C3 = 205. 46 C4 = 271.04 C5 = 269.27∆t ∆2 = 53.108 ∆3 = 49. 564 ∆4 = 58.06 ∆5 = 58.01

Profit end of Day 0.40 −4. 0 1. 32

Profit at the end of Day 2 (or beginning of Day 3).

• Asset at the beginning of Day 2: MV (2) = ∆2S2−C2 = 53.108×39.25−232.82 = 1851. 669

• Asset at the end of Day 2 before rebalancing: MV BR (3) = ∆2S3 − C3 =53.108× 38.75− 205. 46 = 1852. 475

• Profit at the end of Day 2: MV BR (3) −MV (2) erh = 1852. 475− 1851.669e0.08/365 = 0.400 1


• Asset at the beginning of Day 3: MV (3) = ∆3S3−C3 = 49. 564×38.75−205. 46 = 1715. 145

• Asset at the end of Day 3 before rebalancing: MV BR (4) = ∆3S4 − C4 =49. 564× 40− 271.04 = 1711. 52

• Profit at the end of Day 3: MV BR (4) −MV (2) erh = 1711. 52 − 1715.145e0.08/365 = −4. 000 96




• Asset at the beginning of Day 4: MV (4) = ∆4S4 − C4 = 58.06 × 40 −271.04 = 2051. 36

• Asset at the end of Day 4 before rebalancing: MV BR (3) = ∆4S5 − C5 =58.06× 40− 269.27 = 2053. 13

• Profit at the end of Day 4: MV BR (5) −MV (4) erh = 2053. 13 − 2051.36e0.08/365 = 1. 320 3

You can verify that the profit at the end of Day 2, 3, 4 calculated abovematches Derivatives Markets Table 13.2. However, in Table 13.2, the profit atthe end of Day 0 is posted in Day 1 column. Similarly, the profit at the end ofDay 1 is posted in Day 1 column. So on and so forth.

13.3 Textbook Table 13.2

Next, we want to reproduce the textbook Table 13.2.Profit at the end of Day t (or the beginning of Day t + 1) can be broken

down into two parts:= MV BR (t+ 1) −MV (t) erh = MV BR (t+ 1) −MV (t) −MV (t) erh +

MV (t)=MV BR (t+ 1)−MV (t)−MV (t)

¡erh − 1

¢=MV BR (t+ 1)−MV (t)| z

capital gain at end of Day t

+ −MV (t)¡erh − 1

¢| z Interest earned at the end of Day t

Define MV BR (t+ 1)−MV (t) = CapitalGain earned at the end of Day tCapitalGain (t) is also equal to:MV BR (t+ 1)−MV (t) = (∆tSt+1 − Ct+1)−(∆tSt − Ct) = ∆t (St+1 − St)−

(Ct+1 − Ct)

Define −MV (t)¡erh − 1

¢as the interest earned at the end of Day t. If the

trader invests money at the beginning of Day t, then MV (t) is positive and−MV (t)

¡erh − 1

¢is negative. The negative interest earned is just the interest

expense incurred by the trader.

Define MV (t) as the investment made at the beginning of Day t. If MV (t)is negative, then it means that the trader receives money.

Example 13.3.1.

Reproduce the textbook Table 13.2We already reproduced the daily profit in Table 13.2. We just need to

reproduce the investment, interest, and the capital gain.


13.3. TEXTBOOK TABLE 13.2 139

Day 0 1 2Time t t = 0 t = 1/365 t = 2/365

Expiry T T0 = 91/365 T1 = 90/365 T2 = 89/365St S0 = 40 S1 = 40.5 S2 = 39.25Ct C0 = 278.04 C1 = 306.21 C2 = 232.82∆t ∆0 = 58.2404 ∆1 = 61.42 ∆2 = 53.108

Investment (beginning of the day) 2051. 56 2051. 56 1851. 669interest earned during the day −0.45 −3. 385 −0.41Capital gain (end of the day) 0.95 −0.478 0.81

Profit (end of the day) 0.50 −3. 863 0.40

Day 0

We already know:MV (0) = ∆0S0 − C0 = 58. 24× 40− 278. 04 = $2051. 56MV BR (1) = ∆0S1 − C1 = 58. 24× 40.5− 100× 3.0621 = 2052. 51The trader’s profit at the end of Day 0 is:MV BR (1)−MV (0) erh = 2052. 51− 2051. 56e0.08×1/365 = 0.50

To find the capital gain and the interest earned at the end of Day 0, we justneed to break down the profit MV BR (1)−MV (0) erh into two parts:

MV BR (1)−MV (0) erh

=MV BR (1)−MV (0)−MV (0) erh +MV (0)=MV BR (1)−MV (0)| z

capital gain

+ −MV (0)¡erh − 1

¢| z interest earned

The capital gain at the end of Day 0:MV BR (1)−MV (0) = 2052. 51− 2051. 56 = 0.95

The interest credited at the end of Day 0:−MV (0)

¡erh − 1

¢= −2051. 56

¡e0.08×1/365 − 1

¢= −0.449 7 = −0.45

Investment at the beginning of Day 0:MV (0) = 2051. 56

Please note the textbook shows the interest credited at the end of Day 0,capital gain earned at the end of Day 0, and daily profit at the end of Day 0 inDay 1 column.

Day 1MV (1) = ∆1S1 − C1 = 61.420× 40.5− 306.21 = 2181. 3MV BR (2) = ∆1S2 − C2 = 61.420× 39.25− 232.82 = 2177. 915The trader’s profit at the end of Day 1 is:MV BR (2)−MV (1) erh = 2177. 915− 2181. 3e0.08×1/365 = −3. 863



To find the capital gain and the interest earned at the end of Day 0, we justneed to break down the profit MV BR (1)−MV (0) erh into two parts:

MV BR (2)−MV (1) erh =MV BR (2)−MV (1)| z capital gain

+ −MV (1)¡erh − 1


Capital gain at the end of Day 1: MV BR (2)−MV (1) = 2177. 915− 2181.3 = −3. 385

Interest earned at the end of Day 1: −MV (1)¡erh − 1

¢= −2181. 3

¡e0.08×1/365 − 1

¢=

−0.478Investment at the beginning of Day 1:MV (1) = ∆1S1 − C1 = 61.420× 40.5− 306.21 = 2181. 3

You should be able to reproduce Table 13.2 for the other days.Day 3 4 5

Time t t = 3/365 t = 4/365 t = 5/365Expiry T T3 = 88/365 T4 = 87/365 T5 = 86/365

St S3 = 38.75 S4 = 40 S5 = 40Ct C3 = 205. 46 C4 = 271.04 C5 = 269.27∆t ∆3 = 49. 564 ∆4 = 58.06 ∆5 = 58.01

Investment (beginning of the day) 1715. 15 2051. 36interest earned during the day −0.38 −0.45Capital gain (end of the day) −3. 63 1. 77

Profit (end of the day) −4. 01 1. 32

13.4 Textbook Table 13.3The stock price of Table 13.3 follows the binomial tree with σ = 0.3.On Day 0, the stock price is S0 = 40On Day 1 the stock moves up 1 σ

S1 = S0erh+σ

√h = 40e0.08/365+0.3

√1/365 = 40.642

Day 2 the stock moves down 1 σ

S2 = S1erh−σ

√h = 40.642e0.08/365−0.3

√1/365 = 40. 018

Day 3 the stock moves down 1 σ

S3 = S2erh−σ

√h = 40. 018e0.08/365−0.3

√1/365 = 39. 403

Day 4 the stock moves down 1 σS4 = S3e

rh−σ√h = 39. 403e0.08/365−0.3

√1/365 = 38. 797

Day 5 the stock moves up 1 σS5 = S4e

rh+σ√h = 38. 797e0.08/365+0.3

√1/365 = 39. 420

If you use the same method for reproducing Table 13.2, you should be ableto reproduce Table 13.3. When reading Table 13.3, remember the interest, thecapital gain, and the daily profit on Day 1 is the interest, the capital gain, andthe daily profit at the end of Day 0 (or the beginning of Day 1). Similarly, the


13.5. MATHEMATICS OF DELTA HEDGING 141

interest, the capital gain, and the daily profit on any other day is the interest,the capital gain, and the daily profit at the end of the previous day or thebeginning of that day.The author of the textbook uses Table 13.3 to show us that if the stock price

moves up or down 1 σ daily, then the trader’s profit is zero.

13.5 Mathematics of Delta hedging

13.5.1 Delta-Gamma-Theta approximation

First, let’s understand the textbook Equation 13.6:

V (St+h, T − t− h) ≈ V (S0, T − t) +∆t + θh+1

2Γt

2 (Textbook 13.6)

Let’s consider an option written at time t expiring on date T (i.e. expiringin T − t years). The value of this option is V (St, T − t), where St is the stockprice at t. Suppose a tiny time interval h (such as 0.00001 second) has passedand we are now standing at t+h. Now the option has a remaining life T − t−hyears and is worth V (St+h, T − t− h), where St+h is the stock price at t + h.Suppose St+h = St + .

Time t t+ h TStock price St St+h = St +Option value V (St, T − t) V (St+h, T − t− h)

We want to estimate V (St+h, T − t− h). By Taylor series, we have:

f (x0 + x, y0 + y) ≈ f (x0, y0) +∂f (x0, y0)

∂xx +

∂f (x0, y0)

∂yy

+1

2

∂2f (x0, y0)

∂x22x +

1

2

∂2f (x0, y0)

∂y22y

Similarly,V (St+h, T − t− h) = V [St+h,− (t+ h) + T ]

≈ V (St, T − t) +∂V (St, T − t)

∂S+

∂V (St, T − t)

∂th

+1

2

∂2V (St, T − t)

∂S22 +

1

2

∂2V (St, T − t)

∂t2h2

However,∂V (St, T − t)

∂S= ∆t

∂V (St, T − t)

∂t= θ

∂2V (St, T − t)

∂S2= Γt

We decide to ignore1

2

∂2V (St, T − t)

∂t2h2 since it’s close to zero. However,

1

2

∂2V (St, T − t)

∂S22 is not close to zero. The reason that

1

2

∂2V (St, T − t)

∂t2h2



is close to zero but1

2

∂2V (St, T − t)

∂S22 is not close to zero will be explained

in Derivatives Markets Chapter 20 when we derive Ito’s Lemma. For now justaccept it.

Now we have:V (St+h, T − t− h) ≈ V (St, T − t) +∆t + θh+

1

2Γt

2

13.5.2 Understanding the market maker’s profit

Suppose a trader sets up a hedging portfolio at time t. The trader’s profit aftera short interval h (i.e. at time t+ h) is:

Profit (t+ h)=MV BR (t+ h)−MV (t) erh

= (∆tSt+h − Ct+h)− (∆tSt − Ct) erh

= (∆tSt+h − Ct+h)− (∆tSt − Ct)¡erh − 1

¢+ (∆tSt − Ct)

= ∆t (St+h − St)− (Ct+h − Ct)− (∆tSt − Ct)¡erh − 1

¢For a small h, using Taylor series, we get erh ≈ 1 + rhProfit (t+ h) = ∆t (St+h − St)− (Ct+h − Ct)− rh (∆tSt − Ct)

According to the textbook Equation 13.6,

Ct+h − Ct = ∆t + θh+1

2Γt

2

Profit (t+ h) = ∆t (St+h − St)−µ∆t + θh+

1

2Γt

2

¶− rh (∆tSt − Ct)

Since St+h = St + , we have:Profit (t+ h)

= ∆t −µ∆t + θh+

1

2Γt

2


= −µθh+

1

2Γt

2


Profit (t+ h) ≈ −µ1

22Γt + θh− rh [∆tSt − Ct]

¶(Textbook 13.7)

However, St+h = Sterh+σ

√h

Using the Taylor series, we have: erh+σ√h = 1+

³rh+ σ

√h´+1

2

³rh+ σ

√h´2+

...

As h→ 0,1

2

³rh+ σ

√h´2and higher order terms all approach zero.

In addition, as h approaches 0,√h is much larger than h. For example,√

0.0001 = 0.01 is much larger than 0.0001.Hence, we can discard rh but keepσ√h.


13.5. MATHEMATICS OF DELTA HEDGING 143

→ St+h = Sterh+σ

√h = St

µ1 +

³rh+ σ

√h´+1

2

³rh+ σ

√h´2+

¶≈ St

³1 + σ

√h´

→ = St+h − St ≈ Stσ√h

→ 2 ≈ S2t σ2h

Plug the above equation in Textbook Equation 13.7, we get:

Profit (t+ h) ≈ −µ1

2S2t σ

2hΓt + θh− rh [∆tSt − Ct]

¶(Textbook 13.9)

From the textbook Table 13.3, we know that if the stock price moves up ordown by 1 σ, the trader’s profit is zero. So

−µ1

2S2t σ

2hΓt + θh− rh [∆tSt − Ct]

¶= 0

This gives us the Black-Scholes PDE:

1

2S2t σ

2Γt + θ − r [∆tSt − Ct] = 0 (Textbook 13.10)

Next, the textbook has the following topics:

• Delta-hedging of American options

• The advantage to frequent rehedging

• Delta-hedging in practice

• Market making as insurance

These topics are minor ideas. I recommend that you skip them.




Chapter 14

Exotic options: I

Any option that is not a plain vanilla call or put is called an exotic option.There are usually no markets in these options and they are purely bought OTC(over-the-counter). They are much less liquid than standard options. They oftenhave discontinuous payoffs and can have huge deltas near expiration which makethem difficult to hedge.Before studying this chapter, make sure you understand the learning objec-

tive.SOA’s learning outcome for this chapter:

• Explain the cash flow characteristics of the following exotic options: Asian,barrier, compound, gap and exchange

If you want to cut corners, you can skip the pricing formula for exotic optionsbecause calculating the exotic option price is out of the scope of the learningoutcome or learning objective.

14.1 Asian option (i.e. average options)

14.1.1 Characteristics

Asian options (also called average options) have payoffs that are based on theaverage price of the underlying asset over the life of the option. The averageprice can be the average stock price or the average strike price. The averagecan be arithmetic or geometric.

• The unique characteristic of an average price option is that the underlyingasset prices are averaged over some predefined time interval.

• Averaging dampens the volatility and therefore average price options areless expensive than standard options

145


146 CHAPTER 14. EXOTIC OPTIONS: I

• Average price options are path-dependent, meaning that the value of theoption at expiration depends on the path by which the stock arrives at itsfinal price. The price path followed by the underlying asset is crucial tothe pricing of the option.

• Average price options are useful in situations where the trader/hedgeris concerned only about the average price of a commodity which theyregularly purchase.

14.1.2 Examples

Examples are based on 1

• A 9-month European average price contract calls for a payoff equal to thedifference between the average price of a barrel of crude oil and a fixedexercise price of USD18. The averaging period is the last two months ofthe contract. The impact of this contract relative to a standard optioncontract is that the volatility is dampened by the averaging of the crude oilprice, and therefore the option price is lower. The holder gains protectionfrom potential price manipulation or sudden price spikes.

• ACanadian exporting firm doing business in the U.S. is exposed to Can$/US$foreign exchange risk every week. For budgeting purposes the treasurermust pick some average exchange rate in which to quote Can$ cash flows(derived from US$ revenue) for the current quarter. Suppose the treasurerchooses an average FX rate of Can$1.29/US$1.00. If the US$ strength-ens, the cash flows will be greater than estimated, but if it weakens, thecompany’s Can$ cash flows are decreased.

Arithmetic average

We record the stock price every h periods from time 0 to time T . There areN = T/h periods. The arithmetic average is:

A (T ) =1

N

nXi=1

Sih

14.1.3 Geometric average

G (T ) = (ShS2h...SNh)1/N

1Based onhttp://www.fintools.com/doc/exotics/exoticsAbout_Average_Options.html


14.2. BARRIER OPTION 147

14.1.4 Payoff at maturity T

LetAVG (T ) represent the average stock price. For arithmetic average, AV G (T ) =

A (T ) =1

N

nXi=1

Sih. For geometric average, AV G (T ) = (ShS2h...SNh)1/N .

Option Payoffaverage price call max [0, AV G (T )−K]average price put max [0,K −AVG (T )]average strike call max [0, ST −AV G (T )]average strike put max [0, AV G (T )− ST ]

14.2 Barrier option

Barrier options are similar to standard options except that they are extinguishedor activated when the underlying asset price reaches a predetermined barrier orboundary price. As with average options, a monitoring frequency is defined aspart of the option which specifies how often the price is checked for breach ofthe barrier. The frequency is normally continuous but could be hourly, daily,etc.

14.2.1 Knock-in option

• "In" option starts its life worthless and becomes active only if the barrierprice is reached

• Technically, this type of contract is not an option until the barrier priceis reached. So if the barrier price is never reached it is as if the contractnever existed.

• down-and-in (spot price starts above the barrier level and has to movedown for the option to become activated.)

• up-and-in (spot price starts below the barrier level and has to move upfor the option to become activated.)

14.2.2 Knock-out option

• "Out" option starts its life active and becomes null and void if the barrierprice is reached

• The option will expire worthless if the asset price exceeds the barrier price

• down-and-out (spot price starts above the barrier level and has to movedown for the option to become null and void)

• up-and-out (spot price starts below the barrier level and has to move upfor the option to be knocked out)



14.2.3 Rebate option

• Barrier options are sometimes accompanied by a rebate, which is a payoffto the option holder in case of a barrier event.

• Rebates can either be paid at the time of the event or at expiration.

14.2.4 Barrier parity

”Knock − in” option+ ”Knock − out” option = Ordinary Option (14.1)

Example 14.2.1.

Explain whyup-and-in option + up-and-out option = Ordinary option.

T=option expiration timetH=time when the stock price first reaches the barrier, where the barrier

price is greater than the current stock priceDuring the time interval [0, tH)

• the up-and-out option is alive

• the up-and-in option is dead

During the time interval [tH , T ]

• the up-and-out option is dead (because the barrier is reached)

• the up-and-in option is alive (because the barrier is reached)

If we have a down-and-in option and a down-and-out option with two optionson the same stock, having the same barrier price and the same strike price, thenat any moment in the interval [0, T ], we’ll always have exactly one ordinaryoption alive. Hence down-and-in option + down-and-out option = Ordinaryoption.Similarly, down-and-in option + down-and-out option = Ordinary option.

14.2.5 Examples

1. A European call option is written on an underlying with spot price $100,and a knockout barrier of $120. This option behaves in every way likea vanilla European call, except if the spot price ever moves above $120,the option "knocks out" and the contract is null and void. Note that theoption does not reactivate if the spot price falls below $120 again. Onceit is out, it’s out for good.


14.3. COMPOUND OPTION 149

2. A bank purchases an at-the-money 9-month Nikkei call option struck at17,000 with a down-and-out barrier price of 16,000. If the price of theNikkei falls to 16,000 or below, during the 9-month period, the bank willno longer have the benefit of Nikkei price appreciation since the call optionwill have been knocked out.

3. An airline is concerned that events in the Middle East might drive upthe price of fuel. An up-and-in call would allow the airline to buy crudeoil futures at a fixed price if some knock-in boundary price is reached.The price of the U&I call would be less than a standard call with thesame expiration and exercise price so it might be viewed as a cost effectivehedging instrument.

14.3 Compound option

Definition

• With a compound option one has the right to buy an ordinary option ata later date

• Compound options are more expensive to purchase than the underlyingoption, as the purchaser has received a price guarantee and effectivelyextended the life of the option

Make sure you understand the textbook Figure 14.2.For a compound call (i.e. call on call) to be valuable, the following two

conditions need to be met:

• St1 > S∗

• St1 > K

Compound option parity

The key formula is DM 14.12:

CallOnCall − PutOnCall + xe−rt1 = BSCall (DM 14.12)

This formula looks scary but is actually easy to remember. We know theput-call parity C +Ke−rT = P + S0. If we treat the standard BSCall as theunderlying asset, then applying the standard put-call parity, we get:

CallOnCall + xe−rt1 = PutOnCall +BSCall

This is DM 14.12.Similarly, we have:CallOnPut+ xe−rt1 = PutOnPut+BSPut



Options on dividend-paying stocks

This section is a minor part of Exam MFE. I recommend that you skip it.However, if you don’t want to skip, here is the main idea.At t1, the value of an American call option CA (St1 , T − t1) is (see the text-

book’s explanation for this formula):

CA (St1 , T − t1) = max [CE (St1 , T − t1) , St1 +D −K] (DM 14.13)

Here CE (St1 , T − t1) is a European call option. Under the put-call parity,we get:

CE (St1 , T − t1) = PE (St1 , T − t1) + St1 −Ke−r(T−t)

SoCA (St1 , T − t1) = max

£PE (St1 , T − t1) + St1 −Ke−r(T−t), St1 +D −K

¤= St1+D−K+max

£PE (St1 , T − t1) + St1 −Ke−r(T−t) − (St1 +D −K) , 0

¤= St1 +D −K +max

£PE (St1 , T − t1)−

¡D −K

¡1− e−r(T−t)

¢¢, 0¤

CA (St1 , T − t1) = St1+D−K+maxhPE (St1 , T − t1)−

³D −K

³1− e−r(T−t)

´´, 0i

(DM 14.14)max

£PE (St1 , T − t1)−

¡D −K

¡1− e−r(T−t)

¢¢, 0¤is payoff of a call on put

with strike price equal to D −K¡1− e−r(T−t)

¢.

Next, the author made two points:Exercising call on put at t1 is the same as not exercising the American call

at t1. If you exercise the call on put, then it must be true that PE (St1 , T − t1)−¡D −K

¡1− e−r(T−t)

¢¢≥ 0 (otherwise you won’t exercise it). Then DM 14.14

becomesCA (St1 , T − t1) = St1 +D−K +PE (St1 , T − t1)−

¡D −K

¡1− e−r(T−t)

¢¢= St1 + PE (St1 , T − t1)−Ke−r(T−t) = CE (St1 , T − t1)

So exercising the call on put at t1 means that not early exercising the Amer-ican call option at t1 (so the American call option becomes a European calloption).

The second point. If you don’t exercise the call on put at t1, then it mustbe true that PE (St1 , T − t1)−

¡D −K

¡1− e−r(T−t)

¢¢< 0.

This gives us CA (St1 , T − t1) = St1 +D−K. This equation means that wereceive dividendD at t1 and immediately exercise the call at t1 (so the Americancall option at t1 is equal to St1 +D −K). So not exercising the call on put att1 is the same as early exercising the American call option at t1.

This is all you need to know about this section.


14.4. GAP OPTION 151

DM Example 14.2 shows you how to use DM 14.14 to calculate the price ofan American call option with a single dividend. When reading this example,please note two things:(1) DM Example 14.2 has errors. Make sure you download the errata. The

DM textbook errata can be found at the SOA website.(2) The call on put price is calculated using the Excel spreadsheet (so don’t

worry about how to manually calculate the call on put).

14.4 Gap option

14.4.1 Definition

• An option in which one strike price K1 determines the size of the payoffand another strike price K2 determines whether or not the payoff is made.

• Example. A call option pays ST − 5 if ST > 8 and pays zero otherwise.The strike price K1 = 5 determines the size of the payoff but a differentstrike price K2 = 8 determines whether or not the payoff is made. K2 iscalled the payment trigger.

• Example. A put option pays 8 − ST if ST > 5 and pays zero otherwise.The strike price K1 = 8 determines the size of the payoff but a differentstrike price K2 = 8 determines whether or not the payoff is made.

• As the footnote in Page 457 of the textbook indicates, a gap call or putoption is really not a option; once the payment trigger is satisfied, theowner of a gap option MUST exercise the option. Hence the premium ofa gap option can be negative.

14.4.2 Pricing formula

To find the price of a gap option, you can modify a standard option’s formulaby changing d1:

C (K1,K2) = Se−δTN (d1)−K1e−rTN (d2)

P (K1,K2) = K1e−rTN (−d2)− Se−δTN (−d1)

d1 =

lnS

K2+

µr − δ +

1

2σ2¶T

σ√T

d2 = d1 − σ√T

14.4.3 How to memorize the pricing formula

• The call payoff is 0 if ST ≤ K2 and ST −K1 if ST > K2. The call optionprice is just the risk-neutral expected discounted value of the payoff. Sothe call price is Se−δTN (d1)−K1e

−rTN (d2). Similarly, the put price isK1e

−rTN (−d2)− Se−δTN (−d1).



You are given the following information on a gap option on a stock

• The gap call expires in 6 months

• The stock is currently selling for $40

• The payment trigger is $50

• The strike price is $60

• The continuously compounded risk-free interest rate is 6% per year

• The continuously compounded dividend yield is 2% per year

• The stock volatility σ = 30%

Calculate

• the price of the gap call option.

• the price of the gap put option

S = 40 K1 = 60 K2 = 50 r = 0.06δ = 0.02 T = 0.5

d1 =

lnS

K2+

µr − δ +

1

2σ2¶T

σ√T

=

ln40

50+

µ0.06− 0.02 + 1

2× 0.32

¶0.5

0.3√0.5

=

−0.851 6d2 = d1 − σ

√T = −0.851 6− 0.3

√0.5 = −1. 063 7

N (d1) = 0.197 2 N (d2) = 0.143 7N (−d2) = 0.856 3 N (−d1) = 0.802 8C = Se−δTN (d1) − K1e

−rTN (d2) = 40e−0.02×0.50.197 2 − 60e−0.06×0.50.143 7 = −0.56

P = K1e−rTN (−d2) − Se−δTN (−d1) = 60e−0.06×0.50.856 3 − 40e−0.02×0.5

0.802 8 = 18. 07

14.5 Exchange option

• Allows the holder of the option to exchange one asset for another

• Pricing formula. Use the standard Black-Scholes formula except changingσ to

pσ2S + σ2K − 2ρσSσK . Also, make sure you know which is the stock

asset and which is the strike asset. If you give up Asset 2 and receiveAsset 1, Asset 1 is the stock and Asset 2 is the strike asset.


14.5. EXCHANGE OPTION 153

Let’s walk through DM Example 14.3.S = 95.92 K = 3.5020× 27.39 = 95. 92 = S (i.e. at-the-money call)δS = 0.75% δK = 1.17% σ = 0.1694 T = 1

d1 =ln95.92e−0.0075×1

95.92e−0.0117×1+1

2× 0.16942 × 1

0.1694√1

= 0.109 5

d2 = d1 − σ√T = 0.109 5− 0.1694

√1 = −0.059 9

N (d1) = 0.543 6 N (d2) = 0.476 1C = Se−δSTN (d1)−K1e

−δKTN (d2)= 95.92e−0.0075×1 × 0.543 6− 95.92e−0.0117×1 × 0.476 1 = 6. 616

This is slightly different from the textbook call price 6.6133 due to rounding.You can verify the call price using the Excel worksheet "Exchange."

Inputs:Underlying AssetPrice 95.92Volatility 20.300%Dividend Yield 0.750%

Strike AssetPrice 95.92Volatility 22.270%Dividend Yield 1.170%

OtherCorrelation 0.6869Time to Expiration (years) 1

Output:Exchange OptionBlack-Scholes

Call PutPrice 6.6144 6.2154

So the exchange call price is 6.6144; the exchange put price is 6.2154.




Chapter 18

Lognormal distribution

This chapter is an easy read. I’m going to highlight the major points.

18.1 Normal distributionThe normal distribution has the following pdf (probability density function)(DM 18.1):

φ (x;μ, σ) =1

σ√2πexp

"−12

µx− μ

σ

¶2#In the standard normal distribution, μ = 0 and σ = 1. Its pdf is:

φ (x; 0, 1) =1√2πexp

µ−12x2¶

Its cdf (cumulative probability density function) is DM 18.2:

P (z 6 a) = N (a) =R a−∞ φ (x; 0, 1) dx =

R a−∞

1√2πexp

µ−12x2¶dx

A frequently used formula is DM 18.3:N (−a) = 1−N (a)

How to convert a normal random variable to the standard normal randomvariable. If x ∼ N

¡μ, σ2

¢, then we can transform x into a standard normal

random variable using DM 18.4:

z =x− μ

σ

We can verify that z ∼ N (0, 1). First, notice that z =1

σx − 1

σμ is normal

(linear combination of normal random variables are also normal). Next,

E (z) = E

µx− μ

σ

¶=1

σ[E (x)− μ] = 0

155


156 CHAPTER 18. LOGNORMAL DISTRIBUTION

V ar (z) = V ar

µx− μ

σ

¶=1

σ2V ar (x− μ) =

1

σ2V ar (x) =

1

σ2× σ2 = 1

So z ∼ N (0, 1)

DM 18.4 can be rewritten as DM 18.7: x = μz + σSum of normal random variables is still normal. For example, if x1 ∼

N¡μ1, σ

21

¢and x2 ∼ N

¡μ2, σ

22

¢, then

ax1 + bx2 ∼ N¡aμ1 + bμ2, aσ

21 + bσ22 + 2abρσ1σ2

¢Next, the textbook briefly explains the central limit theorem, a concept you

most likely already know.

18.2 Lognormal distributionIf x is normal, then y = ex is lognormal. If you say y is lognormal, it meansthat ln y is normal.Let R (0, t) represent the not-annualized continuously compounded return

over the interval [0, t], the stock price at time t is (DM 18.12):St = S0e

R(0,t)

This leads to DM 18.11: R (0, t) = lnStS0

Next comes an important formula:

If x ∼ N¡m, v2

¢, then E (ex) = em+0.5v

2

and V ar (ex) = e2m+v2³ev

2 − 1´.

You don’t need to know how to derive these formulas. Just memorize them.

18.3 Lognormal model of stock pricesDefine:

• α, the expected (annualized) continuously compounded return earned bythe stock

• δ, the (annualized) continuously compounded dividend yield

• σ, the stock’s volatility

• z, the standard normal random variable

Then the (not annualized) continuously compounded return earned during[0, t] is normally distributed with mean

¡α− δ − 0.5σ2

¢t and standard deviation

σ√t (DM 18.18):

lnStS0∼ N

¡¡α− δ − 0.5σ2

¢t, σ2t

¢Using DM 18.7, we can rewrite the normal random variable with mean¡

α− δ − 0.5σ2¢t and standard deviation σ

√t as:


18.4. LOGNORMAL PROBABILITY CALCULATION 157

x =¡α− δ − 0.5σ2

¢t+ σ

√tz

Then we have DM 18.19:

lnStS0= x =

¡α− δ − 0.5σ2

¢t+ σ

√tz

The stock’s price at time t is (DM 18.20):

St = S0ex = S0 exp

£¡α− δ − 0.5σ2

¢t+ σ

√tz¤

Using DM 18.13, we get DM 18.22:E (St) = S0E

¡exp

£¡α− δ − 0.5σ2

¢t+ σ

√tz¤¢= S0 exp

¡¡α− δ − 0.5σ2

¢t+ 0.5σ2t

¢=

S0e(α−δ)t

18.4 Lognormal probability calculation

Let’s go through some formulas. First, let’s look at DM 18.23:

P (St < K) = N

"lnK − lnS0 −

¡α− δ − 0.5σ2

¢t

σ√t

#= N

µ−∧d2

¶

where∧d2 = −

lnK − lnS0 −¡α− δ − 0.5σ2

¢t

σ√t

=ln

S0K+¡α− δ − 0.5σ2

¢t

σ√t

Here’s how to derive it. lnSt ∼ N£lnS0 +

¡α− δ − 0.5σ2

¢t, σ2t

¤The real world probability of St < K is:

P (St < K) = P (lnSt < lnK) = N

"lnK −

¡lnS0 +

¡α− δ − 0.5σ2

¢t¢

σ√t

#

In the above formula, if we set α = r, we’ll get the risk neutral probabilityof St < K:

P ∗ (St < K) = N

"lnK −

¡lnS0 +

¡r − δ − 0.5σ2

¢t¢

σ√t

#= N (−d2)

Next, let’s calculate the real world probability of P (St > K). Please notethat P (St = K) = 0. This is because St is a continuous random variable; theprobability for a continuous random variable to take on a specific value is zero.So P (St > K) = 1 − P (St < K) − P (St = K) = 1 − P (St < K). The realworld probability of St > K is:

P (St > K) = 1−Nµ−∧d2

¶= 1−

∙1−N

µ∧d2

¶¸= N

µ∧d2

¶= N

⎛⎜⎝ ln S0K +¡α− δ − 0.5σ2

¢t

σ√t

⎞⎟⎠



If we set α = r, we’ll get the risk neutral probability of P (St > K):

P ∗ (St > K) = N (d2) = N

⎛⎜⎝ ln S0K +¡r − δ − 0.5σ2

¢t

σ√t

⎞⎟⎠18.4.1 Lognormal confidence interval

The textbook’s explanation is a bit confusing. Here is how to build a confidenceinterval of the stock price.First, let’s consider building a 95% confidence interval for the standard nor-

mal random variable z. We need to find a value a positive value b such thatP (−b < z < b) = 0.95.

P (−b < z < b) = N (b)−N (−b) = N (b)− [1−N (b)] = 2N (b)− 1 = 0.95=⇒ N (b) =

1 + 0.95

2= 0.975

b = N−1 (0.975) = 1.96.Then the 95% confidence interval is −1.96 < z < 1.96.

Alternatively, we can find −b.P (−b < z < b) = N (b)−N (−b) = [1−N (−b)]−N (−b) = 1− 2N (−b) =

0.95

Then N (−b) = 1− 0.952

= 0.025

−b = N−1 (0.025) = −1.96So the 95% confidence interval is −1.96 < z < 1.96.

Let’s generalize the result.The p confidence interval of the standard normal random variable z is

N−1µ1− p

2

¶< z < N−1

µ1 + p

2

¶.

For example, if p = 0.95, then the confidence interval is:

N−1µ1− 0.952

¶< z < N−1

µ1 + 0.95

2

¶, which is N−1 (0.025) < z <

N−1 (0.975) , which is −1.96 < z < 1.96.

We can also specify the confidence interval as 1−p. For example, if we wantto build a 95% confidence interval, we just set p = 5%.The (1− p) confidence interval of the standard normal random variable z

is N−1µ1− (1− p)

2

¶< z < N−1

µ1 + (1− p)

2

¶, which is N−1

³p2

´< z <

N−1³1− p

2

´.

For example, if p = 5%, then the 1− 5% = 95% confidence interval is:



N−1µ0.05

2

¶< z < N−1

µ1− 0.05

2

¶, which isN−1 (0.025) < z < N−1 (0.975),

which is −1.96 < z < 1.96.

Next, let’s consider building the 95% confidence interval for normal randomvariable x ∼ N

¡μ, σ2

¢. Since x = μ+ zσ, the p confidence interval of x is:

μ+ σN−1µ1− p

2

¶< x < μ+ σN−1

µ1 + p

2

¶.

For example, the 95% confidence interval of x ∼ N¡1, 22

¢is: 1+2 (−1.96) <

x < 1 + 2 (1.96), which is −2. 92 < x < 4. 92.

If we specify the confidence interval in terms of 1−p, then the 1−p confidenceinterval is:

μ+ σN−1³p2

´< x < μ+ σN−1

³1− p

2

Ýou don’t need to memorize messy formulas such as μ+ σN−1

µ1− p

2

¶<

x < μ + σN−1µ1 + p

2

¶or μ + σN−1

³p2

´< x < μ + σN−1

³1− p

2

´. The

following is a simple way to find a confidence interval for x ∼ N¡μ, σ2

¢.

• Step 1. Find the confidence interval −b < z < b for the standard normalrandom variable z. Just solve the equation P (−b < z < b) = N (b) −N (−b) = 2N (b)− 1 =the confidence interval.

• Step 2. After finding b, the confidence interval for x is μ−bσ < x < μ+bσ

Example. Find the 90% confidence interval for x ∼ N¡1, 22

¢. First, we find

the 90% confidence interval for the standard normal random variable z. Wesolve the equation:

P (−b < z < b) = N (b)−N (−b) = N (b)− [1−N (b)] = 2N (b)− 1 = 0.9So N (b) =

1 + 0.9

2= 0.95 and b = 1.645.

Then the 90% confidence interval for z is −1.645 < z < 1.645; for x is1− 1.645× 2 < x < 1 + 1.645× 2, which is −2. 29 < x < 4. 29.Similarly, the 99% confidence interval is calculated as follows:

2N (b)− 1 = 0.99, 2N (b) = 1 + 0.99

2= 0.995, b = 2.576

The 99% confidence interval for z is −2.576 < z < 2.576; for x ∼ N¡1, 22

¢is 1− 2.576× 2 < x < 1 + 2.576× 2, which is −4. 152 < x < 6. 152.

Now let’s walk through DM example 18.6. Here we need to find the 95%confidence interval for the stock price St. The inputs are:

• S0 = 100

• t = 2



• α = 0.1

• σ = 0.3

• δ = 0

We know that lnStS0∼ N

£¡a− δ − 0.5σ2

¢t, σ2t

¤.¡

a− δ − 0.5σ2¢t =

¡0.1− 0− 0.5× 0.32

¢2 = 0.11

σ2t = 0.32 (2) = 0.18

=⇒ lnS2100

∼ N (0.11, 0.18)

How to find the 95% confidence interval for S2.Step 1 What’s the 95% confidence interval for z ∼ N (0, 1)?Answer: −1.96 < z < 1.96.

Step 2 What’s the 95% confidence interval for lnSt100

∼ N (0.11, 0.18)?

Answer: 0.11− 1.96√0.18 < ln

S2100

< 0.11 + 1.96√0.18,

or −0.721 56 < ln S2100

< 0.941 56.

Step 3 Take exponentiation of −0.721 56 < ln S2100

< 0.941 56

exp (−0.721 56) < expµln

S2100

¶< exp (0.941 56)

exp (−0.721 56) = 0.485 994 exp (0.941 56) = 2. 563 98 exp

µln

S2100

¶=

S2100

=⇒ 0.485 994 <S2100

< 2. 563 98 48.60 < S2 < 256.40

So the 95% confidence interval for S2 is (48.60, 256.40) .

Now let’s walk through Row 1 (i.e. 1 day horizon) in DM Table 18.1. Theinputs are:

• S0 = 50

• t = 1/365 (i.e. 1 day)

• α = 0.15

• δ = 0

• σ = 0.3



We need to find the confidence interval that corresponds to −1 < z < 1and the confidence interval that corresponds to −2 < z < 2. First, let consider−1 < z < 1.

lnStS0∼ N

£¡a− δ − 0.5σ2

¢t, σ2t

¤ln

S1/36550

∼ N

µ0.15− 0− 0.5× 0.32

365,0.32

365

¶= N

µ2. 876 71× 10−4, 0.3

2

365

¶What’s the confidence interval for the standard normal random variable z?Answer: −1 < z < 1.

What’s the corresponding confidence interval for lnS1/36550

?

Answer:

2. 876 71× 10−4 − 1×r0.32

365< ln

S1/36550

< 2. 876 71× 10−4 + 1×r0.32

365

−1. 541 505× 10−2 < lnS1/36550

< 1. 599 04× 10−2

50e−1. 541 505×10−2

< S1/365 < 50e1. 599 04×10−2

49. 24 < S1/365 < 50. 81

So the confidence interval for S1/365 is 49. 24 < S1/365 < 50. 81.

What confidence interval is that?P (−1 < z < 1) = N (1)−N (−1) = 2N (1)− 1 = 2× 0.841 34− 1 = 0.682 7

So the confidence interval is 68.27%. So there’s 68.27% chance that S1/365will fall in the range (49. 24, 50. 81).

Similarly, we can calculate the confidence interval that corresponds to −2 <z < 2.

2. 876 71× 10−4 − 2×r0.32

365< ln

S1/36550

< 2. 876 71× 10−4 + 2×r0.32

365

−3. 111 776 × 10−2 < lnS1/36550

< 3. 169 311× 10−2

50e−3. 111 776×10−2

< S1/365 < 50e3. 169 311×10−2

48. 47 < S1/365 < 51. 61

So the confidence interval for S1/365 is 48. 47 < S1/365 < 51. 61.What confidence interval is that?P (−2 < z < 2) = 2N (2)− 1 = 2 (0.977 25)− 1 = 0.954 5So the 95.45% confidence interval for S1/365 is 48. 47 < S1/365 < 51. 61.



18.4.2 Conditional expected prices

One key formula is DM 18.27. The partial expectation of St on the conditionSt < K is:RK

0St × g (St;S0) dSt = E (St)N

µ−∧d1

¶= S0e

(α−δ)tN

µ−∧d1

¶∧d1 =

lnS0K+¡α− δ + 0.5σ2

¢t

σ√t

g (St;S0) is the probability density function of St. Here St;S0 indicates thatthe initial stock price for St is S0.

DM 18.27 is a little difficult to derive. I recommend that you don’t botherhow to prove DM 18.27. Just memorize it. Once you memorize DM 18.27, youcan derive DM 18.28:

E (St|St < K) =

RK0

St × g (St;S0) dSt

P (St < K)=

S0e(α−δ)tN

µ−∧d1

¶N

µ−∧d2

¶Once you memorize DM 18.27, you can also derive DM 18.29:R∞K

St × g (St;S0) dSt =R∞0

St × g (St;S0) dSt −RK0

St × g (St;S0) dStR∞0

St × g (St;S0) dSt = E (St) = S0e(α−δ)tRK

0St × g (St;S0) dSt = E (St)N

µ−∧d1

¶= S0e

(α−δ)tN

µ−∧d1

¶

=⇒R∞K

St × g (St;S0) dSt = S0e(α−δ)t

∙1−N

µ−∧d1

¶¸= S0e

(α−δ)tµ1−

∙1−N

µ∧d1

¶¸¶= S0e

(α−δ)tN

µ∧d1

¶Then you can derive DM 18.30:

E (St|St > K) =

R∞K

St × g (St;S0) dSt

P (St > K)=

S0e(α−δ)t

∙1−N

µ−∧d1

¶¸1−N

µ−∧d2

¶ =

S0e(α−δ)tN

µ∧d1

¶N

µ∧d2

¶

18.4.3 Black-Scholes formula

This section derives the Black-Scholes formula with simple math. The Black-Scholes formula was originally derived using stochastic calculus and partial dif-ferential equation. Many years after the Black-Scholes formula was published,someone came up with this simple proof (hindsight is always 20-20).


18.5. ESTIMATINGTHE PARAMETERS OF A LOGNORMALDISTRIBUTION163

The European call option price in the risk-neutral world, is:C = e−rt

R∞K(St −K)×g∗ (St;S0) dSt = e−rtE∗ (St −K|St > K)P ∗ (St > K)

Here any term with the ∗ sign indicates the term is valuated in the risk-neutral world (i.e. by setting α = r).

E∗ (St −K|St > K)P ∗ (St > K) = E∗ (St|St > K)P ∗ (St > K)−E∗ (K|St > K)P ∗ (St > K)

E∗ (St|St > K) =S0e

(r−δ)tN (d1)

N (d2)P ∗ (St > K) = N (d2)

=⇒ E∗ (St|St > K)P ∗ (St > K) = S0e(r−δ)tN (d1)

E∗ (K|St > K) = K=⇒ E∗ (K|St > K)P ∗ (St > K) = KN (d2)

=⇒ C = e−rt£S0e

(r−δ)tN (d1)−KN (d2)¤= S0e

−δtN (d1)−Ke−rtN (d2)

Similarly, the European put price is:P = Ke−rtN (−d2)− S0e

−δtN (−d1)

18.5 Estimating the parameters of a lognormal

distribution

Consider the stock price over the time interval [t− h, t].

St = St−he(α−δ−0.5σ2)h+σ

√hz

where z is the standard normal random variable withE (z) = 0 and V ar (Z) = 1

=⇒ lnSt

St−h=¡α− δ − 0.5σ2

¢h+ σ

√hz

=⇒ E

µln

StSt−h

¶= E

h¡α− δ − 0.5σ2

¢h+ σ

√hzi=¡α− δ − 0.5σ2

¢h+

σ√hE (z) =

¡α− δ − 0.5σ2

¢h

=⇒ V ar

µln

StSt−h

¶= V ar

h¡α− δ − 0.5σ2

¢h+ σ

√hzi

Since¡α− δ − 0.5σ2

¢h is a constant, we have:

V arh¡α− δ − 0.5σ2

¢h+ σ

√hzi=³σ√h´2

V ar (z) = σ2h

Now let’s go through DM Table 18.2.



week S lnStSt−1

µln

StSt−1

¶21 1002 105.04 0.049171 0.00241783 105.76 0.006831 0.00004674 108.93 0.029533 0.00087225 102.5 −0.060843 0.00370186 104.8 0.022191 0.00049247 104.13 −0.006414 0.0000411

sum 0 0.040470 0.0075721

We have:

E

µln

StSt−h

¶=¡α− δ − 0.5σ2

¢h =

0.040470

6= 0.006 745

=⇒ α = δ + 0.5σ2 +0.006 745

h

In the above equations,α is the stock’s expected (annual) continuously compounded returnδ is the stock’s (annual) continuously compounded dividend yield (δ = 0 in

this problem)σ is the stock’s (annual) volatilityh = 1/52 (52 weeks in one year)

V ar

µln

StSt−h

¶= σ2h =

n

n− 1

⎛⎜⎜⎝ 1n ×Pµln

StSt−1

¶2−

⎛⎜⎜⎝Pln

StSt−1n

⎞⎟⎟⎠2⎞⎟⎟⎠ =

6

5

µ1

6× 0.0075721− 0.006 7452

¶= 0.001 459 83

=⇒ σ2 =0.001 459 83

h

=⇒ α = δ + 0.5σ2 +0.006 745

h= 0 + 0.5× 0.001 459 83

1/52+0.006 745

1/52=

0.388 7

The textbook talks about making two adjustments before calculating α. Ifound this approach confusing. I recommend that you use my approach.

=⇒ σ =

r0.001 459 83

h=

r0.001 459 83

1/52= 0.275 5

So the stock’s expected return per year is α = 38.8 7% and its annual volatil-ity is σ = 0.275 5.

Please note that Eµln

StSt−h

¶=1

n

Pln

StSt−h

=1

n

P(lnSt − lnSt−h) =

1

n(lnSn − lnS0) =

1

nln

SnS0=¡α− δ − 0.5σ2

¢h


18.6. HOW ARE ASSET PRICES DISTRIBUTED 165

In this problem, we could have estimated

E

µln

StSt−h

¶=¡α− δ − 0.5σ2

¢h =

1

nln

SnS0=1

6ln104.13

100= 0.006745

In addition, we have:

α− δ − 0.5σ2 = 1

nhln

SnS0=1

Tln

SnS0

where T = nh is the length of the observation.

We see that when we estimate α − δ − 0.5σ2, only the following 3 factorsmatter:

• the length of the observation T

• the stock’s starting price S0

• the stock’s ending price Sn

The stock prices between S0 and Sn are irrelevant. And for a given T ,increasing the number of observations n doesn’t affect our estimate of α− δ −0.5σ2 (because as n goes up, h goes down, and T = nh is a constant). Frequentobservations don’t improve our estimate of α − δ − 0.5σ2. To improve ourestimate of α− δ − 0.5σ2, we need to increase T .

When we estimate σ, the in-between stock prices do matter and more fre-quent observations do improve our estimate.

18.6 How are asset prices distributedThe lognormal stock price model assumes that the stock returns are normallydistributed. Are stock returns really normally distributed?

18.6.1 Histogram

One method to assess whether stock returns are normally distributed is to plotthe continuously compounded returns as a histogram. Look at DM Figure 18.4.

The histograms in Figure 18.4 don’t appear normal. The textbook offerstwo explanations:

• Stock prices can jump discretely from time to time

• Stock returns are normally distributed, but the variance of the returnchanges over time.

That’s all your need to know about this section.



18.6.2 Normal probability plots

Another method to check whether stock returns are normally distributed is todraw a normal probability plot.To see what a normal probability plot is, let’s go through an example. We

know that the 99% confidence interval for a standard normal random variable is[−2.58, 2.58]. Suppose we take the following samples from the range [−2.58, 2.58]with the step equal to 0.01:−2.58,−2.57,−2.56,−2.55, ..., 0, 0.01, 0.02, ..., 2.57, 2.58

The for each number xi sampled, we calculate yi = P (z < xi) = N (xi). Ifwe plot the data pairs (xi, yi), what do we get? Answer: we’ll get roughly astraight line.

You can experiment this using Excel. Here is a screen shot:A B

1 xi yi = P (Z < xi)2 −2.58 0.004943 −2.57 0.0050854 −2.56 0.005234... ... ...260 0 0.5261 0.01 0.503989262 0.02 0.507978... ... ...517 2.57 0.994915518 2.58 0.99506

Sample formulas.Cell A2 = −2.58 A3 = A2 + 0.01 ... A518 = A517 + 0.01B2 = NORMDIST (A2, 0, 1, 1) B3 = NORMDIST (A3, 0, 1, 1) ... B518 =

NORMDIST (A518, 0, 1, 1)

Next, plot the range "A2:B518" as xy diagram in Excel. The plot you gotshould look like the 2nd diagram in DM Figure 18.2 (see DM page 589). Youcan see that from diagram from z = −2.58 to z = 2.58 is roughly a straight line.The point of this Excel experiment: if you plot samples from a normal

distribution in a normal probability plot, you’ll get roughly a straight-line. Ifyou plot samples from an unknown distribution in a normal probability plot andget roughly a straight line, then the samples are from approximately normaldistribution.Steps on how to construct a normal probability plot:

1. Sort the samples from smallest to biggest. Number the samples as i =1, 2, 3, ..., n. So you samples are x1, x2, ..., xn.


18.7. SAMPLE PROBLEMS 167

2. Assign the cumulative probability to xi as yi = P (x < xi) =i− 0.5

n.Here

0.5 is the continuity adjustment. Calculate the corresponding zi = N−1 (yi).

3. Graph the data pairs (xi, yi) or (xi, zi)

If the graph is roughly a straight line, then x1, x2, ..., xn are approximatelynormal.Here’s a table to summarize the 3 steps for building a normal probability

plot:

index xi yi = P (x < xi) =i− 0.5

nzi = N−1 (yi)

1 x1 y1 =0.5

nz1 = N−1

µ0.5

n

¶2 x2 y2 =

1.5

nz2 = N−1

µ1.5

n

¶3 x3 y3 =

2.5

nz3 = N−1

µ2.5

n

¶... ... ... ...

n xn yn =n− 0.5

nzn = N−1

µn− 0.5

n

¶Let’s walk through DM Example 18.9.

i xi yi = P (x < xi) =i− 0.5

nzi = N−1 (yi)

1 30.5

5= 0.1 N−1 (0.1) = −1.2816

2 41.5

5= 0.3 N−1 (0.3) = −0.5244

3 52.5

5= 0.5 N−1 (0.5) = 0

4 73.5

5= 0.7 N−1 (0.7) = 0.5244

5 114.5

5= 0.9 N−1 (0.9) = 1.2816

Next, plot the data pairs (xi, yi) or (xi, zi). The result is the top two dia-grams in DM Figure 18.6.

18.7 Sample problems

Problem 1

• Stock prices follow the lognormal distribution

• S0 = 100



• α = 0.12

• δ = 0.06

• σ = 0.25

• T = 2

Construct the 95% confidence interval of ST .

Solution

First we find the 95% confidence interval for the standard normal randomvariable z. We need to b such that P (−b < z < b) = 0.95.

P (−b < z < b) = N (b)−N (−b) = N (b)− [1−N (b)] = 2N (b)− 1 = 0.95N (b) =

1 + 0.95

2= 0.975 b = N−1 (0.975) = 1. 96

So the 95% confidence interval for z is (−1.96, 1.96).

lnSTS0∼ N

£¡a− δ − 0.5σ2

¢T, σ2T

¤¡a− δ − 0.5σ2

¢T =

¡0.12− 0.06− 0.5× 0.252

¢2 = 0.057 5

σ2T = 0.252 × 2

The 95% confidence interval for lnSTS0

is:

0.057 5− 1.96√0.252 × 2 = −0.635 46

0.057 5 + 1.96√0.252 × 2 = 0.750 46

The 95% confidence interval for ST is:100e−0.635 46 = 52. 97 100e0.750 46 = 211. 80

The 95% confidence interval for ST is (52. 97, 211. 80)

Problem 2 (Spring 2007 Exam C #4)You are given the following information for a stock with current price 0.25:

• The price of the stock is lognormally distributed with continuously com-pounded expected annual rate of return α = 0.15.

• The dividend yield of the stock is zero.

• The volatility of the stock is σ = 0.35 .



Using the procedure described in the McDonald text, determine the upperbound of the 90% confidence interval for the price of the stock in 6 months.Solution90% confidence interval for a standard normal random variable is (−1.645, 1.645).ln

STS0∼ N

£¡a− δ − 0.5σ2

¢T, σ2T

¤¡a− δ − 0.5σ2

¢T =

¡0.15− 0− 0.5× 0.352

¢0.5 = 0.044 375

σ2T = 0.352 × 0.5

90% confidence interval for lnSTS0is¡0.044 375− 1.645×

√0.352 × 0.5, 0.044 375 + 1.645×

√0.352 × 0.5

¢.

So the upper bound for ST is:0.25e0.044 375+1.645×

√0.352×0.5 = 0.392 7

Problem 3A European call option and a European put option are written on the same

stock. You are given:

• S0 = 100

• K = 105

• α = 0.08

• δ = 0

• σ = 0.3

• T = 0.5

Calculate the probability that the call option will be exercised.Calculate the probability that the put option will be exercised.

SolutionThe call will be exercised is ST > K. We just need to use DM 18.24.

P (ST > K) = N

µ∧d2

¶= N

⎛⎜⎝ ln S0K +¡α− δ − 0.5σ2

¢t

σ√t

⎞⎟⎠= N

⎛⎜⎝ ln 100105 +¡0.08− 0− 0.5× 0.32

¢0.5

0.3√0.5

⎞⎟⎠= N (−0.147 5 ) = 1−N (0.147 5 ) = 1−NormalDist (0.147 5 ) = 0.441 4

The put option will be exercise is ST < K

P (ST < K) = N

µ−∧d2

¶= 1−N

µ∧d2

¶= 1− 0.441 4 = 0.558 6



Alternatively,

P (ST < K) = 1− P (ST > K) = 0.558 6Problem 4

• S0 = 50

• α = 0.1

• δ = 0

• σ = 0.4

Calculate the conditional expected value of the stock at T = 3 given ST > 75.Calculate the conditional expected value of the stock at T = 3 given ST < 75.

Solution

Need to use DM 18.28 and 18.29.

E (St|St < K) =

S0e(α−δ)tN

µ−∧d1

¶N

µ−∧d2

¶ E (St|St > K) =

S0e(α−δ)tN

µ∧d1

¶N

µ∧d2

¶

∧d1 =

lnS0K+¡α− δ + 0.5σ2

¢t

σ√t

=ln50

75+¡0.1− 0 + 0.5× 0.42

¢3

0.4√3

= 0.194 2

∧d2 =

lnS0K+¡α− δ − 0.5σ2

¢t

σ√t

=ln50

75+¡0.1− 0− 0.5× 0.42

¢3

0.4√3

= −0.498 6

E (St|St < K) =50e(0.1−0)3N (−0.194 2)

N (0.498 6)=50e(0.1−0)30.423 0

0.6910= 41. 32

E (St|St > K) =50e(0.1−0)3N (0.194 2)

N (−0.498 6) ==50e(0.1−0)3 (1− 0.423 0)

1− 0.6910 = 126.

03Problem 5Stock prices follow lognormal distribution. You are given:

• S0 = 100

• α = 0.08

• δ = 0

• σ = 0.3

• T = 5



Calculate the 75 percentile of ST . Calculate the median of ST .Solution75 percentile of a standard normal random variable z isN−1 (0.75) = 0.674 5.

lnSTS0∼ N

£¡a− δ − 0.5σ2

¢T, σ2T

¤¡a− δ − 0.5σ2

¢T =

¡0.08− 0− 0.5× 0.32

¢5 = 0.175

σ2T = 0.32 × 5

75 percentile of lnSTS0

is

x = 0.175 + 0.674 5×√0.32 × 5 = 0.627 47

75 percentile of ST is:100ex = 100e0.627 47 = 187. 29

Median is 50 percentile, which corresponds to z = 0 and x = 0.175.Median of ST is: 100e0.175 = 119. 12

Problem 6 (Spring 2007 C #34)The price of a stock in seven consecutive months is:

Month Price1 542 563 484 555 606 587 62

Based on the procedure described in the McDonald text, calculate the an-nualized expected return of the stock.(A) Less than 0.28(B) At least 0.28, but less than 0.29(C) At least 0.29, but less than 0.30(D) At least 0.30, but less than 0.31(E) At least 0.31

Solution



month price x = lnStSt−1

x2

1 542 56 0.036368 0.0013233 48 −0.154151 0.0237624 55 0.136132 0.0185325 60 0.087011 0.0075716 58 −0.033902 0.0011497 62 0.066691 0.004448

sum 0.138150 0.056785

E

µln

StSt−h

¶=¡α− δ − 0.5σ2

¢h =

0.138150

6= 0.023 025

=⇒ α = δ + 0.5σ2 +0.023 025

h

In the above equations,α is the stock’s expected (annual) continuously compounded returnδ is the stock’s (annual) continuously compounded dividend yield (δ = 0 in

this problem)σ is the stock’s (annual) volatilityh = 1/12 (12 months in one year)

V ar

µln

StSt−h

¶= σ2h =

n

n− 1

⎛⎜⎜⎝ 1n ×Pµln

StSt−1

¶2−

⎛⎜⎜⎝Pln

StSt−1n

⎞⎟⎟⎠2⎞⎟⎟⎠ =

6

5

µ1

6× 0.056785− 0.023 0252

¶= 0.010 720 8

σ2 =0.010 720 8

h

=⇒ α = δ + 0.5σ2 +0.023 025

h= 0 + 0.5 × 0.010 720 8

1/12+0.023 025

1/12=

0.340 624 8

The answer is E.


Chapter 19

Monte Carlo valuation

In this chapter, I’m going to walk you through some examples. If you understandmy examples, you’ll know enough for the exam.

19.1 Example 1 Estimate E (ez)

Let’s forget about option pricing for now and focus on a random variable X.Suppose we want to find out E (X), the mean of X, but we don’t have an easyformula to do so. For example, we have a random variable X = ez where z is astandard normal random variable with mean 0 and variance 1.

E (X) =R∞−∞ x (z) f (z) dz =

R∞−∞ exp (z)

1√2πexp

¡−0.5z2

¢dz

Suppose we haven’t heard of DM 18.13 E (ex) = exp¡m+ 0.5v2

¢.

We don’t know how to do the integrationR∞−∞ exp (z)

1√2πexp

¡−0.5z2

¢dz.

How can we calculate E (X)?One approach is to create n instances of z , calculate the corresponding

value of X = ez, and find the sample mean μ =1

n

µnPi=1

Xi

¶. Based on the cen-

tral limit theorem, μ =1

n

µnPi=1

Xi

¶is normally distributed with mean E (μ) =

1

nE

µnPi=1

Xi

¶=1

n(nE (X)) = E (X) and variance V ar (μ) =

1

n2V ar

µnPi=1

Xi

¶=

1

n2×nV ar (X) =

V ar (X)

n. Since E (μ) = E (X), we can calculate the mean of

the sample mean E (μ) and use it to approximate the population mean E (X).

Let’s come back to the example of estimating E (ez). I’m going to produce 10sample means and the use the average of these 10 sample means as an estimateto E (ez). To produce each sample mean, I’m going to randomly draw 10, 000z0s (so I have 10, 000 values of ez to produce each sample mean). Another way

173


174 CHAPTER 19. MONTE CARLO VALUATION

to describe this process is to say that I’m going to have 10 trials with each trialhaving 10, 000 samples.I’m going to do this in Microsoft Excel. The following table shows how to

calculate the sample mean for one trial.

A B C1 simulation i zi Xi = ezi

2 1 −0.042457 0.9584323 2 0.253383 1.2883764 3 0.017632 1.0177885 4 −1.167590 0.3111166 5 0.273333 1.3143387 6 −0.206665 0.8132928 7 0.305234 1.356942

...... ... ... ...10001 10000 −1.506862 0.22160410002 Total 16, 479.27961010003 sample mean 1.647928

Sample formulas in Excel:B2=NORMINV(RAND(),0,1) = −0.042457 C2 = e−0.042457 = 0.958 432

Let’s walk through B2. Rand() is Excel’s formula for a random draw in auniform distribution over (0, 1). Norminv(p, μ, σ) is Excel’s formula for the in-verse normal distribution. For example, for a standard normal random variablez ∼ N (0, 1), P (z ≤ 1.96) = N (1.96) = 0.975. Then Norminv(0.975, 0, 1) =N−1 (0.975) = 1.96. Another example. For a normal random variable Z ∼N¡2, 0.32

¢, P (Z ≤ 2) = 0.5. Then Norminv(0.5, 2, 0.3) = 2.

NORMINV(RAND(),0,1) works this way. First, Rand() generates a numberfrom a uniform distribution over (0, 1). Say this random number is 0.4831. ThenExcel finds that P (z ≤ −0.042457) = 0.4831. Hence a =NORMINV(0.4831,0,1) =−0.042457.

Please note that Excel has a similar formula normsinv for calculating theinverse normal random variable. The "s" in normsinv stands for standard. Sonormsinv produces the inverse standard normal distribution, while norminv pro-duces the inverse normal distribution with mean μ and standard deviation σ. Inother words, normsinv(p) =norminv(p, μ = 0, σ = 1). So normsinv(0.975) =norminv(0.975, 0, 1) =1.96.Similarly, the formula for B3 and C3 are:

B3=NORMINV(RAND(),0,1) = 0.253383 C3 = e0.253383 = 1. 288 376B10001=NORMINV(RAND(),0,1) = −1.506862 C10001 = e−1.506862 =

0.221 604

C10002=sum(C2:C10001)=16, 479.279610


19.1. EXAMPLE 1 ESTIMATE E¡EZ¢

175

The sample mean is

μ =1

10000

µnPi=1

Xi

¶=16, 479.279610

10000= 1.647928

Next, I’m going to produce 9 more trials. Here is the snapshot of Excel:

Trial 1 Trial 2 ... Trial 10i zi Xi = ezi zi Xi = ezi zi Xi = ezi

1 −0.042457 0.958432 −0.985719 0.373171 −0.783098 0.4569882 0.253383 1.288376 −1.324269 0.265997 0.186120 1.2045673 0.017632 1.017788 −0.157924 0.853915 0.504305 1.655835...10000 −1.506862 0.221604 −1.673002 0.187683 1.177804 3.247236PX 16, 479 16, 512 16, 554

μ 1.6479 1.6512 1.6554

Now I have 10 sample means (k = 10):Trial μ μ2

1 1.6479 2.7155742 1.6512 2.7264613 1.6163 2.6124264 1.6325 2.6650565 1.6821 2.829466 1.6626 2.7642397 1.6656 2.7742238 1.5994 2.558089 1.6459 2.70898710 1.6554 2.740349total 16.4589 27.09486

E (X) = E (μ) ' 1

10

Pμi =

1.6479 + 1.6512 + ...+ 1.6554

10= 1.64589

So E (X) ' 1.64589. The correct value of E (X) according to DM 18.13 isE (X) = e0.5 = 1. 648 721.We can also construct the confidence interval of E (μ). The estimated vari-

ance of the sample mean μ is

V ar (μ) 'P¡

μi − μ2¢

k − 1 =

Pμ2i − k (

Pμi)

2

k − 1 =k

k − 1

Ã1

k

Pμ2i −

µ1

k

Pμi

¶2!=

10

9

¡2.709486− 1.645892

¢= 0.000 591 2

pV ar (μ) '

√0.000 591 2 = 0.02 431

The 95% confidence interval is [1. 598 24, 1. 693 54]. This is calculated asfollows:



First, we calculate the critical value z∗ where P (z ≤ z∗) =1− 0.952

=

0.025 . This gives us z∗ = −1.96.Then lower bound of the 95% confidence interval is1.64589− 1.96 (0.02 431) = 1. 598 242 4

Then upper bound of the 95% confidence interval is1.64589 + 1.96 (0.02 431) = 1. 693 537 6So we are 95% certain that the sample meanE (μ) is in the range [1. 598 24, 1. 693 54].

SinceE (X) = E (μ), we are 95% certain that E (X) is in the range [1. 598 24, 1. 693 54].

The 99% confidence interval is calculated as follows:

First, we calculate the critical value z∗ where P (z ≤ z∗) =1− 0.992

=

0.005 . This gives us z∗ = −2.58.Then lower bound of the 99% confidence interval is1.64589− 2.58 (0.02 431) = 1. 583 170 2Then upper bound of the 99% confidence interval is1.64589 + 2.58 (0.02 431) = 1. 708 609 8So we are 99% certain that E (X) is in the range [1. 583 2, 1. 708 6]

Some other key points in this example.This method to generate a normal random variable is called the inversion

method (DM textbook page 622 and 623). This method works as follows:

• Step 1. Generate a random number p from the uniform distribution over(0, 1). Say p = 0.4831.

• Step 2. Ask "For a normal random variable Z with mean μ and standarddeviation σ, what’s the number a such that P (Z ≤ a) = p?" In this ex-ample, we find P (z ≤ a) = 0.4831. Then a = −0.042457 is the simulatednormal random variable.

We can also generate the normal random variable as the sum of 12 uniformlydistributed random variables minus 6 (DM textbook 622):∼Z = u1+u2+ ...+u12−6 where ui is an independent identically distributed

uniform random variable over (0, 1).E (ui) =

R 10uf (t) du=

R 10udu = 1/2 E

¡u2i¢=R 10u2f (t) du=

R 10u2du =

1/3

V ar (ui) =1

3−µ1

2

¶2=1

12

Based the central limit theorem, u1+u2+ ...+u12 is approximately normal.The normal random variable with a constant −6 added is still normal. Hence∼Z is normal.


19.2. EXAMPLE 2 ESTIMATE π 177

E

µ∼Z

¶= E (u1 + u2 + ...+ u12)− 6 = 12×

1

2− 6 = 0

V ar

µ∼Z

¶= V ar (u1 + u2 + ...+ u12) = 12×

1

12= 1

So∼Z roughly a standard normal random variable.

The following snapshot shows how to create 5 standard normal random vari-ables in Excel:

ui trial 1 trial 2 trial 3 trial 4 trial 51 0.6316 0.3308 0.0062 0.3634 0.32072 0.0247 0.6775 0.0296 0.8678 0.73613 0.4223 0.3489 0.8446 0.4283 0.83084 0.1326 0.5315 0.3436 0.3937 0.56585 0.1944 0.6660 0.2083 0.4484 0.60796 0.5996 0.3129 0.5807 0.3446 0.28827 0.7959 0.7749 0.9899 0.5553 0.92428 0.5468 0.2398 0.7499 0.5410 0.79009 0.1719 0.0645 0.0679 0.8638 0.039410 0.0964 0.4656 0.1779 0.8405 0.571811 0.1059 0.9432 0.3354 0.6266 0.490212 0.6106 0.0310 0.6455 0.8713 0.7856sum 4.3328 5.3868 4.9794 7.1447 6.9508sum-6 -1.6672 -0.6132 -1.0206 1.1447 0.9508Each uniform random variable is created using Excel’ formula Rand().

The last row show the 5 standard normal random variable created.For example, the first standard normal random variable is created as follows:(0.6316 + 0.0247 + 0.4223 + ...+ 0.6106)− 6 = −1.6672

This method is more cumbersome than the inversion method.

19.2 Example 2 Estimate π

We decide to estimate π using the Monte Carlo simulation (we can treat π asa random variable that happens to be a constant). We’ll estimate π using aclassic "throwing the dart" method.Imagine a square whose size is two units. The center of this square is the

origin (0, 0). Inside this square sits a unit circle x2 + y2 = 1. The center ofthis circle is also (0, 0). If we randomly throw a dart at the square, what’s theprobability that the dart falls within the unit circle?



-1.0 -0.8 -0.6 -0.4 -0.2 0.2 0.4 0.6 0.8 1.0

-1.0

-0.8

-0.6

-0.4

-0.2

0.2

0.4

0.6

0.8

1.0

x

y

Of all the darts falling into the square (we ignore the darts falling out of thesquare), the probability that the dart falls within the circle is the area of the

circle divided by the area of the square, which is P =π¡12¢

22=

π

4.

We can also focus on the first quadrant. Of all the darts falling in square in

the first quadrant, the probability that they fall in the circle is P =

1

4π¡12¢

12=

π

4. Then

∧π = 4P .

Next, we need to simulate darts falling in the first quadrant. Once again,we’ll do the simulation in Excel. We plan to produce 20 estimates of π. Foreach estimate, we’ll use 10, 000 simulations. Here’s how to produce one trial(one trial=10, 000 simulations).

• Create the first point (x1, y1), where 0 < x1 < 1 and 0 < y1 < 1. We cancreate x1 and y1 by randomly drawing two numbers from U ∼ (0, 1), auniform distribution over [0, 1].

• If x21 + y21 ≤ 1, then (x1, y1) falls in the circle in the first quadrant.

• Similarly, create the 2nd point (x2, y2) by randomly drawing two numbersfrom U ∼ (0, 1). Determine whether (x2, y2) falls in the circle in the firstquadrant.

• ......


19.2. EXAMPLE 2 ESTIMATE π 179

• Create n-th point (xn, yn) by randomly drawing two numbers from U ∼[0, 1]. Determine whether (xn, yn) falls in the circle in the first quadrant.Set n = 65, 000.

• Count m, the total number of points that fall in the circle.

• P ≈m

n. Then

∧π = 4P .

Here’s the snapshot of Excel.A B C D E

1 throw i xi yi x2i + y2i Fall in the circle? (1=Yes, 0=No)2 1 0.2854 0.1694 0.1102 13 2 0.5147 0.2787 0.3426 14 3 0.1915 0.2699 0.1095 15 4 0.6501 0.8420 1.1317 06 5 0.0007 0.6580 0.4330 17 6 0.1520 0.9772 0.9780 18 7 0.3179 0.5713 0.4274 1... ... ... ... ... ...

10, 001 10, 000 0.5560 0.5679 0.6317 1

The number of simulations: n = 10, 000.We find that the number of darts falling in the circle in the first quadrant:

m = 7, 894

P ≈m

n=7, 894

10, 000= 0.7894

∧π = 4 (0.7894) = 3. 157 6

Sample formulas in the above Excel spreadsheet are:B2=rand()=0.2854 C2=rand()=0.1694D2=B2^2+C2^2=0.1102 E2=if(D2>=1,1,0)=1

B3=rand()=0.5147 C3=rand()=0.2787D3=B3^2+C3^2=0.3426 E3=if(D3>=1,1,0)=1m=SUM(E2:E10001)=7, 894We produce 19 more trials. The total number of trials is k = 20. Here is the

snapshot of Excel:

trial mi pi∧πi

³ ∧πi

´21 7,894 0.7894 3.1576 9.970437762 7,818 0.7818 3.1272 9.779379843 7,830 0.783 3.1320 9.809424004 7,752 0.7752 3.1008 9.614960645 7,879 0.7879 3.1516 9.932582566 7,879 0.7879 3.1516 9.932582567 7,778 0.7778 3.1112 9.67956544



8 7,878 0.7878 3.1512 9.930061449 7,891 0.7891 3.1564 9.9628609610 7,896 0.7896 3.1584 9.9754905611 7,844 0.7844 3.1376 9.8445337612 7,880 0.788 3.1520 9.9351040013 7,920 0.792 3.1680 10.0362240014 7,821 0.7821 3.1284 9.7868865615 7,825 0.7825 3.1300 9.7969000016 7,870 0.787 3.1480 9.9099040017 7,886 0.7886 3.1544 9.9502393618 7,822 0.7822 3.1288 9.7893894419 7,839 0.7839 3.1356 9.8319873620 7,762 0.7762 3.1048 9.63978304total 62.7856 197.10829728

The estimated mean of the sample mean is

π ≈ E³∧π´=

P ∧πik

=62.7856

20= 3. 139 28

The estimated variance of the sample mean is:

V ar³∧π´' k

k − 1

Ã1

k

Pμ2i −

µ1

k

Pμi

¶2!=20

19

µ197.10829728

20− 3. 139 282

¶=

0.0003536 27pV ar (μ) '

√0.0003536 27 = 0.01880 5

The 95% confidence interval is [3. 102 4, 1. 3. 176 1]. This is calculated asfollows:3. 139 28− 1.96 (0.01880 5) = 3. 102 422 23. 139 28 + 1.96 (0.01880 5) = 3. 176 137 8

The 99% confidence interval is [3. 090 8, 3. 187 8]. This is calculated as fol-lows:3. 139 28− 2.58 (0.01880 5) = 3. 090 763 1Then upper bound of the 99% confidence interval is3. 139 28 + 2.58 (0.01880 5) = 3. 187 796 9

Now you have some ideas about the Monte Carlo simulation. Let’s look atthe textbook.

19.3 Example 3 Estimate the price of Europeancall or put options

Monte Carlo simulation can be used to price European options, especially whenthere’s no simple formula for the option price such as the arithmetic Asianoption.


19.3. EXAMPLE 3 ESTIMATE THE PRICE OF EUROPEAN CALLOR PUTOPTIONS181

First, let’s use the Monte Carlo method to calculate the price of a Europeancall option and put option. The inputs are (See DM page 377 Example 12.1):

• S0 = 41

• K = 40

• σ = 0.3

• r = 8%

• δ = 0

• T = 0.25 (3 months)

SolutionWe’ll use DM Equation 19.6 to calculate the option price:

V (S0, 0) =1

ne−rT

nPt=1

V¡SiT , T

¢= e−rT

nPt=1

1

nV¡SiT , T

¢nPt=1

1

nV¡SiT , T

¢is the average terminal payoff. So the price of the European

option is just the discounted value of the terminal payoff.

We’re are going to produce 5 sample means. For each sample mean, we’lluse 10 simulations. So we have 5 trials with 20 simulations per trial. For eachtrial, we need to generate the stock’s terminal price at T = 0.25.We generate the terminal stock price using DM 19.3:

ST = S0 exph¡α− δ − 0.5σ2

¢T + σ

√TZi

Please note that the above formula produces the real world stock price atT . If we use the real world terminal stock price to calculate the option price,you have to use the real world discount rate, which is path-dependent (see DMTable 19.1). Path dependent discount rates are difficult to calculate. To avoidthe difficulty of finding the path-dependent discount rates, we’ll want to live inthe risk neutral world where everything earns the risk-free rate. To move intothe risk neutral world, we just set α = r and change DM 19.3 into:

ST = S0 exph¡r − δ − 0.5σ2

¢T + σ

√TZi

S0.25 = 41 exp¡¡0.08− 0.5× 0.32

¢0.25 + 0.3

√0.25Z

¢= 41 exp (0.15Z + 0.008 75)

Next, we’ll produce 10 simulations of Z and S0.25. The following is thesnapshot of Excel simulation:

index Z S0.25 Call payoff put payoff1 0.67830 45.7901 5.7901 0.00002 0.65327 45.6185 5.6185 0.00003 1.21158 49.6034 9.6034 0.0000



4 1.50347 51.8234 11.8234 0.00005 0.62495 45.4251 5.4251 0.00006 -1.89286 31.1369 0.0000 8.86317 0.57282 45.0712 5.0712 0.00008 -0.84887 36.4154 0.0000 3.58469 -0.34767 39.2586 0.0000 0.741410 -0.34540 39.2720 0.0000 0.7280Total 429.4146 43.3317 13.9171Sample calculations.If Z = 0.67830, S0.25 = 41 exp (0.15× 0.67830 + 0.008 75) = 45. 790 1The call payoff is 45. 790 1− 40 = 5. 790 1The put payoff is zero.

The average call payoff at T is:43.3317

10= 4. 333 17

So the call price is 4. 333 17e−rT = 4. 333 17e−0.08×0.25 = 4. 247 4

The average put payoff at T is:13.9171

10= 1. 391 71

So the put price is 1. 391 71e−0.08×0.25 = 1. 364 2

Now I’m going to repeat this process 4 more times. This is my final result:trials call price put price1 4. 247 4 1. 364 22 3.9132 1.23913 0.5436 2.48754 1.6850 1.83425 2.4911 1.9250

The mean of the sample mean for the call price is:4. 247 4 + 3.9132 + 0.5436 + 1.6850 + 2.4911

5= 2. 58

Compare this with the correct price using the Black-Scholes formula (seeDM page 377): C = 3.40The mean of the sample mean for the put price is:1. 364 2 + 1.2391 + 2.4875 + 1.8342 + 1.9250

5= 1. 77

Compare this with the correct price using the Black-Scholes formula (seeDM page 377): P = 1.61The prices from the Monte Carlo simulations are off because the number of

simulations per trial is small.

After running 10 trials (5,000 simulations per trial), I got the following:trials call call^2 put put^21 3.4132 11.6502 1.5976 2.5523


19.3. EXAMPLE 3 ESTIMATE THE PRICE OF EUROPEAN CALLOR PUTOPTIONS183

2 3.3931 11.5134 1.6003 2.56113 3.4085 11.6182 1.6007 2.56224 3.4002 11.5614 1.6249 2.64025 3.4019 11.5729 1.5815 2.50126 3.4224 11.7129 1.6056 2.57787 3.4089 11.6206 1.6158 2.61088 3.4140 11.6551 1.6039 2.57259 3.3998 11.5587 1.5879 2.521410 3.4018 11.5725 1.6216 2.6297

Sum 34.0639 116.03585 16.0398 25.7292So the mean of the sample mean for call price is:34.0639

10= 3. 40639

The variance of the sample mean for the call price is:10

9

µ116.03585

10− 3. 406 392

¶= 1. 024 1× 10−4

√1. 024 1× 10−4 = 0.01

So the 95% confidence interval for the call price is [3. 39, 1. 3. 43]. This iscalculated as follows:3. 40639− 1.96 (0.01) = 3. 393. 40639 + 1.96 (0.01) = 3. 43

So the mean of the sample mean for put price is:16.0398

10= 1. 603 98

The variance of the sample mean for the put price is:10

9

µ25.7292

10− 1. 603 982

¶= 0.000 186 844

√0.000 186 844 = 0.014

So the 95% confidence interval for the call price is [1.58, 1.63]. This is cal-culated as follows:1. 603 98− 1.96 (0.014) = 1. 581. 603 98 + 1.96 (0.014) = 1. 63

By the way, there’s no need for us to use Monte Carlo summations to cal-culate a European call or put option. The Black-Scholes formula can producethe price of a European call or put option. Here we calculate European optionprices to illustrate how to use the Monte Carlo simulation.Please note that using the Monte Carlo simulation to calculate the American

option price is covered in DM 1936, which is out of the scope of exam MFE.



19.4 Example 4 Arithmetic and geometric op-tions

Let’s use the Monte Carlo method to calculate the price of an arithmetic calloption and an arithmetic put option. The inputs are (See DM page 629 Table19.3):

• S0 = 40

• K = 40

• σ = 0.3

• r = 8%

• δ = 0

• T = 0.25 (3 months)

• The average stock price is the average stock prices at the end of Month 1,Month2, and Month 3

Solution

The monthly interval is h = T/3 = 0.25/3The stock prices (the real world price) at the end of Month 1, 2, and 3 are:

end of Month 1: Sh = S0 exp³¡α− δ − 0.5σ2

¢h+ σ

√hz1

énd of Month 2: S2h = Sh exp

³¡α− δ − 0.5σ2

¢h+ σ

√hz2

énd of Month 3: S3h = ST = S2h exp

³¡α− δ − 0.5σ2

¢h+ σ

√hz3

´where z1, z2, and z3 are three separate random draws of the standard normal

distribution.

Since our goal is to calculate the option price, we don’t need the real worldstock price. We just need the risk-neutral stock price. We change the stock’sexpected return α into the risk free rate r. The risk neutral stock prices at theend of Month 1, 2, and 3 are:

Sh = S0 exp³¡r − δ − 0.5σ2

¢h+ σ

√hz1

´S2h = Sh exp

³¡r − δ − 0.5σ2

¢h+ σ

√hz2

´S3h = S2h exp

³¡r − δ − 0.5σ2

¢h+ σ

√hz3

´The average stock price is S =

Sh + S2h + S3h3

The arithmetic call price is: C = e−rT max¡S −K, 0

¢The arithmetic call price is: P = e−rT max

¡K − S, 0

¢


19.4. EXAMPLE 4 ARITHMETIC AND GEOMETRIC OPTIONS 185

The following snapshot of Excel shows one trial (5000 simulations per trial)of the call/put price:

i z1 Sh z2 S2h z3 S3h S C pay P pay1 0.8424 43.1529 0.258 44.2568 −2.4041 36.0434 41.1510 1.1510 0.00002 0.066 40.3468 −1.4415 35.7157 −2.0925 29.8830 35.3152 0.0000 4.68483 0.5357 42.0218 0.1255 42.6051 −0.8295 39.7677 41.4649 1.4649 0.00004 0.0014 40.1217 1.0027 43.8893 −0.8446 40.9128 41.6413 1.6413 0.00005 −1.598 34.9321 −1.5906 30.5258 1.2832 34.2134 33.2238 0.0000 6.77626 1.2843 44.8364 −0.6896 42.3605 −0.3961 41.0516 42.7495 2.7495 0.00007 −0.0712 39.8702 0.2363 40.8134 0.0147 40.9848 40.5561 0.5561 0.00008 −1.0075 36.7649 −0.5005 35.3082 0.6539 37.4745 36.5159 0.0000 3.4841... ... ... ... ... ... ... ... ... ...

5000 −0.0168 40.0585 −1.2203 36.1464 0.6609 38.3874 38.1974 0.0000 1.8026

Sample calculations for the first row.

First, we draw z1 = 0.8424, z2 = 0.258,and z3 = −2.4041 from the standardnormal distribution.

Sh = 40 exp

Ã¡0.08− 0− 0.5× 0.32

¢ 0.253+ 0.3

r0.25

3× 0.8424

!= 40 exp (0.075 87) =

43. 152 9

S2h = 43. 152 9 exp

Ã¡0.08− 0− 0.5× 0.32

¢ 0.253+ 0.3

r0.25

3× 0.258

!=

43. 152 9 exp¡2. 526 01× 10−2

¢= 44. 256 8

S3h = 44. 256 8 exp

Ã¡0.08− 0− 0.5× 0.32

¢ 0.253+ 0.3

r0.25

3× (−2.4041)

!=

44. 256 8 exp (−0.205 28) = 36. 0434S =

43. 152 9 + 44. 256 8 + 36. 0434

3= 41. 151 0

index Call payoff put payoff C P C2 P 2

1 1.1510 0.0000 1.1282 0.0000 1.2728 0.00002 0.0000 4.6848 0.0000 4.5920 0.0000 21.08653 1.4649 0.0000 1.4359 0.0000 2.0618 0.00004 1.6413 0.0000 1.6088 0.0000 2.5882 0.00005 0.0000 6.7762 0.0000 6.6420 0.0000 44.11626 2.7495 0.0000 2.6951 0.0000 7.2636 0.00007 0.5561 0.0000 0.5451 0.0000 0.2971 0.00008 0.0000 3.4841 0.0000 3.4151 0.0000 11.6629... ... ... ... ... ... ...

5000 0.0000 1.8026 0.0000 1.7669 0.0000 3.1219sum 10, 111.9410 7, 390.6006 9, 911.7119 7, 244.2548 61, 373.4417 32, 916.8904

Sample calculations for the first row.



Call price: e−rT×call payoff= e−0.08(0.25) × 1.1510 = 1. 128 2Put payoff: e−rT×put payoff= e−0.08(0.25) × 0.0000 = 0

C2 = 1. 128 22 = 1. 272 8 P 2 = 02 = 0

The estimated option prices are calculated as follows.

index Call payoff put payoff C P C2 P 2

sum 10, 111.9410 7, 390.6006 9, 911.7119 7, 244.2548 61, 373.4417 32, 916.8904

The average call price of these 5000 simulations:

e−rT × 1

5000

Pcall payoff= e−0.08(0.25) × 10111.9410

5000= 1. 982 34

or1

5000

PC =

9911.7119

5000= 1. 982 34

The average put price of these 5000 simulations:

e−rT × 1

5000

Pput payoff= e−0.08(0.25) × 7390.6006

5000= 1. 448 9

or1

5000

PP =

7244.2548

5000= 1. 448 9

The estimated variance of the call price per simulation:

n

n− 1

Ã1

n×P

C2 −µP

C

n

¶2!=

5000

5000− 1

µ1

5000× 61373.4417− 1. 982 342

¶=

8. 346 7

The estimated variance of the put price per simulation:

n

n− 1

Ã1

n×P

P 2 −µP

P

n

¶2!=

5000

5000− 1

Ã1

5000× 32916.8904−

µ7244.2548

5000

¶2!= 4. 485 1

The following is the snapshot of Excel for 30 trials (5000 simulations pertrial) for the arithmetic European call and put prices:



i C P C2 P 2

1 1.9823 1.4489 3.929513 2.0993112 1.9409 1.4486 3.767093 2.0984423 1.9467 1.4920 3.789641 2.2260644 2.0098 1.4211 4.039296 2.0195255 1.9894 1.4447 3.957712 2.0871586 1.9515 1.4886 3.808352 2.2159307 1.9825 1.4813 3.930306 2.1942508 2.0462 1.4408 4.186934 2.0759059 2.1270 1.3565 4.524129 1.84009210 1.9583 1.4675 3.834939 2.15355611 1.9878 1.4691 3.951349 2.15825512 1.9664 1.4496 3.866729 2.10134013 2.0147 1.3785 4.059016 1.90026214 1.9664 1.4327 3.866729 2.05262915 1.9915 1.4253 3.966072 2.03148016 1.9504 1.4592 3.804060 2.12926517 1.9414 1.4800 3.769034 2.19040018 1.9804 1.4186 3.921984 2.01242619 1.9666 1.4724 3.867516 2.16796220 2.0276 1.3779 4.111162 1.89860821 1.9940 1.4217 3.976036 2.02123122 1.9307 1.4815 3.727602 2.19484223 1.9221 1.4903 3.694468 2.22099424 1.9710 1.4491 3.884841 2.09989125 1.9185 1.4526 3.680642 2.11004726 1.9529 1.4638 3.813818 2.14271027 1.9791 1.4538 3.916837 2.11353428 1.9776 1.4676 3.910902 2.15385029 1.9713 1.5092 3.886024 2.27768530 1.9689 1.4640 3.876567 2.143296

sum 59.3139 43.5069 117.3193 63.1309

For the remaining part of the calculation, all you need to know is the fol-lowing:

Trial C P C2 P 2

sum 59.3139 43.5069 117.3193 63.1309

The estimated arithmetic call price is:

E¡C¢=59.3139

30= 1. 977 13

The estimated variance of the call price (per trial) is:



V ar¡C¢=

n

n− 1

⎛⎜⎜⎜⎝ 1n × n=30Pi=1

C2

i −

⎛⎜⎜⎝n=30Pi=1

Ci

n

⎞⎟⎟⎠2⎞⎟⎟⎟⎠ =

30

30− 1

µ1

30× 117.3193− 1. 977 12

¶=

0.001 778The standard deviation is:

√0.001 778 = 0.04 217

We can also estimate the variance of the call price as follows:For each trial, the average call price of the 5, 000 simulations is used to

estimate the call price. C =1

5, 000

5,000Pi=1

Ci.

So the variance of the call price per trial is:

V ar¡C¢=

1

5, 0002

µV ar

5,000Pi=1

Ci

¶=

1

5, 0002×5, 000V ar (C) = 1

5000V ar (C) =

8. 346 7

5000= 0.001 669

The standard deviation is:√0.001 669 = 0.0406

Please note that the two methods of estimating the variance of the call priceper trial often produce close, but not identical, results. The results are notidentical because the estimated variance per trial changes depending how manytrial you have and how many simulations per trial.

The estimated arithmetic put price is:

E¡P¢=43.5069

30= 1. 450 23

The estimated variance of the put price (per trial) is:

V ar¡P¢=

n

n− 1

⎛⎜⎜⎜⎝ 1n × n=30Pi=1

P2

i −

⎛⎜⎜⎝n=30Pi=1

P i

n

⎞⎟⎟⎠2⎞⎟⎟⎟⎠ =

30

30− 1

µ1

30× 63.1309− 1. 450 232

¶=

0.001 237 5The standard deviation is:

√0.001 237 5 = 0.03 518

We can also estimate the variance of the put price as follows:For each trial, the average call price of the 5, 000 simulations is used to

estimate the call price. P =1

5, 000

5,000Pi=1

Pi.

V ar¡P¢=

1

5, 0002

µV ar

5,000Pi=1

Pi

¶=

1

5, 0002×5, 000V ar (P ) = 1

5000V ar (P ) =

4. 4668

5000= 0.000 89

The standard deviation is:√0.000 89 = 0.0298

Now let’s use Monte Carlo simulation to estimate the geometric call and putprices. I’m going to use the same inputs:



• S0 = 40

• K = 40

• σ = 0.3

• r = 8%

• δ = 0

• T = 0.25 (3 months)

• The average stock price is the average stock prices at the end of Month 1,Month2, and Month 3

In addition, I’m going to use the same random draws of standard normalrandom variables z1, z2, and z3 as used for simulating the arithmetic call/putprices. Though I don’t have to use the z1, z2, and z3 used for estimating thearithmetic call/put prices, reuse saves me time. Here is a snapshot of Exceldoing one trial:

i z1 Sh z2 S2h z3 S3h S C pay P pay1 0.8424 43.1529 0.258 44.2568 −2.4041 36.0434 40.9831 0.9831 02 0.066 40.3468 −1.4415 35.7157 −2.0925 29.8830 35.0508 0 4.94923 0.5357 42.0218 0.1255 42.6051 −0.8295 39.7677 41.4466 1.4466 04 0.0014 40.1217 1.0027 43.8893 −0.8446 40.9128 41.6101 1.6101 05 −1.598 34.9321 −1.5906 30.5258 1.2832 34.2134 33.1662 0 6.83386 1.2843 44.8364 −0.6896 42.3605 −0.3961 41.0516 42.7209 2.7209 07 −0.0712 39.8702 0.2363 40.8134 0.0147 40.9848 40.5532 0.5532 08 −1.0075 36.7649 −0.5005 35.3082 0.6539 37.4745 36.5047 0 3.4953... ... ... ... ... ... ... ... .... ...

5000 −0.0168 40.0585 −1.2203 36.1464 0.6609 38.3874 38. 163 6 0 1. 836 4sum 9, 911.3146 7, 528.5134

Sample calculations.The 1st simulation.S = (43.1529× 44.2568× 36.0434)1/3 = 40. 983 1Call payoff: 40. 983 1− 40 = 0.983 1Put payoff: 0

The 5000-th simulation:i z1 Sh z2 S2h z3 S3h S C pay P pay

5000 −0.0168 40.0585 −1.2203 36.1464 0.6609 38.3874 38. 163 6 0 1. 836 4

S = (40.0585× 36.1464× 38.3874)1/3 = 38. 163 6Call payoff: 0Put payoff: 40− 38. 163 6 = 1. 836 4



The next snapshot:i call payoff put payoff C P C2 P 2

1 0.9831 0 0.9636 0 0.9285 02 0 4.9492 0 4.8512 0 23.53413 1.4466 0 1.418 0 2.0107 04 1.6101 0 1.5782 0 2.4907 05 0 6.8338 0 6.6985 0 44.86996 2.7209 0 2.667 0 7.1129 07 0.5532 0 0.5422 0 0.2940 08 0 3.4953 0 3.4261 0 11.7382... ... ... ... ...

5000 0 1. 836 4 0 1. 800 0 0 3.24sum 9, 911.3146 7, 528.5134 9, 715.0581 7, 379.4385 59, 337.3153 33, 852.0593

Sample calculation.The 1st simulation.Call price: 0.9831e−0.08×0.25 = 0.963 6

Put price: 0× e−0.08×0.25 = 0

The 5000-th simulation.Call price: 0

Put price: 1. 836 4e−0.08×0.25 = 1. 800 0The average call price of these 5000 simulations:

e−rT × 1

5000

Pcall payoff= e−0.08(0.25) × 10111.9410

5000= 1. 982 34

or1

5000

PC =

9911.7119

5000= 1. 982 34

The following calculation uses this part of the table:i call payoff put payoff C P C2 P 2

sum 9, 911.3146 7, 528.5134 9, 715.0581 7, 379.4385 59, 337.3153 33, 852.0593

The average call price of these 5000 simulations:

e−rT × 1

5000

Pcall payoff= e−0.08(0.25) × 9911.3146

5000= 1. 943 0

or1

5000

PC =

9715.0581

5000= 1. 943 0

The estimated variance of the call price per simulation is:n

n− 1

Ã1

n×P

C2 −µP

C

n

¶2!=

5000

5000− 1

µ1

5000× 59337.3153− 1. 943 02

¶=

8. 093 8



The average put price of these 5000 simulations:

e−rT × 1

5000

Pput payoff= e−0.08(0.25) × 7528.5134

5000= 1. 475 9

or1

5000

PP =

7379.4385

5000= 1. 475 9

The estimated variance of the put price per simulation is:5000

5000− 1

µ1

5000× 33852.0593− 1. 475 92

¶= 4. 593 0

Next is the snapshot of the 30 trials (1 trial=5,000 simulations):

i C P C2 P 2

1 1. 943 0 1.4759 3.775249 2.1782812 1.9001 1.4754 3.610380 2.1768053 1.9080 1.5200 3.640464 2.3104004 1.9698 1.4476 3.880112 2.0955465 1.9506 1.4727 3.804840 2.1688456 1.9112 1.5171 3.652685 2.3015927 1.9436 1.5091 3.777581 2.2773838 2.0058 1.4675 4.023234 2.1535569 2.0854 1.3830 4.348893 1.91268910 1.9191 1.4958 3.682945 2.23741811 1.9487 1.4959 3.797432 2.23771712 1.9281 1.4775 3.717570 2.18300613 1.9748 1.4058 3.899835 1.97627414 1.9272 1.4608 3.714100 2.13393715 1.9519 1.4523 3.809914 2.10917516 1.9116 1.4863 3.654215 2.20908817 1.9039 1.5072 3.624835 2.27165218 1.9424 1.4453 3.772918 2.08889219 1.9274 1.5007 3.714871 2.25210020 1.9867 1.4049 3.946977 1.97374421 1.9535 1.4483 3.816162 2.09757322 1.8924 1.5100 3.581178 2.28010023 1.8845 1.5182 3.551340 2.30493124 1.9328 1.4771 3.735716 2.18182425 1.8804 1.4809 3.535904 2.19306526 1.9133 1.4916 3.660717 2.22487127 1.9399 1.4814 3.763212 2.19454628 1.9398 1.4947 3.762824 2.23412829 1.9318 1.5376 3.731851 2.36421430 1.9279 1.4921 3.716798 2.226362

sum 58.1356 44.3327 112.7048 65.5497



trial C P C2 P 2

sum 58.1356 44.3327 112.7048 65.5497

The estimated call price is:58.1356

30= 1. 937 85

The estimated put price is:44.3327

30= 1. 477 76

The estimated variance of the call price per trial:30

30− 1

µ1

30× 112.7048− 1. 937 852

¶= 0.001618

The standard deviation is:√0.001618 = 0.040 22

The estimated variance of the call per trial can also be calculated as follows:For each trial, the average call price of the 5, 000 simulations is used to

estimate the call price. C =1

5, 000

5,000Pi=1

Ci.

So the variance of the call price per trial is:

V ar¡C¢=

1

5, 0002

µV ar

5,000Pi=1

Ci

¶=

1

5000V ar (C) =

8. 093 8

5000= 0.001 619

The standard deviation is:√0.001 619 = 0.04 02

The estimated variance of the put price per trial:30

30− 1

µ1

30× 65.5497− 1. 477 762

¶= 0.001 257 3

The standard deviation is:√0.001 257 3 = 0.03545 8

The estimated variance of the call per trial can also be calculated as follows:For each trial, the average put price of the 5, 000 simulations is used to

estimate the put price. P =1

5, 000

5,000Pi=1

Pi.

So the variance of the put price per trial is:

V ar¡P¢=

1

5, 0002

µV ar

5,000Pi=1

Pi

¶=

1

5000V ar (P ) =

4. 593 0

5000= 0.000 918 6

The standard deviation is:√0.000 918 6 = 0.0303

Here’s a question. We know that the estimated standard variance of the putprice per simulation is 4. 593 0 (the standard deviation is

√4. 593 0 = 2. 143 1).

If we want the standard deviation of the put price per trial is 0.02, how manysimulations should be performed in one trial? Here is how to find it.

V ar¡C¢=1

n2

µV ar

nPi=1

Ci

¶=1

n2× nV ar (C) =

V ar (C)

n


19.5. EFFICIENT MONTE CARLO VALUATION 193

Set

rV ar (C)

n= 0.02. We have:

r4. 593 0

n= 0.02 or n =

4.5930

0.022= 11482.

5We need to have about 11,500 simulations per trial.

19.5 Efficient Monte Carlo valuation

19.5.1 Control variance method

Suppose we have n simulations per trial. These n simulations produce n optionprices V1, V2, ..., Vn. We can use the sample mean of these option prices toestimate the true option price. So the true option is estimated as:

V =V1 + V2 + ...+ Vn

n

The variance of V is:

V ar¡V¢= V ar

µV1 + V2 + ...+ Vn

n

¶=

nV ar (V )

n2=

V ar (V )

n

The standard deviation of the sample mean is σV =σV√n, where n is the num-

ber of the simulations and σV is the standard deviation of the option price persimulation. To decrease σV , we need to increase n by doing more simulations.Doing more simulations costs time.However, there are techniques out there to reduce σV without increasing n.

One method is called the control variate method. It goes like this.Suppose we have two random similar variables X and Y . We need to calcu-

late E (X) and E (Y ). We can calculate E (X) easily because there’s a formulafor E (X). E (Y ), on the other hand, doesn’t have a formula and needs to becalculated through the Monte Carlo simulation. Since X and Y are similar, weexpect that our errors in estimating E (X) are similar to our errors in estimatingE (Y ):

E (X)− ∧μX ≈ E (Y )− ∧μYIn the above formula, E (X) and E (Y ) are the true means of X and Y ;

∧μXand

∧μY are the estimated means of X and Y based on the Monte Carlo

simulation.Since our goal is to find E (Y ), we arrange the above formula into:

E (Y ) ≈ ∧μY +hE (X)− ∧μX

i=³∧μY −

∧μX

´+E (X)

So we can use³∧μY −

∧μX

´+ E (X) to estimate E (Y ). We define this new

estimate as∧μ∗Y =

³∧μY −

∧μX

´+E (X).



Why is the new estimate E (Y ) ≈ ∧μ∗Y =

³∧μY −

∧μX

´+ E (X) better than

the old estimate E (Y ) ≈ ∧μY ? It turns out the new estimate often has lowervariance if

∧μXand

∧μY are positively correlated.

The variance of the new estimate is:V ar

³∧μ∗Y

´= V ar

h³∧μY −

∧μX

´+E (X)

i= V ar

³∧μY −

∧μX

´= V ar

³∧μX

´+

V ar³∧μY

´− 2Cov

³∧μX ,

∧μY

´The above formula holds because E (X) is a constant.

Now let’s go through an example. There’s no formula for the arithmeticEuropean call option price; there’s a formula for the geometric European calloption price. So we’ll use the Monte Carlo simulation to estimate the arithmeticEuropean call option. We’ll use the geometric European price as the controlvariate (i.e. a dummy variable).

Let AC=arithmetic European call price GC=geometric European callprice.

∧μ∗AC =

∧μAC −

∧μGC +E (GC)

We’ll find∧μACand

∧μGCthrough Monte Carlo simulation. We’ll calculate

E (GC) using the equation DM 14.18 and DM 14.19 (see Derivatives MarketsAppendix 14.A). Once again, the inputs are:

• S0 = 40

• K = 40

• σ = 0.3

• r = 8%

• δ = 0

• T = 0.25 (3 months)

• The average stock price is the average stock prices at the end of Month 1,Month2, and Month 3 (so N = 3)

First, we’ll calculate the geometric European call option price. By the way,the appendix 14.5 is not on the syllabus of the exam MFE. You can ignore mycalculation and just accept that the geometric call option price is 1. 938 5.Using DM 14.18, we get:

δ∗ =1

2

µ0.08× 2

3+¡0 + 0.5× 0.32

¢ 43− 0.3

2

32× 4× 7

6

¶= 0.03 333

Using DM 14.19, we have:


19.5. EFFICIENT MONTE CARLO VALUATION 195

σ∗ =0.3

3

r4× 76

= 0.216 025

C (S,K, σ∗, r, T, δ∗) = Se−δ∗TN (d1)−Ke−rTN (d2)

d1 =

lnS

K+

µr − δ∗ +

1

2(σ∗)2

¶T

σ√T

=

ln40

40+

µ0.08− 0.03 333 + 1

2× 0.216 0252

¶0.25

0.216 025√0.25

=

0.162 026

N (d1) = NormalDist (0.162 026) = 0.564 357

d2 = d1 − σ∗√T = 0.162 026− 0.216 025

√0.25 = 0.054 013 5

N (d2) = NormalDist (0.054 013 5) = 0.521 538

C = 40e−0.03333×0.25 × 0.564 357− 40e−0.08×0.25 × 0.521 538 = 1. 938 5

So the geometric call option price is 1. 938 5.

Now we have:∧μ∗AC =

∧μAC −

∧μGC + 1. 938 5

Next, we perform 30 trials (1 trial =5,000 simulations).



i AC GC AC −GC (AC −GC)2

1 1.9823 1.9430 0.0393 0.0015442 1.9409 1.9001 0.0408 0.0016653 1.9467 1.9080 0.0387 0.0014984 2.0098 1.9698 0.0400 0.0016005 1.9894 1.9506 0.0388 0.0015056 1.9515 1.9112 0.0403 0.0016247 1.9825 1.9436 0.0389 0.0015138 2.0462 2.0058 0.0404 0.0016329 2.1270 2.0854 0.0416 0.00173110 1.9583 1.9191 0.0392 0.00153711 1.9878 1.9487 0.0391 0.00152912 1.9664 1.9281 0.0383 0.00146713 2.0147 1.9748 0.0399 0.00159214 1.9664 1.9272 0.0392 0.00153715 1.9915 1.9519 0.0396 0.00156816 1.9504 1.9116 0.0388 0.00150517 1.9414 1.9039 0.0375 0.00140618 1.9804 1.9424 0.0380 0.00144419 1.9666 1.9274 0.0392 0.00153720 2.0276 1.9867 0.0409 0.00167321 1.9940 1.9535 0.0405 0.00164022 1.9307 1.8924 0.0383 0.00146723 1.9221 1.8845 0.0376 0.00141424 1.9710 1.9328 0.0382 0.00145925 1.9185 1.8804 0.0381 0.00145226 1.9529 1.9133 0.0396 0.00156827 1.9791 1.9399 0.0392 0.00153728 1.9776 1.9398 0.0378 0.00142929 1.9713 1.9318 0.0395 0.00156030 1.9689 1.9279 0.0410 0.001681

sum 59.3139 58.1356 1.1783 0.046313

So

i AC GC AC −GC (AC −GC)2

sum 59.3139 58.1356 1.1783 0.046313∧μAC =

59.3139

30= 1. 977 13

∧μGC =

58.1356

30= 1. 937 85

Hence the updated estimated price of the geometric European call option is:∧μ∗AC =

∧μAC −

∧μGC + 1. 938 5 = 1. 977 13− 1. 937 85 + 1. 938 5 = 1. 977 78

Alternatively,


19.6. ANTITHETIC VARIATE METHOD 197

∧μAC −

∧μGC =

1.1783

30= 0.03927 67

∧μ∗AC = 0.03927 67 + 1. 938 5 = 1. 977 78

The variance of∧μ∗AC is:

V ar³∧μ∗AC

´= V ar

³∧μAC −

∧μGC + 1. 938 5

´= V ar

³∧μAC −

∧μGC

´=

30

30− 1

µ1

30× 0.046313− 0.03927 672

¶=

1. 145 7× 10−6

In contrast, the estimated variance of the geometric European call pricewithout using the control variate method is, as calculated before,

V ar³∧μGC

´= 0.001618

We can see that V ar³∧μ∗AC

ís much smaller than V ar

³∧μAC

´.

Boyle points out that∧μ∗Y =

³∧μY −

∧μX

´+ E (X) doesn’t always produce

lower variance than the variance of the original estimate∧μY . Boyle recommends

the following new estimate:∧μ∗Y =

∧μY + β

³E (X)− ∧μX

´

If we use the textbook notation, the above new estimate is DM 19.10:A∗ = A+ β

¡G−G

¢Since G is constant (here G is equivalent to E (X)), we have DM 19.11:

V ar (A∗) = V ar£A+ β

¡G−G

¢¤= V ar

¡A− βG

¢= V ar

¡A¢+β2V ar

¡G¢−

2βCov¡A,G

¢The textbook says that V ar (A∗) is minimized if we set β =

Cov¡A,G

¢V ar

¡G¢ .

How do we find β? Typically, we perform a small number of Monte Carlosimulations, run regression of DM 19.10, and estimate β. Then we apply theestimated β to DM 19.10, run more simulations, and calculate A∗.This is all you need to know about the Boyle’s improved control variate

method.

19.6 Antithetic variate method

Antithetic variate method is simple. Suppose we want to estimate E (X) andhave generated random numbers from a symmetric distribution such as a stan-dard normal random variables. We have n random draws z1, z2,...,zn from the



standard normal distributions. Since z1, z2,...,zn are random draws from thestandard normal distribution, −z1, -z2,...,−zn are also random draws from thestandard normal distribution. Next, we calculate two samples means, X1 (usingz1, z2,...,zn ) and X2 (using −z1, -z2,...,−zn). Then we can estimated E (X) asthe average of X1 and X2:

E (X) ≈X1 +X2

2This method is called the antithetic variate method.Here’s an example. We want to estimate E (ez) where z is the standard

normal random variable. We have generated the following 10 standard normalrandom variables:

i zi1 −0.01832 2.04783 −0.48494 −0.65835 −0.46666 −0.37577 1.13928 2.15669 0.109610 −1.5389

This is how to estimate E (ez) using the antithetic variate method:i zi Xi = exp (zi) −zi Yi = exp (−zi)1 −0.0183 0.981866 0.0183 1.0184682 2.0478 7.750831 −2.0478 0.1290183 −0.4849 0.615759 0.4849 1.6240134 −0.6583 0.517731 0.6583 1.9315065 −0.4666 0.627131 0.4666 1.5945636 −0.3757 0.686808 0.3757 1.4560107 1.1392 3.124268 −1.1392 0.3200758 2.1566 8.641706 −2.1566 0.1157189 0.1096 1.115832 −0.1096 0.89619310 −1.5389 0.214617 1.5389 4.659462

sum 24.276548 13.745027

X =24.276548

10= 2. 427 654 8

Y =13.745027

10= 1. 374 502 7

E (ez) ≈2. 427 654 8 + 1. 374 502 7

2= 1. 901 078 75


19.7. STRATIFIED SAMPLING 199

19.7 Stratified sampling

In stratified sampling, we divide the population into several non-overlappinggroup. Each group is called a strata. Then we take a sample from each group.Example. We divide the range [0, 1] into 100 groups: [0, 0.01), [0.01, 0.02),

[0.02, 0.03),...,[0.99, 1). Next, we draw 100 numbers u1, u2, ..., u100 from theuniform distribution [0, 1]. All these 100 numbers are divided by 100. So we

haveu1100

,u2100

,u3100

, ...,u100100

.Next, we addi− 1100

to the i-th numberui100

. Now

we have:

u1100

,u2100

+ 0.01,u3100

+ 0.02, ...,u100100

+ 0.99

Since 0 ≤ ui < 1, 0 ≤ui100

< 0.01.

Sou1100

falls in the group [0, 0.01).

Similarly, 0.01 ≤ u2100

+0.01 < 0.02. Sou2100

+0.01 falls into group [0.01, 0.02).

Similarly, 0.02 ≤ u3100

+0.02 < 0.03. Sou3100

+0.01 falls into group [0.02, 0.03).

So on and so forth.The end result is that we take one sample from [0, 0.01), one sample from

[0.01, 0.02), ..., and one sample from [0.99, 1).This is all you need to know about the stratified sample for the purpose of

passing Exam MFE.

19.7.1 Importance sampling

In the importance sampling, we perform simulations from a conditional distri-bution, not from the original distribution. For example, we want to estimatethe price of an option that is deep out of the money. If we perform simulationsfrom the original distribution, then most of the simulated payoffs will be zero.This is a waste of our time. To make the simulations more efficient, we’ll drawrandom numbers from the conditional distribution where the payoff is not zero.This is all you need to know about the importance sampling.The textbook also mentioned Latin hypercube sampling and low discrepancy

sequences. Since the textbook merely mentioned these terms without providingmuch explanation, I don’t think SOA expects you to know much about term.Skip these terms and move on.

19.8 Sample problems

Problem 1



You are simulating the standard normal random variable z by taking randomdraws from a uniform distribution over (0, 1). Let a represent the simulatedvalue of z. Calculate P (z ≤ a). You are given:

i ui1 0.37632 0.13493 0.4144 0.04055 0.52256 0.04237 0.20418 0.92829 0.679210 0.336811 0.153512 0.157

SolutionPui = 0.3763+0.1349+0.414+0.0405+0.5225+0.0423+0.2041+0.9282+

0.6792 + 0.3368 + 0.1535 + 0.157 = 3. 989 3

The simulated value is a =P

ui − 6 = 3.9893− 6 = −2. 010 7P (z ≤ a) = N (−2. 010 7) = 1−N (2. 010 7) = 0.022

Problem 2

You are simulating E (ez) by taking random 10 draws from a uniform dis-tribution over (0, 1). Calculate the simulated value of E (ez). You are given:

i ui1 0.8782 0.7623 0.0694 0.85 0.0486 0.227 0.1788 0.6619 0.19110 0.258

Solution



i ui zi = N−1 (ui) xi = ezi

1 0.878 1.17 3.2219932 0.762 0.71 2.0339913 0.069 −1.48 0.2276384 0.8 0.84 2.3163675 0.048 −1.66 0.1901396 0.22 −0.77 0.4630137 0.178 −0.92 0.3985198 0.661 0.42 1.5219629 0.191 −0.87 0.41895210 0.258 −0.65 0.522046

sum 11.314619

The estimated value is E (ez) =11.314619

10= 1. 131 461 9

Sample calculation. u1 = 0.878. Look at the normal table. You see thatroughly P (z < 1.17) = 0.878 so z1 = 1.17.

By the way, I’m using Excel to find zi = N−1 (ui) for me so I don’t have tolook at the normal table. So if you can’t match my z, that’s OK. For example,for u1 = 0.878, you might get z1 = 1.16 or z1 = 1.165. However, your finalestimated E (ez) should be close to mine.

x1 = e1.17 = 3. 221 993

u3 = 0.069. Since 0.069 < 0.5, you know that z3 = N−1 (u3) < 0. Youcan’t directly look up z3 from the normal table (the normal table lists only thepositive z values). Use the formula N (−z3) = 1−N (z3) = 1− 0.069 = 0.931 .From the normal table you see that P (z < 1.48) = 0.931. So −z3 = 1.48 orz3 = −1.48.

Problem 3

You are simulating the real-world price ST . Inputs are:

• Stock prices are lognormally distributed

• S0 = 100

• α = 0.06

• r = 0.08

• δ = 0.02

• T = 0.5

• σ = 0.3



•

i ui ∼ U (0, 1)1 0.65152 0.88393 0.76214 0.39225 0.1748

Calculate the average of the simulated real-world stock prices at T .

Solution

The real world price at T is:

ST = S0 exp³¡α− δ − 0.5σ2

¢T + σ

√Tz´= 100e(0.06−0.02−0.5×0.3

2)0.5+0.3√0.5z =

100e−0.002 5+0.3√0.5z

i u1 ∼ U (0, 1) zi = N−1 (u1) ST = 100e−0.002 5+0.3

√0.5z

1 0.6515 0.39 100e−0.002 5+0.3√0.50.39 = 108. 353 8

2 0.8839 1.19 128. 394 53 0.7621 0.71 115.96454 0.3922 −0.27 94.19765 0.1748 −0.94 81.7173

Total 528.6276

The average of the simulated stock prices is:528.6276

5= 105. 73

Please note that the risk free rate r is not needed for solving this problem.

Problem 4

You are simulating the price of an arithmetic average stock price Europeancall option and put option. You are given:

• Stock prices are lognormally distributed

• S0 = 40

• K = 40

• α = 0.06

• r = 0.08

• δ = 0

• T = 1

• σ = 0.3



• Stock prices at the end of Month 4, 8, and 12 are averaged

• The 3 random draws from a uniform distribution (0, 1) are 0.4828, 0.6177,and 0.9345, which are used to simulate the stock price at the end of Month4, 8, and 12 respectively.

Calculate the simulated values of this arithmetic Asian call option and putoption.

Solution

The option price is:

e−rT max

µS∗1/3 + S∗2/3 + S∗1

3−K, 0

¶S∗t is the risk-neutral stock price at t.

S∗t = St−h exp³¡r − δ − 0.5σ2

¢h+ σ

√hz´= St−he(

0.08−0−0.5×0.32)h+0.3√hz

u1 z1 S∗1/3 u2 z2 S∗2/3 u3 z3 S∗10.4828 −0.04 40.1900 0.6177 0.3 42.8303 0.9345 1.51 56.2864

Sample calculation.

S∗1/3 = 40e(0.08−0−0.5×0.32)1/3+0.3

√1/3(−0.04) = 40. 1900

S∗2/3 = 40.1900e(0.08−0−0.5×0.32)1/3+0.3

√1/3(0.3) = 42. 830 3

S∗1 = 42.8303e(0.08−0−0.5×0.32)1/3+0.3

√1/3(1.51) = 56. 286 4

Average stock price:40. 1900 + 42.8303 + 56.2864

3= 46. 435 6

The call payoff: max (46. 435 6− 40, 0) = 6. 435 6The put payoff: max (40− 46. 435 6, 0) = 0

The simulated value of the call price is: e−0.08×16. 435 6 = 5. 940 8

The simulated value of the put price is: e−0.08×10 = 0

Problem 5 (spring 2007 Exam C #19)The price of a non dividend-paying stock is to be estimated using simulation.

It is known that:

• The price St follows the lognormal distribution

• S0 = 50, α = 0.15, and σ = 0.30.



Using the following uniform (0, 1) random numbers and the inversion method,three prices for two years from the current date are simulated 0.9830, 0.0384, 0.7794.Calculate the mean of the three simulated prices.(A) Less than 75 (B) At least 75, but less than 85 (C) At least 85,

but less than 95 (D) At least 95, but less than 115 (E) At least 115

Solution

ST = S0 exp³¡α− δ − 0.5σ2

¢T + σ

√Tz´= 50e(0.15−0−0.5×0.3

2)2+0.3√2z =

50e0.21+0.3√2z

i u1 ∼ U (0, 1) zi = N−1 (u1) ST = 50e0.21+0.3

√2z

1 0.9830 2. 12 50e0.21+0.3√2×2.12 = 151. 63

2 0.0384 −1. 77 50e0.21+0.3√2×(−1. 77) = 29. 11

3 0.7794 0.77 50e0.21+0.3√2×(0. 77) = 85. 52

Total 151. 63 + 29. 11 + 85. 52 = 266. 26

The average of the simulated prices is266. 26

3= 88. 75. The answer is C.


Chapter 20

Brownian motion and Ito’sLemma

20.1 IntroductionAccording to Wikipedia, Brownian motion (named in honor of the botanistRobert Brown) is either the random movement of particles suspended in a fluidor the mathematical model used to describe such random movements, oftencalled a Wiener process.In 1827, while examining pollen grains suspended in water under a micro-

scope, Brown observed minute particles in the pollen grains executing a contin-uous jittery motion. He observed the same motion in particles of dust, enablinghim to rule out the hypothesis that the motion was due to pollen being alive.Although he did not provide a theory to explain the motion, the phenomenonis now known as Brownian motion in his honor.Brownian motion is a useful tool for modeling the stock price. The price

of a stock is constantly hit by random events, just as a particle in the water isconstantly hit by water molecules. In fact, the Brownian motion is to stochasticprocesses as the standard normal distribution to random variables.Brownian motion is an abstract concept. The first step toward learning the

Brownian motion is to see it and experiment it. You can easily find Brownianmotion simulations in the internet. Here are some simulations in the internet:

• http://www.phy.ntnu.edu.tw/java/gas2d/gas2d.html

• http://www.aip.org/history/einstein/brownian.htm

• http://www.stat.umn.edu/~charlie/Stoch/brown.html

• http://www.matter.org.uk/Schools/Content/BrownianMotion

There’s a large body of knowledge on the internet about the Brownian mo-tion. For example, you can check out Wikipedia’s explanation at:

205


206 CHAPTER 20. BROWNIAN MOTION AND ITO’S LEMMA

http://en.wikipedia.org/wiki/Brownian_motion

To help you experiment with the Brownian motion, I designed a spreadsheettitled "Simulate Brownian Motion." Download this spreadsheet.My spreadsheet uses the following Excel functions:

• Rand() returns a random number equal to or greater than 0 but less than 1.In other words, Rand() simulates a random variable uniformly distributedover [0, 1).

• NORMINV(probability,mean,standard_dev) returns the inverse of the nor-mal cumulative distribution for the specified mean and standard deviation.

20.1.1 Big picture

When studying the Brownian motion and Ito’s lemma, remember the followingbig picture. The Black-Scholes option pricing formula Equation 12.1 relies onthe Black-Scholes PDE (Equation 12.24). Equation 12.24 assumes the followingprice model for a risk-free asset and a risky asset (i.e. stock):The price of a risk free asset (i.e. the savings account or a bond) is

B (t) = ert (20.1)

The price of a risky asset (i.e. stock) at time t is:

S (t) = S (0) eα−1

2σ2 t+σ

√tY (t)

(20.2)

In the above equation:

• S (0) is the stock price at time zero

• S (t) is the stock price at time t

• α is the expected (annualized continuously compounded) return of the

stock. In other words, E [S (t)] = S (0) eαt or α =1

tlnE

∙S (t)

S (0)

¸• r is the (annualized continuously compounded) risk-free interest rate.

• σ is the stock’s volatility.

• Y (t) is a random draw of a standard normal random variable. Y (t1) andY (t2) are independent for t1 6= t2

Equation 20.2 is another form of Equation 20.1 in the textbook:

dS (t)

S (t)= αdt+ σdZ (t) (Textbook 20.1)


20.2. BROWNIAN MOTION 207

20.2 Brownian motion

20.2.1 Stochastic process

A stochastic process is a family of random variables indexed by time. Forexample, the temperature out side your house X (t) is a stochastic process. Lett = 0 represent now and t = 1 represent the next time (such as next hour, nextday, next week). Then for a series of time points t = 1, 2, 3, ... there is a familyof random temperatures X (0), X (1), X (2), X (3) ....This is the major difference between a stochastic process and a deterministic

process. In a stochastic process you’ll see a series of random variables; foreach time t there’s a corresponding random variable X (t). In contrast, in adeterministic process, you’ll see only one random variable.

20.2.2 Definition of Brownian motion

Consider a particle that jumps, at discrete times, up or down along the verticalline. At t = 0 the particle is at position zero. After each h-long time period,the particle jumps up or down by a constant distance of k and with equalprobability of 0.5. That is, at t = h, 2h, 3h, ..., nh, the particle either movesup by k or moves down by k, with up and down movements having an equalprobability of 0.5. Let Z (t) represent the height of the article from the positionzero at time t. Clearly Z (0) = 0. We like to find Z (T ), the height of the articleat time T = nh.

The particle’s height

4k3k

2k 2kk k

0 0 0−k −k

−2k −k−3k

−4ktime 0 h 2h 3h 4h

Let’s walk through the above table. At t = 0 the particle is at the positionzero. At t = h, the particle’s height is either k or −k. At t = 2h, the k nodeeither goes up to 2k or goes down to 0. Similarly, the −k node either goes upto 0 or goes down to −2k. So on and so forth.The jump at t = h is: Z (h)− Z (0) = Y (h) kHere Y (h) is a direction indicator. If Y (h) = 1, then the particle moves up

by k; if Y (h) = −1, then the particle moves down by k:Y (h) =

½1 Probability 0.5−1 Probability 0.5



The jump at t = 2h is: Z (2h)− Z (h) = Y (2h) k

Y (2h) =


The jump at t = nh = T is: Z (nh)− Z [(n− 1)h] = Y (nh) kHere Y (nh) is a direction indicator:

Y (nh) =


The direction indicators Y (h), Y (2h), Y (3h), ..., Y (nh) are independentidentically distributed binomial random variables. For i = 1 to n,

E (ih) = (1) 0.5 + (−1) 0.5 = 0Eh(ih)2

i= (1)2 0.5 + (−1)2 0.5 = 1

V ar (ih) = Eh(ih)

2i−E2 (ih) = 1

The particle’s height at time T is:Z (T ) = [ Z (h)− Z (0)]+[ Z (2h)− Z (h)]+[ Z (3h)− Z (2h)]+...+[ Z (nh = T )− Z [(n− 1)h]]

= Y (h) k + Y (2h) k + Y (3h) k + ...+ Y (nh) k= [Y (h) + Y (2h) + Y (3h) + ...+ Y (nh)] k

According to the central limit theorem, Z (T ) is approximately normal withmean and variance as:

E [Z (T )] = kE [Y (h) + Y (2h) + Y (3h) + ...+ Y (nh)] = 0V ar [Z (T )] = k2V ar [Y (h) + Y (2h) + Y (3h) + ...+ Y (nh)]

= k2n = k2T

h=

k2

hT

We want V ar [Z (T )] =k2

hT to exist (i.e. not to become infinite) as n→∞.

To achieve this, we setk2

hto a positive constant:

k2

h= c. To make our

model simple, we set c = 1. Hencek2

h= 1 k =

√h

Now we have:

Z (ih)− Z [(i− 1)h] = Y (ih)√h (20.3)

Now Z (T ) is approximately normal with mean zero and variancek2

hT = T :

Z (T ) ∼ N (0, T )

Please note that another way to specify the model is treat Y (ih) as a randomdraw of a standard normal random variable (instead of a binomial randomvariable):



Z (ih)− Z [(i− 1)h] = Y (ih)√h and Y (ih) ∼ N (0, 1) (20.4)

My spreadsheet for simulating the Brownian motion uses both Equation 20.3and Equation 20.4.

The random process Z (t) as n → ∞ is called the Brownian motion or theWiener process.Next, let’s formally define the Brownian motion.

Definition 20.2.1.

A stochastic process Z (t) is a Brownian motion or a Wiener process if

1. Z (0) = 0 . Brownian motion starts at zero (this is merely for our conve-nience).

2. Z (t+ h)−Z (t) is normally distributed with mean 0 and variance h. Thismeans that the increments over a time interval h is normally distributedwith mean 0 and variance h. This stands true no matter how small or bigh is.

3. Z (t+ s1)− Z (t) is independent of Z (t)− Z (t− s2) where s1, s2 > 0.

4. Z (t) is continuous.

20.2.3 Martingale

The following part is based on Wikipedia.A martingale is a stochastic process (i.e., a sequence of random variables)

such that the conditional expected value of an observation at some time t, givenall the observations up to some earlier time s, is equal to the observation at thatearlier time s.Originally, martingale referred to a betting strategy popular in 18th century

France. The rule of the game is that the gambler wins his stake if a coin comesup heads and loses it if the coin comes up tails. The martingale strategy hadthe gambler double his bet after every loss, so that the first win would recoverall previous losses plus win a profit equal to the original stake. Since eventuallya gambler will win at least once, the martingale betting strategy was thoughtto be sure way of winning. In reality, however, the exponential growth of thebets would eventually bankrupt the gambler.A stochastic process (i.e., a sequence of random variables) X (t) is a martin-

gale if the following holds:E [X (t) |X (s)] = X (s) for t > s.

It can be shown that the Brownian motion Z (t) is martingale.



For t > s, we have:E [Z (t) |Z (s)]= E [Z (t)− Z (s) + Z (s) |Z (s)]= E [Z (t)− Z (s) |Z (s)] +E [Z (s) |Z (s)]

E [Z (s) |Z (s)] = Z (s)Z (t)− Z (s) is independent of Z (s).→ E [Z (t)− Z (s) |Z (s)] = E [Z (t)− Z (s)] = 0→ E [Z (t) |Z (s)] = E [Z (t)− Z (s) |Z (s)] +E [Z (s) |Z (s)] = Z (s)Hence Z (t) is martingale.E [Z (t) |Z (s)] = Z (s) means that the best estimate of the future value of a

Brownian motion is its current value.

20.2.4 Properties of Brownian motion

1. The Brownian motion is continuous everywhere yet differentiable nowhere.

2. The first-order variation is infinite: limn→∞

|Z (h)−Z (0) |+ |Z (2h)−Z (h) |+...+ |Z (nh)− Z [(n− 1)h] |→∞

3. The second-order variation (called quadratic variation) is equal to thelength of the time interval: lim

n→∞[Z (h)− Z (0)]

2+ [Z (2h)− Z (h)]

2+

...+ (Z (nh)− Z [(n− 1)h])2 = T

4. Cov [Z (s) , Z (t)] = min (s, t). The covariance of two Brownian motions isthe shorter time interval.

5. The higher moments of Z (t) is:E [Zn (t)] =

½0 if n is oddtn/2 (n− 1) (n− 3) ...1 if n is even

Let’s look at the first property.If you look at the simulation of Brownian motion over the internet, you’ll

find that the Brownian motion is always continuous yet it’s not differentiableanywhere. Can you imagine that a function is continuously anywhere yet differ-entiable nowhere? If I hadn’t studied the Brownian motion, I would have neverthought that such a function exists.We can explain the non-differentiality using the following equation (it’s text-

book Equation 20.4):

dZ (t) = Y (t)√dt (20.5)

The above equation is another form of Equation 20.3 or 20.4:dZ (t) = Z (t+ dt)− Z (t) = Y (t)

√dt

The textbook explains in the footnote that you can treat Y (t) as a binomial



random variable or a standard normal random variable. Either way, dZ (t) is anormal random variable.From Equation 20.5, we have:dZ (t)

dt=

Y (t)√dt→∞ as dt→ 0

Let’s look at the 2nd property. The following is an intuitive but not rigorousproof.

According to the definition of the Brownian motion, Z (h)−Z (0), Z (2h)−Z (h), ..., Z (nh)− Z [(n− 1)h] are independent identically distributed normalrandom variable with mean 0 and variance h. On average, |Z [(i+ 1)h]−Z (ih) |approaches E|X| where X ∼ N (0, h). Let f (x) represent the probability den-sity function of X, then

E|X| =R∞−∞ |x|f(x) dx > 0 (since |x| ≥ 0, it’s mean must be positive)

Then limn→∞

(|Z (h)− Z (0) |+ |Z (2h)− Z (h) |+ ...+ |Z (nh)− Z [(n− 1)h] |)

approaches nE|X|, which approaches∞ (since n→∞, a positive constant E|X|times n must also approach infinity.)Property 2 should be easy to understand. Since ∆Z (t) = Y (t)

√∆t, for a

tiny interval ∆t,√∆t is much large than ∆t. For example,

√0.000001

0.000001= 1, 000.

Hence during any short interval, the Brownian motion can move up or down byan infinitely large amount.In contrast, for a continuously differentiable function y = f (t), we have:∆y = f 0 (t)∆t→ 0 as ∆t→ 0

Let’s get an intuitive feel of the 3rd property. On average, Z [(i+ 1)h] −Z (ih)2 approaches E

¡X2¢.

E¡X2¢= E2 (X) + V ar (X) = 02 + h = h

limn→∞

[Z (h)− Z (0)]2+ [Z (2h)− Z (h)]

2+ ...+ (Z (nh)− Z [(n− 1)h])2 ap-

proachesh+ h+ ...+ h = nh = TIn contrast, the 2nd order variation of a differentiable function is zero. For

example, we can find the 2nd order variation of the function y = x is zero.Divide the interval [0, T ] into [0, h],[h, 2h],...,[(n− 1)h, nh = T ]

limn→∞

(h− 0)2+(2h− h)2+ ...+[nh− (n− 1)h]2 = nh2 = limn→∞

n

µT

n

¶2=

0It can be proven that the quadratic variation of any continuously differen-

tiable function is zero. The quadratic variation of any continuously differentiablefunction is zero because such a function is roughly linear at any point. For acontinuously differentiable function y = f (t), we have:∆y = f 0 (t)∆t → (∆y)2 = [f 0 (t)]2 (∆t)2

For a tiny interval∆t, (∆t)2 → 0must faster than∆t→ 0. Hence (∆y)2 → 0



In contrast, for a Brownian motion we have∆Z (t) = Y (t)√∆t and [∆Z (t)]2 =

[Y (t)]2∆t→ ∆t. HenceP[∆Z (t)]2 →

P∆t = T

Brownian motion Property 3 reconfirms the idea that Brownian motion isnot differentiable anywhere. If it’s differentiable, then its second order variationwould be zero.

Tip 20.2.1. Just memorize Equation 20.5 dZ (t) = Y (t)√dt. This equation

tells you that the Brownian motion Z (t) is not differentiable anywhere, its sec-ond order variation is t, and its first order variation is infinite.

Tip 20.2.2. To get an intuitive feel of the equation dZ (t) = Y (t)√dt, imagine

you are looking at the Brownian motion under a magnifying class. If you zoomin on the Brownian motion by shrinking the time interval dt, no matter howmuch you reduce dt, you’ll see a jigsaw. In comparison, if you zoom in on acontinuously differentiable function such as y = t2, you’ll see a straight line.

Please note that Derivatives Markets explains Property 2 and 3 using thefollowing formula:

Z [(i+ 1)h]− Z (ih) = Y [(i+ 1)h]√h

Since Y [(i+ 1)h] is a binomial random variable having a value of ±1, then|Z [(i+ 1)h]− Z (ih) | =

√h and Z [(i+ 1)h]− Z (ih) = h. Hence

limn→∞

(|Z (h)− Z (0) |+ |Z (2h)− Z (h) |+ ...+ |Z (nh)− Z [(n− 1)h] |) = n√h→

∞limn→∞

[Z (h)− Z (0)]2+ [Z (2h)− Z (h)]

2+ ...+(Z (nh)− Z [(n− 1)h])2 =

nh = TThe problem with this explanation is that it works if we treat Y [(i+ 1)h]

as binomial random variable whose value is ±1. Such explanation won’t work ifwe treat Y [(i+ 1)h] as a random draw of a standard normal random variable.The explanation I provided here works no matter if you treat Y [(i+ 1)h] as abinomial random variable or a standard normal random variable.Let’s look at Property 4. Suppose s ≤ tCov [Z (s) , Z (t)] = CovZ (s) , Z (s) + [Z (t)− Z (s)]Using the formula Cov (a, b+ c) = Cov (a, b) + Cov (a, c), we get:CovZ (s) , Z (s)+[Z (t)− Z (s)] = CovZ (s) , Z (s)+CovZ (s) , [Z (t)− Z (s)]CovZ (s) , Z (s) = V ar [Z (s)] = sSince Z (s) and [Z (t)− Z (s)] are independent (Brownian motion definition

Point #3), CovZ (s) , [Z (t)− Z (s)] = 0→ Cov [Z (s) , Z (t)] = s = min (s, t)Property 5 is based on the moment formula for a standard normal random

variable φ

E (φn) =

½0 if n is odd(n− 1) (n− 3) ...1 if n is even

(20.6)

We can find the n-th moment of a random variable X using the momentgenerating function (MGF):



E (Xn) =

∙dn

dtnMX (t)

¸t=0

(20.7)

The MGF of a normal random variable X with mean μ and the standarddeviation σ is:

MX (t) = E¡etX

¢= e

μt+1

2σ2t2

(20.8)

→Mφ (t) = e

1

2t2

Using Equation 20.7, you can verify that Equation 20.6 holds.Since Z (t) is a normal random variable with mean 0 and variance t, then

Z (t)√tis a standard normal random variable. Hence

E [Zn (t)] =

½0 if n is oddtn/2 (n− 1) (n− 3) ...1 if n is even

(20.9)

Example 20.2.1. Calculate P [Z (3) > 1]

Z (3) is a normal random variable with mean 0 and variance 3.

P [Z (3) > 1] = 1 − P [Z (3) ≤ 1] = 1 − Φµ1− 0√3

¶= 1 − Φ (0.577 35) =

0.281 9

Example 20.2.2. Calculate P [Z (1) ≤ 0 ∩ Z (2) ≤ 0]

Z (1) is a normal random variable with mean 0 and variance 1. LetX = Z (1)

Z (2) = Z (1) + [Z (2)− Z (1)]Z (2)− Z (1) is a normal random variable with mean 0 and variance 1.Z (2) = Z (1) + [Z (2)− Z (1)] ≤ 0 → [Z (2)− Z (1)] ≤ −Z (1)Let Y = Z (2)− Z (1).X and Y are independent.P [Z (1) ≤ 0 ∩ Z (2) ≤ 0] = P (X ≤ 0 ∩ Y ≤ −X)To have X ≤ 0 ∩ Y ≤ −X, we first fix X at a tiny interval (x, x+ dx)

where −∞ < x < 0. Next, we set Y < −x. Then we are guaranteed to haveX ≤ 0∩Y ≤ −X. Let f (x) and Φ (x) represent the probability density function(pdf) and the cumulative density function (cdf) of a standard normal randomvariable.

P (X ≤ 0 ∩ Y ≤ −X) =R 0−∞ P (x < X < x+ dx)P (Y < −x) =

R 0−∞ [f (x) dx]P (Y < −x)

However, f (x) dx = dΦ (x) and P (Y < −x) = Φ (−x) = 1− Φ (x)R 0−∞ [f (x) dx]P (−x) =

R 0−∞ [dΦ (x)] [1− Φ (x)] =

R 0−∞ [1− Φ (x)] dΦ (x) =R 0

−∞ [1− Φ (x)] dΦ (x)

=

∙Φ (x)− 1

2Φ2 (x)

¸0−∞



= [Φ (0)− Φ (−∞)]− 12

£Φ2 (0)− Φ2 (−∞)

¤=

∙1

2− 0¸−"1

2

µ1

2

¶2− 02

#

=1

2− 12

µ1

2

¶2=3

8

Example 20.2.3. Calculate Cov [Z (5) , Z (2)]

Cov [Z (5) , Z (2)] = min (5, 2) = 2

Example 20.2.4. Calculate E£Z4 (t)

¤.

E£Z4 (t)

¤= t4/2 (4− 1) = 3t2

20.2.5 Arithmetic Brownian motion andGeometric Brown-ian motion

A standard Brownian motion is normally distributed with mean 0 and variance1. Now we want to extend the Brownian motion to allow for non-zero mean andan arbitrary variance. Define a stochastic process:

X (t+ h)−X (t) = αh+ σY (t+ h)√h

Breaking down [0, T ] into small intervals [0, h],[h, 2h],...,[(n− 1)h, nh = T ],we have:

X (T )−X (0) =Pn

i=1

hαh+ σY (ih)

√hi= αT + σ

Pni=1

hY (ih)

√hi

As n→∞,Pn

i=1

hY (ih)

√hi→ Z (T )

X (T )−X (0) = αT + σZ (T ) (20.10)

dX (t) = αdt+ σdZ (t) (20.11)

Equation 20.10 and 20.11 are called arithmetic Brownian motion. α is theinstantaneous mean per unit of time; σ is the instantaneous standard deviationper unit of time.Equation 20.10 and 20.11 indicate thatX (T )−X (0) is normally distributed.

Its mean and variance are:E [X (T )−X (0)] = E [αT + σZ (T )] = αT +σE [Z (T )] = αT +σ×0 = αTV ar [X (T )−X (0)] = V ar [αT + σZ (T )] = V ar [σZ (T )] = σ2V ar [Z (T )] =

σ2T

The textbook lists the major properties and weaknesses of Equation 20.11.Major properties:

1. X (t) is normally distributed.



2. We can change the variance by changing the parameter σ.

3. We can change the mean by changing the parameter α. Now the mean isno longer zero if α 6= 0. And we have E [X (T )] − E [X (0)] = αT . Thismeans that after time T , the stock price drifts away from the price at timezero.

Major weaknesses:

1. The stock priceX (t) can be negative. SinceX (t) is normally distributed,−∞ <X (t) < ∞. Equation 20.11 allows a negative stock price. Of course, thestock price can’t become negative.

2. The expected change of the stock price does not depend on the stock price.In reality, the expected change of the stock price should be proportionalto the stock price. The higher the stock price, the higher the expectedchange. So we like to have E [dX (t)] = αX (t). We need to modifyEquation 20.11 to allow E [dX (t)] = αX (t).

3. The variance of the stock price does not depend on the stock price. Inreality, the variance should be proportional to the stock price. So we needto modify Equation 20.11 to allow σ [X (t) , t] = σX (t).

Major Weakness #2 and #3 can also be stated this way. Equation 20.11 can

be rewritten asdX (t)

X (t)=

α

X (t)dt +

σ

X (t)dZ (t). This indicates that

dX (t)

X (t),

the percentage return on the stock depends on the stock price X (t). However,in reality, we think that the stock return on average shouldn’t depend on thestock price. In other words, instead of Equation 20.11, we like to see

dX (t)

X (t)= αdt+ σdZ (t) (20.12)

or

dX (t) = X (t)αdt+ σX (t) dZ (t) (20.13)

Equation 20.12 and 20.13 are called the geometric Brownian motion.

20.2.6 Ornstein-Uhlenbeck process

We can modify Equation 20.11 to allow for mean reversion. It’s reasonable forus to assume that the stock price or the interest rate will revert to the mean.For example, if the stock price is too high, then it might go down; if the stockprice is too low, it might go up. We modify the drift term in Equation 20.11:

dX (t) = λ [α−X (t)] dt+ σdZ (T ) (20.14)



If α = 0, 20.14 is called the Ornstein-Uhlenbeck process:

dX (t) = −λX (t) dt+ σdZ (T ) (20.15)

The Ornstein-Uhlenbeck process is the most widely used mean revertingstochastic process in financial modeling.

20.3 Definition of the stochastic calculusWe can rewrite Equation 20.12 as:R T

0dX (t) =

R T0X (t)αdt+

R T0σX (t) dZ (t)

orX (T )−X (0) =R T0X (t)αdt+

R T0σX (t) dZt = α

R T0X (t) dt+σ

R T0X (t) dZt

But what’s the meaning ofR T0X (t) dZ (t)? Or generally, what’s the meaning

ofR bag (t) dZ (t)?

To answer this question, let’s take a step back and find out the meaning ofa simple deterministic calculus

R 10x2dx.R 1

0x2dx is the area of the function x2 bounded by x = 0 and x = 1. To find

this area, we divide the interval [0, 1] into n intervals [0, h], [h, 2h], [2h, 3h], ...,[(n− 1)h, nh = 1]. Then we approximate the area with the sum of n rectangles.The area of function x2 over the interval [(i− 1)h, ih] is roughly the area of therectangular with height [(i− 1)h]2 and width ih− (i− 1)h = h.R ih

(i−1)h x2dx ≈ [(i− 1)h]2 h = (i− 1)2 h3R 1

0x2dx = lim

n→∞

Pni=1

R ih(i−1)h x

2dx = limn→∞

Pni=1 (i− 1)

2 h3 = limn→∞

02h +h2h+ (2h)

2h+ ...+ [(n− 1)h]2 h

02h+ h2h+ (2h)2 h+ ...+ [(n− 1)h]2 h= h3

h02 + 12 + 22 + ...+ (n− 1)2

iUsing the famous formula:

12 + 22 + 32 + ...+ n2 =n (n+ 1) (2n+ 1)

6

h3h02 + 12 + 22 + ...+ (n− 1)2

i= h3

(n− 1) (n) (2n− 1)6

Since h =1

n, we have:

R 10x2dx = lim

n→∞

1

n3(n− 1) (n) (2n− 1)

6

= limn→∞

∙1

6

µ1− 1

n

¶(1)

µ2− 1

n

¶¸=1

6

There are other ways to approximateR ih(i−1)h x

2dx. For example, the area of

function x2 over the interval [(i− 1)h, ih] is roughly the area of the rectangularwith height (ih)2 and width ih− (i− 1)h = h.R ih

(i−1)h x2dx ≈ (ih)2 h = i2h3


20.3. DEFINITION OF THE STOCHASTIC CALCULUS 217

R 10x2dx = lim

n→∞

Pni=1

R ih(i−1)h x

2dx = limn→∞

Pni=1 i

2h3 = limn→∞

n (n+ 1) (2n+ 1)

6h3

= limn→∞

n (n+ 1) (2n+ 1)

6

1

n3=1

6The 3rd way to approximate

R ih(i−1)h x

2dx is to take the average height of

the rectangular. So the area of function x2 over the interval [(i− 1)h, ih] is

roughly the area of the rectangular with height(i− 1)2 h2 + (ih)2

2and width

ih− (i− 1)h = h.(i− 1)2 h2 + (ih)2

2=(i− 1)2 + i2

2h2R ih

(i−1)h x2dx ≈

"(i− 1)2 + i2

2h2

#h

R 10x2dx = lim

n→∞

Pni=1

R ih(i−1)h x

2dx = limn→∞

Pni=1

"(i− 1)2 + i2

2h2

#h

=1

2limn→∞

Pni=1

h(i− 1)2 h2

ih+

1

2limn→∞

Pni=1

£i2h2

¤h

=1

2

µ1

6

¶+1

2

µ1

6

¶=1

6It seems natural that we extend this logic of deterministic integration to a

define a stochastic integration.Suppose we partition [a, b] into a = t0 < t1 < t2 < ... < tn = b. We can

make the partition intervals [t0,t1], [t1,t2], ..., [tn−1,tn] have the same length

tk+1 − tk =b− a

n. We can also have the partition intervals [t0,t1], [t1,t2], ...,

[tn−1,tn] have different lengths.R bag (t) dZ (t) = lim

n→∞

Pn−1k=0 g (tk) [Z (tk+1)− Z (tk)]

Definition 20.3.1.

Suppose g (t) is a simple process, meaning that g (t) is piecewise-constantbut may have jumps at a = t0 < t1 < t2 < ... < tn = b. If

R baE£g2 (t)

¤dt <∞,

then the stochastic integralR bag (t) dZ (t) is defined asZ b

a

g (t) dZ (t) =n−1Xk=0

g (tk) [Z (tk+1)− Z (tk)] (20.16)

In the above definition,R baE£g2 (t)

¤dt < ∞ is the sufficient condition forPn−1

k=0 g (tk) [Z (tk+1)− Z (tk)] to exist.

Example 20.3.1. CalculateR T0dZ (t)

Solution.



Here g (t) = 1.R T0E¡12¢dt = T <∞. So

R T0dZ (t) exists.R T

0dZ (t) =

Pn−1k=0 [Z (tk+1)− Z (tk)] = Z (T )

Example 20.3.2.

X (t) =

⎧⎨⎩ 1 if 0 ≤ t ≤ 12 if 1 < t ≤ 23 if 2 < t ≤ 3

CalculateR 30X (t) dZ (t)

Solution.

Dividend [0, 3] into (0, 1), (1, 2), and (2, 3). Then X (t) is constant duringthe interval and jumps at t = 1 and t = 2. HenceR 3

0X (t) dZ (t)

= X (0) [Z (1)− Z (0)] +X (1) [Z (2)− Z (1)] +X (2) [Z (3)− Z (2)]= 1 [Z (1)− Z (0)] + 2 [Z (2)− Z (1)] + 3 [Z (3)− Z (2)]= 1Z (1) + 2 [Z (2)− Z (1)] + 3 [Z (3)− Z (2)]= 3Z (3)− Z (2)− Z (1)

If g (tk) is not a simple process, then we define the stochastic integral asfollows.If E lim

n→∞(X − a)2 = 0, we say that X approach a in mean square.

Explain why the sample mean approaches the population mean in meansquare.Suppose we take n random samples X1, X2,...,Xn from the population X.

Let μ represent the population mean. Then the sample mean is1

n

Pnk=1Xk. It

can be shown that

E limn→∞

µ1

n

Pnk=1Xk − μ

¶2= 0

To see why, notice

E

µ1

n

Pnk=1Xk

¶=1

nE (Pn

k=1Xk) =1

nnμ = μ

E

µ1

n

Pnk=1Xk − μ

¶2= V ar

µ1

n

Pnk=1Xk

¶=1

n2V ar (

Pnk=1Xk)

=1

n2nV ar (X) =

V ar (X)

n

→ E limn→∞

µ1

n

Pnk=1Xk − μ

¶2= 0

So the sample mean approaches the population mean in mean square.



Definition 20.3.2.

Let a = t0 < t1 < t2 < ... < tn = b represent a partition of [a, b]. Define

random variable In =Pn−1

k=0 g (tk) [Z (tk+1)− Z (tk)], where g (tk) is a simpleor a complex process and g (tk) and Z (tk+1) − Z (tk) are independent. IfR baE£g2 (t)

¤dt <∞ and the mean squared difference between In and U is zero

as n→∞:

E limn→∞

(In − U)2= 0 (20.17)

Then we say

•R bag (t) dZ (t) = U

• In converges toR bag (t) dZ (t) in mean square

The above definition holds whether g (t) is a simple or complex process.Please note that in the term g (tk) [Z (tk+1)− Z (tk)], g is evaluated at the

left of the interval [tk, tk+1]:R bag (t) dZ (t) = lim

n→∞

Pn−1k=0 g (tk) [Z (tk+1)− Z (tk)]R b

ag (t) dZ (t) 6= lim

n→∞

Pn−1k=0 g (tk+1) [Z (tk+1)− Z (tk)]R b

ag (t) dZ (t) 6= lim

n→∞

Pn−1k=0

g (tk) + g (tk+1)

2[Z (tk+1)− Z (tk)]

This is different from the deterministic calculusR ih(i−1)h x

2dx, which can beapproximated using 3 heights:

• The left height [(i− 1)h]2

• The right height (ih)2

• The average height (i− 1)2 h2 + (ih)2

2

In addition, we require that g (tk) and Z (tk+1)−Z (tk) are independent. Inother words, we require that g (tk) not depend on the future Brownian incre-ment Z (tk+1) − Z (tk). The requirements that we evaluate g at the left of theinterval [tk, tk+1] and that g (tk) and Z (tk+1) − Z (tk) are independent agreewith our intuition. We calculating g (tk) [Z (tk+1)− Z (tk)], g (tk) is based onthe information available to us during [0, tk] and is independent of the futureBrownian increment Z (tk+1)− Z (tk).

Example 20.3.3. CalculateR T0[dZ (t)]

2.



R T0[dZ (t)]

2=R T0g (t) [dZ (t)]

2 where g (t) = 1Partition [0, T ] into [0, h], [h, 2h],...,[(n− 1)h, nh = T ].Let In =

Pn−1k=0Z [(k + 1)h]− [Z (kh)]2 =

Pnk=1Z (kh)− [Z (k − 1)h]2

Z (kh)− [Z (k − 1)h] is a normal random variable with mean 0 and varianceh. Using Equation 20.9, we have

EZ (kh)− [Z (k − 1)h]2 = hIn =

Pnk=1Z (kh)− [Z (k − 1)h]2

Since E (In) = nh = T , we guess thatR T0[dZ (t)]

2= T

The difficult part is to verify that the mean square error is zero as n→∞:limn→∞

Eh(Pn

k=1Z (kh)− [Z (k − 1)h])2 − T

i2= 0

Let ∆k = Z (kh)− [Z (k − 1)h]→ (

Pnk=1Z (kh)− [Z (k − 1)h])

2= ∆21 +∆

22 + ...+∆2n

→h(Pn

k=1Z (kh)− [Z (k − 1)h])2 − T

i2=£¡∆21 +∆

22 + ...+∆2n

¢− T

¤2=¡∆21 +∆

22 + ...+∆2n

¢2+T 2−2T

¡∆21 +∆

22 + ...+∆2n

¢¡∆21 +∆

22 + ...+∆2n

¢2= ∆41 +∆

42 + ...∆4n + 2∆

21∆

22 + 2∆

21∆

23 + ...

→ E¡∆21 +∆

22 + ...+∆2n

¢2= E

¡∆41 +∆

42 + ...∆4n

¢+2E

¡∆21∆

22 +∆

21∆

23 + ...

¢∆k is normal with mean 0 and variance h→ E

¡∆4k¢= 3h2

→ E¡∆41 +∆

42 + ...∆4n

¢= 3nh2

∆i and ∆j where i 6= j are two independent normal random variables (Point3 of the Brownian motion definition)→ E

¡∆2i∆

2j

¢= E

¡∆2i¢E¡∆2j¢= h× h = h2

There are1

2n (n− 1) pairs of ∆i and ∆j where i 6= j

→ 2E¡∆21∆

22 +∆

21∆

23 + ...

¢= 2× 1

2n (n− 1)h2 = n (n− 1)h2

E¡∆21 +∆

22 + ...+∆2n

¢2= 3nh2 + n (n− 1)h2

E£2T¡∆21 +∆

22 + ...+∆2n

¢¤= 2TE

¡∆21 +∆

22 + ...+∆2n

¢= 2T (nh) = 2T 2

→ Eh(Pn

k=1Z (kh)− [Z (k − 1)h])2 − T

i2= 3nh2 + n (n− 1)h2 + T 2 − 2T = [3n+ n (n− 1)]h2 − T 2

However, h =T

n

→ [3n+ n (n− 1)]h2−T 2 = [3n+ n (n− 1)]µT

n

¶2−T 2 = 3n+ n (n− 1)

n2T 2−

T 2

As n→∞, 3n+ n (n− 1)n2

→ 13n+ n (n− 1)

n2T 2 − T 2 → 0



limn→∞

Eh(Pn

k=1Z (kh)− [Z (k − 1)h])2 − T

i2= 0

HenceR T0[dZ (t)]2 = T .

Example 20.3.4. CalculateR T0Z (t) dZ (t)

Partition [0, T ] into [0, h], [h, 2h],...,[(n− 1)h, nh = T ].Let In =

Pn−1k=0 Z (kh) Z [(k + 1)h]−Z (kh) =

Pnk=1 Z [(k − 1)h] Z (kh)−

Z [(k − 1)h]

Use the formula: ab =(a+ b)

2 −¡a2 + b2

¢2

Let a = Z [(k − 1)h] b = Z (kh)− Z [(k − 1)h]a+ b = Z (kh)Pn

k=1 Z [(k − 1)h] Z (kh)− Z [(k − 1)h]

=Pn

k=1

[Z (kh)]2 − Z [(k − 1)h]2 − Z (kh)− Z [(k − 1)h]2

2

=1

2

Pnk=1[Z (kh)]

2 − Z [(k − 1)h]2− 12

Pnk=1Z (kh)− Z [(k − 1)h]2

=1

2[Z (nh)]

2 − 12

Pnk=1Z (kh)− Z [(k − 1)h]2

=1

2[Z (T )]

2 − 12

Pnk=1Z (kh)− Z [(k − 1)h]2

→ In =1

2[Z (T )]

2 − 12

Pnk=1Z (kh)− Z [(k − 1)h]2

limn→∞

Pnk=1 Z [(k − 1)h] Z (kh)− Z [(k − 1)h] = lim

n→∞In

=1

2[Z (T )]

2 − 12limn→∞

Pnk=1Z (kh)− Z [(k − 1)h]2

From the previous example, we know thatPn

k=1Z (kh)−Z [(k − 1)h]2 ap-proaches

R T0[dZ (t)]2 = T in the mean square. So we guess that

R T0Z (t) dZ (t) =

1

2[Z (T )]2 − 1

2T

Next, we need to prove that E limn→∞

∙In −

µ1

2[Z (T )]

2 − 12T

¶¸2= 0

In −µ1

2[Z (T )]

2 − 12T

¶=1

2T − 1

2

Pnk=1Z (kh)− Z [(k − 1)h]2

In − E (In)

=1

2[Z (T )]2 − 1

2

Pnk=1Z (kh)− Z [(k − 1)h]2 −

µ1

2[Z (T )]2 − 1

2T

¶=1

2

¡T −

Pnk=1Z (kh)− Z [(k − 1)h]2

¢From the previous example, we found thatlimn→∞

E¡T −

Pnk=1Z (kh)− Z [(k − 1)h]2

¢2= 0

→ E limn→∞

∙In −

µ1

2[Z (T )]2 − 1

2T

¶¸2= 0



So we have

Z T

0

Z (t) dZ (t) =1

2[Z (T )]

2 − 12T (20.18)

Equation 20.18 is surprising. In the deterministic calculus, we haveR T0xdx =

1

2T 2.

Equation 20.18 has an extra term −12T . This is why this extra term is

needed. Taking expectation of Equation 20.18:

EhR T0Z (t) dZ (t)

i=1

2E [Z (T )]

2 − 12T

As to be explained later, EhR T0Z (t) dZ (t)

i= 0. We already know that

E [Z (T )]2= T . Hence E

hR T0Z (t) dZ (t)

i=1

2E [Z (T )]

2 − 12T = 0. The extra

term −12T is needed so the expectations of both sides of Equation 20.18 are

equal.

20.4 Properties of the stochastic calculus

1. Lineality.R T0[c1g (t) + c2h (t)] dZ (t) = c1

R T0g (t) dZ (t)+c2

R T0h (t) dZ (t)

2. Zero mean property. IfR T0E£X2 (t)

¤dt <∞, then E

³R T0X (t) dZ (t)

´=

0

The proof is complex. However, for a simple process g (t) and h (t), Property#1 holds due to the definition of the stochastic integral. For a simple processg (t), Property #2 can be easily established.R T

0g (t) dZ (t) =

Pn−1k=0 g (tk) [Z (tk+1)− Z (tk)]

g (tk) is constant during each partition interval and independent of Z (tk+1)−Z (tk). Then

Eg (tk) [Z (tk+1)− Z (tk)] = E [g (tk)]E [Z (tk+1)− Z (tk)] = E [g (tk)] 0 =0

Let’s apply Property 2 toR T0Z (t) dZ (t). Since

R T0E£Z2 (t)

¤dt =

R T0tdt =

1

2T 2 <∞, we have E

hR T0Z (t) dZ (t)

i= 0


20.5. ITO’S LEMMA 223

20.5 Ito’s lemma

20.5.1 Multiplication rules

dZ dtdZ dt 0dt 0 0

The above table means:[dZ (t)]2 = dt dZ (t) dt = 0 (dt)2 = 0

Example 20.5.1. Explain why [dZ (t)]2 = dt

dZ (t) is a normal random variable with mean 0 and variance dt (see thefootnote of Derivatives Markets Page 652). Hence E [dZ (t)]2 = dt

E [dZ (t)− dt]2= V ar [dZ (t)] = dt→ 0

Hence dt approach [dZ (t)]2 in mean square. So [dZ (t)]2 = dt

Example 20.5.2. Explain why dZ (t) dt = 0

E [dZ (t) dt] = E [dZ (t)] dt = 0 (here dt is treated as a constant)E [dZ (t) dt− 0]2 = E [dZ (t)]

2(dt)

2= (dt)

3 → 0Hence 0 approaches dZ (t) dt in mean square. dZ (t) dt = 0.

Example 20.5.3. Explain why (dt)2 = 0

(dt)2 doesn’t contain any Brownian motion term dZ. So we need to calculate

(dt)2 according to the deterministic calculus. In the deterministic calculus,(dt)

2 → 0 as dt→ 0. Hence (dt)2 = 0.The textbook Derivatives Markets also gives the following formula:

dZ × dZ0= ρdt (20.19)

The above formula will be explained later.

20.5.2 Ito’s lemma

In essence, Ito’s lemma is a Taylor series applied to Brownian motion.

Suppose that a stock has an expected instant return∧α [S (t) , t], dividend

yield∧δS (t) , t, and instant volatility

∧σ [S (t) , t] follows geometric Brownian mo-

tion:

dS (t) =

µ∧α−

∧δ

¶dt+

∧σdZ (t) (20.20)

Here∧α,∧δ, and

∧σ are function of the stock price S (t) and time t.



Let C [S (t) , t] represent the value of a call or put option. We want to findout the change of the option value given there’s a small change of the stock priceand a small change of time. Using Taylor series, we have:

dC [S (t) , t] =∂C

∂SdS+

∂C

∂tdt+

1

2

∂2C

∂S2(dS)

2+1

2

∂2C

∂t2(dt)

2+

∂2C

∂S∂tdSdt (20.21)

[dS (t)]2=

∙µ∧α−

∧δ

¶dt+

∧σdZ (t)

¸2=

µ∧α−

∧δ

¶2(dt)

2+2

µ∧α−

∧δ

¶∧σdZ (t) dt+

∧σ2[dZ (t)]

2

dSdt =

µ∧α−

∧δ

¶(dt)

2+∧σdZ (t) dt

Using the multiplication rules: (dt)2 = 0 [dZ (t)]2= dt dZ (t) dt = 0

[dS (t)]2 =∧σ2dt dSdt = 0

Now we have:

dC [S (t) , t] =∂C

∂SdS +

∂C

∂tdt+

1

2

∂2C

∂S2(dS)

2=

∂C

∂SdS +

∂C

∂tdt+

1

2

∂2C

∂S2∧σ2dt

(20.22)Next, apply Equation 20.20 to Equation ??:

→ dC [S (t) , t] =∂C

∂S

∙µ∧α−

∧δ

¶dt+

∧σdZ (t)

¸+

∂C

∂tdt+

1

2

∂2C

∂S2∧σ2dt

=

∙µ∧α−

∧δ

¶∂C

∂S+1

2

∧σ2 ∂2C

∂S2+

∂C

∂t

¸dt+

∂C

∂S

∧σdZ (t)

dC [S (t) , t] =

∙µ∧α−

∧δ

¶∂C

∂S+1

2

∧σ2 ∂2C

∂S2+

∂C

∂t

¸dt+

∂C

∂S

∧σdZ (t) (20.23)

Equation 20.23 is called the Ito’s lemma.

Tip 20.5.1. Don’t bother memorizing Equation 20.23. Just derive Equation20.23 on the spot. First, write down the Taylor series Equation 20.22. Next,apply the multiplication rules. Then you’ll get Equation 20.23.

If S (t) follows a geometric Brownian motion, we have:

• ∧α [S (t) , t] = αS (t)

•∧δ [S (t) , t] = δS (t)

• ∧σ [S (t) , t] = σS (t)

Equation 20.23 becomes:

dC [S (t) , t] =

∙(α− δ)S

∂C

∂S+1

2σ2S2

∂2C

∂S2+

∂C

∂t

¸dt+

∂C

∂SσSdZ (t) (20.24)


20.6. GEOMETRIC BROWNIAN MOTION REVISITED 225

Tip 20.5.2. Don’t bother memorizing Equation 20.24. Just derive Equation20.24 on the spot.

Please note that∂C

∂S= ∆

∂2C

∂S2= Γ

∂C

∂t= θ (option Greeks)

Then Equation 20.22 becomes:

dC [S (t) , t] = ∆dS + θdt+1

2Γ (dS)

2 (20.25)

Consider a tiny interval dt ≈ h. We have:dC [S (t) , t] ≈ C [S (t+ h) , t+ h]− C [S (t) , t]

dS ≈ S (t+ h)− S (t)Equation 20.25 becomes:

C [S (t+ h) , t+ h]−C [S (t) , t] ≈ ∆ [S (t+ h)− S (t)]+θh+1

2Γ [S (t+ h)− S (t)]

2

(20.26)

Equation 20.26 is just the textbook Equation 13.6 (Derivatives Markets page426).If S (t) were deterministic (i.e. if Equation 20.20 didn’t have dZ (t) term),

then∂2C

∂S2→ 0 and Equation 20.22 would be:

dC [S (t) , t] =∂C

∂SdS +

∂C

∂tdt

20.6 Geometric Brownian motion revisited

There are two minor concepts under geometric Brownian motion. SOA caneasily write a question on these concepts. So let’s study them.

20.6.1 Relative importance of drift and noise term

Consider a discrete geometric Brownian motion:

X (t+ h)−X (t) = αX (t)h| z deterministic component

+ σX (t)Y (t)√h| z

random component

One phrase you need to know is called "the ratio of the standard deviationto the drift" (the drift is actually the deterministic component). The ratio ofthe standard deviation to the drift is defined as:

σX (t)√h

αX (t)h=

σ

α√h

(20.27)



20.6.2 Correlated Ito processes

Let W1 (t) and W2 (t) represent two independent Brownian motions. Suppose

Z (t) =W1 (t) (20.28)

Z0(t) = ρW1 (t) +

p1− ρ2W2 (t) (20.29)

Please note that Z (t) is normally distributed with mean 0 and variance tbecause W1 (t) is normally distributed with mean 0 and variance t.You might wonder why Equation 20.29 has constants ρ and

p1− ρ2. These

two constants are needed to make Z0(t) normally distributed with variance t.

Because W1 (t) and W2 (t) are two independent normal random variables, thelinear combination ρW1 (t) +

p1− ρ2W2 (t) is also normally distributed.

EhZ0(t)i= E

hρW1 (t) +

p1− ρ2W2 (t)

i= E [ρW1 (t)] +E

hp1− ρ2W2 (t)

i= ρE [W1 (t)] +

p1− ρ2E [W2 (t)]

= ρ0 +p1− ρ20 = 0

V arhZ0(t)i= V ar

hρW1 (t) +

p1− ρ2W2 (t)

i= V ar [ρW1 (t)] + V ar

hp1− ρ2W2 (t)

i= ρ2V ar [W1 (t)] +

¡1− ρ2

¢V ar [W2 (t)]

= ρ2t+¡1− ρ2

¢t = t

The covariance between Z (t) and Z0(t) is: Cov

hZ (t) , Z

0(t)i

Using the standard formula: Cov (X,Y ) = E (XY )−E (X)E (Y )

→ CovhZ (t) , Z

0(t)i= E

hZ (t)Z

0(t)i−E [Z (t)]E

hZ0(t)i

= EhZ (t)Z

0(t)i− 0× 0 = E

hZ (t)Z

0(t)i

Z (t)Z0(t) =W1 (t)

hρW1 (t) +

p1− ρ2W2 (t)

i= ρ [W1 (t)]

2+p1− ρ2W1 (t)W2 (t)

EhZ (t)Z

0(t)i= E

³ρ [W1 (t)]

2´+E

hp1− ρ2W1 (t)W2 (t)

i= ρE [W1 (t)]

2+p1− ρ2E [W1 (t)W2 (t)]

E [W1 (t)]2 = t

E [W1 (t)W2 (t)] = E [W1 (t)]E [W2 (t)] = 0× 0 = 0

EhZ (t)Z

0(t)i= ρt (20.30)



The textbook says calls CovhZ (t) , Z

0(t)i= E

hZ (t)Z

0(t)i= ρt the corre-

lation between Z (t) and Z0(t). However, the term correlation between X and

Y typically means the following:

ρX,Y =Cov (X,Y )

σXσYIf we use the typical definition, the correlation between Z (t) and Z

0(t) is:

CovhZ (t) , Z

0(t)i

σZ(t)σZ0 (t)

=ρt√t√t= ρ

Since Derivatives Markets is the textbook, we have to adopt its definitionthat the correlation between Z (t) and Z

0(t) is ρt.

dZ (t) and dZ0(t) are both normal random variables with mean 0 and vari-

ance dt. Applying Equation 20.30 and replacing t with dt, we have:

EhdZ (t) dZ

0(t)i= ρdt (20.31)

Finally, let’s explain why Equation 20.19 dZ × dZ0= ρdt holds.

EhdZ (t) dZ

0(t)i= ρdt

EhdZ (t) dZ

0(t)− ρdt

i2= V ar

hdZ (t) dZ

0(t)i= E

hdZ (t) dZ

0(t)i2−E2

hdZ (t) dZ

0(t)i

dZ (t) dZ0(t)

= dW1 (t)× dhρW1 (t) +

p1− ρ2W2 (t)

i= dW1 (t)×

hρdW1 (t) +

p1− ρ2dW2 (t)

i= ρ [dW1 (t)]

2 +p1− ρ2dW1 (t) dW2 (t)h

dZ (t) dZ0(t)i2

= ρ2 [dW1 (t)]4+¡1− ρ2

¢[dW1 (t)]

2[dW2 (t)]

2+2ρ

p1− ρ2 [dW1 (t)]

3dW2 (t)

EhdZ (t) dZ

0(t)i2

= ρ2E [dW1 (t)]4+¡1− ρ2

¢E³[dW1 (t)]

2 [dW2 (t)]2´+2ρ

p1− ρ2E

³[dW1 (t)]

3 dW2 (t)´

= ρ2E [dW1 (t)]4+¡1− ρ2

¢E [dW1 (t)]

2E [dW2 (t)]2+2ρ

p1− ρ2E [dW1 (t)]

3E [dW2 (t)]

= ρ23 (dt)2 +¡1− ρ2

¢dt× dt+ 2ρ

p1− ρ20× 0

= ρ23 (dt)2+¡1− ρ2

¢dt× dt

=£3ρ2 +

¡1− ρ2

¢¤(dt)2 =

¡2ρ2 + 1

¢(dt)2

EhdZ (t) dZ

0(t)i= ρdt

EhdZ (t) dZ

0(t)i2−E2

hdZ (t) dZ

0(t)i

=¡2ρ2 + 1

¢(dt)

2 − ρ2 (dt)2=¡ρ2 + 1

¢(dt)

2



(dt)2= 0

→ EhdZ (t) dZ

0(t)i2−E2

hdZ (t) dZ

0(t)i= 0

→ EhdZ (t) dZ

0(t)− ρdt

i2= V ar

hdZ (t) dZ

0(t)i= 0

Hence dZ (t) dZ0(t) approach ρdt in mean square. Hence

dZ × dZ0= ρdt (20.32)

Suppose the stock price S (t) follows a geometric Brownian motion:dS (t)

S (t)= αdt+ σdZ (t)

1. Apply Ito’s lemma to lnS (t) and derive that lnS (t) is a normal randomvariable with mean lnS (0) +

£¡α− 0.5σ2

¢t¤and variance σ2t.

2. Derive that the mean of S (t) is E [S (t)] = S (0) eαt

As explained in my book and my solution manual, you don’t need to mem-orize Ito’s lemma. Just use the Taylor expansion but keep the (dZ)2 term.First, use Taylor expansion:

d lnS =∂ lnS

∂SdS +

∂ lnS

∂tdt+

1

2

∂2 lnS

∂S2(dS)

2

In the deterministic calculus, you just write d lnS =∂ lnS

∂SdS +

∂ lnS

∂tdt.

However, in the stochastic calculus, we can’t ignore (dZ)2 since (dZ)2 = dt.In this problem, d lnS is a linear function of dZ, so:

(dS)2= [Sαdt+ σdZ]

2= S2

hα2 (dt)

2+ 2ασdtdZ + σ2 (dZ)

2i

Using the multiplication rule (DM 20.17 a, b, c), we get:

(dS)2= S2

hα2 (dt)

2+ 2αdtσdZ + σ2 (dZ)

2i= S2

£α2 × 0 + 2ασ × 0 + σ2dt

¤=

S2σ2dt

So we have to keep the term1

2

∂2 lnS

∂S2(dS)

2.

Anyway, Taylor expansion gives us:

d lnS =∂ lnS

∂SdS +

∂ lnS

∂tdt+

1

2

∂2 lnS

∂S2S2σ2dt

Next,∂ lnS

∂S=1

S

∂2 lnS

∂S= − 1

S2∂ lnS

∂t= 0

→ d lnS =dS

S− 12σ2dt = αdt+ σdZ − 1

2σ2dt =

µα− 1

2σ2¶dt+ σdZ

During a tiny time interval [t, t+ dt], the change of lnS is d lnS.d lnS is a normal random variable. This is why.First, dZ ∼ N (0, dt) .According to DM 20.4, dZ = Y

√dt. So dZ is a normal

random variable with mean 0 and variance dt.



During the fixed interval [t, t+ dt], dt is a constant. Since a constant plus arandom variable is also a random variable,

d lnS =

µα− 1

2σ2¶dt+σdZ is a normal random variable with mean

µα− 1

2σ2¶dt

and variance σ2dt

Now consider the time interval [0, t].

lnS (t) = lnS (0) +R t0d lnS = lnS (0) +

R t0

µα− 1

2σ2¶ds+

R t0σdZ

= lnS (0) +

µα− 1

2σ2¶R t

0ds+ σ

R t0dZ = lnS (0) +

¡α− 0.5σ2

¢t+ σZ

Z is a standard normal random variable with mean 0 and variance 1, thatis, Z ∼ N

¡0, σ2t

¢. Hence σZ is a normal random variable with mean 0 and

variance σ2t. And lnS (t) is normal with mean lnS (0) +¡α− 0.5σ2

¢t and

variance σ2t:

lnS (t) ∼ N£lnS (0) +

¡α− 0.5σ2

¢t, σ2t

¤

Next, from lnS (t) = lnS (0) +¡α− 0.5σ2

¢t+ σZ, we get:

elnS(t) = elnS(0)+(α−0.5σ2)t+σZ

→ S (t) = S (0) e(α−0.5σ2)t+σZ =

hS (0) e(α−0.5σ

2)tieσZ

→ E [S (t)] =hS (0) e(α−0.5σ

2)tiE¡eσZ

¢

We can use DM Equation 18.13 to calculate E¡eσZ

¢. DM Equation 18.13

says that if x is normal with mean m and variance v2, that is x ∼ N¡m, v2

¢,

then

E (ex) = em+0.5v2

(DM 18.13)

→ E¡eσZ

¢= e0+0.5σ

2t

→ E [S (t)] =hS (0) e(α−0.5σ

2)tie0+0.5σ

2

= S (0) eαt

Alternative method to calculateE [S (t)]. Since lnS (t) ∼ N£S (0) +

¡α− 0.5σ2

¢t, σ2t

¤,

using DM 18.13, we have:

E [S (t)] = eS(0)+(α−0.5σ2)t+0.5σ2t = S (0) eαt



I want you to memorize the following results (these results are used over andover in Exam MFE):

dZ ∼ N (0, dt) (20.33)

Z ∼ N (0, t) (20.34)

x ∼ N¡m, v2

¢→ E (ex) = em+0.5v

2

(20.35)

S (t) is geometric Brownian motion → dS (t)

S (t)= αdt+ σdZ (t) (20.36)

dS (t)

S (t)= αdt+ σdZ (t) → d lnS (t) =

¡α− 0.5σ2

¢dt+ σdZ (20.37)

dS (t)

S (t)= αdt+σdZ (t) → lnS (t) ∼ N

£lnS (0) +

¡α− 0.5σ2

¢t, σ2t

¤(20.38)

dS (t)

S (t)= αdt+ σdZ (t) → S (t) = S (0) e(α−0.5σ

2)t+σZ (20.39)

dS (t)

S (t)= αdt+ σdZ (t) → E [S (t)] = S (0) eαt (20.40)

20.7 Sharpe ratioAn asset’s Sharpe ratio is equal to the asset’s risk premium α − r divided bythe asset’s volatility σ:

SR =α− r

σ(20.41)

If two non-dividend paying assets are perfectly corrected (i.e. they are drivenby the same Brownian motion Z (t)), then their Sharpe ratios are equal. Let’sderive this.The price processes of Asset 1 and Asset 2 are:

dS1 = α1S1dt+ σ1S1dZ (20.42)

dS2 = α2S2dt+ σ2S2dZ (20.43)

We can form a riskless portfolio by removing the random Brownian motiondZ. Rewrite Equation 20.42 and 20.43 as:


20.7. SHARPE RATIO 231

µ1

σ1S1

¶dS1 =

α1σ1

dt+ dZ (20.44)µ1

σ2S2

¶dS2 =

α2σ2

dt+ dZ (20.45)

Suppose at time zero we buy N1 =1

σ1S1units of Asset 1 and short sell

N2 =1

σ2S2units of Asset 2. If we hold one unit of Asset 1 at time zero, then

after a tiny interval dt, the value of Asset 1 increases by the amount dS1. If

we hold N1 =1

σ1S1units of Asset 1 at time zero, then after dt the value of

N1 =1

σ1S1units of Asset 1 will increase by

µ1

σ1S1

¶dS1 =

α1σ1

dt+ dZ. Notice

that the increase of the value of Asset 1 has a random component dZ, where dZis a normal random variable with mean 0 and variance dt.Similarly, if we short sell N2 =

1

σ2S2units of Asset 2 at time zero, after dt,

the value of Asset 2 will increase byµ

1

σ2S2

¶dS2 =

α2σ2

dt+dZ. The increase of

the value of Asset 2 has a random component dZ, where dZ is a normal randomvariable with mean 0 and variance dt.Suppose at time zero we simultaneously buy N1units of Asset 1 and short

sell N2 units of Asset 2. Then at dt, we close our position by selling N1 unitsof Asset 1 in the open market and buying N2 units of Asset 2 from the openmarket.The cash flow at time zero:

• We pay N1S1 =

µ1

σ1S1

¶S1 =

1

σ1dollar to buy N1units of Asset 1.

• We receive N2S2 =1

σ2dollars for short selling N2 units of Asset 2

• The net cost is 1

σ1− 1

σ2dollars. If

1

σ1− 1

σ2< 0, then we receive

−µ1

σ1− 1

σ2

¶net cash. To avoid tying up our capital, we go to a bank

and borrow1

σ1− 1

σ2dollars. If

1

σ1− 1

σ2< 0, then we lend −

µ1

σ1− 1

σ2

¶.

• Our net cash outgo is zero.

Our payoff at time dt

• At time dt, we sell off N1 units of Asset 1 in the open market for the price

of S1 + dS1, receiving N1 (S1 + dS1) = N1S1 +N1dS1 =1

σ1+

α1σ1

dt+ dZ



• At time dt we buy N2 units of Asset 2 in the open market at the price

S2 + dS2. We pay N2 (S2 + dS2) = N2S2 +N2dS2 =1

σ2+

α2σ2

dt+ dZ

• At time dt, we pay back the bank both the principal and the accrued

interest. The sum of the principal and the interest isµ1

σ1− 1

σ2

¶erdt

Our profit at dt is:

Profit =

µ1

σ1+

α1σ1

dt+ dZ

¶−µ1

σ2+

α2σ2

dt+ dZ

¶−µ1

σ1− 1

σ2

¶erdt (20.46)

If1

σ1− 1

σ2< 0, we’ll lend −

µ1

σ1− 1

σ2

¶at t = 0. Then at dt, we’ll receive

−µ1

σ1− 1

σ2

¶erdt from the borrower.

The Taylor expansion:

erdt = 1 + rdt+r2

2!(dt)

2+

r3

3!(dt)

3+ ... = 1 + rdt

Equation 20.46 can be rewritten as:µ1

σ1+

α1σ1

dt

¶−µ1

σ2+

α2σ2

dt

¶−µ1

σ1− 1

σ2

¶erdt

=

µ1

σ1− 1

σ2

¶+

µα1σ1− α2

σ2

¶dt−

µ1

σ1− 1

σ2

¶erdt

=

µ1

σ1− 1

σ2

¶+

µα1σ1− α2

σ2

¶dt−

µ1

σ1− 1

σ2

¶−µ1

σ1− 1

σ2

¶rdt

=

µα1σ1− α2

σ2

¶dt−

µ1

σ1− 1

σ2

¶rdt

=

µα1 − r

σ1− α2 − r

σ2

¶dt

= (SR1 − SR2) dt

Profit =

µα1 − r

σ1− α2 − r

σ2

¶dt = (SR1 − SR2) dt (20.47)

Equation 20.46 and 20.47 don’t have the random term dZ (t), indicating thatthe profit is surely made.Our cash outgo is zero at t = 0, but we’ll have the profit indicated by 20.47

at dt.

Wealth 0 −→µα1 − r

σ1− α2 − r

σ2

¶dt = (SR1 − SR2) dt

Time 0 −→ dt

To avoid arbitrage, the profit needs to be zero:

→ α1 − r

σ1=

α2 − r

σ2SR1 = SR2


20.8. RISK NEUTRAL PROCESS 233

20.8 Risk neutral processThis section is difficult because the author tired to put too many complex con-cepts in this small section. It seems that the author was in a big hurry to finishthis chapter. The author mentioned many concepts (such as martingale, Gir-sanov’s theorem) but he didn’t really explain them, leaving us hanging in theair asking why.The only thing worth studying is how to transform a standard geometric

Brownian motion into a risk neutral process.The standard geometric Brownian motion is:dS (t)

S (t)= (α− δ) dt+ σdZ (t)

This is how to transform:dS (t)

S (t)= (α− δ) dt+ σdZ (t)

= (r − δ) dt+ σdZ (t) + (α− δ) dt

= (r − δ) dt+ σ

∙dZ (t) +

α− δ

σdt

¸Define d

∼Z (t) = dZ (t) +

α− δ

σdt

→ dS (t)

S (t)= (α− δ) dt+ σdZ (t) = (r − δ) dt+ σd

∼Z (t)

Just learn this transformation and move on.

20.9 Valuing a claim on Sa

In a call or put, the payoff at T is a linear function of the stock price S (T ).What if the payoff is linear? For example, what if the payoff at T is S (T ) raisedto some power a?

20.9.1 Process followed by Sa

Though the textbook uses Sa, I’m going to use SA. The reason is explainedlater.Suppose the stock price follows geometric Brownian motion:dS

S= (α− δ) dt+ σdZ

This is why I use SA. If you use Sa, the letter a in Sa is very similar to

the α (alpha) in the equationdS

S= (α− δ) dt + σdZ. This can easily lead to

confusion and mistake. As a matter of fact, when I was deriving the formula forE (Sa), I couldn’t match the textbook’s formula. It took me a while to figureout that I accidentally switched a and α.We need to determine the process followed by payoff at T is SA.



dSA =∂SA

∂SdS +

1

2

∂2SA

∂S2(dS)

2= ASA−1dS +

1

2A (A− 1)SA−2 (dS)2

However, (dS)2 = [(α− δ) dt+ σdZ]2S2 =

¡σ2dt

¢S2 = σ2S2dt

→ dSA = ASA−1dS +1

2A (A− 1)SA−2σ2S2dt

= ASA−1dS +1

2A (A− 1)SAσ2dt

→ dSA

SA= A

dS

S+1

2A (A− 1)σ2dt

= A (α− δ) dt+AσdZ + 0.5A (A− 1)σ2dt=£A (α− δ) + 0.5A (A− 1)σ2

¤dt+AσdZ

We see that SA follows a geometric Brownian motion with drift A (α− δ)+0.5A (A− 1)σ2 and risk component AσdZ.

20.9.2 Formula for SA (t) and E£SA (t)

¤Next, we apply Equation 20.37. Replace α with A (α− δ)+0.5A (A− 1)σ2 andσ with Aσ.

d lnSA (t) =£A (α− δ) + 0.5A (A− 1)σ2 − 0.5A2σ2

¤dt+AσdZ

=£A (α− δ) dt− 0.5Aσ2

¤dt+AσdZ

Using Equation 20.38, we get:lnSA (t) ∼ N

£lnSA (0) +

£A (α− δ)− 0.5Aσ2

¤t,A2σ2t

¤Using Equation 20.39:SA (t) = SA (0) e[A(α−δ)−0.5Aσ

2]t+AσZ

Using Equation 20.40 just replacing α with A (α− δ) + 0.5A (A− 1)σ2

E£SA (t)

¤= SA (0) e[A(α−δ)+0.5A(A−1)σ

2]t

We can also derive E£SA (t)

¤using SA (t) = SA (0) e[A(α−δ)−0.5Aσ

2]t+AσZ

E£SA (t)

¤= E

³SA (0) e[A(α−δ)−0.5Aσ

2]t+AσZ´

= SA (0)E³e[A(α−δ)−0.5Aσ

2]t+AσZ´

= SA (0)E³e[A(α−δ)−0.5Aσ

2]teAσZ´

= SA (0) e[A(α−δ)−0.5Aσ2]tE

¡eAσZ

¢AσZ is a normal random variable with mean 0 and variance V ar [AσZ (t)] =

A2σ2V ar [Z (t)] = A2σ2t. Using Equation 20.35, we get:E¡eAσZ

¢= e0.5A

2σ2t

→ E£SA (t)

¤= SA (0) e[A(α−δ)−0.5Aσ

2]te0.5A2σ2t = SA (0) e[A(α−δ)+0.5A(A−1)σ

2]t


20.9. VALUING A CLAIM ON SA 235

Alternative method to calculate E£SA (t)

¤. Using Equation 20.39, we get:

dS

S= (α− δ) dt+ σdZ → S (t) = S (0) e(α−δ−0.5σ

2)t+σZ

→ SA (t) = SA (0) eA(α−δ−0.5σ2)t+AσZ

→ E£SA (t)

¤= E

³SA (0) eA(α−δ−0.5σ

2)t+AσZ´

= SA (0) eA(α−δ−0.5σ2)tE

¡eAσZ

¢= SA (0) e[A(α−δ)−0.5Aσ

2]te0.5A2σ2t

= SA (0) e[A(α−δ)+0.5A(A−1)σ2]t

20.9.3 Expected return of a claim on SA (t)

Consider two geometric Brownian motions S (t) and SA (t).dS

S= (α− δ) dt+ σdZ

where α is the expected return on a claim on S (t) and δ is the continuousdividend yield earned by S (t)

dSA

SA=£A (α− δ) + 0.5A (A− 1)σ2

¤dt+AσdZ

= (γ − δ∗) dt+AσdZwhere γ is the expected return on a claim on SA (t) and δ∗ is the continuous

dividend yield earned by SA (t). The textbook calls δ∗ the lease rate.

These two processes have the same risk dZ. Consequently, they have thesame Sharpe ratio:

α− r

σ=

γ − r

Aσ→ γ = r +A (α− r)

where r is the continuously compounded risk-free interest rate.

20.9.4 Specific examples

The textbook keeps mentioning Jensen’s inequality. So let’s first talk aboutJensen’s inequality. Jensen’s inequality is in the appendix C of the textbook.You can also find information at http://en.wikipedia.org/wiki/Convex_functionJensen’s inequality says

1. if f (x) is convex, then for any probability distribution, we have E [f (x)] ≥f [E (x)]

2. if f (x) is concave, then for any probability distribution, we haveE [f (x)] ≤f [E (x)]



What’s a convex function? What’s a concave function?

If at any point you draw a tangent line, the function f (x) stays above thetangent line, then f (x) is a convex function.

If at any point you draw a tangent line, the function f (x) stays below thetangent line, then f (x) is a concave function.

A twice differentiable function of one variable is convex on an interval if andonly if its second derivative is non-negative there.

A twice differentiable function of one variable is concave on an interval ifand only if its second derivative is negative there.

Here’s a simple explanation of Jensen’s inequality.



Consider y = x2. Sined2y

dx2= 2 > 0, y = x2 is a convex function. Suppose

we take two points A (1, 1) and B (4, 16). The mid point of the line AB is

C

µ1 + 4

2= 2.5,

1 + 16

2= 8.5

¶. The fact that y = x2 is a convex function means

that y = x2 curves up. Then it follows that the point C must be above the point

D¡2.5, 2.52 = 6. 25

¢. From the graph below, we clearly sees that

12 + 42

2= 8.5

(which is the height of the point C) is greater thanµ1 + 4

2

¶2= 6.25 (which

is the height of the point D). This is an example where the mean of a convexfunction is greater than the function of the average.

0 1 2 3 4 50

5

10

15

20

25

x

y

A

B

C

D

12 + 42

2>

µ1 + 4

2

¶2, an example of E [f (x)] ≥ f [E (x)].



Next, let’s consider a concave function y =√x. Consider two points A (1, 1)

and B (4, 2). The mid point of AB is C

Ã1 + 4

2= 2.5,

√1 +√4

2= 1.5

!. Since

y =√x curves down, then it follows that C must be lower thanD

µ1 + 4

2= 2.5,

√2.5 = 1. 58

¶.

From the graph below, we clearly sees that

√1 +√4

2= 1.5 (which is the height

of the point C) is less than

r1 + 4

2= 1. 58 (which is the height of the point

D). This is an example where the mean of a concave function is less than thefunction of the average.

0 1 2 3 4 50.0

0.5

1.0

1.5

2.0

x

y

A

B

CD

√1 +√4

2<

r1 + 4

2, an example of E [f (x)] ≤ f [E (x)]

By now you should have intuitive feel of Jensen’s inequality. Let’s move on.



The textbook considers the following examples: A = −1, 0, 1.If A = 1, the time 0 value of the claim S (T ) at T is:V (0) = FP

0,T [S (T )] = e−rTS (0) e[(r−δ)]T = S (0) e−δT

This is just DM Equation 5.4.

If A = 0, then the claim is just one dollar: S0 (T ) = 1→ V (0) = FP

0,T [1] = e−rT

So the time 0 value of getting $1 at T is e−rT . V (0) is equal $1 discountedback to time 0 at the risk-free rate.

If A = 2, the time 0 value of the claim S2 (T ) at T is:

V (0) = FP0,T

£S2 (T )

¤= e−rTS2 (0) e[2(r−δ)+σ

2]T

As a general rule, the forward price of any asset at T is just the prepaidforward price accumulating at the risk-rate from time 0 to T :

F0,T = FP0,T e

rT

We use the risk-free rate r in the above equation. To get an asset at T , weas a buyer can either pay the seller FP

0,T at time 0 or pay the seller F0,T at timeT . To avoid the arbitrage, the two payments should differ only in timing. SoF0,T = FP

0,T erT .

→ F0,T£S2 (T )

¤= FP

0,T

£S2 (T )

¤erT = S2 (0) e[2(r−δ)+σ

2]T

F0,T [S (T )] = FP0,T [S (T )] e

rT = S (0) e−δT erT = S (0) e(r−δ)T

→ F0,T£S2 (T )

¤= S2 (0) e[2(r−δ)+σ

2]T = (F0,T [S (T )])2 eσ

2T

Since eσ2T ≥ 1, we have F0,T

£S2 (T )

¤≥ (F0,T [S (T )])2. This agrees with

Jensen’s inequality. Roughly speaking 1, F0,T = E (ST ).

S2 is twice differentiable andd2

dSS2 = 2 > 0. Hence S2 is convex.

According to Jensen’s inequality, we have E£S2 (T )

¤≥ (E [S (T )])2. This

leads to F0,T£S2 (T )

¤= E

£S2 (T )

¤≥ (F0,T [S (T )])2 = (E [S (T )])2.

If A = −1, the time 0 value of the claim 1

S (T )at T is:

V (0) = FP0,T

∙1

S (T )

¸= e−rT

1

S (0)e[−(r−δ)+σ

2]T = e−rT1

S (0) e(r−δ)Teσ

2T

→ F0,T

∙1

S (T )

¸= FP

0,T

∙1

S (T )

¸erT =

1

S (0) e(r−δ)Teσ

2T

However, F0,T [S (T )] = S (0) e(r−δ)T

→ F0,T

∙1

S (T )

¸=

1

F0,T [S (T )]eσ

2T ≥ 1

F0,T [S (T )]

1Experts don’t agree whether F0,T = E (ST ). Some say the forward price is the unbiasedestimate of the expected future spot price, that is, F0,T = E (ST ). Others disagree. However,it’s safe to see that F0,T is very close to E (ST ).



Since1

S (T )is concave, according to Jensen’s inequality, we have:

F0,T

∙1

S (T )

¸= E

∙1

S (T )

¸≥µF0,T

∙1

S (T )

¸¶2=

µE

∙1

S (T )

¸¶2.

S−1 is twice differentiable andd2

dS2S−1 =

2

S3> 0. Hence S2 is convex.

According to Jensen’s inequality, we have E∙

1

S (T )

¸≥µE

∙1

S (T )

¸¶2. This

leads to F0,T

∙1

S (T )

¸= E

∙1

S (T )

¸≥µF0,T

∙1

S (T )

¸¶2=

µE

∙1

S (T )

¸¶2.

If A = 0.5, the time 0 value of the claimpS (T ) at T is:

V (0) = FP0,T

hpS (T )

i= e−rT

pS (0)e[0.5(r−δ)+0.5×0.5(0.5−1)σ

2]T

= e−rTpS (0)e[0.5(r−δ)−0.125σ

2]T

→ F0,T

hpS (T )

i=pS (0)e[0.5(r−δ)−0.125σ

2]T =pS (0)e0.5(r−δ)T

=pS (0) e(r−δ)T e−0.125σ

2T =pF0,T [S (T )]e

−0.125σ2T

Since e−0.125σ2T ≤ 1, we see that F0,T

hpS (T )

i≤pF0,T [S (T )]

This agrees with Jensen’s inequality.pS (T ) is twice differentiable.

d2

dS2√S = −4S−1.5 < 0. Hence

pS (T ) is

concave.→ F0,T

hpS (T )

i= E

hpS (T )

i≤pF0,T [S (T )] =

hpE (S (T ))

i

Example 20.9.1.

The price of a stock follows a geometric Brownian motion:dS (t)

S (t)= (0.1− 0.04) dt+ 0.3dZ

The current price of the stock is 10.The continuously compounded dividend yield is 0.04 per year.The continuously compounded risk-free rate is 0.06 per year.The seller and the buyer enter a forward contract. The contract requires the

seller to pay the buyer S2 (5) five years from now.

Calculate

• the prepaid forward price

• the forward price

• γ

• δ∗



• The probability that S2 (5) ≤ 160.

Solution.

dS (t)

S (t)= (0.1− 0.04) dt+ 0.3dZ

FP0,T

£SA (T )

¤= e−rTSA (0) e[A(r−δ)+0.5A(A−1)σ

2]T

FP0,5

£S2 (5)

¤= e−0.06×5102e(2(0.06−0.04)+0.5×2(2−1)×0.3

2)5 = 141. 906 8

F0,5£S2 (5)

¤= FP

0,5

£S2 (5)

¤e0.06×5 = 141. 906 8e0.06×5 = 191. 554 1

γ = r +A (α− r) = 0.06 + 2 (0.1− 0.06) = 0.14δ∗ = γ −

£A (α− δ) + 0.5A (A− 1)σ2

¤= 0.14−

¡2 (0.1− 0.04) + 0.5× 2 (2− 1)× 0.32

¢= −0.07

It’s OK for δ∗ to be negative.lnSA (t) ∼ N

£lnSA (0) +

£A (α− δ)− 0.5Aσ2

¤t, A2σ2

¤lnS2 (5) ∼ N

£ln 102 +

£2 (0.1− 0.04)− 0.5× 2× 0.32

¤5, 22 × 0.32 × 5

¤¡2 (0.1− 0.04)− 0.5× 2× 0.32

¢5 = 0.15

ln 102 +¡2 (0.1− 0.04)− 0.5× 2× 0.32

¢5 = ln 100 + 0.15 = 4. 755 17

P£S2 (5) ≤ 160

¤= P

£lnS2 (5) ≤ ln 160

¤= Φ (z)

z =ln 160− 4. 755 17√22 × 0.32 × 5

= 0.238 5

Φ (z) = 0.594 3 → P£S2 (5) ≤ 160

¤= 0.594 3

Example 20.9.2.


S (t)= (0.12− 0.05) dt+ 0.25dZ


seller to pay the buyer1

S (4)four years from now.

Calculate



• γ



• δ∗

• The probability that 1

S (4)≤ 0.03.

Solution.dS (t)

S (t)= (0.12− 0.05) dt+ 0.25dZ

FP0,T

£SA (T )

¤= e−rTSA (0) e[A(r−δ)+0.5A(A−1)σ

2]T

FP0,4

∙1

S (4)

¸= e−0.07×4

1

20e(−1(0.07−0.05)+0.5×(−1)(−1−1)×0.25

2)4 = 0.04479

F0,4

∙1

S (4)

¸= FP

0,4

∙1

S (4)

¸e0.07×4 = 0.04479e0.07×4 = 0.05 926

γ = r +A (α− r) = 0.07− 1 (0.12− 0.07) = 0.02δ∗ = γ −

£A (α− δ) + 0.5A (A− 1)σ2

¤= 0.02−

¡−1 (0.12− 0.05) + 0.5× (−1) (−1− 1)× 0.252

¢= 0.027 5

lnSA (t) ∼ N£lnSA (0) +

£A (α− δ)− 0.5Aσ2

¤t,A2σ2

¤ln

1

S (4)∼ N

∙ln1

20+£−1 (0.12− 0.05)− 0.5× (−1)× 0.252

¤4, (−1)2 × 0.252 × 4

¸¡−1 (0.12− 0.05)− 0.5× (−1)× 0.252

¢4 = −0.155

ln1

20+£−1 (0.12− 0.05)− 0.5× (−1)× 0.252

¤4 = ln

1

20−0.155 = −3. 150 7

P

∙1

S (4)≤ 0.03

¸= P

∙ln

1

S (4)≤ ln 0.03

¸= Φ (z)

z =ln 0.03− (−3. 150 7)q(−1)2 × 0.252 × 4

= −0.711 7

Φ (z) = 0.238 3 → P

∙1

S (4)≤ 0.03

¸= 0.238 3

Example 20.9.3.


S (t)= (0.15− 0.04) dt+ 0.35dZ


seller to pay the buyerpS (6) six years from now.

Calculate





• γ

• δ∗

• The probability thatpS (6) ≤ 4.

Solution.

dS (t)

S (t)= (0.15− 0.04) dt+ 0.35dZ

FP0,T

£SA (T )

¤= e−rTSA (0) e[A(r−δ)+0.5A(A−1)σ

2]T

FP0,6

hpS (6)

i= e−0.08×6

√10e(0.5(0.08−0.04)+0.5×0.5(0.5−1)×0.35

2)6 = 2. 012 6

F0,6

hpS (6)

i= FP

0,6

hpS (6)

ie0.08×6 = 2. 012 6e0.08×6 = 3. 252 5

γ = r +A (α− r) = 0.08 + 0.5 (0.15− 0.08) = 0.115δ∗ = γ −

£A (α− δ) + 0.5A (A− 1)σ2

¤= 0.115 −

¡0.5 (0.15− 0.04) + 0.5× 0.5 (0.5− 1)× 0.352

¢= 0.075 3

lnSA (t) ∼ N£lnSA (0) +

£A (α− δ)− 0.5Aσ2

¤t, A2σ2

¤lnpS (6) ∼ N

£ln√10 +

£0.5 (0.15− 0.04)− 0.5× 0.5× 0.352

¤5, 0.52 × 0.352 × 6

¤¡0.5 (0.15− 0.04)− 0.5× 0.5× 0.352

¢6 = 0.146 25

ln√10 +

£0.5 (0.15− 0.04)− 0.5× 0.5× 0.352

¤5 = ln

√10 + 0.146 25 = 1.

297 5

Php

S (6) ≤ 4i= P

hlnpS (6) ≤ ln 4

i= Φ (z)

z =ln 4− 1. 297 5√0.52 × 0.352 × 6

= 0.207 1

Φ (z) = 0.582 0 → Php

S (6) ≤ 4i= 0.582 0




Chapter 21

Black-Scholes equation

21.1 Differential equations and valuation undercertainty

21.1.1 Valuation equation

Let’s consider a risk-free world where all assets just earn the risk free interestrate. Support at time t we spend S (t) to buy one share of a stock. Then at t+h,we receive D (t+ h)h amount of the dividend, where D (t+ h) represents thedividend accumulated per unit of time during [t, t+ h]. At t+ h after receivingD (t+ h)h dividend, we sell the stock and receive S (t+ h). The total amountof money we have at t+h is D (t+ h)h+S (t+ h). Had we put S (t) amount ofmoney in a savings account, we would have S (t) (1 + rh) amount of money att+h, where rh represent the (not-annualized) risk-free interest rate per h period.To avoid arbitrage, we need to have S (t) (1 + rh) = D (t+ h)h + S (t+ h).Rearranging this equation, we get:

S (t) =D (t+ h)h+ S (t+ h)

1 + rh(DM 21.1)

S (t+ h)− S (t) +D (t+ h)h = rhS (t) (DM 21.2)

dS (t)

dt= lim

h→0

S (t+ h)− S (t)

h= lim

h→0

hrhhS (t)−D (t+ h)

i= rS (t)−D (t)

(DM 21.3)

In Equation 21.3, r =1

h× rh represents the annualized continuously com-

pounded interest rate per year. rh is the interest rate per h period; the total

number of h lengths in a year is1

h. Hence r =

1

h× rh represents the annualized

continuously compounded interest rate per year.

245


246 CHAPTER 21. BLACK-SCHOLES EQUATION

21.1.2 Bonds

For a zero-coupon bond, D (t) = 0. DM Equation 21.3 becomesdS (t)

dt= rS (t).

This gives us

S (t) = S (T ) e−r(T−t) (DM 21.4)

Since the boundary condition is S (T ) = 1, we have S (t) = e−r(T−t). Thisequation says that $1 at T is worth e−r(T−t) at time t.

21.1.3 Dividend paying stock

At time t, you spend S (t) and buy one share of a stock. This gives you twothings:

• at time T you can sell the stock and get S (T ), which is worth S (T ) e−r(T−t)at t

• you accumulate dividend at a continuous rate of D (s), where D (s) is theinstant dividend earned per unit of time at time s. The total value ofthe continuous dividend earned during the interval [s, s+ ds] is D (s) ds,which is worth [D (s) ds] e−r(s−t) = D (s) e−r(s−t)ds at time t. The presentvalue at time t of the total continuous dividend earned during the interval[t, T ] is

R TtD (s) e−r(s−t)ds The PV of this continuous flow of dividend is

DaT−t|r = D1− e−r(T−t)

r

To avoid arbitrage, we have S (t) = S (T ) e−r(T−t) +R TtD (s) e−r(s−t)ds.

Please note that if D (s) = D is a constant, thenR TtD (s) e−r(s−t)ds =

DR Tte−r(s−t)ds = DaT−t|r = D

1− e−r(T−t)

r.

Anyway,R TtD (s) e−r(s−t)ds is a continuous annuity. If you have trouble

understandingR TtD (s) e−r(s−t)ds, refer to your FM book.

21.2 Black-Scholes equation

21.2.1 How to derive Black-Scholes equation

This section derives DM Equation 21.11.

Vt + 0.5σ2S2VSS + (r − δ)SVS − rV = 0 (DM 21.11)

This is the outline of how to derive this formula.At time t

• We buy one option on the stock. We pay V


21.2. BLACK-SCHOLES EQUATION 247

• We buy N shares of the stock (a negative N means short selling stocks).We pay NS .

• We deposit W means into a savings account. We pay W .

To have zero-financing, we set our total initial cost to zero

I = V +NS +W = 0 (DM 21.7)

Next, let’s consider the change of I during [t, t+ dt]:

dI = dV +N (dS + δSdt) + dW (DM 21.8)

dI is the interest earned on the option during [t, t+ dt]. Since W is investedin a savings account, we have dW = rWdt. This says that the interest earnedon W during [t, t+ dt] is rWdt. Notice that the change of S is dS + δSdt (thesum of the change of the stock price dS and the dividend received δSdt). ApplyIt’o lemma:

dV = Vtdt+ VSdS + 0.5σ2S2VSSdt

→ dI = Vtdt + VSdS + 0.5σ2S2VSSdt + N (dS + δSdt) + dW = Vtdt +

(VS +N) dS + 0.5σ2S2VSSdt+NδSdt+ dW

Set N = −VS . Then dS term becomes zero and W = − (V − VSS).Because our initial cost is zero, the interest we earned dI should be zero.

dI = Vtdt+ 0.5σ2S2VSSdt− VSδSdt− r (V − VSS) dt = 0

This leads to DM 21.11.Vt + 0.5σ

2S2VSS + (r − δ)SVS − rV = 0

21.2.2 Verifying the formula for a derivative

Simple PV calculation

Verification that the price of a zero-coupon bond satisfies the Black-Scholesequation DM 21.11.

$1 at time T is worth V (t) = e−r(T−t) at t.=⇒ VS = VSS = 0 Vt = re−r(T−t) = rV=⇒ Vt + 0.5σ

2S2VSS + (r − δ)SVS − rV = 0

Verification that the price of a prepaid forward contract satisfies the Black-Scholes equation DM 21.11.

V (S, t) = S (t) e−δ(T−t)

=⇒ VS = e−δ(T−t) VSS = 0 Vt = δS (t) e−δ(T−t)

=⇒ Vt+0.5σ2S2VSS+(r − δ)SVS−rV = δSe−δ(T−t)+(r − δ)Se−δ(T−t)−

rSe−δ(T−t) = 0



Call option

The textbook explains that the price of a European call option satisfies (1) theboundary condition and, (2) DM 21.11.

V = S (t) e−δ(T−t)N (d1)−Ke−r(T−t)N (d2)

Verification that the call price formula meets the boundary condition.The boundary condition is that at the call expiration date T the call is worth

S (T )−K (T ) if S (T ) > K (T ) and zero otherwise.

d1 =

lnS (t)

K+

µr − δ +

1

2σ2¶(T − t)

σ√T − t

d2 = d1 − σ√T − t

If t approaches T , then d1 =

lnS (T )

K+

µr − δ +

1

2σ2¶(T − t)

σ√T − t

, d2 = d1,

and

• If S (T ) > K, thenS (T )

K> 1, ln

S (T )

K> 0, d1 =

lnS (T )

Kσ√T − t

+

µr − δ +

1

2σ2¶√

T − t

σ=

+∞ , N (d1) = N (d2) = 1, V = S − T.

• If S (T ) < K, then lnS (T )

K< 0, d1 = −∞ , N (d1) = N (d2) = 0 and

V = 0

By the way, we don’t need to worry about S (T ) = K because the proba-bility of S (T ) = K is zero. The probability that a continuous random variabletakes on a fixed value is zero. When we talk about the probability regarding acontinuous random variable X, we talk about the probability that X falls in arange [a, b], not the probability that X takes on a single value. If the probabilitythat X takes a single value is not zero, then the total probability that a < X < bwill be infinite because there are infinite number of single values in the range[a, b].We have proved that the call price satisfies the boundary condition.

For the verification that the call price satisfies DM 21.11, see my solution toDM Problem 21.5, 21.6, and 21.7.We can also verify that the European put price satisfies DM 21.11. The put

price isV = Ke−r(T−t)N (−d2)− S (t) e−δ(T−t)N (−d1)Using the formula N (−x) = 1−N (x), we can rewrite the put price asV = Ke−r(T−t) [1−N (d2)] − S (t) e−δ(T−t) [1−N (d1)] = Ke−r(T−t) −

S (t) e−δ(T−t) + S (t) e−δ(T−t)N (d1)−Ke−r(T−t)N (d2)



You can also derive the above equation using the put-call parity. Notice thatS (t) e−δ(T−t)N (d1)−Ke−r(T−t)N (d2) is the call price.Each of the three terms,Ke−r(T−t), S (t) e−δ(T−t), and S (t) e−δ(T−t)N (d1)−

Ke−r(T−t)N (d2), satisfies the BS PDE. Hence the put price satisfies the BSPDE.I won’t prove that the put price satisfies the boundary condition V (t = T ) =

K − S (T ) if K > S (T ) and zero otherwise. You can easily prove this yourself.

Key formula to remember:If t → T , then d1 = d2 and

• If S (T ) > K, N (d1) = N (d2) = 1

• If S (T ) < K, N (d1) = N (d2) = 0

All or nothing option

The textbook points out that S (t) e−δ(T−t)N (d1) and e−r(T−t)N (d2) each sat-isfy the BS PDE (see my solution to DMProblem 21.5, 21.6). So S (t) e−δ(T−t)N (d1)can be a price of a derivative; e−r(T−t)N (d2) can be a price of another deriv-ative. What derivatives are priced as S (t) e−δ(T−t)N (d1) and e−r(T−t)N (d2)respectively?

Let’s check the boundary condition. As t→ T ,

• S (t) e−δ(T−t)N (d1)→ S (T ) if S (T ) > K

• S (t) e−δ(T−t)N (d1)→ 0 if S (T ) < K

S (t) e−δ(T−t)N (d1)must be the price of an option that pays S (T ) if S (T ) >K and zero otherwise. Such an option is an asset-or-nothing option.Similarly, as t→ T

• e−r(T−t)N (d2)→ 1 if S (T ) > K

• e−r(T−t)N (d2)→ 0 if S (T ) < K

e−r(T−t)N (d2) must be the price of an option that pays 1 if S (T ) > K andzero otherwise. Such an option is an cash-or-nothing option.

We can break down a European call option into one asset-or-nothing optionand several cash-or-nothing options. Buying a European call option is equivalentto buying one asset-or-nothing option and selling K units of cash-or-nothingoption (you can verify that they have the same payoff). Hence the price of theEuropean call option is S (t) e−δ(T−t)N (d1)−Ke−r(T−t)N (d2).Similarly, we can break down a gap option (gap option is explained in Chap-

ter 14). In a gap call, the payoff is S (T )−K1 if S (T ) > K2. This is equivalent



to buying one asset-or-nothing call and selling K1 units of cash-or-nothing op-tion. So the price of this gap call is S (t) e−δ(T−t)N (d1) − K1e

−r(T−t)N (d2),

where d1 =ln¡Ste−δ(T−t)/K2e

−r(T−t)¢+ 0.5σ2 (T − t)

σ√T − t

and d2 = d1−σ√T − t.

• P ∗ (ST > K) = N (d2). This is the risk-neutral probability that the callwill be exercised (i.e. call will finish in the money)

• P ∗ (ST < K) = 1−N (d2) = N (−d2). This is the risk-neutral probabilitythat the call will NOT be exercised.

• Ke−r(T−t)N (d2) is the strike price multiplied by the risk-neutral proba-bility that the strike price will ever be paid. This is the expected value ofthe strike price *if* the call will be exercised

• S (t) e−δ(T−t)N (d1) is the expected value of the stock price *if* the callwill be exercised (i.e. if ST > K)

• N (d1) is the expected fractional share of the stock *if* the call will beexercised

• If an option pays one stock when ST > K and zero otherwise, then thisoption is worth S (t) e−δ(T−t)N (d1)

• If an option pays $1 when ST > K and zero otherwise, then this option isworth PV of $1 (which is e−r(T−t)) multiplied by P ∗ (ST > K)

21.2.3 Black-Scholes equation and equilibrium returns

This section derives the BS PDE using the idea that the actual expected returnon an option should be equal to the equilibrium expected return.The expected continuously compounded return on an option is αoption =

E (dV )

V dt. Here dV is the increase of the option value (i.e. the interest earned

on the option) during [t, t+ dt]. At time t, if you spend V dollars and buy anoption, then during the tiny interval [t, t+ dt], your option value will go up by

dV . The (not annualized) return earned on the option per dt period isdV

V.

Since the number of dt periods in one year is1

dt, the annualized (continuously

compounded) rate of return per $1 invested in the option isdV

V× 1

dt=

dV

V dt.

The expected return on the option is EµdV

V dt

¶=

E (dV )

V dt. Here V and dt are

treated as constants. dV is a random variable.

To calculate E (dV ), we use Ito’s lemma:



dV =£0.5σ2S2VSS + (α− δ)SVS + Vt

¤dt+ SVSσdZ

→ E [dV ] = E¡£0.5σ2S2VSS + (α− δ)SVS + Vt

¤dt¢+E (SVSσdZ)

=£0.5σ2S2VSS + (α− δ)SVS + Vt

¤dt+ SVSσE (dZ)

We know that E (dZ) = 0. Recall dZ (t) is a normal random variable withmean 0 and variance dt (see the footnote of Derivatives Markets Page 652).

Hence αoption =E (dV )

V dt=0.5σ2S2VSS + (α− δ)SVS + Vt

V

The (not annualized) unexpected return on the option isSVSσdZ

V. Here we

didn’t divide SVSσdZ by dt because the unexpected return is not annualized.

DefineSVSσ

V= σoption .

Hence the (not annualized) unexpected return on the option is σoptiondZ.The Ito’s lemma can be written as:

dV

V= αoptiondt+ σoptiondZ

By the way, it seems that the textbook has a typo in DM 21.18:

E (dV )

V− dV

V=

SVSσdZ

V(DM 21.18)

The correct formula should be:dV

V− E (dV )

V=

SVSσdZ

VConsider two assets, an stock and an option on the stock.

The process of the stock price:dS

S= αdt+ σdZ (DM 20.1)

The process of the option price:dV

V= αoptiondt+ σoptiondZ

These two processes are driven by the same dZ. According to Chapter 20,the stock and the option on the stock must have the same Sharpe ratio. Hence

α− r

σ=

αoption − r

σoption

=⇒ α− r

σ=

αoption − rSVSσ

V

α− r =V

SVS(αoption − r)

=⇒ α− r =V

SVS

µ0.5σ2S2VSS + (α− δ)SVS + Vt

V− r

¶

=⇒ (α− r)SVS = 0.5σ2S2VSS + (α− δ)SVS + Vt − rV

=⇒ Vt + 0.5σ2S2VSS + (r − δ)SVS − rV = 0 (BS PDE)



By the way, according to DM 12.8, the option elasticity is Ω =SVSV. Hence

σoption =SVSσ

V= σΩ (DM 21.20). DM 21.20 is slightly different from DM

12.9:σoption = σ|Ω| (DM 12.9)σoption = σΩ (DM 21.20)Obviously, the author of Derivatives Markets changed his definition of σoption .

In Chapter 12, σoption is non negative; in Chapter 21, σoption can be positive,zero, or negative.

21.3 Risk-neutral pricing

This section repeats the old idea of risk neutral pricing. Under risk neutralpricing, we can set the expected return on the stock α to r and get the correctprice of a derivative.This section contains many formulas. Make sure you understand the meaning

of each formula.The textbook lists the following equations:

dS

S= (r − δ) dt+ σd

˜

Z (DM 21.28)

d

dtE∗ (dV ) = Vt + 0.5σ

2S2VSS + (r − δ)SVS (DM 21.30)

d

dtE∗ (dV ) = rV (DM 21.31)

These equations are risk-neutral version of similar formulas.

Next, the textbook introduced a new symbol f (ST : St). f (ST : St) is theconditional probability of the stock’s terminal price ST given that the pricetoday is St.You’ll want to memorize the following formulas:P (ST > K) =

R∞K

f (ST : St) dST (this holds whether f (ST : St) is risk-neutral probability or the true probability)

The price of a derivative is Vt = e−r(T−t)E∗ (VT )The price of a European call is

Vt = e−r(T−t)E∗ (VT ) = e−r(T−t)RK00× f∗ (ST : St) dST+e

−r(T−t) R∞K[S (T )−K]

f∗ (ST : St) dST = e−r(T−t)R∞K[S (T )−K] f∗ (ST : St) dST

Please note that the price of the call at expiration is VT = 0 if S (T ) < Kand S (T )−K if S (T ) > K.

Similarly, the price of a European put isVt = e−r(T−t)E∗ (VT ) = e−r(T−t)

RK0[K − S (T )]× f∗ (ST : St) dST


Chapter 22

Exotic options: II

This is an easy chapter.

22.1 All-or-nothing options

Cash-or-nothing option.

• A cash call pays $1 at T is ST > K and $0 if ST ≤ K

• A cash put pays $1 at T is ST < K and $0 if ST > K

Asset-or-nothing option.

• An asset-or-nothing call pays ST at T is ST > K and $0 if ST ≤ K

• An asset-or-nothing put pays ST at T is ST < K and $0 if ST > K

Price of cash-or-nothing option.

• The price of a cash call option is e−r(T−t)N (d2), whereN (d2) = P ∗ (ST > K)(i.e. the risk neutral probability of ST > K)

• The price of a cash put option is e−r(T−t) [1−N (d2)] = e−r(T−t)N (−d2),where N (−d2) = P ∗ (ST < K) (i.e. the risk neutral probability of ST <K)

Price of asset-or-nothing option.

• The price of an asset call option is Ste−δ(T−t)N (d1)

• The price of an asset put option is Ste−δ(T−t) [1−N (d1)] = Ste−δ(T−t)N (−d1)

253


254 CHAPTER 22. EXOTIC OPTIONS: II

Price of a European call and put

• Buying an ordinary European call option is equivalent to buying oneasset-or-nothing call and selling K cash-or-nothing calls. The call price isSte−r(T−t)N (d1)−Ke−r(T−t)N (d2)

• Buying an ordinary European put option is equivalent to buying K cash-or-nothing puts and selling one asset-or-nothing put. The put price isKe−r(T−t)N (−d2)− Ste

−r(T−t)N (−d1)

Price of a gap call and put

• Buying a gap call option is equivalent to buying one asset-or-nothing callwith strike price K2 and selling K1 cash-or-nothing calls with strike priceK2. The gap call price is Ste−r(T−t)N (d1)−K1e

−r(T−t)N (d2) .

• Buying a gap put option is equivalent to buying K1 asset-or-nothing putswith strike price K2 and selling one asset-or-nothing put with strike priceK2. The gap put price is K1e

−r(T−t)N (−d2)− Ste−r(T−t)N (−d1) .

• For both a gap call and put, d1 =ln¡Ste−δ(T−t)/K2e

−r(T−t)¢+ 0.5σ2 (T − t)

σ√T − t

and d2 = d1 − σ√T − t.

Delta-hedging all-or-nothing optionsIt’s difficult to hedge an all-or-nothing option because the payoff is not con-

tinuous. The textbook explains this well. Refer to the textbook.


Chapter 23

Volatility

The concept of implied volatility is explained in DM section 12.5. The impliedvolatility of an option is the volatility implied by the market price of the optionbased on the Black-Scholes formula. If you plug the implied volatility into theBlack-Scholes formula, the formula should produce the option price that is equalto the current market price of the option.The Black-Scholes formula assumes that a given stock has a constant volatil-

ity. Under the BS formula, the stock’s volatility is a inherent characteristic ofthe stock; it doesn’t depend on other factors (such as the stick price K and theoption’s expiry T ). In reality, however, the implied volatility of a stock dependson K and T .The volatility smile is a long-observed pattern where at-the-money options

tend to have lower implied volatilities than in- or out-of-the-money options.

• The implied volatility is lowest when the option is at the money

• The implied volatility gets higher and higher as the option gets more andmore in-the-money

• The implied volatility gets higher and higher as the option gets more andmore out-of-the money.

If we plot the implied volatility in a 2-D plane (setting the implied volatilityσ as Y and the strike price K as X), the diagram looks like a letter U (the letterU looks like a smile).To see a diagram of the volatility smile, refer to

• http://www.optiontradingpedia.com/volatility_smile.htm

• http://en.wikipedia.org/wiki/Volatility_smile

Volatility surface is a 3-D diagram of the implied volatility σ as a functionof the strike price K and time to maturity T .

255


256 CHAPTER 23. VOLATILITY

To account for the factor that the implied volatility depends on K and T ,

we can rewrite DM 20.25dS (t)

S (t)= (α− δ) dt + σdZ (t) as DM 23.1

dS (t)

S (t)=

(α− δ) dt+σ (St,Xt,t) dZ (t). In DM 23.1, the instant volatility σ (St,Xt,t) dZ (t)is a function of the stock price St, another factor Xt,, and the time t.

Historical volatility. Please note that DM 23.2ˆσ2

=1

h

∙1

n− 1nPi=1

ε2i

¸is not

new. This formula is already used in DM Chapter 11 "Estimating Volatility." In

DM 23.2,1

n− 1nPi=1

ε2i is the estimated variance per h period. Since the number

of h periods in one year is1

h, the variance per year is

1

h

∙1

n− 1nPi=1

ε2i

¸.

This is all you need to know about Chapter 23.


Chapter 24

Interest rate models

24.1 Market-making and bond pricing

24.1.1 Review of duration and convexity

Duration and convexity are explained in Derivatives Markets Section 7.3. Sec-tion 7.3 is excluded from the MFE syllabus. However, Derivatives Markets page

781 mentions the duration hedging:

...hedging a bond portfolio based on duration does not result in a perfecthedge. Recall that the duration of a zero-coupon bond is the bond’s time tomaturity...

Because Chapter 24 mentions duration and Chapter 24 is on the syllabus,SOA may test your knowledge about duration hedging. So read DM Section7.3 and take a quick review of duration and convexity. Next, let’s solve a fewproblems.

Example 24.1.1.

The yield to maturity of a 5-year zero-coupon bond is 6%. The face amountof the bond is 100.Calculate

• the value of the bond

• the Macaulay duration

• the modified duration

• the convexity

257


258 CHAPTER 24. INTEREST RATE MODELS

In addition, calculate the bond value if the yield to maturity

• moves up to 7%

• moves down to 5%.

Solution.

The bond value is P0 = 100× 1.06−5 = 74. 725 8The (Macaulay) duration of a zero-coupon bond is its maturity T . So the

(Macaulay) duration isDMac = T = 5 years.The modified duration is

Dmod =DMac

1 + yield=

5

1.06= 4. 716 98 years

The convexity of a zero-coupon bond is:

C =T (T + 1)

(1 + y)2=5× 61.062

= 26. 699 893

If the yield to maturity is now 7%, the bond value is:P1 = 100× 1.07−5 = 71. 298 6

We can also use duration and convexity to approximate the bond value underthe new yield.The new bond value is

P1 = P0 +∆P = P0

µ1 +

∆P

P0

¶∆P

P0= −Dmod∆y+

1

2C (∆y)2 = −4. 716 98×0.01+1

2×26. 699 893×(0.01)2 =

−0.045 8

→ P1 = 74. 725 8 (1− 0.045 8) = 71. 303 3This is very close to the correct amount of 71. 298 6.

If the yield to maturity is now 5%, the bond value is:P1 = 100× 1.05−5 = 78. 352 6We can also use duration and convexity to approximate the bond value under

the new yield.∆P

P0= −Dmod∆y +

1

2C (∆y)

2= −4. 716 98 × (−0.01) + 1

2× 26. 699 893 ×

(−0.01)2 = 0.048 5

→ P1 = 74. 725 8 (1 + 0.048 5) = 78. 350 0This is very close to the correct amount of 78. 352 6.

Example 24.1.2.


24.1. MARKET-MAKING AND BOND PRICING 259

The yield to maturity of a 4-year zero-coupon bond is 8%. The face amountof the bond is 100.Calculate

• the value of the bond

• the Macaulay duration

• the modified duration

• the convexity

In addition, calculate the bond value if the yield to maturity

• moves up to 9%

• moves down to 7%.

Solution.

The bond value is P0 = 100× 1.08−4 = 73. 5030DMac = T = 4 years.

Dmod =DMac

1 + yield=

4

1.08= 3. 703 7 years

C =T (T + 1)

(1 + y)2 =

4× 51.072

= 17. 468 8


the new yield.The new bond value is

P1 = P0 +∆P = P0

µ1 +

∆P

P0

¶∆P

P0= −Dmod∆y+

1

2C (∆y)

2= −3. 703 7× 0.01+ 1

2× 17. 468 8× (0.01)2 =

−0.036 2→ P1 = 73. 5030 (1− 0.036 2) = 70. 842 2This is very close to the correct amount of 70. 842 5.


the new yield.∆P

P0= −Dmod∆y +

1

2C (∆y)2 = −3. 703 7 × (−0.01) + 1

2× 17. 468 8 ×

(−0.01)2 = 0.037 9

→ P1 = 73. 5030 (1 + 0.037 9) = 76. 288 8This is very close to the correct amount of 76. 289 5.



Example 24.1.3.

Portfolio A consists of one unit of 4-year zero coupon bond.Portfolio B consists of x units of 1-year zero coupon bond and y units of

8-year zero coupon bond. Portfolio B is used to duration hedge Portfolio A.Each of the three zero-coupon bonds above has 100 face amount.

The current annual effective interest rate is 5% for all three bonds.

Assume that the yield curve changes by a uniform amount if there’s a change.Demonstrate why duration-hedging leads to arbitrage under two scenarios:

• the annual effective interest rate moves up to 6% immediately after thethree bonds are issued.

• the annual effective interest rate instantly moves down to 4% immediatelyafter the three bonds are issued.

Solution.

To duration hedge Portfolio A, we need to satisfy two conditions:

• Portfolio A and B have the same present value

• Portfolio A and B have the same duration

So we set up the following equations:100

1.051x+

100

1.058y =

100

1.054100

1.051x

100

1.051x+

100

1.058y× 1 +

100

1.058y

100

1.051x+

100

1.058y× 8 = 4

This is the meaning of the second equation. Portfolio B consists of twobonds. If a portfolio consists of multiple bonds, then the portfolio’s duration isjust the weighted average of the each bond’s duration, with weight equal to thepresent value of each bond.

The 2nd equation can be simplified asµ100

1.051x

¶1+

µ100

1.058y

¶8 = 4× 100

1.054

100

1.051x+

100

1.058y =

100

1.054µ100

1.051x

¶1 +

µ100

1.058y

¶8 = 4× 100

1.054

→ x = 0.493 62 , y = 0.520 93



So Portfolio B needs to consist of 0.493 62 unit of 1-year zero coupon bondand 0.520 93 units of 8-year zero coupon bond.

If the interest rate moves up to 6%:

Portfolio A is worth:100

1.064= 79. 2094

Portfolio B is worth:100

1.061× 0.493 62 + 100

1.068× 0.520 93 = 79. 251 7

To arbitrage, at t = 0 (when the interest rate is 5%), we buy low and sellhigh:

• buy Portfolio B. We pay 100

1.051x+

100

1.058y =

100

1.054= 82. 270 2

• sell Portfolio A. We receive 100

1.054= 82. 270 2

So our net cost is zero.

Then instantly later at t = 0+, the interest rate moves up to 6%. ThenPortfolio B (our asset) is worth 79. 251 7; Portfolio A (our liability) is worth 79.2094.Our net profit is 79. 251 7− 79. 2094 = 0.042 3If the interest rate moves down to 4%:

Portfolio A is worth:100

1.044= 85. 480 4

Portfolio B is worth:100

1.041× 0.493 62 + 100

1.048× 0.520 93 = 85. 527 3

To arbitrage, at t = 0 (when the interest rate is 5%), we buy low and sellhigh:

• buy Portfolio B. We pay 100

1.051x+

100

1.058y =

100

1.054= 82. 270 2

• sell Portfolio A. We receive 100

1.054= 82. 270 2

So our net cost is zero.

Then instantly later at t = 0+, the interest rate moves down to 4%. ThenPortfolio B (our asset) is worth 85. 527 3; Portfolio A (our liability) is worth 85.480 4.Our net profit is 85. 527 3− 85. 480 4 = 0.046 9

So no matter the interest rate moves up to 6% or moves down to 4%, we canalways make free money by buying Portfolio B and selling Portfolio A.



Why is Portfolio B better than Portfolio A? It turns out B has high convexity.

Convexity of A:

CA =TA (TA + 1)

(1 + y)2=4× 51.052

= 18. 1406

Convexity of B is just the weighted average convexities of the two bonds,with weights being the present value of the bond.

CB =

1× 21.052

×µ100

1.051× 0.493 62

¶+8× 91.052

×µ100

1.058× 0.520 93

¶100

1.051× 0.493 62 + 100

1.058× 0.520 93

=

1× 21.052

×µ100

1.051× 0.493 62

¶+8× 91.052

×µ100

1.058× 0.520 93

¶100

1.054

= 29. 024 9

CB > CA

If the interest rate moves up by ∆y, then the present value of a portfolio is:

P1 = P0 +∆P = P0

µ1 +

∆P

P0

¶∆PA

PA0

= −DAmod∆y +

1

2CA (∆y)2

∆PB

PB0

= −DBmod∆y +

1

2CB (∆y)2

However, DAmod = DB

mod =4

1.05= 3. 809 5 CB > CA

→ ∆PB

PB0

>∆PA

PA0

Since PA0 = PB

0 → ∆PB > ∆PA

→ PB1 > PA

1 So Portfolio B always worth more than Portfolio A undera flat yield curve.For example, if the interest rate moves up to 6%, then∆PA

PA0

= −3. 809 5× 0.01 + 12× 18. 1406× 0.012 = −0.037 2

∆PB

PB0

= −3. 809 5× 0.01 + 12× 29. 024 9× 0.012 = −0.036 6

PA1 =

100

1.054(1− 0.037 2) = 79. 2098

PB1 =

100

1.054(1− 0.036 6) = 79. 259 2

PB1 − PA

1 = 79. 259 2− 79. 2098 = 0.049 4Key point to remember:



1. Two portfolios can have the same present value, the same duration, butdifferent convexities.

2. Under the assumption of the parallel shift of a flat yield curve, we canalways make free money by buying the high-convexity portfolio and sellthe low-convexity portfolio.

3. The parallel shift of a flat yield curve assumption leads to arbitrage.

Information about three zero-coupon bonds:Maturity (Yrs) Face2 1004 1007 100

The current interest rate is 7% for all three bonds. Assume a parallel shiftof a flat yield curve. Design an arbitrage strategy.

We need to form 2 portfolios. These two portfolios have the same PV, thesame duration, but different convexity. We can make free money by buying thehigh convexity portfolio and selling the low convexity portfolio.The low convexity portfolio is the 4-year bond (Portfolio A).The high convexity portfolio consists of x unit of 2-year bond and y unit

of 7-year bond (Portfolio B). This is called a barbell. A barbell bond portfoliocombines short maturities (low duration) with long maturities (high duration)for a blended, moderate maturity (moderate duration)

Portfolio Maturity (Yrs) PV Duration Convexity Units

A 4100

1.074= 76. 29 4

4× 51.072

= 17. 468 8 1

B 2100

1.072= 87. 34 2

2× 31.072

= 5. 240 6 x

B 7100

1.077= 62. 27 7

7× 81.072

= 48. 912 6 y

100

1.072x+

100

1.077y =

100

1.074µ100

1.072x

¶2 +

µ100

1.077y

¶7 = 4× 100

1.074

x = 0.524 1, y = 0.490 0

At time 0, we buy Portfolio B (which consists of 0.524 1 unit of 2-year bondand 0.490 0 unit of 7-year bond). Simultaneously, we sell Portfolio A.The convexity of Portfolio A is: CA = 17. 468 8

The convexity of Portfolio B is:



CB =

2× 31.072

× 100

1.072× 0.524 1 + 7× 8

1.072× 0.490 0× 100

1.077100

1.074

= 22. 708 9

For example, if the new interest rate is 7.25% for all bonds with differentmaturities, then

PB1 =

100

1.07252× 0.524 1 + 0.490 0× 100

1.07257= 75. 584 1

PA1 =

100

1.07254= 75. 580 9

PB1 > PA

1

Our profit is 75. 584 1− 75. 580 9 = 0.003 2

If the new interest rate is 6.5% for all bonds with different maturities, then

PB1 =

100

1.0652× 0.524 1 + 0.490 0× 100

1.0657= 77. 739 6

PA1 =

100

1.0654= 77. 732 3

PB1 > PA

1

Our profit is 77. 739 6− 77. 732 3 = 0.007 3

Either way, we make money.By now you should see that the parallel shift of a flat yield curve is a bad

model.

24.1.2 Interest rate is not so simple

We all know what an interest rate is, yet a derivative on interest rate is surpris-ingly difficult. To get a sense of the difficulty, suppose we want to calculate theprice of a European call on a 1-year zero-coupon bond that pays $100 one yearfrom now. Here are the inputs:

• The call expires in one year

• The strike price is $100

• The continuous risks-free rate is r = 6% per year

• The current price of the bond is 100e−0.06 = 94. 18.

• The volatility of the bond return is σ = 30%

Using the Black-Scholes formula, we find:d1 = 0.15 d2 = −0.15 N (d1) = 0.5596 N (d2) = 0.4404→ C = 94. 18× 0.5596− 94. 18× 0.4404 = 11. 23



Everything looks fine. However, after further thinking, you realize that thecall price C should be zero. At T = 1, the bond pays $100. Hence the 100-strikecall value is zero. Why should anyone buy a 100-strike call on an asset worth$100 at call expiration?What went wrong? It turns out that we can’t use the Black-Scholes formula

to calculate the price of an interest rate derivative:

• The Black-Scholes option formula assume that the term structure of theinterest rate is flat and deterministic (see DM page 379). However, If theinterest rate is known and constant, there won’t be any need for interestrate derivatives. This is similar to the idea that if the stock price is knownand constant, there won’t be any need for call or put option.

• The standard deviation of the return σ is a constant. However, the stan-dard deviation of the return σ is not constant. Unlike a stock, a bond hasa finite maturity. At the maturity date, the bond value is its face amount.Hence the standard deviation of a bond’s return decreases as the bondapproaches its maturity.

The key point. It’s much harder to calculate the price of an option interestrate because interest rate is not a tradeable underlying asset. For call and puton stocks, we can buy or short sell stocks to set up the hedge portfolio. However,we can’t go out and buy a 5% interest rate to hedge an option on interest rate.Now let’s go to the textbook.

24.1.3 Impossible bond pricing model

A bond is a derivative on interest rate. We normally don’t think this way, buta bond derives its value from interest rate. If the market interest rate goes up,the bond value goes down; if the market interest rate goes down, the bond valuegoes up.

So our starting point is to set up a stochastic random variable called theshort interest rate. The textbook keeps using the phrase "short interestrate" or "short rate" without giving a clear definition. Here is the definition:

Definition 24.1.1. A short interest rate or short rate is just the instantaneousinterest rate r (t) over a short (hence the name "short rate") interval [t, t+ dt].

First, we assume that the short interest rate r (t) follows the Ito’s process:

dr = α (r) dt+ σ (r) dZ (24.1)

Next, the text book explains that we can’t assume r (t) follows a flat yieldcurve ("impossible bond pricing model").A flat yield curve means that the interest rate is independent of time. To

have a flat yield curve, the following two conditions are met:



• The initial interest rate r is a constant regarding time, that is, r (t) = t

• If the interest rate changes, the change is also independent of time (i.e.parallel shift of the yield curve)

If the continuously compounded interest rate r is constant, then the presentvalue at time t of $1 to be received at T is:

P (t, T ) = e−r(T−t) (24.2)

For example, if a zero coupon bond pays us $1 at T = 2, the PV of this bondat time zero is P (0, 2) = e−2r. The PV of this bond at t = 1 is P (1, 2) = e−r.Next, let’s review the textbook’s proof and find out why Equation 24.1 and

24.2 won’t work together.

Suppose we want to delta hedge a bond. Whereas delta hedging a call meansbuying ∆ shares of a stock, delta hedging a bond means buying ∆ units of abond with a different maturity date.Suppose at time t we buy a bond maturing at T2. "A bond maturing at T2"

just means that we, the bond holder, will receive $1 at T2. So the price of thisbond at time t is P (t, T2) = e−r(T2−t).To delta hedge this bond, at t, we buy ∆ bonds maturing at T1 (please note

the textbook uses N instead of ∆). The cost is ∆P (t, T1) = ∆e−r(T1−t)

The total cost of buying two bonds is ∆P (t, T1) + P (t, T2) = ∆e−r(T1−t) +

e−r(T2−t). To avoid tying up our capital, at time t we borrow ∆e−r(T1−t) +e−r(T2−t) from a bank to finance the purchase of two bonds. So at time t, ourportfolio is:

• buy 1 bond maturing at T2

• buy ∆ bonds maturing at T1

• borrow e−r(T2−t) +∆e−r(T1−t) from a bank

Our net position is zero:I = ∆P (t, T1) + P (t, T2) +W = 0 where W = − [∆P (t, T1) + P (t, T2)] =

−£∆e−r(T1−t) + e−r(T2−t)

¤. As seen before, if we borrow money, we use a neg-

ative number. If we lend money, we use the a positive number. This is why Wis negative.In the next instant dt, the interest rate moves up to r + dr. Now change of

our portfolio value is:dI = ∆× dP (t, T1) + dP (t, T2) + dWUsing Ito’s lemma, we have:dP (t, T1) =

∂P (t,T1)∂t dt+ ∂P (t,T1)

∂r dr + 12∂2P (t,T1)

∂r2 (dr)2

∂P (t,T1)∂t = ∂

∂te−r(T1−t) = rP (t, T1)

∂P (t,T1)∂r = ∂

∂re−r(T1−t) = −P (t, T1) (T1 − t)



∂2P (t,T1)∂r2 = ∂2

∂r2 e−r(T1−t) = er(t−T1) (T1 − t)

2= P (t, T1) (T1 − t)

2

(dr)2= (αdt+ σdZ)

2= σ2 (dZ)

2= σ2dt

→ dP (t, T1) = rP (t, T1) dt− P (t, T1) (T1 − t) dr + 12P (t, T1) (T1 − t)2 σ2dt

Similarly,dP (t, T2) = rP (t, T2) dt− P (t, T2) (T2 − t) dr + 1

2P (t, T2) (T2 − t)2σ2dt

However, dW = rWdt. To see why, notice that dW is the interest earned onW during the interval [t, t+ dt]. During [t, t+ dt], the continuous interest rater can be treated as a discrete interest rate. The interest earned is justthe initial capital × interest rate × length of the interval = rWdtNow we have DM Equation 24.6:

dI = ∆hrP (t, T1) dt− P (t, T1) (T1 − t) dr + 1

2P (t, T1) (T1 − t)2σ2dt

i+hrP (t, T2) dt− P (t, T2) (T2 − t) dr + 1

2P (t, T2) (T2 − t)2 σ2dti

+rWdt

We’ll want to have dI = 0. If dI = 0, then the value of our portfolio won’tchange during the same during [t, t+ dt]. To make dI = 0, we first choose ∆such that the dr term is zero.

→ ∆ [−P (t, T1) (T1 − t)] + [−P (t, T2) (T2 − t)] = 0

→ ∆ = − (T2−t)P (t,T2)(T1−t)P (t,T1)

This negative delta means that we need to sell (T2−t)P (t,T2)(T1−t)P (t,T1) units of bondmaturing at T1.

By the way, ∆ = − (T2−t)P (t,T2)(T1−t)P (t,T1) is similar to DM Equation 7.13 (DerivativesMarkets page 227):

N = −D1B1 (y1) / (1 + y1)

D2B2 (y2) / (1 + y2)(DM 7.13)

So buying ∆ bonds is really duration-hedging.Next, we want to set the dt term to zero:

∆hrP (t, T1) +

12P (t, T1) (T1 − t)

2σ2i+hrP (t, T2) +

12P (t, T2) (T2 − t)

2σ2i+

rW = 0

→ r [∆P (t, T1) + P (t, T2) +W ]+ 12∆P (t, T1) (T1 − t)

2σ2+1

2P (t, T2) (T2 − t)2σ2 =

0However,∆P (t, T1) + P (t, T2) +W = 0

→ 12∆P (t, T1) (T1 − t)

2σ2 + 1

2P (t, T2) (T2 − t)2σ2 = 0

→ −12(T2−t)P (t,T2)(T1−t)P (t,T1)P (t, T1) (T1 − t)2 σ2 + 1

2P (t, T2) (T2 − t)2 σ2 = 0

→ −12 (T2 − t)P (t, T2) (T1 − t)σ2 + 12P (t, T2) (T2 − t)

2σ2 = 0

→ −12 (T2 − t)P (t, T2) [(T1 − t)− (T2 − t)]σ2 = 0



→ −12 (T2 − t)P (t, T2) (T1 − T2)σ2 = 0

→ T1 = T2 ∆ = −1∆ = −1 means that sell one bond.If we buy a bond maturing in T2 years, the only way to hedge the risk is to

sell this bond! So one moment you buy a bond. The next moment you sell it.Your net position is zero. Of course you are hedged against all risks, but thishedging is really doing nothing type of hedging.This tells us that the bond pricing model based on Equation 24.1 and 24.2

are impossible. In our words, Equation 24.1 is OK. But Equation 24.2 is bad.This confirms that a parallel shift of a flat yield curve is a bad assumption forpricing a bond.

24.1.4 Equilibrium equation for bonds

Since it’s bad to assume that r is a flat yield curve, we switch our gears andassume that r isn’t a flat yield curve. Now we just assume that Equation 24.1holds.

dP (r, t, T ) =∂P

∂rdr +

∂P

∂tdt+

1

2

∂2P

∂t2(dr)2

(dr)2= [α (r) dt+ σ (r) dZ]

2= σ2 (r) (dZ)

2= σ2 (r) dt

→ dP (r, t, T ) =∂P

∂rdr+

∂P

∂tdt+

1

2

∂2P

∂t2σ2 (r) dt =

∂P

∂r[α (r) dt+ σ (r) dZ]+

∂P

∂tdt+

1

2

∂2P

∂t2σ2 (r) dt

dP (r, t, T ) =

∙α (r)

∂P

∂r+1

2

∂2P

∂t2σ2 (r) +

∂P

∂t

¸dt+

∂P

∂rσ (r) dZ (24.3)

Define

α (r, t, T ) =1

P (r, t, T )

∙α (r)

∂P

∂r+1

2

∂2P

∂t2σ2 (r) +

∂P

∂t

¸(24.4)

q (r, t, T ) =1

P (r, t, T )

∂P

∂rσ (r) (24.5)

Now Equation 24.3 becomes:

dP (r, t, T )

P (r, t, T )= α (r, t, T ) dt+ q (r, t, T ) dZ (24.6)



dP (r, t, T )

P (r, t, T )is the bond’s return. Equation 24.6 says that the bond’s return

is the sum of a drift term α (r, t, T ) dt and a random component q (r, t, T ) dZ.

Please also note that generally q (r, t, T ) is negative. This is because∂P

∂ris

negative. If r goes up, the bond price goes down; if r goes down, P goes up.Next, we are going to derive DM Equation 24.16:

α1 (r, t, T1)− r

q1 (r, t, T1)=

α2 (r, t, T2)− r

q2 (r, t, T2)(24.7)

Suppose we have two bonds, Bond #1 and Bond #2 maturing at T1 and T2respectively. Each bond follows DM Equation 24.6:

dP (r, t, T1)

P (r, t, T1)= α1 (r, t, T1) dt+ q1 (r, t, T1) dZ

dP (r, t, T2)

P (r, t, T2)= α2 (r, t, T2) dt+ q2 (r, t, T2) dZ

At t = 0, we form a self-financing portfolio consisting of

• buying ∆ units of Bond #1

• buying 1 Bond #2

• borrowing ∆P (r, t, T1)+P (r, t, T2) from a bank at the riskless short rater

The value of this portfolio is I

dI = ∆P (r, t, T1)dP (r, t, T1)

P (r, t, T1)+ P (r, t, T2)

dP (r, t, T2)

P (r, t, T2)− [∆P (r, t, T1) + P (r, t, T2)] rdt

= ∆P (r, t, T1) [α1 (r, t, T1) dt+ q1 (r, t, T1) dZ]+P (r, t, T2) [α2 (r, t, T2) dt+ q2 (r, t, T2) dZ]−[∆P (r, t, T1) + P (r, t, T2)] rdt= ∆P (r, t, T1) [α1 (r, t, T1)− r] dt+ P (r, t, T2) [α2 (r, t, T2)− r] dt+ [∆P (r, t, T1) q1 (r, t, T1) + P (r, t, T2) q2 (r, t, T2)] dZ

Choose∆ such that∆P (r, t, T1) q1 (r, t, T1)+P (r, t, T2) q2 (r, t, T2) = 0. Thisremoves the stochastic random term dZ. Now dI is deterministic. Since theportfolio is self-financing and riskless, it earns zero interest rate. Hence∆P (r, t, T1) [α1 (r, t, T1)− r] dt+ P (r, t, T2) [α2 (r, t, T2)− r] dt = 0

→ −P (r, t, T2) q2 (r, t, T2)P (r, t, T1) q1 (r, t, T1)

P (r, t, T1) [α1 (r, t, T1)− r] dt+P (r, t, T2) [α2 (r, t, T2)− r] dt =

0

→ −q2 (r, t, T2)q1 (r, t, T1)

[α1 (r, t, T1)− r] dt+ [α2 (r, t, T2)− r] dt = 0

→ α1 (r, t, T1)− r

q1 (r, t, T1)=

α2 (r, t, T2)− r

q2 (r, t, T2)



Define the following term as Sharp ratio:

α (r, t, T )− r

q (r, t, T )= φ (r, t) (24.8)

Apply Equation 24.4 and 24.6 to 24.8:1

P (r, t, T )

∙α (r)

∂P

∂r+1

2

∂2P

∂t2σ2 (r) +

∂P

∂t

¸− r

1

P (r, t, T )

∂P

∂rσ (r)

= φ (r, t)

→∙α (r)

∂P

∂r+1

2

∂2P

∂t2σ2 (r) +

∂P

∂t

¸− rP = φ (r, t)

∂P

∂rσ (r)

1

2σ2 (r)

∂2P

∂r2+ [α (r)− σ (r)φ (r, t)]

∂P

∂r+

∂P

∂t− rP = 0 (24.9)

Any interest-dependent securities (not just zero coupon bonds) must satisfyEquation 24.9.

To solve the bond price P (r, t, T ), we need to use Equation 24.9 togetherwith the following boundary condition:

P (r, T, T ) = 1 (24.10)

The solution to Equation 24.9 and 24.10 is:

P (r, t, T ) = E∗t

Ãexp

"−Z T

t

r (s) ds

#!= E∗t (exp [−R (t, T )]) (24.11)

where E∗means that the expectation is based on the risk neutral probabilityand

R (t, T ) =

Z T

t

r (s) ds (24.12)

Equation 24.11 is the general formula for a bond price.

24.1.5 Delta-Gamma approximation for bonds

The key formula is:

1

dtE∗ (dP ) = rP (24.13)

Don’t worry about how to prove Equation 24.13. Just memorize its meaning.Equation 24.13 says that under the risk neutral distribution, the bond is pricedto earn a risk-free rate.


24.2. EQUILIBRIUM SHORT-RATE BOND PRICE MODELS 271

24.2 Equilibrium short-rate bond price models

24.2.1 Arithmetic Brownian motion (i.e. Merton model)

A simple model is to assume that the short rate r (t) follows arithmetic Brownianmotion:

dr (t) = αdt+ σdZ (24.14)

In addition, we assume φ (r, t) = φAdvantage:

• The model is simple.

Disadvantages:

• r (t) can go negative. Since r (t) is normally distributed with mean r0+αtand variance σ2t, r (t) can be negative. A negative interest rate won’tmake any sense.

• r (t) is not mean reverting.

• V ar [r (t)] = σ2t. This is undesirable because the volatility of the interestshould depend on the interest rate. If the interest rate is high, then thevolatility is high.

We can derive the bond price under the Merton model. Under this model,Equation 24.9 now becomes:

12σ

2 ∂2P∂r2 + [α− σφ] ∂P∂r +

∂P∂t − rP = 0

We guess the solution is P (r, t, T ) = A (T − t) e−B(T−t)r(t) where A (T − t)and B (T − t) are two functions of T − t.

∂P

∂r= −ABe−Br ∂2P

∂r2= AB2e−Br

∂P

∂t= rAB

0e−Br −A0e−Br

→ 12σ

2AB2e−Br − [α− σφ]ABe−Br + rAB0e−Br −A0e−Br − rAe−Br = 0

1

2σ2AB2e−Br − [α− σ]ABe−Br −A0e−Br = rA

³e−Br −B

0e−Br

´(24.15)

Equation 24.15 should hold for any r. The only way to make it work for anyr is:

12σ

2AB2 − [α− σφ]AB −A0 = 0

A³1−B

0´= 0

Using the boundary condition is P (r, T, T ) = 1, we get A (0) e−B(0)r = 1.For this equation to hold for any r, we need to have A (0) = 1 and B (0) = 0.

Hence from A³1−B

0´= 0, we get:



B0= 1 → B = T − t

From 12σ

2AB2 − [α− σφ]AB −A0 = 0. we get:A0

A= 1

2σ2B2 − [α− σφ]B = 1

2σ2 (T − t)

2 − [α− σφ] (T − t)

→ d lnA (T − t) = 12σ

2 (T − t)2 − (α− σφ) (T − t)

→ lnA (T − t) = 16σ

2 (T − t)3 − 1

2(α− σφ) (T − t)

2+ C where C is a con-

stant.

→ A = A (0) exp

∙16σ

2 (T − t)3 − 12(α− σφ) (T − t)2

¸= exp

∙16σ

2 (T − t)3 − 12(α− σφ) (T − t)2

¸Finally, we have:

P (r, t, T ) = exp

∙16σ

2 (T − t)3 − 12(α− σφ) (T − t)2

¸e−(T−t)r(t)

You don’t need to memorize the above formula. Just understand how toderive the formula in case SOA or CAS gives you a tough problem.

24.2.2 Rendleman-Bartter model

This model assume that r (t) follows a geometric Brownian motion:

dr (t)

r (t)= αdt+ σdZ (24.16)

Advantage:

• r (t) can’t be negative

• V ar [r (t)] = r2 (t)σ2t. The variance increases if r (t) increases. This isdesirable because the volatility of the interest is high if the interest rateis high.

Disadvantage:

• not mean-reverting

24.2.3 Vasicek model

According to Wikipedia, Vasicek’s model was the first one to capture mean re-version, an essential characteristic of the interest rate that sets it apart fromother financial prices. Stock prices can rise indefinitely. However, interest ratescannot. Excessively high interest rate would hamper economic activity, prompt-ing a decrease in interest rates. Similarly, interest rates can not decrease indef-initely. If interest rates are too low, few people are willing to lend their money.This tends to push up the interest rate. As a result, interest rates move in alimited range, showing a tendency to revert to a long run value.


24.2. EQUILIBRIUM SHORT-RATE BOND PRICE MODELS 273

The model is:

dr (t) = α (b− r) dt+ σdZ (24.17)

The variance of r (t) is σ2tAdvantage:

• mean reverting

Disadvantage:

• can produce a negative interest rate

• V ar [r (t)] = σ2t. So the variance doesn’t increase if r (t) increases.

We can also derive the bond price P (t, T ) under the Vasicek model. How-ever, the derivation is far more complex than in the Merton model. Here is theoutline of the derivation.Under the Vasicek model, Equation 24.9 becomes:12σ

2 ∂2P∂r2 + [α (b− r)− σφ] ∂P∂r +

∂P∂t − rP = 0

subject to the boundary condition P (r, T, T ) = 1Once again, we guess the solution is P (r, t, T ) = A (T − t) e−B(T−t)r(t)

→ ∂P

∂r= −ABe−Br ∂2P

∂r2= AB2e−Br

∂P

∂t= rAB

0e−Br −A0e−Br

→ 12σ

2AB2e−Br−[α (b− r)− σφ]ABe−Br+rAB0e−Br−A0e−Br−rAe−Br =

0

→ 12σ

2AB2 − [α (b− r)− σφ]AB + rAB0 −A0 − rA = 0

→ 12σ

2AB2 − (αb− σφ)AB + σφAB −A0 =³A−AB

0 − aAB´r

For the above equation to hold for any r, we need to have:A−AB

0 − αAB = 012σ

2AB2 − (αb− σφ)AB + σφAB −A0 = 0

A−AB0 − αAB = 0 → 1−B

0 − αB = 0The boundary condition is B (0) = 0

This gives us: B =1− e−α(T−t)

αEquation 1

2σ2AB2 − (αb− σφ)AB + σφAB − A0 = 0 is hard to solve.

Someone solved this equation for us. The result is DM Equation 24.26. Youdon’t need to memorize the solution for A or B. Just memorize P (r, t, T ) =A (T − t) e−B(T−t)r(t).

Example 24.2.1. May 2007 SOA MFE #13

Let P (r, t, T ) denote the price at time t of $1 to be paid with certainty attime T , t ≤ T , if the short rate at time t is equal to r. For a Vasicek modelyou are given:

P (0.04, 0, 2) = 0.9445



P (0.05, 1, 3) = 0.9321P (r∗, 2, 4) = 0.8960Calculate r∗.The price of any bond must satisfy Equation 24.9. We guess that the solution

to Equation 24.9 is P (r, t, T ) = A (T − t) e−B(T−t)r(t)

P (0.04, 0, 2) = 0.9445 → A (2) e−B(2)0.04 = 0.9445P (0.05, 1, 3) = 0.9321 → A (2) e−B(2)0.05 = 0.9321

→ A (2) e−B(2)0.04

A (2) e−B(2)0.05=0.9445

0.9321eB(2)0.01 =

0.9445

0.9321

→ eB(2) =

µ0.9445

0.9321

¶ 1

0.01=

µ0.9445

0.9321

¶100= 3. 749 26

→ A (2) = 0.9445eB(2)0.04 = 0.9445£eB(2)

¤0.04= 0.9445

¡3. 749 260.04

¢=

0.995 77P (r∗, 2, 4) = 0.8960

→ A (2) e−B(2)r∗= A (2)

£eB(2)

¤−r∗= 0.8960

→ 0.995 77¡3. 749 26−r

∗¢= 0.8960

→¡3. 749 26−r

∗¢=0.8960

0.995 77= 0.899 81

→ −r∗ ln 3. 749 26 = ln 0.899 81→ r∗ = − ln 0.899 81

ln 3. 749 26= 0.07988 4 = 0.08

Please note r∗ can be solved without using any specifics of the Vasicek model.As a matter of fact, this problem can be rewritten as:Let P (r, t, T ) denote the price at time t of $1 to be paid with certainty at

time T , t ≤ T , if the short rate at time t is equal to r. You are given:P (0.04, 0, 2) = 0.9445P (0.05, 1, 3) = 0.9321P (r∗, 2, 4) = 0.8960Calculate r∗.

24.2.4 CIR model

The model assumes that the short interest rate r (t) follows the stochastic dif-ferential equation:

dr (t) = α (b− r) dt+ σpr (t)dZ (24.18)

In CIR model, V ar [r (t)] = σ2r (t) t.The drift factor α (b− r) in the CIR model is the same as the drift factor

in the Vasicek model. It ensures mean reversion of the interest rate towardsthe long run value b, with speed of adjustment governed by the strictly positiveparameter a.The standard deviation factor, σ

pr (t), corrects the main drawback of Va-

sicek’s model, ensuring that the interest rate cannot become negative. Thus,at low values of the interest rate, the standard deviation becomes close to zero,


24.3. BOND OPTIONS, CAPS, AND THE BLACK MODEL 275

canceling the effect of the random shock on the interest rate. Consequently,when the interest rate gets close to zero, its evolution becomes dominated bythe drift factor, which pushes the rate upwards.Advantage:

• Mean-reverting

• Not allow a negative interest rate

• Variance of V ar [r (t)] = σ2r (t) t. The higher the r (t), the higher theV ar [r (t)]. This is a desirable feature.

To solve the bond price under CIR model, again we guess the solution isP (r, t, T ) = A (T − t) e−B(T−t)r(t). The solution is listed in DM page 788. Youdon’t need to memorize the solution. Just memorize P (r, t, T ) = A (T − t) e−B(T−t)r(t).

24.3 Bond options, caps, and the Black model

24.3.1 Black formula

Notations:

• Pt (T, T + s). This is the price agreed upon at time t, which will be paid atT in order to receive $1 at T + s. Simply put, Pt (T, T + s) is the presentvalue of $1 discounted from T + s to T using the interest rate availableat t. For example, at time zero we know that the annual interest ratefrom t = 1 to t = 3 is 10% compounded continuously. Then P0 (1, 1 + 2)is just PV of $1 discounted from t = 3 to t = 1 using 10% continuouslycompounded interest rate. So P0 (1, 1 + 2) = e−0.1(2) = 0.818 73

• PT (T, T + s). This is PV $1 discounted from T + s to T . Generally, wesimplify the symbol PT (T, T + s) as P (T, T + s)

Consider a call option with strike price K, expiring at time T , on a zero-coupon bond paying $1 at time T +s. If you buy this call option, then at T youhave the right to buy a bond maturing at time T + s for the guaranteed priceK. Since time T cost of a bond maturing in T + s is PV of $1 discounted fromT + s to T , buying a bond maturing at time T + s for the guaranteed price Kreally means "give up K and receive PV of $1 discounted from T +s to T ." ThePV of $1 discounted from T + s to T is PT (T, T + s). So the call payoff is

Call Payoff = max [0, PT (T, T + s)−K] (DM 24.30)

Next, the textbook has a difficult formula:

Ft,T [P (T, T + s)] =P (t, T + s)

P (t, T )(DM 24.31)



To understand the meaning of DM 24.31, use an example. Suppose t = 0,T = 1, and s = 2. In addition, assume that the interest rate is always 10%compounded continuously.

• Ft,T [P (T, T + s)] = F0,1 [P (1, 3)] is the price agreed upon at t = 0 to bepaid at T = 1 in order to receive $1 at T + s = 3. In other words, if youpay F0,1 [P (1, 3)] at T = 1, you should receive PV of $1 discounted fromT + s = 3 to T = 1. To avoid arbitrage, F0,1 [P (1, 3)] = e−0.1(2) = e−0.02

• P (t, T + s) = P (0, 3) is PV of $1 discounted from T + s = 3 to time zero.P (0, 3) = e−0.1(3) = e−0.03

• P (t, T ) = P (0, 1) is PV of $1 discounted from T = 1 to time zero.P (0, 1) = e−0.1

Clearly we see that e−0.02 = e−0.03

e−0.1 .Next, the textbook gives us the price of the call on a bond. Consider a call

option is written at time zero with strike price K, expiring at time T , on azero-coupon bond paying $1 at time T + s. If you buy this call option at timezero, then at T you have the right to pay K and receive the PV of $1 discountedfrom T + s to T .

To find the call price, we’ll use DM Equation 12.5:C = FP

0,T (S)N (d1)− FP0,T (K)N (d2)

d1=ln

FP0,T (S)

FP0,T

(K)+0.5σ2T

σ√T

d2 = d1 − σ√T

This is how I memorize DM Equation 12.5:C =Time zero cost of what you get at T ×N (d1)−Time zero cost of what

you give at T ×N (d2)d1=

³ln Time zero cost of what you get at T

Time zero cost of what you give at T + 0.5σ2T´/σ√T

d2 = d1 − σ√T

Let’s apply DM Equation 12.5 to call on bond.

• What we get at T is P (T, T + s), PV of $1 discounted from T + s to T

• Time zero cost of P (T, T + s) is P (0, T + s), PV of $1 discounted fromT + s to time 0

• What we give at T is K

• Time zero cost of what we give at T is KP (0, T ), PV of K discountedfrom T to time 0


24.3. BOND OPTIONS, CAPS, AND THE BLACK MODEL 277

C =Time zero cost of what you get at T ×N (d1)−Time zero cost of what

you give at T ×N (d2)

d1=³ln Time zero cost of what you get at T


d2 = d1 − σ√T

This gives us the price of the call on a bond:

C = P (0, T + s)N (d1)− P (0, T )KN (d2)

d1=ln P(0,T+s)

P(0,T)+0.5σ2T

σ√T

d2 = d1 − σ√T

We can also calculate the price of the put on the bond:P = P (0, T )KN (−d2)− P (0, T + s)N (−d1)

Please note that DM Equation 24.32 uses the following formula:C = P (0, T ) [FN (d1)−KN (d2)]

d1=ln F

K+0.5σ2T

σ√T

This is how to derive these formulas. Notice, F = F0,T [P (T, T + s)] =P (0,T+s)P (0,T )

→ C = P (0, T + s)N (d1)− P (0, T )KN (d2)= FP (0, T )N (d1)− P (0, T )KN (d2)= P (0, T ) [FN (d1)−KN (d2)]

Example 24.3.1. SOA May 2007 MFE #7

You are given the following information:

Bond maturity (years) 1 2Zero-coupon bond price 0.9434 0.8817A European call option, that expires in 1 year, gives you the right to purchase

a 1-year bond for 0.9259.The bond forward price is lognormally distributed with volatility σ = 0.05.Using the Black formula, calculate the price of the call option.

Solution.

If you buy this option, then at T = 1, you can pay K = 0.9259 and buy a1-year bond. This 1-year bond will give you $1 at time T + s = 2. .



• The value of this bond at T = 1 is PV of $1 discounted from T + s = 2 toT = 1.Time zero cost of PV of $1 discounted from T + s = 2 to T = 1 isjust PV of $1 discounted from T + s = 2 to t = 0. The cost is 0.8817

• Time zero cost of the strike price K at T = 1 is just PV of K discountedfrom T = 1 to time zero. So PV (K) = K × 0.9434 = 0.9259× 0.9434

C =Time zero cost of what you get at T ×N (d1)−Time zero cost of whatyou give at T ×N (d2)



→ d1 =ln 0.8817

0.9259×0.9434+0.5×0.052×1

0.05√1

= 0.212 0

→ d2 = d1 − σ√T = 0.212 0− 0.05

√1 = 0.162 0

N (d1) = 0.583 9 N (d2) = 0.564 3C = 0.8817× 0.583 9− 0.9259× 0.9434× 0.564 3 = 0.021 9

We can also calculate the put price.P =Time zero cost of what you give at T ×N (−d2)−Time zero cost of what

you get at T ×N (−d1)P = 0.9259× 0.9434× (1− 0.564 3)− 0.8817× (1− 0.583 9) = 0.013 7

If you want to use the formula C = P (0, T ) [FN (d1)−KN (d2)], this ishow:

• P (0, T ) = P (0, 1) is PV of $1 discounted from T = 1 to time zero. SoP (0, 1) = 0.9434

• F = F0,T [P (T, T + s)] = F0,1 [P (1, 2)] is the forward price of 1-year bond.This is the price agreed up at time zero, paid at T = 1 in order to receive$1 at T + s = 2. Using Equation DM 24/31, we have F0,1 [P (1, 2)] =P (0,2)P (0,1) =

0.88170.9434

d1=ln F

K+0.5σ2T

σ√T

=ln

0.88170.94340.9259 +0.5×0.05

2×10.05√1

= 0.212 0

d2 = 0.212 0− 0.05√1 = 0.162 0

N (d1) = 0.583 9 N (d2) = 0.564 3

C = P (0, T ) [FN (d1)−KN (d2)]= 0.9434

¡0.88170.9434 × 0.583 9− 0.9259× 0.564 3

¢= 0.0220

P = P (0, T ) [KN (d2)− FN (d1)]= 0.9434

¡0.9259× (1− 0.564 3)− 0.8817

0.9434 × (1− 0.583 9)¢

= 0.013 7


24.4. BINOMIAL INTEREST RATE MODEL 279

24.3.2 Interest rate caplet

Notation

• Rt (T, T + s). The (not annualized) interest rate pre-agreed upon at timet where t ≤ T that applies to the future time interval [T, T + s].

• RT (T, T + s). The (not annualized) interest rate agreed upon at time Tthat applies to the time interval [T, T + s].

• Caplet. A caplet gives the buyer the right to buy the time-T marketinterest rate RT (T, T + s) by paying a fixed strike interest rate KR. IfKR ≥ RT (T, T + s), the caplet expires worthless. The payoff of the capletat T + s is max [0, RT (T, T + s)−KR]. The payoff of the caplet at T ismax [0, RT (T, T + s)−KR]

1 +RT (T, T + s)

To calculate the price of the caplet, we first modify the payoff:max [0, RT (T, T + s)−KR]

1 +RT (T, T + s)= (1 +KR)max

∙0,

RT (T, T + s)−KR

(1 +RT (T, T + s)) (1 +KR)

¸=

(1 +KR)max

∙0,

1

1 +KR− 1

1 +RT (T, T + s)

¸= (1 +KR)max

∙0,

1

1 +KR− PT (T, T + s)

¸max

∙0,

1

1 +KR− PT (T, T + s)

¸is the payoff of a put on a bond. This put

gives the buyer the right, at T , to sell a bond that matures at T + s for aguaranteed price 1

1+KR. Let P represent the price of this put. Hence the price

of the caplet is (1 +KR)P .

24.4 Binomial interest rate model

Binomial interest rate tree is really simple. Unfortunately, the author of thetextbook uses too many math symbols and formulas, making this section hardto read. What I will do here is to walk you through a few examples. If youunderstand these examples, you are fine.

Example 24.4.1. (DM Example 24.3)

The following is the 3-period interest rate tree (DM Figure 24.3)t = 0 t = 1 t = 2

0.180.14

0.100.10

0.100.06

0.02



Make sure you understand the above table. The 10% interest rate at t = 0applies to the interval [t = 0, t = 1]. The 14% and 6% interest rates apply tothe interval [t = 1, t = 2]. The interest rates 0.18, 0.10, and 0.02 at t = 2 applyto the interval [t = 2, t = 3].The price of the 1-year bond is just the PV of $1 discounted from t = 1 to

t = 0. Hence P (0, 1) = e−0.1

The price of the 2-year bond is just the PV of $1 discounted from t = 2 tot = 0. What’s tricky is that we have two interest rates during [t = 1, t = 2].

• If the path is u, then $1 at t = 2 travels back to t = 0 through 2 interestrates: 0.14, and 0.1. The PV of $1 discounted from t = 2 to t = 1 ise−0.14. The PV of e−0.14 discounted from t = 1 to t = 0 is e−0.14e−0.11 =e−(0.14+0.1). So the 2-year bond is worth e−(0.14+0.1) at t = 0

• If the path is d, then $1 at t = 2 travels back to t = 0 through 2 interestrates: 0.06 and 0.1. The 2-year bond is worth e−(0.06+0.1) at t = 0

Since the risk-neutral probability of up or down is 50%, then the 2-year bondis worth the following at t = 0:

P (0, 2) = 0.5e−(0.14+0.1) + 0.5e−(0.06+0.1) = 0.819 4The term 0.5e−(0.14+0.1) + 0.5e−(0.06+0.1) can be rewritten as:0.5e−(0.14+0.1) + 0.5e−(0.06+0.1) = E∗

he−

2i=0 rih

iwith h = 1

The price of the 3-year bond is just the PV of $1 discounted from t = 3 tot = 0.

• If the path is uu, then $1 at t = 3 travels back to t = 0 through 3 interestrates: 0.18, 0.14, and 0.1. The PV of $1 discounted from t = 3 to t = 0 ise−(0.18+0.14+0.1). So the 3-year bond is worth e−(0.18+0.14+0.1) at t = 0

• If the path is ud, then $1 at t = 3 travels back to t = 0 through 3 interestrates: 0.1, 0.14, and 0.1. The 3-year bond is worth e−(0.1+0.14+0.1) at t = 0

• If the path is du, then $1 at t = 3 travels back to t = 0 through 3 interestrates: 0.1, 0.06, and 0.1. The 3-year bond is worth e−(0.1+0.06+0.1) at t = 0

• If the path is dd, then $1 at t = 3 travels back to t = 0 through 3 interestrates: 0.02, 0.06, and 0.1. The 3-year bond is worth e−(0.02+0.06+0.1) att = 0

Since the risk-neutral probability for each is 0.52 = 0.25, the price of a 3-yearbond is the following at t = 0:

P (0, 3) = 0.25¡e−(0.18+0.14+0.1) + e−(0.1+0.14+0.1) + e−(0.1+0.06+0.1) + e−(0.02+0.06+0.1)

¢=

0.743 8

The term 0.25¡e−(0.18+0.14+0.1) + e−(0.1+0.14+0.1) + e−(0.1+0.06+0.1) + e−(0.02+0.06+0.1)

¢can be rewritten as:


24.4. BINOMIAL INTEREST RATE MODEL 281

0.25¡e−(0.18+0.14+0.1) + e−(0.1+0.14+0.1) + e−(0.1+0.06+0.1) + e−(0.02+0.06+0.1)

¢=

E∗he−

3i=0 rih

iThe general formula is DM 24.44:

P (0, n) = E∗he−

ni=0 rih

i(DM 24.31)

Option pricing example

Example 24.4.2. (SOA May 2007 #9)

You use a binomial interest rate model to evaluate a 7.5% interest rate capon a $100 three-year loan. You are given:(i) The interest rates for the binomial tree are as follows:

• r0 = 6%

• ru = 7.704%

• rd = 4.673%

• ruu = 9.892%

• rud = rdu = 6.000%

• rdd = 3.639%

(ii) All interest rates are annual effective rates.(iii) The risk-neutral probability that the annual effective interest rate moves

up or down is 0.5.(iv) The loan interest payments are made annually.Using the binomial interest rate model, calculate the value of this interest

rate cap.t = 0 t = 1 t = 2

9.892%7.704%

6.000%6%

6.000%4.673%

3.639%

Once again, the interest rate at t in the above table applies to the period[t, t+ 1]. For example, the 6% rate applies to [0, 1] (i.e. Year 1).First, let’s understand what’s an interest rate cap. Imagine you borrowed

$100 from a bank. Your interest accrued on the loan each year is not basedon a fixed interest rate (such as 8%), but is based on the then market interestrate. For example, if the market interest rate is 6% in Year 1, then your interestpayment at the end of Year 1 is 100× 0.06 = 6.0.



As a borrower, you are worried that the market interest may go up. Forexample, if the Year 1 interest rate is 20%, then your interest payment at theend of Year 1 is 100× 0.2 = 20.How can you reduce your risk? One thing you can do is to buy an interest

cap. Suppose you buy an interest rate cap of 7.5%. This is what happens:

• If the market interest rate is at or below 7.5%, then the cap doesn’t kickin. So you get nothing from the cap

• if the market interest rate is above 7.5%, then the party who sold youthe cap will pay you the excess of the market interest rate over the caprate. For example, if the market interest rate in Year 1 is 10%, then theseller of the cap will pay you 100 (0.1− 0.075) = 2. 5 at the end of Year1. You still own the bank 100 × 0.1 = 10 at the end of Year 1, but youpay 100 × 0.075 = 7. 5 out of your own pocket. The cap seller pays you100 (0.1− 0.075) = 2. 5. So together you get 7. 5 + 2.5 = 10. You mail a$10 check to the bank.

In summary, if you buy an interest capped of 7.5%, then the interest onyour loan is capped at 7.5% regardless of the market interest rate (because anyexcess of the market rate over the cap 7.5% is paid by the cap seller).Now you know what an interest cap is, let’s solve this problem.

The first year market rate 6% is below the cap rate. At the end of Year 1,you get nothing from the cap

If the 2nd year market rate is 7.704%, then at the end of year 2 the capseller pays you 100 (0.07704− 0.075) = 0.204 . Notice that 0.204 occurs att = 2. To help keep track of payments, we’ll discount this payment to t = 1.

The discounted value is0.204

1 + 0.07704at t = 1.

If the 2nd year market rate is 4.673%, then the payment at the end of year2 is zero.

Let’s calculate the cap payoff in Year 3. Of the 4 interest rates during[t = 2, t = 3] , only when the market interest rate is 9.892% do we get a payoffof 100 (0.09892− 0.075) = 2. 392. This payment occurs at t = 3. The PV of

this payment at t = 2 is2. 392

1 + 0.09892.

Now we can draw a payoff table:


24.5. BLACK-DERMAN-TOY MODEL 283

t = 0 t = 1 t = 2

9.892% Payoff:100 (0.09892− 0.075)

1 + 0.09892

7.704% Payoff:100 (0.07704− 0.075)

1 + 0.077046.000%

6%6.000%

4.673%3.639%

The PV of payoff100 (0.07704− 0.075)

1 + 0.07704= 0.189 4 1 at t = 1 discounted to

t = 0 is100 (0.07704− 0.075)(1 + 0.09892) (1 + 0.06)

. The risk neutral probability of reaching u

node is 0.5.

The PV of payoff100 (0.09892− 0.075)

1 + 0.09892= 2. 176 68 at t = 2 discounted to

t = 0 is100 (0.09892− 0.075)

(1 + 0.09892) (1 + 0.07704) (1 + 0.06). The risk neutral probability of

reaching uu node is 0.52 = 0.25

Hence risk neutral based expected present value of the cap payoff is:100 (0.07704− 0.075)(1 + 0.07704) (1 + 0.06)

×0.5+ 100 (0.09892− 0.075)(1 + 0.09892) (1 + 0.07704) (1 + 0.06)

×0.52 =0.565 99 = 0.57

So the price of the cap is $0.57

24.5 Black-Derman-Toy model

The BDT model is a procedure to produce a discrete binomial interest ratetree that matches the observed term structure of interest rates. Once again,to quickly explain the essence of the BDT model, I’m walk you through anexample. If you understand this example, you are ready for the exam.Suppose we gathered the following data from the market (DM Table 24.2):Maturity n (Yrs) YTM Bond Price Volatility in Yr 1 on (n− 1)-Yr bond1 10% 0.90912 11% 0.8116 10%3 12% 0.7118 15%4 12.5% 0.6243 14%

We want to produce an interest rate tree that will match the above table.What should we do?First, let’s go through some terms.



YTM (yield to maturity) is the discrete annual effective interest rate earnedby a bond. The formula is

P (0, T ) = (1 + Y TM)−T

For example, a 2-year bond in the table is worth 0.8116. This means thatPV of $1 discounted from t = 2 to t = 0 is 0.8116. To find YTM, we solve thefollowing equation:0.8116 = (1 + i)

−2 → i = 0.11So the YTM for this 2-year bond is 11%

The volatility in Yr 1 on (n− 1) bond is the standard deviation of the naturallog of the YTM on an (n− 1)-year bond issued at t = 1 and maturity at timen. For example, 10% volatility in the table means that standard deviation ofthe natural log of the YTM on a 1-year bond issued at t = 1 maturing in t = 2is 10%. This concept will be clear to you later.

Next, next use the BDT model to build an interest rate tree. We’ll first findthe interest rate ru and rd, the two possible interest rates for Year 2 (i.e. forthe time interval [t = 1, t = 2]).

t = 0 t = h = 1ru

r0 = 10%rd

Once again, rt is the interest for the interval [t, t+ 1]. For example, r0 = 10%is the interest rate for Year 1. ru and rd are the two interest rates for Year 2(i.e. from t = 1 to t = 2).We assume the risk neutral probability of up and down is 0.5. We want to

find ru and rd to satisfy the condition that standard deviation of the naturallog of the YTM on a 1-year bond issued at t = 1 maturing in t = 2 is 10%.The price of a 1-year bond issued at t = 1 maturing in t = 2 is

P (1, 2) =1

1 + ruor

1

1 + rdThe YTM is:

(1 + Y TMu)−1=

1

1 + ru→ Y TMu = ru

or (1 + Y TMd)−1=

1

1 + rd→ Y TMd = rd

We know that the standard deviation of lnY TMu = ln ru and lnY TMd =ln rd is 10%.

The mean of the log of the YTM on a 1-year bond issued at t = 1 maturingin t = 2 rate is:

E (ln r1) = 0.5 ln ru + 0.5 ln rd = 0.5 (ln ru + ln rd)



We use the variance formula V ar (X) = E [X −E (X)]2 to calculate the

variance of lnY TMu = ln ru and lnY TMd = ln rd:0.5 [ln ru −E (ln r1)]

2 + 0.5 [ln rd −E (ln r1)]2

= 0.5 (ln ru − 0.5 ln ru − 0.5 ln rd)2 + 0.5 (ln rd − 0.5 ln ru − 0.5 ln rd)2

= 0.5 (0.5 ln ru − 0.5 ln rd)2 + 0.5 (0.5 ln rd − 0.5 ln ru)2

= 0.5 (0.5 ln ru − 0.5 ln rd)2 + 0.5 (0.5 ln ru − 0.5 ln rd)2

= (0.5 ln ru − 0.5 ln rd)2 =µ0.5 ln

rurd

¶2The standard deviation of lnY TMu = ln ru and lnY TMd = ln rd is σ1 =

10%.→ 0.5 ln

rurd= σ1 ln

rurd= 2σ1 ru = rde

2σ1 = rde0.2

Now we see that matching volatility requires ru = rde2σ. This relationship

holds for every node.

Next, we want to reproduce the 2-year bond price of 0.8116.If the Year 2 interest rate is ru, then the PV of $1 discounted from t = 2 to

t = 0 is1

(1 + ru) (1 + r0)If the Year 2 interest rate is rd, then the PV of $1 discounted from t = 2 to

t = 0 is1

(1 + rd) (1 + r0)The risk-neutral probability of up and down is 0.5. The expected 2-year

bond price is:

0.5× 1

(1 + ru) (1 + r0)+ 0.5× 1

(1 + rd) (1 + r0)

To match the observed price P (0, 2) = 0.8116, we have:

0.5× 1

(1 + ru) (1 + r0)+ 0.5× 1

(1 + rd) (1 + r0)= P (0, 2)

To sum up, we have two equations:⎧⎨⎩ ru = rde2σ1

0.5× 1

(1 + ru) (1 + r0)+ 0.5× 1

(1 + rd) (1 + r0)= P (0, 2)

Or⎧⎨⎩ ru = rde0.2

0.5× 1

(1 + ru) (1 + 0.1)+ 0.5× 1

(1 + rd) (1 + 0.1)= 0.8116

→ 1

1 + rd+

1

1 + rde0.2= 2× 0.8116 (1 + 0.1)

→ rd = 0.108 265 = 10.83%



→ ru = 0.108 265e0.2 = 0.132 2 = 13.22%

We can use the same procedure to calculate ruu, rud, rdu, rdd. To simplifyour model, we arbitrary set rud = rdu (i.e. we assume the tree is recombining).

t = 0 t = h = 1 t = 2h = 2ruu = rdde

4σ2

ru = 13.22%r0 = 10% rud = rdu = rdde

2σ2

rd = 10.83%rdd

We want to choose rdd such that the standard deviation of the natural logof the YTM on a 2-year bond issued at t = 1 and maturing at t = 3 is 15%.

First, we calculate P (1, 3), the price of a 2-year bond issued at t = 1 andmaturing at t = 3.If the 2nd year interest rate is ru = 13.22%, then the 3rd year rate is either

ruu or rud with equal risk-neutral probability of 0.5. Then the expected PV of$1 discounted from t = 3 to t = 1 is

P (1, 3, ru) = 0.5×1

(1 + ru) (1 + ruu)+ 0.5× 1

(1 + ru) (1 + rud)

=0.5

1.1322

µ1

1 + rdde4σ2+

1

1 + rdde2σ2

¶The YTM can solved as follows:

P (1, 3, ru) = (1 + i)−2 → i = P (1, 3, ru)−0.5 − 1

If the 2nd year interest rate is rd = 10.83%, then the 3rd year rate is eitherrdu or rdd with equal risk-neutral probability of 0.5. Then the expected PV of$1 discounted from t = 3 to t = 1 is

P (1, 3, rd) = 0.5×1

(1 + rd) (1 + rdu)+ 0.5× 1

(1 + rd) (1 + rdd)

=0.5

1.1083

µ1

1 + rdde2σ2+

1

1 + rdd

¶The YTM can solved as follows:

P (1, 3, rd) = (1 + i)−2 → i = P (1, 3, rd)

−0.5 − 1

→ 0.15 = 0.5 lnP (1, 3, ru)

−0.5 − 1P (1, 3, rd)

−0.5 − 1

= 0.5 ln

∙0.5

1.1322

µ1

1 + rdde4σ2+

1

1 + rdde2σ2

¶¸−0.5− 1∙

0.5

1.1083

µ1

1 + rdde2σ2+

1

1 + rdd

¶¸−0.5− 1

By the way, please note that σ2 is not 10%.



We should also match the 3-year bond price P (0, 3) = 0.7118The PV of $1 at t = 3 discounted to t = 0 is

• 1

(1 + r0) (1 + ru) (1 + ruu)if the path is 0→ u→ uu

• 1

(1 + r0) (1 + ru) (1 + rud)if the path is 0→ u→ ud

• 1

(1 + r0) (1 + rd) (1 + rdu)if the path is 0→ d→ du

• 1

(1 + r0) (1 + rd) (1 + rdd)if the path is 0→ d→ dd

Each path has a risk neutral probability of 0.25. Now we have:

0.25

(1 + r0) (1 + ru) (1 + ruu)+

0.25

(1 + r0) (1 + ru) (1 + rud)

+0.25

(1 + r0) (1 + rd) (1 + rdu)+

0.25

(1 + r0) (1 + rd) (1 + rdd)= P (0, 3)

0.25

(1.1) (1.1322) (1 + rdde4σ)+

0.25

(1.1) (1.1322) (1 + rdde2σ)

+0.25

(1.1) (1.1083) (1 + rdde2σ)+

0.25

(1.1) (1.1083) (1 + rdd)= 0.7118

Now our equations are:

0.15 = 0.5 ln

∙0.5

1.1322

µ1

1 + rdde4σ2+

1

1 + rdde2σ2

¶¸−0.5− 1∙

0.5

1.1083

µ1

1 + rdde2σ2+

1

1 + rdd

¶¸−0.5− 1

0.25

(1.1) (1.1322) (1 + rdde4σ2)+

0.25

(1.1) (1.1322) (1 + rdde2σ2)

+0.25

(1.1) (1.1083) (1 + rdde2σ2)+

0.25

(1.1) (1.1083) (1 + rdd)= 0.7118

These equations are hard to solve manually. Special software is needed tosolve them.

You can verify that the solutions are: rdd = 0.0925 σ2 = 0.195 0

0.5 ln

µ0.5

1.1322

µ1

1 + 0.0925e4×0.1950+

1

1 + 0.0925e2×0.1950

¶¶−0.5− 1µ

0.5

1.1083

µ1

1 + 0.0925e2×0.1950+

1

1 + 0.0925

¶¶−0.5− 1

= 0.149 97 =

0.15



0.25

(1.1) (1.1322) (1 + 0.0925e4×0.1950)+

0.25

(1.1) (1.1322) (1 + 0.0925e2×0.1950)

+0.25

(1.1) (1.1083) (1 + 0.0925e2×0.1950)+

0.25

(1.1) (1.1083) (1 + 0.0925)= 0.711 755 = 0.7118

Then the year 3 interest rates are:rdd = 0.0925rdu = rud = 0.0925e

2×0.1950 = 0.136 6ruu = 0.0925e

4×0.1950 = 0.201 8

We can use the same logic and calculate the Year 4 interest rates. However,I’m not going to do the calculation because the calculation is overly intensive.

Make sure you can reproduce the Yr 2 and Yr 3 rates. Yr 2 rates can beeasily reproduced. SOA or CAS can ask you to calculate the Year 2 rates usingthe BDT model. Yr 3 rates are harder. A full calculation of Year 3 rates byhand is difficult. However, SOA or CAS can give you some partial informationon Year 3 rates and ask you to calculate the rest.


Solution to Derivatives Markets: SOA ExamMFE and CAS Exam 3 FE

Yufeng Guo

July 14, 2009

Contents

Introduction vii

9 Parity and other option relationships 1

10 Binomial option pricing I 23

11 Binomial option pricing II 91

12 Black-Scholes formula 107

13 Market making and delta hedging 125

14 Exotic options: I 149

18 Lognormal distribution 161

19 Monte Carlo simulation 177

20 Brownian motion and Ito’s lemma 187

21 The Black-Scholes equation 193

22 Exotic options: II 203

23 Volatility 205

24 Interest rate models 209

iii

Preface

This is Guo’s solution to Derivatives Markets (2nd edition ISBN 0-321-28030-X)for SOA MFE or CAS Exam 3 FE. Unlike the official solution manual publishedby Addison-Wesley, this solution manual provides solutions to both the even-numbered and odd-numbered problems for the chapters that are on the SOAExam MFE and CAS Exam 3 FE syllabus. Problems that are out of the scopeof the SOA Exam MFE and CAS Exam 3 FE syllabus are excluded.

Please report any errors to [email protected].

This book is the exclusive property of Yufeng Guo. Redistribution of thisbook in any form is prohibited.

v

Introduction

Recommendations on using this solution manual:

1. Obviously, you’ll need to buy Derivatives Markets (2nd edition) to see theproblems.

2. Make sure you download the textbook errata from http://www.kellogg.northwestern.edu/faculty/mcdonald/htm/typos2e_01.html

vii

Chapter 9

Parity and other optionrelationships

Problem 9.1.

S0 = 32 T = 6/12 = 0.5 K = 35C = 2.27 r = 0.04 δ = 0.06

C + PV (K) = P + S0e−δT

2.27 + 35e−0.04(0.5) = P + 32e−0.06(0.5) P = 5. 522 7

Problem 9.2.

S0 = 32 T = 6/12 = 0.5 K = 30C = 4.29 P = 2.64 r = 0.04C + PV (K) = P + S0 − PV (Div)4.29 + 30e−0.04(0.5) = 2.64 + 32− PV (Div)PV (Div) = 0.944

Problem 9.3.

S0 = 800 r = 0.05 δ = 0T = 1 K = 815 C = 75 P = 45a. Buy stock+ sell call+buy put=buy PV (K)C + PV (K) = P + S0→ PV (K = 815) = S0|z

buy stock

+ −C|zsell call

+ P|zbuy put

= 800 + (−75) + 45 = 770

So the position is equivalent to depositing 770 in a savings account (or buy-ing a bond with present value equal to 770) and receiving 815 one year later.770eR = 815 R = 0.056 8

1

CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS

So we earn 5.68%.

b. Buying a stock, selling a call, and buying a put is the same as depositingPV (K) in the savings account. As a result, we should just earn the risk freeinterest rate r = 0.05. However, we actually earn R = 0.056 8 > r. To arbitrage,we "borrow low and earn high." We borrow 770 from a bank at 0.05%. We usethe borrowed 770 to finance buying a stock, selling a call, and buying a put.Notice that the net cost of buying a stock, selling a call, and buying a put is770.One year later, we receive 770eR = 815. We pay the bank 770e0.05 = 809.

48. Our profit is 815− 809. 48 = 5. 52 per transaction.If we do n such transactions, we’ll earn 5. 52n profit.

Alternative answer: we can burrow at 5% (continuously compounding) andlend at 5.6 8% (continuously compounding), earning a risk free 0.68%. So ifwe borrow $1 at time zero, our risk free profit at time one is e0.0568 − e0.05 =0.00717 3; if we borrow $770 at time zero, our risk free profit at time one is0.00717 3×770 = 5. 52. If we borrow n dollars at time zero, we’ll earn 0.00717 3ndollars at time one.c. To avoid arbitrage, we need to have:PV (K = 815) = S0|z

buy stock

+ −C|zsell call

+ P|zbuy put

= 815e−0.05 = 775. 25

→ C − P = S0 − PV (K) = 800− 775. 25 = 24. 75

d. C − P = S0 − PV (K) = 800−Ke−rT = 800−Ke−0.05

If K = 780 C − P = 800− 780e−0.05 = 58. 041If K = 800 C − P = 800− 800e−0.05 = 39. 016If K = 820 C − P = 800− 820e−0.05 = 19. 992If K = 840 C − P = 800− 840e−0.05 = 0.967

Problem 9.4.

To solve this type of problems, just use the standard put-call parity.To avoid calculation errors, clearly identify the underlying asset.The underlying asset is €1. We want to find the dollar cost of a put option

on this underlying.The typical put-call parity:C + PV (K) = P + S0e

−δT

C, K, P , and S0 should all be expressed in dollars. S0 is the current (dollarprice) of the underlying. So S0 = $0.95.

C = $0.0571 K = $0.93δ is the internal growth rate of the underlying asset (i.e. €1). Hence δ = 0.04Since K is expressed in dollars, PV (K) needs to be calculated using the

dollar risk free interest r = 0.06.0.0571 + 0.93e−0.06(1) = P + 0.95e−0.04(1) P = $0.02 02

www.actuary88.com c°Yufeng Guo 2


Problem 9.5.

As I explained in my study guide, don’t bother memorizing the followingcomplex formula:

C$ (x0,K, T ) = x0KPf

µ1

x0,1

K,T

¶Just use my approach to solve this type of problems.Convert information to symbols:

The exchange rate is 95 yen per euro. Y 95 =€1 or Y 1 =€1

95

Yen-denominated put on 1 euro with strike price Y100 has a premium Y8.763→ (€1→ Y 100)0 =Y8.763

What’s the strike price of a euro-denominated call on 1 yen? €K → 1Y

Calculate the price of a euro-denominated call on 1 yen with strike price €K(€K → 1Y )0 = €?

€1→ Y 100 → €1

100→ Y 1

The strike price of the corresponding euro-denominated yen call isK =€1

100=€0.01µ

€1

100→ Y 1

¶0

=1

100× (€1→ Y 100)0 =

1

100(Y 8.763)

Since Y 1 =€1

95, we have:

1

100(Y 8.763) =

1

100(8.763)

µ€1

95

¶=€9. 224 2× 10−4

→µ€1

100→ Y 1

¶0

=€9. 224 2× 10−4

So the price of a euro-denominated call on 1 yen with strike price K =€1

100is €9. 224 2× 10−4



Problem 9.6.

The underlying asset is €1. The standard put-call parity is:C + PV (K) = P + S0e

−δT

C, K, P , and S0 should all be expressed in dollars. S0 is the current (dollarprice) of the underlying.

δ is the internal growth rate of the underlying asset (i.e. €1).We’ll solve Part b first.

b. 0.0404 + 0.9e−0.05(0.5) = 0.0141 + S0e−0.035(0.5) S0 = $0.920 04

So the current price of the underlying (i.e. €1) is S0 = $0.920 04. In otherwords, the currency exchange rate is $0.920 04 =€1

a. According to the textbook Equation 5.7, the forward price is:F0,T = S0e

−δT erT = 0.920 04e−0.035(0.5)e0.05(0.5) = $0.926 97

Problem 9.7.

The underlying asset is one yen.a. C +Ke−rT = P + S0e

−δT

0.0006 + 0.009e−0.05(1) = P + 0.009e−0.01(1)

0.0006 + 0.008561 = P + 0.008 91 P = $0.00025

b. There are two puts out there. One is the synthetically created put usingthe formula:

P = C +Ke−rT − S0e−δT

The other is the put in the market selling for the price for $0.0004.

To arbitrage, build a put a low cost and sell it at a high price. At t = 0, we:

• Sell the expensive put for $0.0004

• Build a cheap put for $0.00025. To build a put, we buy a call, depositKe−rT in a savings account, and sell e−δT unit of Yen.

T = 1 T = 1t = 0 ST < 0.009 ST ≥ 0.009

Sell expensive put 0.0004 ST − 0.009 0Buy call −0.0006 0 ST − 0.009Deposit Ke−rT in savings −0.009e−0.05(1) 0.009 0.009

Short sell e−δT unit of Yen 0.009e−0.01(1) ST STTotal $0.00015 0 0

0.0004− 0.0006− 0.009e−0.05(1) + 0.009e−0.01(1) = $0.00015



At t = 0, we receive $0.00015 yet we don’t incur any liabilities at T = 1 (sowe receive $0.00015 free money at t = 0).

c. At-the-money means K = S0 (i.e. the strike price is equal to the currentexchange rate).Dollar-denominated at-the-money yen call sells for $0.0006. To translate this

into symbols, notice that under the call option, the call holder can give $0.009and get Y 1.

"Give $0.009 and get Y 1" is represented by ($0.009→ Y 1). This option’spremium at time zero is $0.0006. Hence we have:($0.009→ Y 1)0 = $0.0006We are asked to find the yen denominated at the money call for $1. Here

the call holder can give c yen and get $1. "Give c yen and get $1" is representedby (Y c→ $1). This option’s premium at time zero is (Y c→ $1)0.First, we need to calculate c, the strike price of the yen denominated dollar

call. Since at time zero $0.009 = Y 1, we have $1 = Y1

0.009. So the at-the-

money yen denominated call on $1 is c =1

0.009. Our task is to find this option’s

premium:µY

1

0.009→ $1

¶0

=?

We’ll find the premium for Y 1 →$0.009, the option of "give 1 yen and get$0.009." Once we find this premium, we’ll scale it and find the premium of "give1

0.009yen and get $1."

We’ll use the general put-call parity:(AT → BT )0 + PV (AT ) = (BT → AT )0 + PV (BT )

($0.009→ Y 1)0 + PV ($0.009) = (Y 1→ $0.009)0 + PV (Y 1)

PV ($0.009) = $0.009e−0.05(1)

Since we are discounting $0.009 at T = 1 to time zero, we use the dollarinterest rate 5%.

PV (Y 1) = $0.009e−0.01(1)

If we discount Y1 from T = 1 to time zero, we get e−0.01(1) yen, which isequal to $0.009e−0.01(1).

So we have:$0.0006+$0.009e−0.05 = (Y 1→ $0.009)0 + $0.009e

−0.01(1)

(Y 1→ $0.009)0 = $2. 506 16× 10−4µ1

0.009Y 1→ $1

¶0

=1

0.009(Y 1→ $0.009)0 = $

2. 506 16× 10−40.009

= $2.

784 62× 10−2 = Y2. 784 62× 10−2

0.009= Y 3. 094



So the yen denominated at the money call for $1 is worth $2. 784 62× 10−2or Y 3. 094.

We are also asked to identify the relationship between the yen denominatedat the money call for $1 and the dollar-denominated yen put. The relationshipis that we use the premium of the latter option to calculate the premium of theformer option.Next, we calculate the premium for the yen denominated at-the-money put

for $1:µ$→ Y

1

0.009

¶0

=1

0.009($0.009→ Y 1)0

=1

0.009× $0.0006 = $ 0.0 666 7

= Y 0.0 666 7× 1

0.009= Y 7. 407 8

So the yen denominated at-the-money put for $1 is worth $ 0.0 666 7 or Y7. 407 8.I recommend that you use my solution approach, which is less prone to errors

than using complex notations and formulas in the textbook.

Problem 9.8.

The textbook Equations 9.13 and 9.14 are violated.This is how to arbitrage on the calls. We have two otherwise identical

calls, one with $50 strike price and the other $55. The $50 strike call is morevaluable than the $55 strike call, but the former is selling less than the latter.To arbitrage, buy low and sell high.We use T to represent the common exercise date. This definition works

whether the two options are American or European. If the two options areAmerican, we’ll find arbitrage opportunities if two American options are ex-ercised simultaneously. If the two options are European, T is the commonexpiration date.

The payoff is:T T T

Transaction t = 0 ST < 50 50 ≤ ST < 55 ST ≥ 55Buy 50 strike call −9 0 ST − 50 ST − 50Sell 55 strike call 10 0 0 − (ST − 55)Total 1 0 ST − 50 ≥ 0 5

At t = 0, we receive $1 free money.At T , we get non negative cash flows (so we may get some free money, but

we certainly don’t owe anybody anything at T ). This is clearly an arbitrage.



This is how to arbitrage on the two puts. We have two otherwise identicalputs, one with $50 strike price and the other $55. The $55 strike put is morevaluable than the $50 strike put, but the former is selling less than the latter.To arbitrage, buy low and sell high.The payoff is:

T T TTransaction t = 0 ST < 50 50 ≤ ST < 55 ST ≥ 55Buy 55 strike put −6 55− ST 55− ST 0Sell 50 strike put 7 − (50− ST ) 0 0Total 1 5 55− ST > 0 0

At t = 0, we receive $1 free money.At T , we get non negative cash flows (so we may get some free money, but

we certainly don’t owe anybody anything at T ). This is clearly an arbitrage.

Problem 9.9.

The textbook Equation 9.15 and 9.16 are violated.We use T to represent the common exercise date. This definition works

whether the two options are American or European. If the two options areAmerican, we’ll find arbitrage opportunities if two American options are ex-ercised simultaneously. If the two options are European, T is the commonexpiration date.This is how to arbitrage on the calls. We have two otherwise identical calls,

one with $50 strike price and the other $55. The premium difference betweenthese two options should not exceed the strike difference 15− 10 = 5. In otherwords, the 50-strike call should sell no more than 10+5. However, the 50-strikecall is currently selling for 16 in the market. To arbitrage, buy low (the 55-strikecall) and sell high (the 50-strike call).

The $50 strike call is more valuable than the $55 strike call, but the formeris selling less than the latter.The payoff is:

T T TTransaction t = 0 ST < 50 50 ≤ ST < 55 ST ≥ 55Buy 55 strike call −10 0 0 ST − 55Sell 50 strike call 16 0 − (ST − 50) − (ST − 50)Total 6 0 − (ST − 50) ≥ −5 −5

So we receive $6 at t = 0. Then at T , our maximum liability is $5. So makeat least $1 free money.

This is how to arbitrage on the puts. We have two otherwise identical calls,one with $50 strike price and the other $55. The premium difference betweenthese two options should not exceed the strike difference 15− 10 = 5. In other



words, the 55-strike put should sell no more than 7 + 5 = 12. However, the55-strike put is currently selling for 14 in the market. To arbitrage, buy low(the 50-strike put) and sell high (the 55-strike put).The payoff is:

T T TTransaction t = 0 ST < 50 50 ≤ ST < 55 ST ≥ 55Buy 50 strike put −14 50− ST 0 0Sell 55 strike put 7 − (55− ST ) − (55− ST ) 0Total 7 −5 − (55− ST ) < −5 0

So we receive $7 at t = 0. Then at T , our maximum liability is $5. So makeat least $2 free money.

Problem 9.10.

Suppose there are 3 options otherwise identical but with different strike priceK1 < K2 < K3 where K2 = λK1 + (1− λ)K2 and 0 < λ < 1.Then the price of the middle strike price K2 must not exceed the price of a

diversified portfolio consisting of λ units of K1-strike option and (1− λ) unitsof K2-strike option:

C [λK1 + (1− λ)K3] ≤ λC (K1) + (1− λ)C (K3)P [λK1 + (1− λ)K3] ≤ λP (K1) + (1− λ)P (K3)

The above conditions are called the convexity of the option price with respectto the strike price. They are equivalent to the textbook Equation 9.17 and 9.18.If the above conditions are violated, arbitrage opportunities exist.

We are given the following 3 calls:Strike K1 = 50 K2 = 55 K3 = 60Call premium 18 14 9.50

λ50 + (1− λ) 60 = 55→ λ = 0.5 0.5 (50) + 0.5 (60) = 55

Let’s check:C [0.5 (50) + 0.5 (60)] = C (55) = 14

0.5C (50) + 0.5C (60) = 0.5 (18) + 0.5 (9.50) = 13. 75C [0.5 (50) + 0.5 (60)] > 0.5C (50) + 0.5C (60)So arbitrage opportunities exist. To arbitrage, we buy low and sell high.The cheap asset is the diversified portfolio consisting of λ units of K1-strike

option and (1− λ) units of K3-strike option. In this problem, the diversifiedportfolio consists of half a 50-strike call and half a 60-strike call.The expensive asset is the 55-strike call.



Since we can’t buy half a call option, we’ll buy 2 units of the portfolio (i.e.buy one 50-strike call and one 60-strike call). Simultaneously,we sell two 55-strike call options.We use T to represent the common exercise date. This definition works

whether the options are American or European. If the options are American,we’ll find arbitrage opportunities if the American options are exercised simulta-neously. If the options are European, T is the common expiration date.The payoff is:

T T T TTransaction t = 0 ST < 50 50 ≤ ST < 55 55 ≤ ST < 60 ST ≥ 60buy two portfoliosbuy a 50-strike call −18 0 ST − 50 ST − 50 ST − 50buy a 60-strike call −9.5 0 0 0 ST − 60Portfolio total −27. 5 0 ST − 50 ST − 50 2ST − 110

Sell two 55-strike calls 2 (14) = 28 0 0 −2 (ST − 55) −2 (ST − 55)Total 0.5 0 ST − 50 ≥ 0 60− ST > 0 0

−27. 5 + 28 = 0.5ST − 50− 2 (ST − 55) = 60− ST2ST − 110− 2 (ST − 55) = 0So we get $0.5 at t = 0, yet we have non negative cash flows at the expiration

date T . This is arbitrage.

The above strategy of buying λ units of K1-strike call, buying (1− λ) unitsof K3-strike call, and selling one unit of K2-strike call is called the butterflyspread.

We are given the following 3 puts:Strike K1 = 50 K2 = 55 K3 = 60Put premium 7 10.75 14.45

λ50 + (1− λ) 60 = 55→ λ = 0.5 0.5 (50) + 0.5 (60) = 55

Let’s check:P [0.5 (50) + 0.5 (60)] = P (55) = 10.75

0.5P (50) + 0.5P (60) = 0.5 (7) + 0.5 (14.45) = 10. 725P [0.5 (50) + 0.5 (60)] > .5P (50) + 0.5P (60)So arbitrage opportunities exist. To arbitrage, we buy low and sell high.The cheap asset is the diversified portfolio consisting of λ units of K1-strike

put and (1− λ) units of K3-strike put. In this problem, the diversified portfolioconsists of half a 50-strike put and half a 60-strike put.The expensive asset is the 55-strike put.



Since we can’t buy half a option, we’ll buy 2 units of the portfolio (i.e. buyone 50-strike put and one 60-strike put). Simultaneously,we sell two 55-strikeput options.The payoff is:

T T T TTransaction t = 0 ST < 50 50 ≤ ST < 55 55 ≤ ST < 60 ST ≥ 60buy two portfoliosbuy a 50-strike put −7 50− ST 0 0 0buy a 60-strike put −14.45 60− ST 60− ST 60− ST 0Portfolio total −21. 45 110− 2ST 60− ST 60− ST 0

Sell two 55-strike puts 2 (10.75) −2 (55− ST ) −2 (55− ST ) 0 0Total 0.05 0 ST − 50 ≥ 0 60− ST > 0 0

−21. 45 + 2 (10.75) = 0.0550− ST + 60− ST = 110− 2ST

−21. 45 + 2 (10.75) = 0.05110− 2ST − 2 (55− ST ) = 060− ST − 2 (55− ST ) = ST − 50So we get $0.05 at t = 0, yet we have non negative cash flows at the expiration

date T . This is arbitrage.The above strategy of buying λ units of K1-strike put, buying (1− λ) units

of K3-strike put, and selling one unit of K2-strike put is also called the butterflyspread.

Problem 9.11.

This is similar to Problem 9.10.We are given the following 3 calls:Strike K1 = 80 K2 = 100 K3 = 105Call premium 22 9 5

80λ+ 105 (1− λ) = 100→ λ = 0.2 0.2 (80) + 0.8 (105) = 100C [0.2 (80) + 0.8 (105)] = C (100) = 9

0.2C (80) + 0.8C (105) = 0.2 (22) + 0.8 (5) = 8. 4C [0.2 (80) + 0.8 (105)] > 0.2C (80) + 0.8C (105)So arbitrage opportunities exist. To arbitrage, we buy low and sell high.The cheap asset is the diversified portfolio consisting of λ units of K1-strike

option and (1− λ) units of K3-strike option. In this problem, the diversifiedportfolio consists of 0.2 unit of 80-strike call and 0.8 unit of 105-strike call.The expensive asset is the 100-strike call.



Since we can’t buy a fraction of a call option, we’ll buy 10 units of the port-folio (i.e. buy two 80-strike calls and eight 105-strike calls). Simultaneously,wesell ten 100-strike call options.We use T to represent the common exercise date. This definition works

whether the options are American or European. If the options are American,we’ll find arbitrage opportunities if the American options are exercised simulta-neously. If the options are European, T is the common expiration date.The payoff is:

T TTransaction t = 0 ST < 80 80 ≤ ST < 100buy ten portfoliosbuy two 80-strike calls −2 (22) 0 2 (ST − 80)buy eight 105-strike calls −8 (5) 0 0Portfolio total −84 0 2 (ST − 80)

Sell ten 100-strike calls 10 (9) 0 0Total 6 0 2 (ST − 80) ≥ 0

T TTransaction t = 0 100 ≤ ST < 105 ST ≥ 105buy ten portfoliosbuy two 80-strike calls −2 (22) 2 (ST − 80) 2 (ST − 80)buy eight 105-strike calls −8 (5) 0 8 (ST − 105)Portfolio total −84 2 (ST − 80) 10ST − 1000

Sell ten 100-strike calls 10 (9) −10 (ST − 100) −10 (ST − 100)Total 6 8 (105− ST ) > 0 0−2 (22)− 8 (5) = −44− 40 = −84

−84 + 10 (9) = −84 + 90 = 62 (ST − 80) + 8 (ST − 105) = 10ST − 10002 (ST − 80)− 10 (ST − 100) = 840− 8ST = 8 (105− ST )10ST − 1000− 10 (ST − 100) = 0So we receive $6 at t = 0, yet we don’t incur any negative cash flows at

expiration T . So we make at least $6 free money.

We are given the following 3 put:Strike K1 = 80 K2 = 100 K3 = 105Put premium 4 21 24.8

80λ+ 105 (1− λ) = 100→ λ = 0.2 0.2 (80) + 0.8 (105) = 100P [0.2 (80) + 0.8 (105)] = P (100) = 21

0.2P (80) + 0.8P (105) = 0.2 (4) + 0.8 (24.8) = 20. 64



P [0.2 (80) + 0.8 (105)] > 0.2P (80) + 0.8P (105)So arbitrage opportunities exist. To arbitrage, we buy low and sell high.The cheap asset is the diversified portfolio consisting of λ units of K1-strike

option and (1− λ) units of K3-strike option. In this problem, the diversifiedportfolio consists of 0.2 unit of 80-strike put and 0.8 unit of 105-strike put.The expensive asset is the 100-strike put.Since we can’t buy half a fraction of an option, we’ll buy 10 units of the port-

folio (i.e. buy two 80-strike puts and eight 105-strike puts). Simultaneously,wesell ten 100-strike put options.The payoff is:

T TTransaction t = 0 ST < 80 80 ≤ ST < 100buy ten portfoliosbuy two 80-strike puts −2 (4) 2 (80− ST ) 0buy eight 105-strike puts −8 (24.8) 8 (105− ST ) 8 (105− ST )Portfolio total −84 1000− 10ST 8 (105− ST )

Sell ten 100-strike puts 10 (21) −10 (100− ST ) −10 (100− ST )Total 3. 6 0 2 (ST − 80) ≥ 0

T TTransaction t = 0 100 ≤ ST < 105 ST ≥ 105buy ten portfoliosbuy two 80-strike puts −2 (4) 0 0buy eight 105-strike puts −8 (24.8) 8 (105− ST ) 0Portfolio total −84 8 (105− ST ) 0

Sell ten 100-strike puts 10 (21) 0 0Total 3. 6 8 (105− ST ) > 0 0

−2 (4)− 8 (24.8) = −206. 42 (80− ST ) + 8 (105− ST ) = 1000− 10ST−206. 4 + 10 (21) = 3. 61000− 10ST − 10 (100− ST ) = 08 (105− ST )− 10 (100− ST ) = 2 (ST − 80)

We receive $3. 6 at t = 0, but we don’t incur any negative cash flows at T .So we make at least $3. 6 free money.

Problem 9.12.

For two European options differing only in strike price, the following condi-tions must be met to avoid arbitrage (see my study guide for explanation):0 ≤ CEur (K1, T )− CEur (K2, T ) ≤ PV (K2 −K1) if K1 < K2



0 ≤ PEur (K2, T )− PEur (K1, T ) ≤ PV (K2 −K1) if K1 < K2

a.Strike K1 = 90 K2 = 95Call premium 10 4

C (K1)− C (K2) = 10− 4 = 6K2 −K1 = 95− 90 = 5C (K1)− C (K2) > K2 −K1 ≥ PV (K2 −K1)Arbitrage opportunities exist.To arbitrage, we buy low and sell high. The cheap call is the 95-strike call;

the expensive call is the 90-strike call.We use T to represent the common exercise date. This definition works

whether the two options are American or European. If the two options areAmerican, we’ll find arbitrage opportunities if two American options are ex-ercised simultaneously. If the two options are European, T is the commonexpiration date.The payoff is:

T T TTransaction t = 0 ST < 90 90 ≤ ST < 95 ST ≥ 95Buy 95 strike call −4 0 0 ST − 95Sell 90 strike call 10 0 − (ST − 90) − (ST − 90)Total 6 0 − (ST − 90) ≥ −5 −5

We receive $6 at t = 0, yet we our max liability at T is −5. So we’ll makeat least $1 free money.

b.T = 2 r = 0.1Strike K1 = 90 K2 = 95Call premium 10 5.25

C (K1)− C (K2) = 10− 5.25 = 4. 75K2 −K1 = 95− 90 = 5PV (K2 −K1) = 5e

−0.1(2) = 4. 094C (K1)− C (K2) > PV (K2 −K1)Arbitrage opportunities exist.

Once again, we buy low and sell high. The cheap call is the 95-strike call;the expensive call is the 90-strike call.The payoff is:

T T TTransaction t = 0 ST < 90 90 ≤ ST < 95 ST ≥ 95Buy 95 strike call −5.25 0 0 ST − 95Sell 90 strike call 10 0 − (ST − 90) − (ST − 90)Deposit 4. 75 in savings −4. 75 4. 75e0.1(2) 4. 75e0.1(2) 4. 75e0.1(2)

Total 0 5. 80 95. 80− ST > 0 0.80



4. 75e0.1(2) = 5. 80− (ST − 90) + 4. 75e0.1(2) = 95. 80− STST − 95− (ST − 90) + 4. 75e0.1(2) = 0.80Our initial cost is zero. However, our payoff is always non-negative. So we

never lose money. This is clearly an arbitrage.It’s important that the two calls are European options. If they are American,

they can be exercised at different dates. Hence the following non-arbitrageconditions work only for European options:0 ≤ CEur (K1, T )− CEur (K2, T ) ≤ PV (K2 −K1) if K1 < K2

0 ≤ PEur (K2, T )− PEur (K1, T ) ≤ PV (K2 −K1) if K1 < K2

c.We are given the following 3 calls:Strike K1 = 90 K2 = 100 K3 = 105Call premium 15 10 6

λ90 + (1− λ) 105 = 100 λ =1

3

→ 1

3(90) +

2

3(105) = 100

C

∙1

3(90) +

2

3(105)

¸= C (100) = 10

1

3C (90) +

2

3C (105) =

1

3(15) +

2

3(6) = 9

C

∙1

3(90) +

2

3(105)

¸>1

3C (90) +

2

3C (105)

Hence arbitrage opportunities exist. To arbitrage, we buy low and sell high.

The cheap asset is the diversified portfolio consisting of1

3unit of 90-strike

call and2

3unit of 105-strike call.

The expensive asset is the 100-strike call.Since we can’t buy a partial option, we’ll buy 3 units of the portfolio (i.e.

buy one 90-strike call and two 105-strike calls). Simultaneously,we sell three100-strike calls.

The payoff at expiration T :

T T T TTransaction t = 0 ST < 90 90 ≤ ST < 100 100 ≤ ST < 105 ST ≥ 105buy 3 portfoliosbuy one 90-strike call −15 0 ST − 90 ST − 90 ST − 90buy two 105-strike calls 2 (−6) 0 0 0 2 (ST − 105)Portfolios total −27 0 ST − 90 ST − 90 3ST − 300

Sell three 100-strike calls 3 (10) 0 0 −3 (ST − 100) −3 (ST − 100)Total 3 0 ST − 90 ≥ 0 2 (105− ST ) > 0 0



−15 + 2 (−6) = −27ST − 90 + 2 (ST − 105) = 3ST − 300−27 + 3 (10) = 3ST − 90− 3 (ST − 100) = 210− 2ST = 2 (105− ST )3ST − 300− 3 (ST − 100) = 0

So we receive $3 at t = 0, but we incur no negative payoff at T . So we’llmake at least $3 free money.

Problem 9.13.

a. If the stock pays dividend, then early exercise of an American call optionmay be optimal.Suppose the stock pays dividend at tD.Time 0 ... ... tD ... ... T

Pro and con for exercising the call early at tD.

• +. If you exercise the call immediately before tD, you’ll receive dividendand earn interest during [tD, T ]

• −. You’ll pay the strike price K at tD, losing interest you could haveearned during [tD, T ]. If the interest rate, however, is zero, you won’t loseany interest.

• −. You throw away the remaining call option during [tD, T ]. Had youwaited, you would have the call option during [tD, T ]

If the accumulated value of the dividend exceeds the value of the remainingcall option, then it’s optimal to exercise the stock at tD.As explained in my study guide, it’s never optimal to exercise an American

put early if the interest rate is zero.

Problem 9.14.

a. The only reason that early exercise might be optimal is that the underlyingasset pays a dividend. If the underlying asset doesn’t pay dividend, then it’snever optimal to exercise an American call early. Since Apple doesn’t paydividend, it’s never optimal to exercise early.

b. The only reason to exercise an American put early is to earn interest onthe strike price. The strike price in this example is one share of AOL stock.Since AOL stocks won’t pay any dividends, there’s no benefit for owning anAOL stock early. Thus it’s never optimal to exercise the put.If the Apple stock price goes to zero and will always stay zero, then there’s

no benefit for delaying exercising the put; there’s no benefit for exercising the



put early either (since AOL stocks won’t pay dividend). Exercising the putearly and exercising the put at maturity have the same value.If, however, the Apple stock price goes to zero now but may go up in the

future, then it’s never optimal to exercise the put early. If you don’t exerciseearly, you leave the door open that in the future the Apple stock price mayexceed the AOL stock price, in which case you just let your put expire worthless.If the Apple stock price won’t exceed the AOL stock price, you can alwaysexercise the put and exchange one Apple stock for one AOL stock. There’s nohurry to exercise the put early.

c. If Apple is expected to pay dividend, then it might be optimal to exercisethe American call early and exchange one AOL stock for one Apple stock.However, as long as the AOL stock won’t pay any dividend, it’s never optimal

to exercise the American put early to exchange one Apple stock for one AOLstock.

Problem 9.15.

This is an example where the strike price grows over time.

If the strike price grows over time, the longer-lived option is at least asvaluable as the shorter lived option. Refer to Derivatives Markets Page 298.We have two European calls:Call #1 K1 = 100e

0.05(1.5) = 107. 788 T1 = 1.5 C1 = 11.50Call #2 K2 = 100e

0.05 = 105. 127 T2 = 1 C2 = 11.924

The longer-lived call is cheaper than the shorter-lived call, leading to arbi-trage opportunities. To arbitrage, we buy low (Call #1) and sell high (Call#2).The payoff at expiration T1 = 1.5 if ST2 < 100e

0.05 = 105. 127

T1 T1Transaction t = 0 T2 ST1 < 100e

0.05(1.5) ST1 ≥ 100e0.05(1.5)Sell Call #2 11.924 0 0 0

buy Call #1 −11.50 0 ST1 − 100e0.05(1.5)Total 0.424 0 ST1 − 100e0.05(1.5) ≥ 0

We receive $0.424 at t = 0, yet our payoff at T1 is always non-negative. Thisis clearly an arbitrage.

The payoff at expiration T1 = 1.5 if ST2 ≥ 100e0.05 = 105. 127

T1 T1Transaction t = 0 T2 ST1 < 100e

0.05(1.5) ST1 ≥ 100e0.05(1.5)Sell Call #2 11.924 100e0.05 − ST2 100e0.05(1.5) − ST1 100e0.05(1.5) − ST1buy Call #1 −11.50 0 ST1 − 100e0.05(1.5)Total 0.424 100e0.05(1.5) − ST1 < 0 0



If ST2 ≥ 100e0.05, then payoff of the sold Call #2 at T2 is 100e0.05 − ST2 .From T2 to T1,

• 100e0.05 grows into¡100e0.05

¢e0.05(T1−T2) =

¡100e0.05

¢e0.05(0.5) = 100e0.05(1.5)

• ST2 becomes ST1 (i.e. the stock price changes from ST2 to ST1)

We receive $0.424 at t = 0, yet our payoff at T1 can be negative. This is notan arbitrage.

So as long as ST2 < 100e0.05 = 105. 127 , there’ll be arbitrage opportunities.

Problem 9.16.

Suppose we do the following at t = 0:

1. Pay Ca to buy a call

2. Lend PV (K) = Ke−rL at rL

3. Sell a put, receiving P b

4. Short sell one stock, receiving Sb0

The net cost is P b + Sb0 − (Ca +Ke−rL).The payoff at T is:

If ST < K If ST ≥ KTransactions t = 0Buy a call −Ca 0 ST −KLend Ke−rL at rL −KerL K KSell a put P b ST −K 0Short sell one stock Sb0 −ST −STTotal P b + Sb0 − (Ca +Ke−rL) 0 0

The payoff is always zero. To avoid arbitrage, we need to haveP b + Sb0 − (Ca +Ke−rL) ≤ 0

Similarly, we can do the following at t = 0:

1. Sell a call, receiving Cb

2. Borrow PV (K) = Ke−rB at rB

3. Buy a put, paying P a

4. Buy one stock, paying Sa0



The net cost is¡Cb +Ke−rB

¢−¡P b + Sb0

¢.

The payoff at T is:If ST < K If ST ≥ K

Transactions t = 0Sell a call Cb 0 K − STBorrow Ke−rB at rB Ke−rB −K −KBuy a put −P a K− ST 0Buy one stock −Sa0 ST STTotal

¡Cb +Ke−rB

¢−¡P b + Sb0

¢0 0

The payoff is always zero. To avoid arbitrage, we need to have¡Cb +Ke−rB

¢−¡P b + Sb0

¢≤ 0

Problem 9.17.

a. According to the put-call parity, the payoff of the following position isalways zero:

1. Buy the call

2. Sell the put

3. Short the stock

4. Lend the present value of the strike price plus dividend

The existence of the bid-ask spread and the borrowing-lending rate differencedoesn’t change the zero payoff of the above position. The above position alwayshas a zero payoff whether there’s a bid-ask spread or a difference between theborrowing rate and the lending rate.If there is no transaction cost such as a bid-ask spread, the initial gain of

the above position is zero. However, if there is a bid-ask spread, then to avoidarbitrage, the initial gain of the above position should be zero or negative.The initial gain of the position is:¡P b + Sb0

¢− [Ca + PVrL (K) + PVrL (Div)]

There’s no arbitrage if¡P b + Sb0

¢− [Ca + PVrL (K) + PVrL (Div)] ≤ 0

In this problem, we are given

• rL = 0.019

• rB = 0.02

• Sb0 = 84.85. We are told to ignore the transaction cost. In addition, weare given that the current stock price is 84.85. So Sb0 = 84.85.

• The dividend is 0.18 on November 8, 2004.



To find the expiration date, you need to know this detail. Puts and callsare called equity options at the Chicago Board of Exchange (CBOE). In CBOE,the expiration date of an equity option is the Saturday immediately followingthe third Friday of the expiration month. (To verify this, go to www.cboe.com.Click on "Products" and read "Production Specifications.")If the expiration month is November, 2004, the third Friday is November 19.

Then the expiration date is November 20.

T =11/20/2004− 10/15/2004

365=36

365= 0.09863

If the expiration month is January, 2005, the third Friday is January 21.Then the expiration date is January 22, 2005.

T =1/22/2005− 10/15/2004

365

Calculate the days between 1/22/2005 and10/15/2004 isn’t easy. Fortu-nately,we can use a calculator. BA II Plus and BA II Plus Professional have"Date" Worksheet. When using Date Worksheet, use the ACT mode. ACTmode calculates the actual days between two dates. If you use the 360 daymode, you are assuming that there are 360 days between two dates.When using the date worksheet, set DT1 (i.e. Date 1) as October 10, 2004

by entering 10.1504; set DT2 (i.e. Date 2) as January 22, 2004 by entering1.2204. The calculator should tell you that DBD=99 (i.e. the days between twodays is 99 days).

So T =1/22/2005− 10/15/2004

365=99

365= 0.271 23

If you have trouble using the date worksheet, refer to the guidebook of BAII Plus or BA II Plus Professional.

The dividend day is tD=11/8/2004− 10/15/2004

365=24

365= 0.06 575

PVrL (Div) = 0.18e−0.06575(0.019) = 0.18PVrL (K) = Ke−0.019T¡P b + Sb0


= P b + 84.85−¡Ca +Ke−0.019T + 0.18

¢K T Ca P b

¡P b + Sb0


75 0.0986 10.3 0.2 0.2 + 84.85−¡10.3 + 75e−0.019×0.0986 + 0.18

¢= −0.29

80 0.0986 5.6 0.6 0.6 + 84.85−¡5.6 + 80e−0.019×0.0986 + 0.18

¢= −0.18

85 0.0986 2.1 2.1 2.1 + 84.85−¡2.1 + 85e−0.019×0.0986 + 0.18

¢= −0.17

90 0.0986 0.35 5.5 5.5 + 84.85−¡0.35 + 90e−0.019×0.0986 + 0.18

¢= −1. 15

75 0.271 2 10.9 0.7 0.7 + 84.85−¡10.9 + 75e−0.019×0.271 2 + 0.18

¢= −0.14

80 0.271 2 6.7 1.45 1.45 + 84.85−¡6.7 + 80e−0.019×0.271 2 + 0.18

¢= −0.17

85 0.271 2 3.4 3.1 3.1 + 84.85−¡3.4 + 85e−0.019×0.271 2 + 0.18

¢= −0.19

90 0.271 2 1.35 6.1 6.1 + 84.85−¡1.35 + 90e−0.019×0.271 2 + 0.18

¢= −0.12

b. According to the put-call parity, the payoff of the following position isalways zero:



1. Sell the call

2. Borrow the present value of the strike price plus dividend

3. Buy the put

4. Buy one stock

If there is transaction cost such as the bid-ask spread, then to avoid arbitrage,the initial gain of the above position is zero. However, if there is a bid-ask spread,the initial gain of the above position can be zero or negative.The initial gain of the position is:Cb + PVrB (K) + PVrB (Div)− (P a + Sa0 )There’s no arbitrage ifCb + PVrB (K) + PVrB (Div)− (P a + Sa0 ) ≤ 0PVrB (Div) = 0.18e−0.06575(0.02) = 0.18PVrL (K) = Ke−0.02T

Cb+PVrB (K)+PVrB (Div)−(P a + Sa0 ) = Cb+Ke−0.02T+0.18−(P a + 84.85)

K T Cb P a Cb + PVrB (K) + PVrB (Div)− (P a + Sa0 )75 0.0986 9.9 0.25 9.9 + 75e−0.02×0.0986 + 0.18− (0.25 + 84.85) = −0.1780 0.0986 5.3 0.7 5.3 + 80e−0.02×0.0986 + 0.18− (0.7 + 84.85) = −0.2385 0.0986 1.9 2.3 1.9 + 85e−0.02×0.0986 + 0.18− (2.3 + 84.85) = −0.2490 0.0986 0.35 5.8 0.35 + 90e−0.02×0.0986 + 0.18− (5.8 + 84.85) = −0.3075 0.271 2 10.5 0.8 10.5 + 75e−0.02×0.271 2 + 0.18− (0.8 + 84.85) = −0.3880 0.271 2 6.5 1.6 6.5 + 80e−0.02×0.271 2 + 0.18− (1.6 + 84.85) = −0.2085 0.271 2 3.2 3.3 3.2 + 85e−0.02×0.271 2 + 0.18− (3.3 + 84.85) = −0.2390 0.271 2 1.2 6.3 1.2 + 90e−0.02×0.271 2 + 0.18− (6.3 + 84.85) = −0.26

Problem 9.18.

Suppose there are 3 options otherwise identical but with different strike priceK1 < K2 < K3 where K2 = λK1 + (1− λ)K2 and 0 < λ < 1.Then the price of the middle strike price K2 must not exceed the price of a

diversified portfolio consisting of λ units of K1-strike option and (1− λ) unitsof K2-strike option:

C [λK1 + (1− λ)K3] ≤ λC (K1) + (1− λ)C (K3)P [λK1 + (1− λ)K3] ≤ λP (K1) + (1− λ)P (K3)

The above conditions are called the convexity of the option price with respectto the strike price. They are equivalent to the textbook Equation 9.17 and 9.18.If the above conditions are violated, arbitrage opportunities exist.

K T Cb Ca

80 0.271 2 6.5 6.785 0.271 2 3.2 3.490 0.271 2 1.2 1.35



85 = λ (80) + (1− λ) (90) → λ = 0.5a.If we buy a 80-strike call, buy a 90-strike call, sell two 85-strike calls

• A 80-strike call and a 90-strike call form a diversified portfolio of calls,which is always as good as two 85-strike calls

• So the cost of buying a 80-strike call and a 90-strike call can never be lessthan the revenue of selling two 85-strike calls

What we pay if we buy a 80-strike call and a 90-strike call: 6.7 + 1.35 = 8.05What we get if we sell two 85-strike calls: 3.2× 2 = 6. 48. 05 > 6. 4So the convexity condition is met.I recommend that you don’t bother memorizing textbook Equation 9.17 and

9.18.

b. If we sell a 80-strike call, sell a 90-strike call, buy two 85-strike calls

• A 80-strike call and a 90-strike call form a diversified portfolio of calls,which is always as good as two 85-strike calls

• So the revenue of selling a 80-strike call and a 90-strike call should neverbe less than the cost of buying two 85-strike calls.

What we get if we sell a 80-strike call and a 90-strike call: 6.5 + 1.2 = 7. 7What we pay if we buy two 85-strike calls: 3.4× 2 = 6. 87. 7 > 6.8So the convexity condition is met.

c. To avoid arbitrage, the following two conditions must be met:C [λK1 + (1− λ)K3] ≤ λC (K1) + (1− λ)C (K3)P [λK1 + (1− λ)K3] ≤ λP (K1) + (1− λ)P (K3)These conditions must be met no matter you are a market-maker or anyone

else buying or selling options, no matter you pay a bid-ask spread or not.




Chapter 10

Binomial option pricing I

Problem 10.1.

The stock price today is S = 100. The stock at T is either

• Su = uS = 1.3× 100 = 130

• Sd = dS = 0.8× 100 = 80

a. For a call, the payoff at T is

• Vu = max (0, Su −K) = max (0, 130− 105) = 25

• Vd = max (0, Sd −K) = max (0, 80− 105) = 0

We hold a replicating portfolio (4, B) at t = 0. This portfolio will havevalue Vu if the stock goes up to Su or Vd if the stock goes down to Sd. We setup the following equations:½

4Su +BerT = Vu4Sd +BerT = Vd

→½4130 +Be0.08(0.5) = 25480 +Be0.08(0.5) = 0

→ B = −38. 431 6 4 = 0.5So the option premium is:V = 4S +B = 0.5× 100 + (−38. 431 6) = 11. 568 4

b. For a put, the payoff at T is either

• Vu = max (0,K − Su) = max (0, 105− 130) = 0

• Vd = max (0, Sd −K) = max (0, 105− 80) = 25

23

CHAPTER 10. BINOMIAL OPTION PRICING I



½4130 +Be0.08(0.5) = 0480 +Be0.08(0.5) = 25

B = 62. 451 3 4 = −0.5So the option premium is:V = 4S +B = −0.5× 100 + 62. 451 3 = 12. 451 3

Problem 10.2.

The stock price today is S = 100.The stock at T

• Su = uS = 1.3× 100 = 130

• Sd = dS = 0.8× 100 = 80

a. Payoff at T is either

• Vu = max (0, Su −K) = max (0, 130− 95) = 35

• Vd = max (0, Sd −K) = max (0, 80− 95) = 0



½4130 +Be0.08(0.5) = 35480 +Be0.08(0.5) = 0

B = −53. 804 2 4 = 0.7

So the option premium is:V = 4S +B = 0.7× 100 + (−53. 804 2) = 16. 195 8

b. There are two calls out there. One can be synthetically built by buying0.7 share of a stock and borrowing $53. 804 2 from a bank. The other is in themarket selling for $17. To arbitrage, we buy low and sell high.

• Buy low. Buy 0.7 share of a stock and borrow $53. 804 2 from a bank. Ourinitial cash outgo is 0.7 × 100 + (−53. 804 2) = 16. 195 8. This grows toeither (0.7) 130−53. 804 2e0.08(0.5) = 35 if the stock goes up or (0.7) 80−53.804 2e0.08(0.5) = 0 if the stock goes down at T .

• Sell high. We sell a call for 17, receiving $17 at time zero. We pay eithermax (0, 130− 95) = 35 if the stock goes up or max (0, 80− 95) = 0 if thestock goes down at T .



The net cash flow at T is zero. The net cash inflow at time zero is 17 −16. 195 8 = 0.804 2. So we receive $0.804 2 at time zero without incurring anyliability at T ; we have made $0.804 2 free money. Millions of investors will copythis arbitraging strategy, which will bid down the call price from 17 to the fairprice of 16. 195 8.

c. There are two calls out there. One can be synthetically built by buying0.7 share of a stock and borrowing $16. 195 8 from a bank. The other is in themarket selling for $15.5. To arbitrage, we buy low and sell high.

• Sell high. Short sell 0.7 share of a stock and deposit $53. 804 2 in a bank.Our initial cash inflow is 0.7 × 100 + (−53. 804 2) = 16. 195 8. At T , weneed to buy back 0.7 share of a stock from the market to close our shortposition. In addition, our bank account grows into 53. 804 2e0.08(0.5). Soour cash inflow at T is either (−0.7) 130 + 53. 804 2e0.08(0.5) = −35 if thestock goes up or (−0.7) 80 + 53. 804 2e0.08(0.5) = 0 if the stock goes down.

• Buy low. We buy a call for 15.5 at time zero, paying $15.5. We get eithermax (0, 130− 95) = 35 if the stock goes up or max (0, 80− 95) = 0 if thestock goes down at T .

The net cash flow at T is zero. The initial cash inflow is 16. 195 8− 15.5 =0.695 8. So we receive $0.695 8 at time zero without incurring any liabilityat T ; we have made $0.695 8 free money. Millions of investors will copy thisarbitraging strategy, which will bid up the call price from 15.5 to the fair priceof 16. 195 8.

Problem 10.3.

The stock price today is S = 100. The stock at T is

• Su = uS = 1.3× 100 = 130

• Sd = dS = 0.8× 100 = 80

a. Payoff at T is

• Vu = max (0,K − Su) = max (0, 95− 130) = 0

• Vd = max (0,K − Sd) = max (0, 95− 80) = 15



½4130 +Be0.08(0.5) = 0480 +Be0.08(0.5) = 15

B = 37. 470 9 4 = −0.3



So the option premium is:V = 4S +B = −0.3× 100 + 37. 470 9 = 7. 470 9

b. There are two puts out there. One can be synthetically built by shortselling 0.3 share of a stock and depositing $37. 470 9 in a bank. The other is inthe market selling for $8. To arbitrage, we buy low and sell high.

• Buy low. Short sell 0.3 share of a stock and deposit $37. 470 9 in a bank.Our initial cash inflow is 0.3× 100− 37. 470 9 = −7. 470 9. At T , we needto buy back 0.3 share of a stock to close our short position. In addition,our initial deposit grows to 37. 470 9e0.08(0.5) = 39. If the stock goes up,our cash outgo at T is (−0.3) 130+37. 470 9e0.08(0.5) = 0; if the stock goesdown to 80 at T , our cash outgo is (−0.3) 80 + 37. 470 9e0.08(0.5) = 15.

• Sell high. We sell a put for 8, receiving $8 at time zero. At T , we payeither max (0, 95− 130) = 0 if the stock goes up or max (0, 95− 80) = 15if the stock goes down.

The net cash flow of "buy low, sell high" is zero at T . The net cash inflow attime zero is 8− 7. 470 9 = 0.529 1. So we receive $0.529 1 at time zero withoutincurring any liability at T ; we have made $0.529 1 free money. Millions ofinvestors will copy this arbitraging strategy, which will bid down the put pricefrom 8 to the fair price of 7. 470 9.

c. There are two puts out there. One can be synthetically built by shortselling 0.3 share of a stock and depositing $37. 470 9 in a bank. The other is inthe market selling for $6. To arbitrage, we buy low and sell high.

• Buy low. We buy a put for $6 at time zero. At T we receive eithermax (0, 95− 130) = 0 if the stock goes up or max (0, 95− 80) = 15 if thestock goes down.

• Sell high. At time zero we buy 0.3 share of a stock and borrow $37. 470 9in a bank. Our initial cash inflow is −0.3× 100 + 37. 470 9 = 7. 470 9. AtT , we sell 0.3 share of a stock, receiving 0.3× 130 = 39 if the stock goesup or receiving 0.3× 80 = 24 if the stock goes down. In addition, we needto pay the bank 37. 470 9e0.08(0.5) = 39. So our net cash inflow at T is39− 39 = 0 if the stock goes up or 24− 39 = −15 if the stock goes down.

The net cash flow of "buy low, sell high" is zero at T . The net cash inflow attime zero is 7. 470 9− 6 = 1. 470 9. So we receive $1. 470 9 at time zero withoutincurring any liability at T ; we have made $1. 470 9 free money. Millions ofinvestors will copy this arbitraging strategy, which will bid up the put pricefrom 6 to the fair price of 7. 470 9.

Problem 10.4.



The problem doesn’t say the option is European or American. Let’s assumethe option is European.

T = 1 n = 2 →period length h = T/2 = 0.5Period 0 1 2

Suu = 100¡1.32

¢= 169

Su = 100 (1.3) = 130 Vuu = 169− 95 = 74Vu =?(∆u, Bu)

Sud = 100 (1.3) (0.8) = 104S = 100 Vud = 104− 95 = 9V =?(∆, B) =?

Sd = 100 (0.8) = 80Vd =? Sdd = 100

¡0.82

¢= 64

(∆d, Bd) Vdd = 0

We start from right to left.Step 1 Calculate (∆u, Bu), the replicating portfolio at Node u which will

produce the payoff Vud and Vdd.½4uSuu +Bue

rh = Vuu4uSud +Bue

rh = Vud→½4u169 +Bue

0.08(0.5) = 744u104 +Bue

0.08(0.5) = 94u = 1 Bu = −91. 2750

The premium at Node u is:Vu = 4uSu +Bu = 1 (130)− 91. 2750 = 38. 725

Step 2 Calculate(∆d, Bd), the replicating portfolio at Node d which willproduce the payoff Vud and Vdd.½

4dSud +Bderh = Vud

4dSdd +Bderh = Vdd

→½4d104 +Bde

0.08(0.5) = 94d64 +Bde

0.08(0.5) = 04d = 0.225 , Bd = −13. 835 4

The premium at Node d is:Vd = 4dSd +Bd = 0.225 (80)− 13. 835 4 = 4. 164 6

Step 3 Calculate(∆, B), the replicating portfolio at time zero, which willproduce the payoff Vu and Vd.½


→½4130 +Be0.08(0.5) = 38. 725480 +Be0.08(0.5) = 4. 164 6

4 = 0.691 2 B = −49. 127 1

The premium at time zero is:V = 4S +B = 0.691 2 (100)− 49. 127 1 = 19. 992 9

The final diagram is:



Period 0 1 2Suu = 100

¡1.32

¢= 169

Su = 100 (1.3) = 130 Vuu = 169− 95 = 74Vu = 38. 725∆u = 1Bu = −$91. 2750

S = 100 Sud = 100 (1.3) (0.8) = 104V = 19. 992 9 Vud = 104− 95 = 9∆ = 0.691 2B = −$49. 127 1

Sd = 100 (0.8) = 80Vd = 4. 164 6 Sdd = 100

¡0.82

¢= 64

∆d = 0.225 Vdd = 0Bd = −$13. 835 4

Problem 10.5.

This question asks us to redo the previous problem by setting the initialstock price to 80, 90, 110, 120, and 130.

• S = 80.


¡1.32

¢= 135. 2

Su = 80 (1.3) = 104 Vuu = 135. 2− 95 = 40. 2Vu =?(∆u, Bu) =?

Sud = 80 (1.3) (0.8) = 83. 2S = 80 Vud = 0V =?(∆, B) =?

Sd = 80 (0.8) = 64Vd =? Sdd = 80

¡0.82

¢= 51. 2

(∆d, Bd) =? Vdd = 0½4u135. 2 +Bue

0.08(0.5) = 40. 24u83. 2 +Bue

0.08(0.5) = 04u = 0.773 08 Bu = −61. 7980

Vu = 4uSu +Bu = 0.773 08 (104)− 61. 7980 = 18. 602 32½4d83. 2 +Bde

0.08(0.5) = 04d51. 2 +Bde

0.08(0.5) = 04d = 0.0 Bd = 0

Vd = 4dSd +Bd = 0 (64)− 0 = 0½4104 +Be0.08(0.5) = 18. 602 32464 +Be0.08(0.5) = 0

4 = 0.465 058 B = −28. 596 7



V = 4S +B = 0.465 058 (80)− 28. 596 7 = 8. 607 94

The final diagram is:Period 0 1 2

Suu = 80¡1.32

¢= 135. 2

Su = 80 (1.3) = 104 Vuu = 135. 2− 95 = 40. 2Vu = 18. 602 32∆u = 0.773 08Bu = −61. 798

Sud = 80 (1.3) (0.8) = 83. 2S = 80 Vud = 0V = 8. 607 94∆ = 0.465 058B = −28. 596 7

Sd = 80 (0.8) = 64Vd = 0 Sdd = 80

¡0.82

¢= 51. 2

∆d = 0 Vdd = 0Bd = 0



• S = 90.


¡1.32

¢= 152. 1

Su = 90 (1.3) = 117 Vuu = 152. 1− 95 = 57. 1Vu =?4u =?Bu =?

S = 90 Sud = 90 (1.3) (0.8) = 93. 6V =? Vud = 0∆ =?B =?

Sd = 90 (0.8) = 72Vd =? Sdd = 90

¡0.82

¢= 57. 6

∆d =? Vdd = 0Bd =?½

4u152. 1 +Bue0.08(0.5) = 57. 1

4u93. 6 +Bue0.08(0.5) = 0

4u = 0.976 068 Bu = −87. 777 7

Vu = 4uSu +Bu = 0.976 068 (117)− 87. 777 7 = 26. 422 26½4d93. 6 +Bde

0.08(0.5) = 04d57. 6 +Bde

0.08(0.5) = 04d = 0.0 Bd = 0

Vd = 4dSd +Bd = 0 (72)− 0 = 0½4117 +Be0.08(0.5) = 26. 422 26472 +Be0.08(0.5) = 0

4 = 0.587 161 B = −40. 617 97

V = 4S +B = 0.587 161 (90)− 40. 617 97 = 12. 226 52


Suu = 90¡1.32

¢= 152. 1

Su = 90 (1.3) = 117 Vuu = 152. 1− 95 = 57. 1Vu = 26. 422 264u = 0.976 068Bu = −87. 777 7

S = 90 Sud = 90 (1.3) (0.8) = 93. 6V = 12. 226 52 Vud = 04 = 0.587 161B = −40. 617 97

Sd = 90 (0.8) = 72Vd = 0 Sdd = 90

¡0.82

¢= 57. 6

∆d = 0 Vdd = 0Bd = 0 0



• S = 110.


¡1.32

¢= 185. 9

Su = 110 (1.3) = 143 Vuu = 185. 9− 95 = 90. 9Vu =?4u =?Bu =?

S = 110 Sud = 110 (1.3) (0.8) = 114. 4V =? Vud = 114. 4− 95 = 19. 4∆ =?B =?

Sd = 110 (0.8) = 88Vd =? Sdd = 110

¡0.82

¢= 70. 4

∆d =? Vdd = 0Bd =?½

4u185. 9 +Bue0.08(0.5) = 90. 9

4u114. 4 +Bue0.08(0.5) = 19. 4

4u = 1 Bu = −91. 2750

Vu = 4uSu +Bu = 1 (143)− 91. 2750 = 51. 725½4d114. 4 +Bde

0.08(0.5) = 19. 44d70. 4 +Bde

0.08(0.5) = 04d = 0.440 909 Bd = −29. 822 9

Vd = 4dSd +Bd = 0.440 909 (88)− 29. 822 9 = 8. 977 092½4143 +Be0.08(0.5) = 51. 725488 +Be0.08(0.5) = 8. 977 092

4 = 0.777 235 B = −57. 089 7

V = 4S +B = 0.777 235 (110)− 57. 089 7 = 28. 406 15


Suu = 110¡1.32

¢= 185. 9

Su = 110 (1.3) = 143 Vuu = 185. 9− 95 = 90. 9Vu = 51. 7254u = 1Bu = −91. 2750

S = 110 Sud = 110 (1.3) (0.8) = 114. 428. 406 15 Vud = 114. 4− 95 = 19. 44 = 0.777 235B = −57. 089 7

Sd = 110 (0.8) = 88Vd = 8. 977 092 Sdd = 110

¡0.82

¢= 70. 4

∆d = 0.440 909 Vdd = 0Bd = −29. 822 9



• S = 120.


¡1.32

¢= 202. 8

Su = 120 (1.3) = 156 Vuu = 202. 8− 95 = 107. 8Vu =?4u =?Bu =?

S = 120 Sud = 120 (1.3) (0.8) = 124. 8V =? Vud = 124. 8− 95 = 29. 8∆ =?B =?

Sd = 120 (0.8) = 96Vd =? Sdd = 120

¡0.82

¢= 76. 8

∆d =? Vdd = 0Bd =?½

4u202. 8 +Bue0.08(0.5) = 107. 8

4u124. 8 +Bue0.08(0.5) = 29. 8

4u = 1 Bu = −91. 275

Vu = 4uSu +Bu = 1 (156)− 91. 275 = 64. 725½4d124. 8 +Bde

0.08(0.5) = 29. 84d76. 8 +Bde

0.08(0.5) = 04d = 0.620 833 Bd = −45. 810 44

Vd = 4dSd +Bd = 0.620 833 (96)− 45. 810 44 = 13. 789 528½4156 +Be0.08(0.5) = 64. 725496 +Be0.08(0.5) = 13. 789 528

4 = 0.848 925 B = −65. 052 4

V = 4S +B = 0.848 925 (120)− 65. 052 4 = 36. 818 6


Suu = 120¡1.32

¢= 202. 8

Su = 120 (1.3) = 156 Vuu = 202. 8− 95 = 107. 8Vu = 64. 7254u = 1

Bu = −91. 275S = 120 Sud = 120 (1.3) (0.8) = 124. 8

V = 36. 818 6 Vud = 124. 8− 95 = 29. 84 = 0.848 925B = −65. 052 4

Sd = 120 (0.8) = 96Vd = 13. 789 528 Sdd = 120

¡0.82

¢= 76. 8

4d = 0.620 833 Vdd = 0Bd = −45. 810 44



• S = 130.


¡1.32

¢= 219. 7

Su = 130 (1.3) = 169 Vuu = 219. 7− 95 = 124. 7Vu =?4u =?Bu =?

S = 130 Sud = 130 (1.3) (0.8) = 135. 2V =? Vud = 135. 2− 95 = 40. 2∆ =?B =?

Sd = 130 (0.8) = 104Vd =? Sdd = 130

¡0.82

¢= 83. 2

∆d =? Vdd = 0Bd =?½

4u219. 7 +Bue0.08(0.5) = 124. 7

4u135. 2 +Bue0.08(0.5) = 40. 2

4u = 1 Bu = −91. 275

Vu = 4uSu +Bu = 1 (169)− 91. 275 = 77. 725½4d135. 2 +Bde

0.08(0.5) = 40. 24d83. 2 +Bde

0.08(0.5) = 04d = 0.773 077 Bd = −61. 7980

Vd = 4dSd +Bd = 0.773 077 (104)− 61. 7980 = 18. 602½4169 +Be0.08(0.5) = 77. 7254104 +Be0.08(0.5) = 18. 602

4 = 0.909 585 B = −73. 015

V = 4S +B = 0.909 585 (130)− 73. 015 = 45. 231The final diagram is:Period 0 1 2

Suu = 130¡1.32

¢= 219. 7

Su = 130 (1.3) = 169 Vuu = 219. 7− 95 = 124. 7Vu = 77. 7254u = 1Bu = −91. 275

S = 130 Sud = 130 (1.3) (0.8) = 135. 2V = 45. 231 Vud = 135. 2− 95 = 40. 24 = 0.909 585B = −73. 015

Sd = 130 (0.8) = 104Vd = 18. 602 Sdd = 130

¡0.82

¢= 83. 2

4d = 0.773 077 Vdd = 0Bd = −61. 7980

Notice that the delta for a call is always positive. A positive delta meansbuying stocks. If you sell a call, the risk you face is that the stock price may go



up to infinity, at which case the call holder will pay only the strike price to buya priceless stock from you. To hedge your risk, you need to already own somestocks at t = 0. If indeed the future stock price goes up at T , your stock willalso go up in value, offsetting your loss in the call.Intuitively, the higher the initial stock price, everything else equal, the higher

the stock price in the future, the higher the payoff of the call. Hence to hedgethe risk, the market maker needs to buy more shares of stocks at t = 0. So thehigher the initial stock price, the higher the initial delta, the more stocks themarket maker needs to buy at t = 0.



Problem 10.6.


¡1.32

¢= 169

Su = 100 (1.3) = 130 Vuu = 0Vu =?4u =?Bu =?

S = 100 Sud = 100 (1.3) (0.8) = 104V =? Vud = 0∆ =?B =?

Sd = 100 (0.8) = 80Vd =? Sdd = 100

¡0.82

¢= 64

∆d =? Vdd = 95− 64 = 31Bd =?½

4u169 +Bue0.08(0.5) = 0

4u104 +Bue0.08(0.5) = 0

4u = 0 Bu = 0

Vu = 4uSu +Bu = 0 (130) + 0 = 0½4d104 +Bde

0.08(0.5) = 04d64 +Bde

0.08(0.5) = 314d = −0.775 Bd = 77. 439 6

Vd = 4dSd +Bd = −0.775 (80) + 77. 439 6 = 15. 439 6½4130 +Be0.08(0.5) = 0480 +Be0.08(0.5) = 15. 439 6

4 = −0.308 792 B = 38. 568 932

V = 4S +B = −0.308 792 (100) + 38. 568 932 = 7. 689 732The final diagram is:Period 0 1 2

Suu = 100¡1.32

¢= 169

Su = 100 (1.3) = 130 Vuu = 0Vu = 04u = 0Bu = 0

S = 100 Sud = 100 (1.3) (0.8) = 104V = 7. 689 732 Vud = 04 = −0.308 792B = 38. 568 932

Sd = 100 (0.8) = 80Vd = 15. 439 6 Sdd = 100

¡0.82

¢= 64

4d = −0.775 Vdd = 95− 64 = 31Bd = 77. 439 6



Problem 10.7.

• S = 80


¡1.32

¢= 135. 2

Su = 80 (1.3) = 104 Vuu = 0Vu =?4u =?Bu =?

S = 80 Sud = 80 (1.3) (0.8) = 83. 2V =? Vud = 95− 83. 2 = 11. 8∆ =?B =?

Sd = 80 (0.8) = 64Vd =? Sdd = 80

¡0.82

¢= 51. 2

∆d =? Vdd = 95− 51. 2 = 43. 8Bd =?½

4u135. 2 +Bue0.08(0.5) = 0

4u83. 2 +Bue0.08(0.5) = 11. 8

4u = −0.226 923 Bu = 29. 477 02

Vu = 4uSu +Bu = −0.226 923 (104) + 29. 477 02 = 5. 877½4d83. 2 +Bde

0.08(0.5) = 11. 84d51. 2 +Bde

0.08(0.5) = 43. 84d = −1 Bd = 91. 275

Vd = 4dSd +Bd = −1 (64) + 91. 275 = 27. 275½4104 +Be0.08(0.5) = 5. 877464 +Be0.08(0.5) = 27. 275

4 = −0.534 95 B = 59. 1

V = 4S +B = −0.534 95 (80) + 59. 1 = 16. 304The final diagram is:Period 0 1 2

Suu = 80¡1.32

¢= 135. 2

Su = 80 (1.3) = 104 Vuu = 0Vu = 5. 8774u = −0.226 923Bu = 29. 477 02

S = 80 Sud = 80 (1.3) (0.8) = 83. 2V = 16. 304 Vud = 95− 83. 2 = 11. 84 = −0.534 95B = 59. 1

Sd = 80 (0.8) = 64Vd = 27. 275 Sdd = 80

¡0.82

¢= 51. 2

4d = −1 Vdd = 95− 51. 2 = 43. 8Bd = 91. 275



• S = 90


¡1.32

¢= 152. 1

Su = 90 (1.3) = 117 Vuu = 0Vu =?4u =?Bu =?

S = 90 Sud = 90 (1.3) (0.8) = 93. 6V =? Vud = 95− 93. 6 = 1. 4∆ =?B =?

Sd = 90 (0.8) = 72Vd =? Sdd = 90

¡0.82

¢= 57. 6

∆d =? Vdd = 95− 57. 6 = 37. 4Bd =?½

4u152. 1 +Bue0.08(0.5) = 0

4u93. 6 +Bue0.08(0.5) = 1. 4

4u = −0.02 393 Bu = 3. 497 27

Vu = 4uSu +Bu = −0.02 393 (117) + 3. 497 27 = 0.697 5½4d93. 6 +Bde

0.08(0.5) = 1. 44d57. 6 +Bde

0.08(0.5) = 37. 44d = −1 Bd = 91. 275

Vd = 4dSd +Bd = −1 (72) + 91. 275 = 19. 275½4117 +Be0.08(0.5) = 0.697 5472 +Be0.08(0.5) = 19. 275

4 = −0.412 833 B = 47. 077 722

V = 4S +B = −0.412 833 (90) + 47. 077 722 = 9. 923The final diagram is:Period 0 1 2

Suu = 90¡1.32

¢= 152. 1

Su = 90 (1.3) = 117 Vuu = 0Vu = 0.697 54u = −0.02 393Bu = 3. 497 27

S = 90 Sud = 90 (1.3) (0.8) = 93. 6V = 9. 923 Vud = 95− 93. 6 = 1. 44 = −0.412 833B = 47. 077 722

Sd = 90 (0.8) = 72Vd = 19. 275 Sdd = 90

¡0.82

¢= 57. 6

4d = −1 Vdd = 95− 57. 6 = 37. 4Bd = 91. 275



• S = 110


¡1.32

¢= 185. 9

Su = 110 (1.3) = 143 Vuu = 0Vu =?4u =?Bu =?

S = 110 Sud = 110 (1.3) (0.8) = 114. 4V =? Vud = 0∆ =?B =?

Sd = 110 (0.8) = 88Vd =? Sdd = 110

¡0.82

¢= 70. 4

∆d =? Vdd = 95− 70. 4 = 24. 6Bd =?½

4u185. 9 +Bue0.08(0.5) = 0

4u114. 4 +Bue0.08(0.5) = 0

4u = 0 Bu = 0

Vu = 4uSu +Bu = 0 (143) + 0 = 0½4d114. 4 +Bde

0.08(0.5) = 04d70. 4 +Bde

0.08(0.5) = 24. 64d = −0.559 091 Bd = 61. 452 1

Vd = 4dSd +Bd = −0.559 091 (88) + 61. 452 1 = 12. 252½4143 +Be0.08(0.5) = 0488 +Be0.08(0.5) = 12. 252

4 = −0.222 76 B = 30. 606 14

V = 4S +B = −0.222 76 (110) + 30. 606 14 = 6. 103The final diagram is:Period 0 1 2

Suu = 110¡1.32

¢= 185. 9

Su = 110 (1.3) = 143 Vuu = 0Vu = 04u = 0Bu = 0

S = 110 Sud = 110 (1.3) (0.8) = 114. 4V = 6. 103 Vud = 04 = −0.222 76B = 30. 606 14

Sd = 110 (0.8) = 88Vd = 12. 252 Sdd = 110

¡0.82

¢= 70. 4

4d = −0.559 091 Vdd = 95− 70. 4 = 24. 6Bd = 61. 452 1



• S = 120


¡1.32

¢= 202. 8

Su = 120 (1.3) = 156 Vuu = 0Vu =?4u =?Bu =?

S = 120 Sud = 120 (1.3) (0.8) = 124. 8V =? Vud = 0∆ =?B =?

Sd = 120 (0.8) = 96Vd =? Sdd = 120

¡0.82

¢= 76. 8

∆d =? Vdd = 95− 76. 8 = 18. 2Bd =?½

4u202. 8 +Bue0.08(0.5) = 0

4u124. 8 +Bue0.08(0.5) = 0

4u = 0 Bu = 0

Vu = 4uSu +Bu = 0 (156) + 0 = 0½4d124. 8 +Bde

0.08(0.5) = 04d76. 8 +Bde

0.08(0.5) = 18. 24d = −0.379 167 Bd = 45. 464 56

Vd = 4dSd +Bd = −0.379 167 (96) + 45. 464 56 = 9. 065½4156 +Be0.08(0.5) = 0496 +Be0.08(0.5) = 9. 065

4 = −0.151 083 B = 22. 644 85

V = 4S +B = −0.151 083 (120) + 22. 644 85 = 4. 514 89


Suu = 120¡1.32

¢= 202. 8

Su = 120 (1.3) = 156 Vuu = 0Vu = 04u = 0Bu = 0

S = 120 Sud = 120 (1.3) (0.8) = 124. 8V = 4. 514 89 Vud = 04 = −0.151 083B = 22. 644 85

Sd = 120 (0.8) = 96Vd = 9. 065 Sdd = 120

¡0.82

¢= 76. 8

4d = −0.379 167 Vdd = 95− 76. 8 = 18. 2Bd = 45. 464 56



• S = 130


¡1.32

¢= 219. 7

Su = 130 (1.3) = 169 Vuu = 0Vu =?4u =?Bu =?

S = 130 Sud = 130 (1.3) (0.8) = 135. 2V =? Vud = 0∆ =?B =?

Sd = 130 (0.8) = 104Vd =? Sdd = 130

¡0.82

¢= 83. 2

∆d =? Vdd = 95− 83. 2 = 11. 8Bd =?½

4u219. 7 +Bue0.08(0.5) = 0

4u135. 2 +Bue0.08(0.5) = 0

4u = 0 Bu = 0

Vu = 4uSu +Bu = 0 (169) + 0 = 0½4d135. 2 +Bde

0.08(0.5) = 04d83. 2 +Bde

0.08(0.5) = 11. 84d = −0.226 923 Bd = 29. 477 0

Vd = 4dSd +Bd = −0.226 923 (104) + 29. 477 0 = 5. 877½4169 +Be0.08(0.5) = 04104 +Be0.08(0.5) = 5. 877

4 = −0.09 041 5 B = 14. 681 05

V = 4S +B = −0.09 041 5 (130) + 14. 681 05 = 2. 927 1The final diagram is:Period 0 1 2

Suu = 130¡1.32

¢= 219. 7

Su = 130 (1.3) = 169 Vuu = 0Vu = 04u = 0Bu = 0

S = 130 Sud = 130 (1.3) (0.8) = 135. 2V = 2. 927 1 Vud = 04 = −0.09 041 5B = 14. 681 05

Sd = 130 (0.8) = 104Vd = 5. 877 Sdd = 130

¡0.82

¢= 83. 2

4d = −0.226 923 Vdd = 95− 83. 2 = 11. 8Bd = 29. 477 0



Notice that the delta for a put is always negative. A negative delta meansshorting sell stocks. If you sell a put, the risk you face is that the stock pricemay go down to zero, at which case the put holder sells you a worthless stock forthe strike price. To hedge your risk, you need to short sell some stocks at t = 0.If indeed the future stock price goes down at T , you can buy back stocks inthe market at low price to close your short position on stocks, earning a profit.Your profit from the short sale can offset your loss in the put.Intuitively, the higher the initial stock price, everything else equal, the higher

the stock price in the future, the lower the payoff of the put. Hence to hedgethe risk, the market maker needs to short sell fewer shares of stocks at t = 0. Sothe higher the initial stock price, the lower the absolute value of the delta for aput, the fewer of the stocks that the market maker needs to short sell initially.

Problem 10.8.


¡1.32

¢= 169

Su = 100 (1.3) = 130 Vuu = 0EVu = 0Vu =?4u =?Bu =?

S = 100 Sud = 100 (1.3) (0.8) = 104EV = 0 Vud = 0V =?∆ =?B =?

Sd = 100 (0.8) = 80EVd = 95− 80 = 15Vd =? Sdd = 100

¡0.82

¢= 64

∆d =? Vdd = 95− 64 = 31Bd =?½

4u169 +Bue0.08(0.5) = 0

4u104 +Bue0.08(0.5) = 0

4u = 0 Bu = 0

Vu = max (4uSu +Bu, EVu) = max [0 (130) + 0, 0] = 0

This is the logic behind the calculation of Vu. The American put can beexercised at the end of Period 1 and end of Period 2.

• If exercised at the end of Period 1, the put is worthEVu = max (0, 95− 130) =0 at the end of Period 1.

• If exercised at Period 2, the put is worth 4uSu +Bu = 0 (130) + 0 = 0 atthe end of Period 1.



We compare the two values and take the greater.½4d104 +Bde

0.08(0.5) = 04d64 +Bde

0.08(0.5) = 314d = −0.775 Bd = 77. 439 6

Vd = max (4dSd +Bd, EVd) = max [−0.775 (80) + 77. 439 6, 15] = 15. 439 6½4130 +Be0.08(0.5) = 0480 +Be0.08(0.5) = 15. 439 6

4 = −0.308 792 B = 38. 568 932

V = max (4S +B,EV ) = max [−0.308 792 (100) + 38. 568 932 , 0] = 7. 689 732


Suu = 100¡1.32

¢= 169

Su = 100 (1.3) = 130 Vuu = 0EVu = 0Vu = 04u = 0Bu = 0

S = 100 Sud = 100 (1.3) (0.8) = 104EV = 0 Vud = 0V = 7. 689 7∆ = −0.308 792B = 38. 568 932

Sd = 100 (0.8) = 80EVd = 95− 80 = 15Vd = 15. 439 6 Sdd = 100

¡0.82

¢= 64

4d = −0.775 Vdd = 95− 64 = 31Bd = 77. 439 6

Problem 10.9.

a.time 0 T

Su = 100 (1.2) = 120Vu = 120− 50 = 70

S = 100V =?

∆ =?, B =?Sd = 100 (1.05) = 105Vd = 105− 50 = 55

Here d = 1.05 > 1. Is there anything wrong with this? No necessarily. Thenon-arbitrage requirement is textbook Equation 10.4: u > e(r−δ)h > dLet’s check. e(r−δ)h = e(0.07696−0)1 = 1. 08u = 1.2 > e(r−δ)h > d = 1.05



So it’s OK to have d > 1 as long as u > e(r−δ)h > d is met.½4120 +Be0.07696(1) = 704105 +Be0.07696(1) = 55

4 = 1 B = −46. 296 3

V = 4S +B = 1 (100)− 46. 296 3 = 53. 703 7

The final diagram is:time 0 T

Su = 100 (1.2) = 120Vu = 120− 50 = 70

S = 100V = 53. 703 74 = 1, B = −46. 296 3

Sd = 100 (1.05) = 105Vd = 105− 50 = 55

b. Now the stock price has a bigger increase and a bigger decrease. Weare not clear whether the call premium will increase or decrease. We have tocalculate the premium.

time 0 TSu = 100 (1.4) = 140Vu = 140− 50 = 90

S = 100V =?∆ =?, B =?

Sd = 100 (0.6) = 60Vd = 60− 50 = 10½

4140 +Be0.07696(1) = 90460 +Be0.07696(1) = 10

4 = 1 B = −46. 296 3

V = 4S +B = 1 (100)− 46. 296 3 = 53. 703 7

The call premium is the same. What’s going on?It turns out that as long Su > Sd > K, the call premium is fixed regardless

of how you choose u and d.time 0 T

Su = uS > KVu = uS −K

SV =?∆ =?, B =?

Sd = dS > KVd = dS −K½

4uS +Berh = uS −K4dS +Berh = dS −K



→ 4 = 1 B = −Ke−rh C = 4S + B = S − Ke−rh = 100 −50e−0.07696(1) = 53. 703 7

We can explain this using the put-call parity. If Su > Sd > K, the put isnever exercised.

C + PV (K) = P + SSince P = 0, we have C = S−PV (K) = S−Ke−rh = 100−50e−0.07696(1) =

53. 703 7

c. I think the question wants us to compare b and c, not a and c.Compared with b, now u is the same but d gets smaller. In fact, now at

Node d, the call expires worthless.

time 0 TSu = 100 (1.4) = 140Vu = 140− 50 = 90

S = 100V =?∆ =?, B =?

Sd = 100 (0.4) = 40Vd = 0

We might think the call option should be worth less since it expires worthlessat Node d.½

4140 +Be0.07696(1) = 90440 +Be0.07696(1) = 0

4 = 0.9 B = −33. 333

V = 4S +B = 0.9 (100)− 33. 333 = 56. 667 > 53. 703 7Why did the call premium goes up to 56. 667? Does this lead to arbitrage?First, I’m going to show you this is not an arbitrage. Now we have two call

options:Call b Call c

time 0 TSu = 100 (1.4) = 140Vu = 140− 50 = 90

S = 100V = 53. 703 74 = 1, B = −46. 296 3

Sd = 100 (0.6) = 60Vd = 60− 50 = 10

time 0 TSu = 100 (1.4) = 140Vu = 140− 50 = 90

S = 100V = 56. 6674 = 0.9, B = −33. 333

Sd = 100 (0.4) = 40Vd = 0

What if we buy Call b and simultaneously sell Call c? Does this lead toarbitrage?The answer is No because these two calls are on two different stocks. If

these two options have the same underlying asset, then we’ll make free money



by "buy low and sell high." However, these two calls are on two different stocks;if the stock is the same, there won’t be two different values of d0s.Call b is on a stock whose up price is 140 and down price is 60. Call c is

on the stock whose up price is 140 and down price is 40. Clearly, the stockunder Call c is more volatile than the stock under Call b. Hence Call c is morevaluable than Call b.We can also explain why Call b is more valuable using the put-call parity.C + PV (K) = P + SFor b, the put is never exercised. Hence P = 0 and C = S − PV (K).For c, the put is exercised Node d. Hence P > 0 and C = P + S − PV (K).So Call c exceeds Call b.

Problem 10.10.

h = T/n = 1/3

u = e(r−δ)h+σ√h = e(0.08−0)1/3+0.3

√1/3 = 1. 221 246

d = e(r−δ)h−σ√h = e(0.08−0)1/3−0.3

√1/3 = 0.863 693


u− d=

e(0.08−0)1/3 − 0.863 6931. 221 246− 0.863 693 = 0.456 806

πd = 1− πu = 1− 0.456 806 = 0.543 194

a. Calculate the call premiumPeriod 2 3

Suuu = 100¡1. 221 2463

¢= 182. 141 7

Suu = 100¡1. 221 2462

¢= 149. 144 2 Vuuu = 182. 141 7− 95 = 87. 141 7

EVuu = 149. 144 2− 95 = 54. 144 2Vuu =? Suud = 100

¡1. 221 2462

¢(0.863 693) = 128. 814 8

Vuud = 128. 814 8− 95 = 33. 814 8Sud = 100 (1. 221 246) (0.863 693) = 105. 478 2EVud = 105. 478 2− 95 = 10. 478 2Vud =?

Sudd = 100 (1. 221 246)¡0.863 6932

¢= 91. 100 8

Sdd = 100¡0.863 6932

¢= 74. 596 6 Vudd = 0

EVdd = 0Vdd =? Sddd = 100

¡0.863 6933

¢= 64. 428 5

Vddd = 0

Let R represent the roll-back value.V Ruu = e−rh (πuVuuu + πdVuud) = e−(0.08)1/3 (0.456 806× 87. 141 7 + 0.543 194× 33. 814 8) =

56. 644 02Vuu = max

¡V Ruu, EVuu

¢= max (56. 644 02, 54. 144 2) = 56. 644 02



V Rud = e−rh (πuVuud + πdVudd) = e−(0.08)1/3 (0.456 806× 33. 814 8 + 0.543 194× 0) =

15. 040 33Vud = max

¡V Rud, EVud

¢= max (15. 040 33, 10. 478 2) = 15. 040 33

V Rdd = e−rh (πuVudd + πdVddd) = e−(0.08)1/3 (0.456 806× 0 + 0.543 194× 0) =

0Vdd = max

¡V Rdd, EVdd

¢= max (0, 0) = 0

Now we have:Period 1 2

Suu = 100¡1. 221 2462

¢= 149. 144 2

EVuu = 149. 144 2− 95 = 54. 144 2Su = 100 (1. 221 246) = 122. 124 6 Vuu = 56. 644 02EVu = 122. 124 6− 95 = 27. 124 6Vu =? Sud = 100 (1. 221 246) (0.863 693) = 105. 478 2

EVud = 105. 478 2− 95 = 10. 478 2Vud = 15. 040 33

Sd = 100 (0.863 693) = 86. 369 3EVd = 0 Sdd = 100

¡0.863 6932

¢= 74. 596 6

Vd =? EVdd = 0Vdd = 0

V Ru = e−rh (πuVuu + πdVud) = e−(0.08)1/3 (0.456 806× 56. 644 02 + 0.543 194× 15. 040 33) =

33. 149 27Vu = max

¡V Ru , EVu

¢= max (33. 149 27, 27. 124 6) = 33. 149 27

V Rd = e−rh (πuVud + πdVdd) = e−(0.08)1/3 (0.456 806× 15. 040 33 + 0.543 194× 0) =

6. 689 72Vd = max

¡V Rd , EVd

¢= max (6. 689 72, 0) = 6. 689 72


Su = 100 (1. 221 246) = 122. 124 6EVu = 122. 124 6− 95 = 27. 124 6Vu = 33. 149 27

S = 100EV = 100− 95 = 5V =? Sd = 100 (0.863 693) = 86. 369 3

EVd = 0Vd = 6. 689 72

V R = e−rh (πuVu + πdVd) = e−(0.08)1/3 (0.456 806× 33. 149 27 + 0.543 194× 6. 689 72) =18. 282 51

V = max¡V R, EV

¢= max (18. 282 51, 5) = 18. 282 51



From the above calculation you can see that the exercise value is nevergreater than the roll back value. Hence the American call option is never exer-cised early; the American call and European call have the same value of $18.282 51.

Generally, if a stock doesn’t pay dividend, then an American call and anotherwise identical European call on this stock are worth the same.

c. Calculate the American put premium.

Period 2 3Suuu = 100

¡1. 221 2463

¢= 182. 141 7

Suu = 100¡1. 221 2462

¢= 149. 144 2 Vuuu = 0

EVuu = 0Vuu =? Suud = 100

¡1. 221 2462

¢(0.863 693) = 128. 814 8

Vuud = 0Sud = 100 (1. 221 246) (0.863 693) = 105. 478 2EVud = 0Vud =?

Sudd = 100 (1. 221 246)¡0.863 6932

¢= 91. 100 8

Sdd = 100¡0.863 6932

¢= 74. 596 6 Vudd = 95− 91. 100 8 = 3. 899 2

EVdd = 95− 74. 596 6 = 20. 403 4Vdd =? Sddd = 100

¡0.863 6933

¢= 64. 428 5

Vddd = 95− 64. 428 5 = 30. 571 5

V Ruu = e−rh (πuVuuu + πdVuud) = e−(0.08)1/3 (0.456 806× 0 + 0.543 194× 0) =

0

Vuu = max¡V Ruu, EVuu

¢= max (0, 0) = 0

V Rud = e−rh (πuVuud + πdVudd) = e−(0.08)1/3 (0.456 806× 0 + 0.543 194× 3. 899 2) =

2. 062 29

Vud = max¡V Rud, EVud

¢= max (2. 062 29, 0) = 2. 062 29

V Rdd = e−rh (πuVudd + πdVddd) = e−(0.08)1/3 (0.456 806× 3. 899 2 + 0.543 194× 30. 571 5) =

17. 903 58

Vdd = max¡V Rdd, EVdd

¢= max (17. 903 58, 20. 403 4) = 20. 403 4

At the dd node, the exercise value is greater than the roll-back value. Theput is exercised at dd. Now we have:



Period 1 2Suu = 100

¡1. 221 2462

¢= 149. 144 2

EVuu = 0Su = 100 (1. 221 246) = 122. 124 6 Vuu = 0EVu = 0Vu =? Sud = 100 (1. 221 246) (0.863 693) = 105. 478 2

EVud = 0Vud = 2. 062 29

Sd = 100 (0.863 693) = 86. 369 3EVd = 95− 86. 369 3 = 8. 630 7 Sdd = 100

¡0.863 6932

¢= 74. 596 6

Vd =? EVdd = 95− 74. 596 6 = 20. 403 4Vdd = 20. 403 4

V Ru = e−rh (πuVuu + πdVud) = e−(0.08)1/3 (0.456 806× 0 + 0.543 194× 2. 062 29) =

1. 090 75

Vu = max¡V Ru , EVu

¢= max (1. 090 75, 0) = 1. 090 75

V Rd = e−rh (πuVud + πdVdd) = e−(0.08)1/3 (0.456 806× 2. 062 29 + 0.543 194× 20. 403 4) =

11. 708 64

Vd = max¡V Rd , EVd

¢= max (11. 708 64, 8. 630 7) = 11. 708 64


Su = 100 (1. 221 246) = 122. 124 6EVu = 0Vu = 1. 090 75

S = 100EV = 0V =? Sd = 100 (0.863 693) = 86. 369 3

EVd = 95− 86. 369 3 = 8. 630 7Vd = 11. 708 64

V R = e−rh (πuVu + πdVd) = e−(0.08)1/3 (0.456 806× 1. 090 75 + 0.543 194× 11. 708 64) =6. 677 85

V = max¡V R, EV

¢= max (6. 677 85, 0) = 6. 677 85

b. Calculate the European put premium. Verify that the put-call parityholds.

Since the option is European, we just calculate the roll-back value.



Period 2 3Vuuu = 0

Vuu =?Vuud = 0

Vud =?Vudd = 3. 899 2

Vdd =?Vddd = 30. 571 5

Vuu = e−rh (πuVuuu + πdVuud) = e−(0.08)1/3 (0.456 806× 0 + 0.543 194× 0) =0

Vud = e−rh (πuVuud + πdVudd) = e−(0.08)1/3 (0.456 806× 0 + 0.543 194× 3. 899 2) =2. 062 29

Vdd = e−rh (πuVudd + πdVddd) = e−(0.08)1/3 (0.456 806× 3. 899 2 + 0.543 194× 30. 571 5) =17. 903 58


Vuu = 0Vu =?

Vud = 2. 062 29Vd =?

Vdd = 17. 903 58

Vu = e−rh (πuVuu + πdVud) = e−(0.08)1/3 (0.456 806× 0 + 0.543 194× 2. 062 29) =1. 090 75

Vd = e−rh (πuVud + πdVdd) = e−(0.08)1/3 (0.456 806× 2. 062 29 + 0.543 194× 17. 903 58) =10. 386 48


Vu = 1. 090 75V =?

Vd = 10. 386 48

V = e−rh (πuVu + πdVd) = e−(0.08)1/3 (0.456 806× 1. 090 75 + 0.543 194× 10. 386 48) =5. 978 6Check whether the put-call parity-holds:C +Ke−rT = 18. 282 51 + 95e−(0.08)1 = 105. 978 6P + S = 5. 978 6 + 100 = 105. 978 6→ C +Ke−rT = P + S

Problem 10.11.

h = T/n = 1/3

u = e(r−δ)h+σ√h = e(0.08−0.08)1/3+0.3

√1/3 = 1. 189 11



d = e(r−δ)h−σ√h = e(0.08−0.08)1/3−0.3

√1/3 = 0.840 965


u− d=

e(0.08−0.08)1/3 − 0.840 9651. 189 11− 0.840 965 = 0.456 807

πd = 1− πu = 1− 0.456 807 = 0.543 193

a.

• Calculate the price of the American call option


¡1. 189 113

¢= 168. 138 1

Suu = 100¡1. 189 112

¢= 141. 398 3 Vuuu = 168. 138 1− 95 = 73. 138 1

EVuu = 141. 398 3− 95 = 46. 398 3Vuu =? Suud = 100

¡1. 189 112

¢(0.840 965) = 118. 9110

Vuud = 118. 9110− 95 = 23. 911Sud = 100 (1. 189 11) (0.840 965) = 100EVud = 100− 95 = 5Vud =?

Sudd = 100 (1. 189 11)¡0.840 9652

¢= 84. 096 5

Sdd = 100¡0.840 9652

¢= 70. 722 2 Vudd = 0


¡0.840 9653

¢= 59. 474 9

Vddd = 0

V Ruu = e−rh (πuVuuu + πdVuud) = e−(0.08)1/3 (0.456 807× 73. 138 1 + 0.543 193× 23. 911) =

45. 177 3


¢= max (45. 177 3, 46. 398 3) = 46. 398 3

The call is early exercised at the uu node.

V Rud = e−rh (πuVuud + πdVudd) = e−(0.08)1/3 (0.456 807× 23. 911 + 0.543 193× 0) =

10. 635 3


¢= max (10. 635 3, 5) = 10. 635 3


0


¢= max (0, 0) = 0



Period 1 2Vuu = 46. 398 3

Su = 100 (1. 189 11) = 118. 911EVu = 118. 911− 95 = 23. 911Vu =?

Vud = 10. 635 3Sd = 100 (0.840 965) = 84. 096 5EVd = 0Vd =?

Vdd = 0


26. 262 3Vu = max

¡V Ru , EVu

¢= max (26. 262 3, 23. 911) = 26. 262 3


4. 730 4Vd = max

¡V Rd , EVd

¢= max (4. 730 4, 0) = 4. 730 4

Period 0 1Vu = 26. 262 3

S = 100EV = 100− 95 = 5Vu =?

Vd = 4. 730 4

V R = e−rh (πuVu + πdVd) = e−(0.08)1/3 (0.456 807× 26. 262 3 + 0.543 193× 4. 730 4) =14. 183 0

V = max¡V R, EV

¢= max (14. 183 0, 5) = 14. 183 0

• Calculate the price of the European call option

Period 0 1 2 3Suuu = 100

¡1. 189 113

¢= 168. 138 1

Vuuu = 168. 138 1− 95 = 73. 138 1Vuu

Suud = 100¡1. 189 112

¢(0.840 965) = 118. 9110

Vu Vuud = 118. 9110− 95 = 23. 911

V VudSudd = 100 (1. 189 11)

¡0.840 9652

¢= 84. 096 5

Vd Vudd = 0Vdd

Sddd = 100¡0.840 9653

¢= 59. 474 9

Vddd = 0



Vuu = e−rh (πuVuuu + πdVuud) = e−(0.08)1/3 (0.456 807× 73. 138 1 + 0.543 193× 23. 911) =45. 177 3

Vud = e−rh (πuVuud + πdVudd) = e−(0.08)1/3 (0.456 807× 23. 911 + 0.543 193× 0) =10. 635 3

Vdd = e−rh (πuVudd + πdVddd) = e−(0.08)1/3 (0.456 807× 0 + 0.543 193× 0) =0

Vu = e−rh (πuVuu + πdVud) = e−(0.08)1/3 (0.456 807× 45. 177 3 + 0.543 193× 10. 635 3) =25. 719 3

Vd = e−rh (πuVud + πdVdd) = e−(0.08)1/3 (0.456 807× 10. 635 3 + 0.543 193× 0) =4. 730 4

V = e−rh (πuVu + πdVd) = e−(0.08)1/3 (0.456 807× 25. 719 3 + 0.543 193× 4. 730 4) =13. 941 5Since the American call is early exercised at Node uu, the price of the Amer-

ican call 14. 183 0 is greater than the price of the European call 13. 941 5.

c. Calculate the American put premium.


¡1. 189 113

¢= 168. 138 1

Suu = 100¡1. 189 112

¢= 141. 398 3 Vuuu = 0


¡1. 189 112

¢(0.840 965) = 118. 9110

Vuud = 0Sud = 100 (1. 189 11) (0.840 965) = 100EVud = 0Vud =?

Sudd = 100 (1. 189 11)¡0.840 9652

¢= 84. 096 5

Sdd = 100¡0.840 9652

¢= 70. 722 2 Vudd = 95− 84. 096 5 = 10. 903 5

EVdd = 95− 70. 722 2 = 24. 277 8Vdd =? Sddd = 100

¡0.840 9653

¢= 59. 474 9

Vddd = 95− 59. 474 9 = 35. 525 1V Ruu = e−rh (πuVuuu + πdVuud) = e−(0.08)1/3 (0.456 807× 0 + 0.543 193× 0) =

0Vuu = max

¡V Ruu, EVuu

¢= max (0, 0) = 0


5. 766 9Vud = max

¡V Rud, EVud

¢= max (5. 766 9, 0) = 5. 766 9


23. 638 9Vdd = max

¡V Rdd, EVdd

¢= max (23. 638 9, 24. 277 8) = 24. 277 8

The American put is early exercised at Node dd.



Period 1 2Vuu = 0

Su = 100 (1. 189 11) = 118. 911EVu = 0Vu =?

Vud = 5. 766 9Sd = 100 (0.840 965) = 84. 096 5EVd = 95− 84. 096 5 = 10. 903 5Vd =?

Vdd = 24. 277 8


3. 050 1Vu = max

¡V Ru , EVu

¢= max (3. 050 1, 0) = 3. 050 1


15. 405 6Vd = max

¡V Rd , EVd

¢= max (15. 405 6, 10. 903 5) = 15. 405 6

Period 0 1Vu = 3. 050 1

S = 100EV = 0Vu =?

Vd = 15. 405 6

V R = e−rh (πuVu + πdVd) = e−(0.08)1/3 (0.456 807× 3. 050 1 + 0.543 193× 15. 405 6) =9. 504 7

V = max¡V R, EV

¢= max (9. 504 7, 0) = 9. 504 7

b. Calculate the European put premium. Verify that the put-call parityholds.


¡1. 189 113

¢= 168. 138 1

Vuuu = 0Vuu

Vu Suud = 100¡1. 189 112

¢(0.840 965) = 118. 9110

Vuud = 0V

VudSudd = 100 (1. 189 11)

¡0.840 9652

¢= 84. 096 5

Vd Vudd = 95− 84. 096 5 = 10. 903 5Vdd

Sddd = 100¡0.840 9653

¢= 59. 474 9

Vddd = 95− 59. 474 9 = 35. 525 1



Vuu = e−rh (πuVuuu + πdVuud) = e−(0.08)1/3 (0.456 807× 0 + 0.543 193× 0) =0

Vud = e−rh (πuVuud + πdVudd) = e−(0.08)1/3 (0.456 807× 0 + 0.543 193× 10. 903 5) =5. 766 9

Vdd = e−rh (πuVudd + πdVddd) = e−(0.08)1/3 (0.456 807× 10. 903 5 + 0.543 193× 35. 525 1) =23. 638 9

Vu = e−rh (πuVuu + πdVud) = e−(0.08)1/3 (0.456 807× 0 + 0.543 193× 5. 766 9) =3. 050 1

Vd = e−rh (πuVud + πdVdd) = e−(0.08)1/3 (0.456 807× 5. 766 9 + 0.543 193× 23. 638 9) =15. 067 6

V = e−rh (πuVu + πdVd) = e−(0.08)1/3 (0.456 807× 3. 050 1 + 0.543 193× 15. 067 6) =9. 325 9

So the European put is worth 9. 325 9.Verify the put-call parity:C +Ke−rT = 13. 941 5 + 95e−(0.08)1 = 101. 637 55P + Se−δT = 9. 325 9 + 100e−(0.08)1 = 101. 637 53Ignoring rounding difference, we get C +Ke−rT = P + Se−δT

The put-call parity holds.

Problem 10.12.

a. h = T/n = 0.5/2 = 0.25

u = e(r−δ)h+σ√h = e(0.08)0.25+0.3

√0.25 = 1. 185 305

d = e(r−δ)h−σ√h = e(0.08)0.25−0.3

√0.25 = 0.878 095


u− d=

e(0.08)0.25 − 0.878 0951. 185 305− 0.878 095 = 0.462 57

πd = 1− πu = 1− 0.462 57 = 0.537 43

b. Since the stock doesn’t pay dividend, the American call and an otherwiseidentical European call have the same value.


¡1. 185 3052

¢= 56. 197 92

Vuu = 56. 197 92− 40 = 16. 197 92Vu =?

Sud = 40 (1. 185 305) (0.878 095) = 41. 632 4V =? Vud = 41. 632 4− 40 = 1. 632 4

Vd =? Sdd = 40¡0.878 0952

¢= 30. 842 03

Vdd = 0



Vu = e−rh (πuVuu + πdVud) = e−(0.08)0.25 (0.462 57× 16. 197 92 + 0.537 43× 1. 632 4) =8. 204 2

Vd = e−rh (πuVud + πdVdd) = e−(0.08)0.25 (0.462 57× 1. 632 4 + 0.537 43× 0) =0.740 1

V = e−rh (πuVu + πdVd) = e−(0.08)0.25 (0.462 57× 8. 204 2 + 0.537 43× 0.740 1) =4. 109 7

So both the American call and the European call are worth 4. 109 7.

We also calculate the European call price using the following shortcut:Node Payoff Risk Neutral Probuu 16. 197 92 π2u = 0.462 57

2

ud 1. 632 4 2πuπd = 2× 0.462 57× 0.537 43dd 0 π2d = 0.537 43

2

V = e−rTP

payoff ×RiskNeutralProb= e−0.08(0.5)

¡16. 197 92× 0.462 572 + 1. 632 4× 2× 0.462 57× 0.537 43 + 0× 0.537 432

¢= 4. 109 8Please note that the above shortcut works only for European options.

c. Even if the stock doesn’t pay dividend, it may be still optimal to exercisean American put early.

• Calculate the premium of the American put


¡1. 185 3052

¢= 56. 197 92

Su = 40 (1. 185 305) = 47. 412 2 Vuu = 0EVu = 0Vu =?

S = 40 Sud = 40 (1. 185 305) (0.878 095) = 41. 632 4EV = 0 Vud = 0V =?

Sd = 40 (0.878 095) = 35. 123 8EVd = 40− 35. 123 8 = 4. 876 2Vd =? Sdd = 40

¡0.878 0952

¢= 30. 842 03

Vdd = 40− 30. 842 03 = 9. 157 97

V Ru = e−rh (πuVuu + πdVud) = e−(0.08)1/3 (0.456 807× 0 + 0.543 193× 0) =

0Vu = max

¡V Ru , EVu

¢= max (0, 0) = 0

V Rd = e−rh (πuVud + πdVdd) = e−(0.08)1/3 (0.456 807× 0 + 0.543 193× 9. 157 97) =

4. 843 6Vd = max

¡V Rd , EVd

¢= max (4. 843 6, 4. 876 2) = 4. 876 2



So the American put is early exercised at Node d.

V R = e−rh (πuVu + πdVd) = e−(0.08)1/3 (0.456 807× 0 + 0.543 193× 4. 876 2) =2. 579 0

V = max¡V R, EV

¢= max (2. 579 0, 0) = 2. 579 0

The American put is worth 2. 579 0.

• Calculate the premium of the European put


¡1. 185 3052

¢= 56. 197 92

Vuu = 0Vu

Sud = 40 (1. 185 305) (0.878 095) = 41. 632 4V Vud = 0

Vd Sdd = 40¡0.878 0952

¢= 30. 842 03

Vdd = 40− 30. 842 03 = 9. 157 97

Vu = e−rh (πuVuu + πdVud) = e−(0.08)1/3 (0.456 807× 0 + 0.543 193× 0) = 0

Vd = e−rh (πuVud + πdVdd) = e−(0.08)1/3 (0.456 807× 0 + 0.543 193× 9. 157 97) =4. 843 6

V = e−rh (πuVu + πdVd) = e−(0.08)1/3 (0.456 807× 0 + 0.543 193× 4. 843 6) =2. 561 8

The European put is worth 2. 561 8.

Problem 10.13.

a. From the previous problem, we know:Period 0 1 2

Suu = 40¡1. 185 3052

¢= 56. 197 92

Su = 40 (1. 185 305) = 47. 412 2 Vuu = 56. 197 92− 40 = 16. 197 92Vu = 8. 204 2

S = 40 (∆u, Bu) Sud = 40 (1. 185 305) (0.878 095) = 41. 632 4V = 4. 109 7 Vud = 41. 632 4− 40 = 1. 632 4(∆, B)

Sd = 40 (0.878 095) = 35. 123 8Vd = 0.740 1 Sdd = 40

¡0.878 0952

¢= 30. 842 03

(∆d, Bd) Vdd = 0

First, let find the replicating portfolio.



• The replicating portfolio at t = 0:½4Su +Berh = Vu4Sd +Berh = Vd½447. 412 2 +Be0.08(0.25) = 8. 204 2435. 123 8 +Be0.08(0.25) = 0.740 1

4 = 0.607 41 B = −20. 186 7

• The replicating portfolio Node u:½4uSuu +Bue

rh = Vuu4uSud +Berh = Vud½4u56. 197 92 +Bue

0.08(0.25) = 16. 197 924u41. 632 4 +Bue

0.08(0.25) = 1. 632 44u = 1 Bu = −39.

207 9

• The replicating portfolio Node d:½4dSud +Bde

rh = Vud4dSdd +Berh = Vdd½4d41. 632 4 +Bde

0.08(0.25) = 1. 632 44d30. 842 03 +Bde

0.08(0.25) = 04d = 0.151 28 Bd = −4.

573 5At t = 0, we

• sell the over-priced call for $5

• build a synthetic call by buying 4 = 0.607 41 share of the stock andborrowing $20. 186 7 from a bank. Cost: 4S + B = 0.607 41 (40) − 20.186 7 = 4. 109 7.

So we receive 5− 4. 109 7 = 0.890 3 at t = 0.

b.

• Suppose the call in the market at time h is fairly priced

To liquidate our position at t = h (i.e. the end of Period 1) yet hedge oursold call, we’ll do the following:

1. Sell our 4 = 0.607 41 share of the stock, receiving 4Sh

2. Pay the bank Berh = 20. 186 7e0.08(0.25) = 20. 594 5



3. Buy the call from the open market for the fair price of Vh, which is eitherVu = 8. 204 2 if the stock price goes up or Vd = 0.740 1 if the stock pricegoes down.

Suppose the stock price goes up to Su and we liquidate our position at Nodeu. We’ll

1. Sell our4 = 0.607 41 share of the stock, receiving4Su = 0.607 41 (47. 412 2) =$28. 798 6

2. Pay the bank Berh = 20. 186 7e0.08(0.25) = $20. 594 5

3. Buy the call from the open market for the fair price of Vu = $8. 204 2

The net receipt is: 28. 798 6 − 20. 594 5 − 8. 204 2 = −0.000 1 = $0 (if weignore rounding errors)So we receive $0.890 3 free money at t = 0 without incurring any liability at

time h.

Suppose the stock price goes up to Sd and we liquidate our position at Nodeu. We’ll

1. Sell our4 = 0.607 41 share of the stock, receiving4Sd = 0.607 41 (35. 123 8) =$21. 334 5

2. Pay the bank Berh = 20. 186 7e0.08(0.25) = $20. 594 5

3. Buy the call from the open market for the fair price of Vd = $0.740 1

The net receipt is: 21. 334 5−20. 594 5−0.740 1 = −0.000 1 = 0 (if we ignorerounding errors)So we receive $0.890 3 free money at t = 0 without incurring any liability at

time h.

• Suppose the call in the market at h is over priced

We’ll not liquidate our position at h. Instead, we rebalance our replicatingportfolio at t = h and liquidate our position at T = 2h. We’ll consider twosituation.

Situation #1 — the stock price goes up to Su = 40 (1. 185 305) = 47. 412 2at time hWe’ll change our replicating portfolio from (∆, B) = (0.607 41, B = −$20. 186 7)

to (∆u, Bu) = (1,−$39. 207 9) at h. We need to buy 1 − 0.607 41 = 0.392 59share of the stock. The cost is 0.392 59 (47. 412 2) = 18. 613 6. We’ll borrow 18.613 6 from the bank to pay for this.



So at time h, our total debt to the bank is: 18. 613 4 + 20. 186 7e0.08(0.25) =39. 207 9. The number of stocks we have is 1. Now at time h, our replicatingportfolio is exactly (∆u, Bu) = (1,−$39. 207 9).Then at T = 2hIf Suu = 40

¡1. 185 3052

¢= 56. 197 92

1. Our written call is exercised against us. We need to pay the call holderVuu = 56. 197 92− 40 = 16. 197 92.

2. We sell ∆u stocks in the market and pay off our loan from the bank. Ournet cash receipt is ∆uSuu +Bue

rh = 1× 56. 197 92− 39. 207 9e0.08(0.25) =16. 197 97.

3. Our net cash flow is zero (ignoring rounding).

If Sud = 40 (1. 185 305) (0.878 095) = 41. 632 4

1. Our written call is exercised against us. We need to pay the call holderVud = 41. 632 4− 40 = 1. 632 4

2. We sell ∆u stocks in the market and pay off our loan from the bank. Ournet cash receipt is ∆uSud + Bue

rh = 1 × 41. 632 4 − 39. 207 9e0.08(0.25) =1. 632 4

3. Our net cash flow is zero.

So we receive $0.890 3 free money at t = 0 without incurring any liability atT .Situation #2 — the stock price goes down to Sd = 40 (0.878 095) = 35.

123 8 at time hWe’ll change our replicating portfolio from (∆, B) = (0.607 41, B = −$20. 186 7)

to (∆d, Bd) = (0.151 28,−$4. 573 5) at h. We need to sell 0.607 4 − 0.151 28 =0.456 12 share of the stock, receiving 0.456 12 (35. 123 8) = $16. 020 7. We im-mediately send a check of $16. 020 7 to the bank to partially pay our debt. Nowour remaining debt to the bank is20. 186 7e0.08(0.25) − 16. 020 7 = 4. 573 8 = 4. 573 5 (ignore rounding)Now at time h, our replicating portfolio is exactly (∆d, Bd) = (0.151 28,−$4. 573 5)Then at T = 2hIf Sdu = 40 (1. 185 305) (0.878 095) = 41. 632 4

1. Our written call is exercised against us. We need to pay the call holderVdu = 41. 632 4− 40 = 1. 632 4

2. We sell ∆d stocks in the market and pay off our loan from the bank. Ournet cash receipt is∆dSdu+Bde

rh = 0.151 28×41. 632 4−4. 573 5e0.08(0.25) =1. 632 3

3. Our net cash flow is zero (ignoring rounding).



If Sdd = 40¡0.878 0952

¢= 30. 842 03

1. Our written call expires worthless.

2. We sell ∆d stocks in the market and pay off our loan from the bank. Ournet cash receipt is∆dSdd+Bde

rh = 0.151 28×30. 842 03−4. 573 5e0.08(0.25) ≈0

3. Our net cash flow is zero.

So we receive $0.890 3 free money at t = 0 without incurring any liability atT .

c. If the call option is under priced at h, we just sell off our replicatingportfolio (i.e. sell ∆ share of the stock and pay back our loan to the bank) andpurchase a call from the market. Because the value of our replicating portfolioat h represents the fair price of a call and the call in the market is lower thanthe fair price, the money we get from selling our replicating portfolio exceedsthe call price in the market. So we’ll make some profit.Now we are standing at time h. The call we buy at time h has life from time

h to T = 2h. The call we initially sold at time zero has life from h to T = 2h.These two calls exactly offset each other so our liability at T is zero.

Problem 10.14.

We can use the general binomial tree formula by setting S = 0.92 andδ = r€ = 0.03.

h = T/n = 0.75/3 = 0.25

u = 1.2 d = 0.9


u− d=

e(0.04−0.03)0.25 − 0.91.2− 0.9 = 0.341 677

πd = 1− πu = 1− 0.341 677 = 0.658 323

We’ll calculate part b first.b. Calculate the American call premium.



Period 2 3Suuu = 0.92

¡1. 23

¢= 1. 589 76

Suu = 0.92¡1. 22

¢= 1. 324 8 Vuuu = 1. 589 76− 0.85 = 0.739 76

EVuu = 1. 324 8− 0.85 = 0.474 8Vuu =? Suud = 0.92

¡1. 22

¢(0.9) = 1. 192 32

Vuud = 1. 192 32− 0.85 = 0.342 32Sud = 0.92 (1. 2) (0.9) = 0.993 6EVud = 0.993 6− 0.85 = 0.143 6Vud =?

Sudd = 0.92 (1. 2)¡0.92

¢= 0.894 24

Sdd = 0.92¡0.92

¢= 0.745 2 Vudd = 0.894 24− 0.85 = 0.044 24

EVdd = 0Vdd =? Sddd = 0.92

¡0.93

¢= 0.670 68

Vddd = 0

V Ruu = e−rh (πuVuuu + πdVuud) = e−(0.04)0.25 (0.341 677× 0.739 76 + 0.658 323 × 0.342 32) =

0.473 359Vuu = max

¡V Ruu, EVuu

¢= max (0.473 359, 0.474 8) = 0.474 8

So the American call is early exercised at the uu node.V Rud = e−rh (πuVuud + πdVudd) = e−(0.04)0.25 (0.341 677× 0.342 32 + 0.658 323 × 0.044 24) =

0.144 633


¢= max (0.144 633, 0.143 6) = 0.144 633

V Rdd = e−rh (πuVudd + πdVddd) = e−(0.04)0.25 (0.341 677× 0.044 24 + 0.658 323 × 0) =

0.01 496 5Vdd = max

¡V Rdd, EVdd

¢= max (0.01 496 5, 0) = 0.014 965

Period 1 2Vuu = 0.474 8

Su = 0.92 (1. 2) = 1. 104EVu = 1. 104− 0.85 = 0.254Vu =?

Vud = 0.144 633Sd = 0.92 (0.9) = 0.828EVd = 0Vd =?

Vdd = 0.01 496 5

V Ru = e−rh (πuVuu + πdVud) = e−(0.04)0.25 (0.341 677× 0.474 8 + 0.658 323 × 0.144 633) =

0.254 882Vu = max

¡V Ru , EVu

¢= max (0.254 882 , 0.254) = 0.254 882

V Rd = e−rh (πuVud + πdVdd) = e−(0.04)0.25 (0.341 677× 0.144 633 + 0.658 323 × 0.01 496 5) =

0.05 8680


¢= max (0.05 8680 , 0) = 0.058 68



Period 0 1Vu = 0.254 882

S = 0.92EV = 0.92− 0.85 = 0.07V =?

Vd = 0.058 68

V R = e−rh (πuVu + πdVu) = e−(0.04)0.25 (0.341 677× 0.254 882 + 0.658 323 × 0.058 68) =0.124 467

V = max¡V R, EV

¢= max (0.124 467 , 0.07) = 0.124 467

So the American call premium is 0.124 467 .

a. Calculate the European call premium.Period 0 1 2 3

Suuu = 0.92¡1. 23

¢= 1. 589 76

Vuuu = 1. 589 76− 0.85 = 0.739 76Vuu

Suud = 0.92¡1. 22

¢(0.9) = 1. 192 32

Vu Vuud = 1. 192 32− 0.85 = 0.342 32

V VudSudd = 0.92 (1. 2)

¡0.92

¢= 0.894 24

Vd Vudd = 0.894 24− 0.85 = 0.044 24Vdd

Sddd = 0.92¡0.93

¢= 0.670 68

Vddd = 0

Vuu = e−rh (πuVuuu + πdVuud) = e−(0.04)0.25 (0.341 677× 0.739 76 + 0.658 323 × 0.342 32) =0.473 359

Vud = e−rh (πuVuud + πdVudd) = e−(0.04)0.25 (0.341 677× 0.342 32 + 0.658 323 × 0.044 24) =0.144 633

Vdd = e−rh (πuVudd + πdVddd) = e−(0.04)0.25 (0.341 677× 0.044 24 + 0.658 323 × 0) =0.01 496 5

Vu = e−rh (πuVuu + πdVud) = e−(0.04)0.25 (0.341 677× 0.473 359 + 0.658 323 × 0.144 633) =0.254 394

V Rd = e−rh (πuVud + πdVdd) = e−(0.04)0.25 (0.341 677× 0.144 633 + 0.658 323 × 0.01 496 5) =

0.05 8680

V = e−rh (πuVu + πdVu) = e−(0.04)0.25 (0.341 677× 0.254 394 + 0.658 323 × 0.058 68) =0.124 302

So the European call premium is 0.124 302 .



Problem 10.15.

We can use the general binomial tree formula by setting S = 0.92 andδ = r€ = 0.03.

h = T/n = 0.75/3 = 0.25

u = 1.2 d = 0.9


u− d=

e(0.04−0.03)0.25 − 0.91.2− 0.9 = 0.341 677

πd = 1− πu = 1− 0.341 677 = 0.658 323

b. Calculate the American put premium.Period 2 3

Suuu = 0.92¡1. 23

¢= 1. 589 76

Suu = 0.92¡1. 22

¢= 1. 324 8 Vuuu = 0

EVuu = 0Vuu =? Suud = 0.92

¡1. 22

¢(0.9) = 1. 192 32

Vuud = 0Sud = 0.92 (1. 2) (0.9) = 0.993 6EVud = 1− 0.993 6 = 0.006 4Vud =?

Sudd = 0.92 (1. 2)¡0.92

¢= 0.894 24

Sdd = 0.92¡0.92

¢= 0.745 2 Vudd = 1− 0.894 24 = 0.105 76

EVdd = 1− 0.745 2 = 0.254 8Vdd =? Sddd = 0.92

¡0.93

¢= 0.670 68

Vddd = 1− 0.670 68 = 0.329 32

V Ruu = e−rh (πuVuuu + πdVuud) = e−(0.04)0.25 (0.341 677× 0 + 0.658 323 × 0) =

0


¢= max (0, 0) = 0

V Rud = e−rh (πuVuud + πdVudd) = e−(0.04)0.25 (0.341 677× 0 + 0.658 323 × 0.105 76) =

0.06893 1


¢= max (0.06893 1, 0.006 4) = 0.068 931

V Rdd = e−rh (πuVudd + πdVddd) = e−(0.04)0.25 (0.341 677× 0.105 76 + 0.658 323 × 0.329 32) =

0.250 418


¢= max (0.250 418, 0.254 8) = 0.254 8

The American put is early exercised at the dd node.



Period 1 2Vuu = 0

Su = 0.92 (1. 2) = 1. 104EVu = 0Vu =?

Vud = 0.068 931Sd = 0.92 (0.9) = 0.828EVd = 1− 0.828 = 0.172Vd =?

Vdd = 0.254 8

V Ru = e−rh (πuVuu + πdVud) = e−(0.04)0.25 (0.341 677× 0 + 0.658 323 × 0.068 931 ) =

0.04 492 7Vu = max

¡V Ru , EVu

¢= max (0.04 492 7 , 0) = 0.04 492 7

V Rd = e−rh (πuVud + πdVdd) = e−(0.04)0.25 (0.341 677× 0.068 931 + 0.658 323 × 0.254 8) =

0.189 389


¢= max (0.189 389 , 0.172) = 0.189 389

Period 0 1Vu = 0.04 492 7

S = 0.92EV = 1− 0.92 = 0.08V =?

Vd = 0.189 389

V R = e−rh (πuVu + πdVu) = e−(0.04)0.25 (0.341 677× 0.04 492 7 + 0.658 323 × 0.189 389) =0.138 636

V = max¡V R, EV

¢= max (0.138 636 , 0.08) = 0.138 636

So the American put premium is 0.138 636 .

a. Calculate the European put premium.Period 0 1 2 3

Suuu = 0.92¡1. 23

¢= 1. 589 76

Vuuu = 0Vuu

Vu Suud = 0.92¡1. 22

¢(0.9) = 1. 192 32

Vuud = 0V Vud

Sudd = 0.92 (1. 2)¡0.92

¢= 0.894 24

Vd Vudd = 1− 0.894 24 = 0.105 76Vdd

Sddd = 0.92¡0.93

¢= 0.670 68

Vddd = 1− 0.670 68 = 0.329 32



Vuu = e−rh (πuVuuu + πdVuud) = e−(0.04)0.25 (0.341 677× 0 + 0.658 323 × 0) =

0Vud = e−rh (πuVuud + πdVudd) = e−(0.04)0.25 (0.341 677× 0 + 0.658 323 × 0.105 76) =

0.06893 1Vdd = e−rh (πuVudd + πdVddd) = e−(0.04)0.25 (0.341 677× 0.105 76 + 0.658 323 × 0.329 32) =

0.250 418

Vu = e−rh (πuVuu + πdVud) = e−(0.04)0.25 (0.341 677× 0 + 0.658 323 × 0.068 931 ) =0.04 492 7

Vd = e−rh (πuVud + πdVdd) = e−(0.04)0.25 (0.341 677× 0.068 931 + 0.658 323 × 0.250 418) =0.186 533

V = e−rh (πuVu + πdVu) = e−(0.04)0.25 (0.341 677× 0.04 492 7 + 0.658 323 × 0.186 533) =0.136 775

So the European put premium is 0.136 775 .

Problem 10.16.

This problem looks scary, but it’s actually simple if you know what’s theunderlying asset.The underlying asset is $1. The call option gives the call holder the right to

purchase $1 with a guaranteed price (i.e. strike price) of 120 Yen. So the strikeprice and the option price are expressed in Yen.To simplify the problem, we can treat the underlying asset $1 as a stock and

translate the original problem into the following:

• Now we are living in Japan and are interested in options on a stock.

• The current stock price is S = 120 Yen.

• The continuously compounded risk free interest rate is rf = 0.01 (remem-ber we are living in Japan).

• The stock’s volatility is σ = 0.1.

• The stock’s continuous dividend yield is δ = 0.05; the dollar interest rateis like the dividend yield.

• T = 1

• h = T/3 = 1/3

Now we can use the standard binomial formula.

u = e(r−δ)h+σ√h = e(0.01−0.05)1/3+0.1

√1/3 = 1. 045 402



d = e(r−δ)h−σ√h = e(0.01−0.05)1/3−0.1

√1/3 = 0.931 398


u− d=

e(0.01−0.05)1/3 − 0.931 3981. 045 402− 0.931 398 = 0.485 57

πd = 1− πu = 1− 0.485 57 = 0.514 43

a. Calculate the price of the American call optionPeriod 2 3

Suuu = 120¡1. 045 4023

¢= 137. 098 0

Suu = 120¡1. 045 4022

¢= 131. 143 8 Vuuu = 137. 098 0− 120 = 17. 098

EVuu = 131. 143 8− 120 = 11. 143 8Vuu =? Suud = 120

¡1. 045 4022

¢(0.931 398) = 122. 147 1

Vuud = 122. 147 1− 120 = 2. 147 1Sud = 120 (1. 045 402) (0.931 398) = 116. 842 2EVud = 0Vud =?

Sudd = 120 (1. 045 402)¡0.931 3982

¢= 108. 826 6

Sdd = 120¡0.931 3982

¢= 104. 100 3 Vudd = 0


¡0.931 3983

¢= 96. 958 8

Vddd = 0

V Ruu = e−rh (πuVuuu + πdVuud) = e−(0.01)1/3 (0.485 57× 17. 098 + 0.514 43× 2. 147 1) =

9. 375 5Vuu = max

¡V Ruu, EVuu

¢= max (9. 375 5, 11. 143 8) = 11. 143 8

So the American call is early exercised at the uu node.

V Rud = e−rh (πuVuud + πdVudd) = e−(0.01)1/3 (0.485 57× 2. 147 1 + 0.514 43× 0) =

1. 039 1


¢= max (1. 039 1, 0) = 1. 039 1


0Vdd = max

¡V Rdd, EVdd

¢= max (0, 0) = 0

Period 1 2Vuu = 11. 143 8

Su = 120 (1. 045 402) = 125. 448 2EVu = 125. 448 2− 120 = 5. 448 2Vu =?

Vud = 1. 039 1Sd = 120 (0.931 398) = 111. 767 8EVd = 0Vd =?

Vdd = 0




5. 925 9Vu = max

¡V Ru , EVu

¢= max (5. 925 9 , 5. 448 2) = 5. 925 9


0.502 9


¢= max (0.502 9 , 0) = 0.502 9

Period 0 1Vu = 5. 925 9

S = 120EV = 0V =?

Vd = 0.502 9

V R = e−rh (πuVu + πdVu) = e−(0.01)1/3 (0.485 57× 5. 925 9 + 0.514 43× 0.502 9) =3. 125 7

V = max¡V R, EV

¢= max (3. 125 7 , 0) = 3. 125 7

So the American call premium is 3. 125 7 Yen.

Calculate the price of the European callPeriod 0 1 2 3

Suuu = 120¡1. 045 4023

¢= 137. 098 0

Vuuu = 137. 098 0− 120 = 17. 098Vuu

Vu Suud = 120¡1. 045 4022

¢(0.931 398) = 122. 147 1

Vuud = 122. 147 1− 120 = 2. 147 1V Vud

Sudd = 120 (1. 045 402)¡0.931 3982

¢= 108. 826 6

Vd Vudd = 0Vdd

Sddd = 120¡0.931 3983

¢= 96. 958 8

Vddd = 0

Vuu = e−rh (πuVuuu + πdVuud) = e−(0.01)1/3 (0.485 57× 17. 098 + 0.514 43× 2. 147 1) =9. 375 5

Vud = e−rh (πuVuud + πdVudd) = e−(0.01)1/3 (0.485 57× 2. 147 1 + 0.514 43× 0) =1. 039 1

Vdd = e−rh (πuVudd + πdVddd) = e−(0.01)1/3 (0.485 57× 0 + 0.514 43× 0) =0

Vu = e−rh (πuVuu + πdVud) = e−(0.01)1/3 (0.485 57× 9. 375 5 + 0.514 43× 1. 039 1) =5. 070 1



Vd = e−rh (πuVud + πdVdd) = e−(0.01)1/3 (0.485 57× 1. 039 1 + 0.514 43× 0) =0.502 9

V = e−rh (πuVu + πdVu) = e−(0.01)1/3 (0.485 57× 5. 070 1 + 0.514 43× 0.502 9) =2. 711 5

So the European call premium is 2. 711 5 Yen.

b. Calculate the price of the American putPeriod 2 3

Suuu = 120¡1. 045 4023

¢= 137. 098 0

Suu = 120¡1. 045 4022

¢= 131. 143 8 Vuuu = 0


¡1. 045 4022

¢(0.931 398) = 122. 147 1

Vuud = 0Sud = 120 (1. 045 402) (0.931 398) = 116. 842 2EVud = 120− 116. 842 2 = 3. 157 8Vud =?

Sudd = 120 (1. 045 402)¡0.931 3982

¢= 108. 826 6

Sdd = 120¡0.931 3982

¢= 104. 100 3 Vudd = 120− 108. 826 6 = 11. 173 4

EVdd = 120− 104. 100 3 = 15. 899 7Vdd =? Sddd = 120

¡0.931 3983

¢= 96. 958 8

Vddd = 120− 96. 958 8 = 23. 041 2

V Ruu = e−rh (πuVuuu + πdVuud) = e−(0.01)1/3 (0.485 57× 0 + 0.514 43× 0) =

0Vuu = max

¡V Ruu, EVuu

¢= max (0, 0) = 0


5. 728 8


¢= max (3. 157 8, 5. 728 8) = 5. 728 8


17. 221 1Vdd = max

¡V Rdd, EVdd

¢= max (17. 221 1, 15. 899 7) = 17. 221 1

Period 1 2Vuu = 0

Su = 120 (1. 045 402) = 125. 448 2EVu = 0Vu =?

Vud = 5. 728 8Sd = 120 (0.931 398) = 111. 767 8EVd = 120 − 111. 767 8 = 8. 232 2Vd =?

Vdd = 17. 221 1




2. 937 3


¢= max (2. 937 3 , 0) = 2. 937 3


11. 602 0


¢= max (11. 602 0 , 8. 232 2) = 11. 602

Period 0 1Vu = 2. 937 3

S = 120EV = 0V =?

Vd = 11. 602

V R = e−rh (πuVu + πdVu) = e−(0.01)1/3 (0.485 57× 2. 937 3 + 0.514 43× 11. 602) =7. 370 1

V = max¡V R, EV

¢= max (7. 370 1 , 0) = 7. 370 1

Since the exercise value is never greater than the roll-back value, the Amer-ican put is not early exercised. Hence the American put and the European puthave the same value.Both the American put and European put are worth 7. 370 1 Yen.

c. Since the underlying asset generates dividend (the Yen interest rate islike the dividend rate), it may be optimal to early exercise an American call.However, even if the asset generates dividend, it may not be optimal to earlyexercise an American put.

Problem 10.17.

Method 1 Apply the stock binomial formula to futuresAs explained in my study guide, we can apply the stock binomial formula

to options on futures if we do the following three things:

• Set the dividend yield equal to the risk free rate (i.e. δ = r)

• 4 =Cu − Cd

Su − Sdinstead of 4 = e−rh

Cu − Cd

Su − Sd

• B = V = e−rhµCu1− d

u− d+ Cd

u− 1u− d


SuCd − SdCu

Su − Sd.



h = 1 year long.


√h = e0.1

√1 = 1. 105 170 9

d = e(r−δ)h−σ√h = e−0.1

√1 = 0.904 837 4


u− d=1− d

u− d=

1− 0.904 837 41. 105 170 9 − 0.904 837 4 = 0.475 021

πd = 1− πu = 1− 0.475 021 = 0.524 979

Period 0 1Su = 300 (1. 105 170 9) = 331. 551 27Vu = 331. 551 27− 290 = 41. 551 27

S = 300EV = 300− 290 = 10V =?(∆, B) =?

Sd = 300 (0.904 837 4) = 271. 451 22Vd = 0

V R = e−rh (πuVu + πdVu) = e−(0.06)1 (0.475 021× 41. 551 27 + 0.524 979 × 0) =18. 588 29

V = max¡V R, EV

¢= max (18. 588 29 , 10) = 18. 588 29

So the call premium is $18. 588 29 at t = 0.

4 =Cu − Cd

Su − Sd=

41. 551 27− 0331. 551 27− 271. 451 22 = 0.691 368

B = V = 18. 588 29

So to form a replicating portfolio at time zero, we buy 0.691 368 unit offutures contract (i.e. enter 0.691 368 unit of futures contract as a buyer) andinvest $18. 588 29 in a savings account.

Method 2Just as in Method 1, we calculateu = eσ

√h = e0.1

√1 = 1. 105 170 9

d = e−σ√h = e−0.1

√1 = 0.904 837 4

Next, we build a futures price tree.Period 0 1

Fu1,1 = 300 (1. 105 170 9) = 331. 551 27

Vu = 331. 551 27− 290 = 41. 551 27F0,1 = 300EV = 300− 290 = 10V =?(∆, B) =?

F d1,1 = 300 (0.904 837 4) = 271. 451 22

Vd = 0



If we buy ∆ futures contract at t = 0, our gain in these futures contracts atT is:

• ∆¡Fu1,1 − F0,1

¢= ∆F0,1 (u− 1) in the u node

• ∆¡F d1,1 − F0,1

¢= ∆F0,1 (d− 1) in the d node

If we put $B in a savings account, we’ll get BerT at T .We want our replicating portfolio and the call option have the same payoff

at T :½∆¡Fu1,1 − F0,1

¢+BerT = Vu

∆¡F d1,1 − F0,1

¢+BerT = Vd

→½∆F0,1 (u− 1) +BerT = Vu∆F0,1 (d− 1) +BerT = Vd

½∆300 (1. 105 170 9− 1) +Be0.06(1) = 41. 551 27∆300 (0.904 837 4− 1) +Be0.06(1) = 0

∆ = 0.691 368 B = 18. 588 29

Since it costs nothing to enter a future contract, the call premium is equalto B:

V = B = 18. 588 29

The statement "Replicating a call option always entails borrowing to buythe underlying asset" is true for calls on stocks. However, it’s not true for callson futures; it costs nothing to enter a futures contract (so you don’t need toborrow money to finance your transaction on futures). If you write a call onfutures, you need to enter ∆ futures contracts and simultaneously deposit thecall premium into a savings account. This way, at call expiration date T , thecombined payoff of your ∆ futures contracts and your deposit in the savingsaccount will replicate the payoff of a call on futures.

Problem 10.18.

We’ll apply the stock binomial formula to futures. We need to do the fol-lowing three things:

• Set the dividend yield equal to the risk free rate (i.e. δ = r)

• 4 =Cu − Cd

Su − Sdinstead of 4 = e−rh

Cu − Cd

Su − Sd

• B = V = e−rhµCu1− d

u− d+ Cd

u− 1u− d


SuCd − SdCu

Su − Sd.



h = 1/3


√h = e0.3

√1/3 = 1. 189 110

d = e(r−δ)h−σ√h = e−0.3

√1/3 = 0.840 965

Notice that ud = 1


u− d=1− d

u− d=

1− 0.840 9651. 189 110 − 0.840 965 = 0.456 807

πd = 1− πu = 1− 0.456 807 = 0.543 193

• calculate the American call premium

Period 2 3Suuu = 1000u

3 = 1681. 380 8Suu = 1000u

2 = 1413. 982 6 Vuuu = 1681. 380 8− 1000 = 681. 380 8EVuu = 1413. 982 6− 1000 = 413. 982 6Vuu =? Suud = 1000u

2d = 1000u = 1189. 1100Vuud = 1189. 11− 1000 = 189. 11

Sud = 1000ud = 1000EVud = 0Vud =?

Sudd = 1000ud2 = 1000d = 840. 965

Sdd = 1000d2 = 707. 222 1 Vudd = 0

EVdd = 0Vdd =? Sddd = 1000d

3 = 1000¡0.840 965 3

¢= 594. 749

Vddd = 0

V Ruu = e−rh (πuVuuu + πdVuud) = e−(0.05)1/3 (0.456 807× 681. 380 8 + 0.543 193 × 189. 11) =

407. 140 2


¢= max (407. 140 2, 413. 982 6) = 413. 982 6

The American call is early exercised at Node uu.V Rud = e−rh (πuVuud + πdVudd) = e−(0.05)1/3 (0.456 807× 189. 11 + 0.543 193 × 0) =

84. 958 9


¢= max (84. 958 9, 0) = 84. 958 9

V Rdd = e−rh (πuVudd + πdVddd) = e−(0.05)1/3 (0.456 807× 0 + 0.543 193 × 0) =

0


¢= max (0, 0) = 0



Period 1 2Vuu = 413. 982 6

Su = 1000 (1. 189 11) = 1189. 11EVu = 1189. 11− 1000 = 189. 11Vu =?

Vud = 84. 958 9Sd = 1000 (0.840 965 ) = 840. 965EVd = 0Vd =?

Vdd = 0

V Ru = e−rh (πuVuu + πdVud) = e−(0.05)1/3 (0.456 807× 413. 982 6 + 0.543 193 × 84. 958 9) =

231. 370 7Vu = max

¡V Ru , EVu

¢= max (231. 370 7 , 189. 11) = 231. 370 7

V Rd = e−rh (πuVud + πdVdd) = e−(0.05)1/3 (0.456 807× 84. 958 9 + 0.543 193 × 0) =

38. 168 4


¢= max (38. 168 4 , 0) = 38. 168 4

Period 0 1Vu = 231. 370 7

S = 1000EV = 0V =?

Vd = 38. 168 4

V R = e−rh (πuVu + πdVu) = e−(0.05)1/3 (0.456 807× 231. 370 7 + 0.543 193 × 38. 168 4) =124. 334 9

V = max¡V R, EV

¢= max (124. 334 9 , 0) = 124. 334 9

So the American call premium is 124. 334 9.

• Calculate the European call premium

Period 0 1 2 3Suuu = 1000u

3 = 1681. 380 8Vuuu = 1681. 380 8− 1000 = 681. 380 8

VuuSuud = 1000u

2d = 1000u = 1189. 1100Vu Vuud = 1189. 11− 1000 = 189. 11

V Vud

Vd Sudd = 1000ud2 = 1000d = 840. 965

Vudd = 0Vdd Sddd = 1000d

3 = 1000¡0.840 965 3

¢= 594. 749

Vddd = 0



Vuu = e−rh (πuVuuu + πdVuud) = e−(0.05)1/3 (0.456 807× 681. 380 8 + 0.543 193 × 189. 11) =407. 140 2

Vud = e−rh (πuVuud + πdVudd) = e−(0.05)1/3 (0.456 807× 189. 11 + 0.543 193 × 0) =84. 958 9

Vdd = e−rh (πuVudd + πdVddd) = 0

Vu = e−rh (πuVuu + πdVud) = e−(0.05)1/3 (0.456 807× 407. 140 2 + 0.543 193 × 84. 958 9) =228. 296 7

Vd = e−rh (πuVud + πdVdd) = e−(0.05)1/3 (0.456 807× 84. 958 9 + 0.543 193 × 0) =38. 168 4

V = e−rh (πuVu + πdVu) = e−(0.05)1/3 (0.456 807× 228. 296 7 + 0.543 193 × 38. 168 4) =122. 953 9

So the European call premium is $122. 953 9.

• Time zero replicating portfolio for the European call

B = V = $122. 953 9


=228. 296 7− 38. 168 4

1000 (1. 189 11)− 1000 (0.840 965 ) = 0.546 118

• calculate the American put premium

Period 2 3Suuu = 1000u

3 = 1681. 380 8Suu = 1000u

2 = 1413. 982 6 Vuuu = 0EVuu = 0Vuu =? Suud = 1000u

2d = 1000u = 1189. 1100Vuud = 0

Sud = 1000ud = 1000EVud = 0Vud =?

Sudd = 1000ud2 = 1000d = 840. 965

Sdd = 1000d2 = 707. 222 1 Vudd = 1000− 840. 965 = 159. 035

EVdd = 1000− 707. 222 1 = 292. 777 9Vdd =? Sddd = 1000d

3 = 1000¡0.840 965 3

¢= 594. 749

Vddd = 1000− 594. 749 = 405. 251

V Ruu = e−rh (πuVuuu + πdVuud) = e−(0.05)1/3 (0.456 807× 0 + 0.543 193 × 0) =

0Vuu = max

¡V Ruu, EVuu

¢= max (0, 0) = 0

V Rud = e−rh (πuVuud + πdVudd) = e−(0.05)1/3 (0.456 807× 0 + 0.543 193 × 159. 035) =

84. 958 9




¢= max (84. 958 9, 0) = 84. 958 9

V Rdd = e−rh (πuVudd + πdVddd) = e−(0.05)1/3 (0.456 807× 159. 035 + 0.543 193 × 405. 251) =

287. 938 62


¢= max (287. 938 62, 292. 777 9) = 292. 777 9

The American put is early exercised at the node dd.

Period 1 2Vuu = 0

Su = 1000 (1. 189 11) = 1189. 11EVu = 0Vu =?

Vud = 84. 958 9Sd = 1000 (0.840 965 ) = 840. 965EVd = 1000− 840. 965 = 159. 035Vd =?

Vdd = 292. 777 9

V Ru = e−rh (πuVuu + πdVud) = e−(0.05)1/3 (0.456 807× 0 + 0.543 193 × 84. 958 9) =

45. 386 3


¢= max (45. 386 3 , 0) = 45. 386 3

V Rd = e−rh (πuVud + πdVdd) = e−(0.05)1/3 (0.456 807× 84. 958 9 + 0.543 193 × 292. 777 9) =

194. 574 6


¢= max (194. 574 6 , 159. 035) = 194. 574 6

Period 0 1Vu = 45. 386 3

S = 1000EV = 0V =?

Vd = 194. 574 6

V R = e−rh (πuVu + πdVu) = e−(0.05)1/3 (0.456 807× 45. 386 3 + 0.543 193 × 194. 574 6) =124. 334 7

V = max¡V R, EV

¢= max (124. 334 7 , 0) = 124. 334 7

So the American put premium is 124. 334 7.

• Calculate the European put premium.



Period 0 1 2 3Suuu = 1000u

3 = 1681. 380 8Vuuu = 0

VuuSuud = 1000u

2d = 1000u = 1189. 1100Vu Vuud = 0

V VudSudd = 1000ud

2 = 1000d = 840. 965Vd Vudd = 1000− 840. 965 = 159. 035

VddSddd = 1000d

3 = 1000¡0.840 965 3

¢= 594. 749

Vddd = 1000− 594. 749 = 405. 251

Vuu = e−rh (πuVuuu + πdVuud) = e−(0.05)1/3 (0.456 807× 0 + 0.543 193 × 0) =0

Vud = e−rh (πuVuud + πdVudd) = e−(0.05)1/3 (0.456 807× 0 + 0.543 193 × 159. 035) =84. 958 9

Vdd = e−rh (πuVudd + πdVddd) = e−(0.05)1/3 (0.456 807× 159. 035 + 0.543 193 × 405. 251) =287. 938 62

Vu = e−rh (πuVuu + πdVud) = e−(0.05)1/3 (0.456 807× 0 + 0.543 193 × 84. 958 9) =45. 386 3

Vd = e−rh (πuVud + πdVdd) = e−(0.05)1/3 (0.456 807× 84. 958 9 + 0.543 193 × 287. 938 62) =191. 989 4

V = e−rh (πuVu + πdVu) = e−(0.05)1/3 (0.456 807× 45. 386 3 + 0.543 193 × 191. 989 4) =122. 9537

So the European call premium is $122. 9537.

• Time zero replicating portfolio for the European call

B = V = $122. 9537


=45. 386 3− 191. 989 4

1000 (1. 189 11)− 1000 (0.840 965 ) = −0.421 1

So at t = 0 we need to enter 0.421 1 futures contract as a seller and deposit122. 9537 in a savings account.

• Why the European call and a European put have the same premium

The standard put-call parity is:C + PV (K) = P + SS is the price of the underlying asset at time zero. For the futures contract,

S is the present value of the forward price:

S = PV (F0,T )In this problem, the forward price and the strike price are equal (i.e. K =

F0,T ). Hence PV (K) = PV (F0,T ) = S. This gives us C = P .



Problem 10.19.

For options on a stock index, you can use the standard binomial formula.u = e(r−δ)h+σ

√h = e(0.05−0.03)1+0.3

√1 = 1. 377 128

d = e(r−δ)h−σ√h = e(0.05−0.03)1−0.3

√1 = 0.755 784


u− d=

e(0.05−0.03)1 − 0.755 7841. 377 128− 0.755 784 = 0.425 557

πd = 1− πu = 1− 0.425 557 = 0.574 443

a. Calculate the price of the European call optionPeriod 0 1 2 3

Suuu = 100¡1. 377 1283

¢= 261. 169 8

Vuuu = 261. 169 8− 95 = 166. 169 8Vuu

Vu Suud = 100¡1. 377 1282

¢(0.755 784) = 143. 333 0

Vuud = 143. 333 0− 95 = 48. 3330V Vud

Sudd = 100 (1. 377 128)¡0.755 7842

¢= 78. 662 9

Vd Vudd = 0Vdd

Sddd = 100¡0.755 7843

¢= 43. 171 1

Vddd = 0

Vuu = e−rh (πuVuuu + πdVuud) = e−(0.05)1 (0.425 557× 166. 169 8 + 0.574 443 × 48. 333) =93. 676 4

Vud = e−rh (πuVuud + πdVudd) = e−(0.05)1 (0.425 557× 48. 3330 + 0.574 443 × 0) =19. 565 3

Vdd = e−rh (πuVudd + πdVddd) = 0

Vu = e−rh (πuVuu + πdVud) = e−(0.05)1 (0.425 557× 93. 676 4 + 0.574 443 × 19. 565 3) =48. 611 4

Vd = e−rh (πuVud + πdVdd) = e−(0.05)1 (0.425 557× 19. 565 3 + 0.574 443 × 0) =7. 920 1

V = e−rh (πuVu + πdVu) = e−(0.05)1 (0.425 557× 48. 611 4 + 0.574 443 × 7. 920 1) =24. 005 8

So the European call premium is $24. 005 8.

b. Calculate the European put premium




¡1. 377 1283

¢= 261. 169 8

Vuuu = 0Vuu

Vu Suud = 100¡1. 377 1282

¢(0.755 784) = 143. 333 0

Vuud = 0V Vud

Sudd = 100 (1. 377 128)¡0.755 7842

¢= 78. 662 9

Vd Vudd = 95− 78. 662 9 = 16. 337 1Vdd

Sddd = 100¡0.755 7843

¢= 43. 171 1

Vddd = 95− 43. 171 1 = 51. 828 9

Vuu = e−rh (πuVuuu + πdVuud) = e−(0.05)1 (0.425 557× 166. 169 8 + 0.574 443 × 48. 333) =0

Vud = e−rh (πuVuud + πdVudd) = e−(0.05)1 (0.425 557× 0 + 0.574 443 × 16. 337 1) =8. 927 0

Vdd = e−rh (πuVudd + πdVddd) = e−(0.05)1 (0.425 557× 16. 337 1 + 0.574 443 × 51. 828 9) =34. 934 0

Vu = e−rh (πuVuu + πdVud) = e−(0.05)1 (0.425 557× 0 + 0.574 443 × 8. 927 0) =4. 8780

Vd = e−rh (πuVud + πdVdd) = e−(0.05)1 (0.425 557× 8. 927 0 + 0.574 443 × 34. 934 0) =22. 702 6

V = e−rh (πuVu + πdVu) = e−(0.05)1 (0.425 557× 4. 8780 + 0.574 443 × 22. 702 6) =14. 379 9

So the European put premium is $14. 379 9.

c. If we switch S and K and switch r and δ and recalculate the option price,what happens?After the switch, we have:

• S = 95

• K = 100

• r = 3%

• δ = 5%

u = e(r−δ)h+σ√h = e(0.03−0.05)1+0.3

√1 = 1. 323 13

d = e(r−δ)h−σ√h = e(0.03−0.05)1−0.3

√1 = 0.726 149


u− d=

e(0.03−0.05)1 − 0.726 1491. 323 13− 0.726 149 = 0.425 557

πd = 1− πu = 1− 0.425 557 = 0.574 443



• Calculate the European call premium


¡1. 323 133

¢= 220. 0550

Vuuu = 220. 0550− 100 = 120. 055Vuu

Suud = 95¡1. 323 132

¢(0.726 149) = 120. 768 7

Vu Vuud = 120. 768 7− 100 = 20. 768 7V Vud

Sudd = 95 (1. 323 13)¡0.726 1492

¢= 66. 279 3

Vd Vudd = 0Vdd

Sddd = 95¡0.726 1493

¢= 36. 374 8

Vddd = 0

Vuu = e−rh (πuVuuu + πdVuud) = e−(0.03)1 (0.425 557× 120. 055 + 0.574 443 × 20. 768 7) =61. 158 1

Vud = e−rh (πuVuud + πdVudd) = e−(0.03)1 (0.425 557× 20. 768 7 + 0.574 443 × 0) =8. 577 1

Vdd = e−rh (πuVudd + πdVddd) = e−(0.03)1 (0.425 557× 0 + 0.574 443 × 0) =0

Vu = e−rh (πuVuu + πdVud) = e−(0.03)1 (0.425 557× 61. 158 1 + 0.574 443 × 8. 577 1) =30. 038 5

Vd = e−rh (πuVud + πdVdd) = e−(0.03)1 (0.425 557× 8. 577 1 + 0.574 443 × 0) =3. 542 2

V = e−rh (πuVu + πdVd) = e−(0.03)1 (0.425 557× 30. 038 5 + 0.574 443 × 3. 542 2) =14. 3799The call premium after the switch is equal to the put premium before the

switch.

• Calculate the European put premium


¡1. 323 133

¢= 220. 0550

Vuuu = 0Vuu

Suud = 95¡1. 323 132

¢(0.726 149) = 120. 768 7

Vu Vuud = 0V Vud

Sudd = 95 (1. 323 13)¡0.726 1492

¢= 66. 279 3

Vd Vudd = 100− 66. 279 3 = 33. 720 7Vdd

Sddd = 95¡0.726 1493

¢= 36. 374 8

Vddd = 100− 36. 374 8 = 63. 625 2



Vuu = e−rh (πuVuuu + πdVuud) = e−(0.03)1 (0.425 557× 0 + 0.574 443 × 0) =0

Vud = e−rh (πuVuud + πdVudd) = e−(0.03)1 (0.425 557× 0 + 0.574 443 × 33. 720 7) =18. 798 1

Vdd = e−rh (πuVudd + πdVddd) = e−(0.03)1 (0.425 557× 33. 720 7 + 0.574 443 × 63. 625 2) =49. 394 8

Vu = e−rh (πuVuu + πdVud) = e−(0.03)1 (0.425 557× 0 + 0.574 443 × 18. 798 1) =10. 479 3

Vd = e−rh (πuVud + πdVdd) = e−(0.03)1 (0.425 557× 18. 798 1 + 0.574 443 × 49. 394 8) =35. 299 1

V = e−rh (πuVu + πdVd) = e−(0.03)1 (0.425 557× 10. 479 3 + 0.574 443 × 35. 299 1) =24. 005 8

The put premium after the switch is equal to the call premium before theswitch.By the way, you can also use the textbook’s spreadsheet "optbasics2" to

calculate the European option premium and verify

• the European call price is $14. 379 9 (which is the European put pricebefore the switch)

• the European put price is $24. 005 8 (which is the European call pricebefore the switch)

What a coincidence, you might wonder. Why? This can be explained usingthe Black-Scholes option formulas:The price of a European call option is:δ

C = Se−δTN (d1)−Ke−rTN (d2) (Textbook 12.1)


P = Ke−rTN (−d2)− Se−δTN (−d1) (Textbook 12.3)

d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

=ln

Se−δT

Ke−rT+1

2σ2T

σ√T

=ln

Se−δT

Ke−rT

σ√T

+1

2σ√T

(Textbook 12.2a)

d2 = d1 − σ√T (Textbook 12.2b)

Before the switch:



C = Se−δTN (d1)−Ke−rTN (d2)P = Ke−rTN (−d2)− Se−δTN (−d1)

d1 =ln

Se−δT

Ke−rT

σ√T

+1

2σ√T

d2 = d1 − σ√T =

lnSe−δT

Ke−rT

σ√T

+1

2σ√T − σ

√T =

lnSe−δT

Ke−rT

σ√T

− 12σ√T

→ C = Se−δTN

⎛⎜⎜⎝ lnSe−δT

Ke−rT

σ√T

+1

2σ√T

⎞⎟⎟⎠−Ke−rTN


Ke−rT

σ√T

− 12σ√T

⎞⎟⎟⎠→ P = Ke−rTN

⎛⎜⎜⎝− lnSe−δT

Ke−rT

σ√T

+1

2σ√T

⎞⎟⎟⎠−Se−δTN⎛⎜⎜⎝− ln

Se−δT

Ke−rT

σ√T

− 12σ√T

⎞⎟⎟⎠After we switch S and K and switch r and δ:C =

£Se−δTN (d1)−Ke−rTN (d2)

¤r←→δ

K←→S= Ke−rTN (d1)− Se−δTN (d2)

P =£Ke−rTN (−d2)− Se−δTN (−d1)

¤r←→δ

K←→S= Se−δTN (−d2)−Ke−rTN (−d1)

d1 =

⎡⎢⎢⎣ lnSe−δT

Ke−rT

σ√T

+1

2σ√T

⎤⎥⎥⎦r←→δ

K←→S

=ln

Ke−rT

Se−δT

σ√T

+1

2σ√T

→ −d1 = −ln

Ke−rT

Se−δT

σ√T

− 12σ√T =

lnSe−δT

Ke−rT

σ√T

− 12σ√T

d2 = d1 − σ√T =

lnKe−rT

Se−δT

σ√T

+1

2σ√T − σ

√T =

lnKe−rT

Se−δT

σ√T

− 12σ√T

→ −d2 = −ln

Ke−rT

Se−δT

σ√T

+1

2σ√T =

− ln Ke−rT

Se−δT

σ√T

+1

2σ√T =

lnSe−δT

Ke−rT

σ√T

+

1

2σ√T

→ C = Ke−rTN

⎛⎜⎜⎝ lnKe−rT

Se−δT

σ√T

+1

2σ√T

⎞⎟⎟⎠−Se−δTN⎛⎜⎜⎝ ln

Ke−rT

Se−δT

σ√T

− 12σ√T

⎞⎟⎟⎠→ P = Se−δTN


Ke−rT

σ√T

+1

2σ√T

⎞⎟⎟⎠−Ke−rTN


Ke−rT

σ√T

− 12σ√T

⎞⎟⎟⎠www.actuary88.com c°Yufeng Guo 81


You can see that the call premium after the switch is equal the put pre-mium before the switch; the put premium after the switch is equal the callpremium before the switch. This conclusion applies to both European optionsand American options. It’s more complex to prove this is true for Americanoption. However, we are not going to worry about the proof.

Intuitively, you can think that the by switching S andK and switching r andδ, we are switching the strike asset and the underlying asset. So after-the-switchput is like a before-the-switch call; after-the-switch call is like a before-the-switchput.

By the way, the payoff of the before-switch call is not equal to the payoff ofthe after-switch put (however the premiums are the same):

Period 3 European call payoff before switch Period 3 European call payoff after switchSuuu = 100

¡1. 377 1283

¢= 261. 169 8 Suuu = 95

¡1. 323 133

¢= 220. 0550

Vuuu = 261. 169 8− 95 = 166. 169 8 Vuuu = 0

Suud = 100¡1. 377 1282

¢(0.755 784) = 143. 333 0 Suud = 95

¡1. 323 132

¢(0.726 149) = 120. 768 7

Vuud = 143. 333 0− 95 = 48. 3330 Vuud = 0

Sudd = 100 (1. 377 128)¡0.755 7842

¢= 78. 662 9 Sudd = 95 (1. 323 13)

¡0.726 1492

¢= 66. 279 3

Vudd = 0 Vudd = 100− 66. 279 3 = 33. 720 7

Sddd = 100¡0.755 7843

¢= 43. 171 1 Sddd = 95

¡0.726 1493

¢= 36. 374 8

Vddd = 0 Vddd = 100− 36. 374 8 = 63. 625 2

the payoff of the before-switch put is not equal to the payoff of the after-switch call (however the premiums are the same):

Period 3 European put payoff before switch Period 3 European call payoff after switchSuuu = 1000u

3 = 1681. 380 8 Suuu = 95¡1. 323 133

¢= 220. 0550

Vuuu = 0 Vuuu = 220. 0550− 100 = 120. 055

Suud = 1000u2d = 1000u = 1189. 1100 Suud = 95

¡1. 323 132

¢(0.726 149) = 120. 768 7

Vuud = 0 Vuud = 120. 768 7− 100 = 20. 768 7

Sudd = 1000ud2 = 1000d = 840. 965 Sudd = 95 (1. 323 13)

¡0.726 1492

¢= 66. 279 3

Vudd = 1000− 840. 965 = 159. 035 Vudd = 0

Sddd = 1000d3 = 1000

¡0.840 965 3

¢= 594. 749 Sddd = 95

¡0.726 1493

¢= 36. 374 8

Vddd = 1000− 594. 749 = 405. 251 Vddd = 0

Problem 10.20.



For options on a stock index, you can use the standard binomial formula.u = e(r−δ)h+σ

√h = e(0.05−0.03)1+0.3

√1 = 1. 377 128

d = e(r−δ)h−σ√h = e(0.05−0.03)1−0.3

√1 = 0.755 784


u− d=

e(0.05−0.03)1 − 0.755 7841. 377 128− 0.755 784 = 0.425 557

πd = 1− πu = 1− 0.425 557 = 0.574 443

a. Calculate the price of the American call optionPeriod 2 3

Suuu = 100¡1. 377 1283

¢= 261. 169 8

Suu = 100¡1. 377 1282

¢= 189. 648 2 Vuuu = 261. 169 8− 95 = 166. 169 8

EVuu = 189. 648 2− 95 = 94. 648 2Vuu =?

Suud = 100¡1. 377 1282

¢(0.755 784) = 143. 333 0

Sud = 100 (1. 377 128) (0.755 784) = 104. 081 1 Vuud = 143. 333 0− 95 = 48. 3330EVud = 104. 081 1− 95 = 9. 081 1Vud =?

Sudd = 100 (1. 377 128)¡0.755 7842

¢= 78. 662 9

Sudd = 100¡0.755 7842

¢= 57. 120 9 Vudd = 0


¡0.755 7843

¢= 43. 171 1

Vddd = 0

V Ruu = e−rh (πuVuuu + πdVuud) = e−(0.05)1 (0.425 557× 166. 169 8 + 0.574 443 × 48. 333) =

93. 676 4Vuu = max

¡V Ruu, EVuu

¢= max (93. 676 4, 94. 648 2) = 94. 648 2

V Rud = e−rh (πuVuud + πdVudd) = e−(0.05)1 (0.425 557× 48. 3330 + 0.574 443 × 0) =

19. 565 3Vud = max

¡V Rud, EVud

¢= max (19. 565 3, 9. 081 1) = 19. 565 3

V Rdd = e−rh (πuVudd + πdVddd) = 0


¢= 0

Period 1 2Vuu = 94. 648 2

Su = 100 (1. 377 128) = 137. 712 8EVu = 137. 712 8− 95 = 42. 712 8Vu =?

Vud = 19. 565 3Sd = 100 (0.755 784) = 75. 578 4EVd = 0Vd =?

Vdd = 0



V Ru = e−rh (πuVuu + πdVud) = e−(0.05)1 (0.425 557× 94. 648 2 + 0.574 443 × 19. 565 3) =

49. 004 8Vu = max

¡V Ru , EVu

¢= 49. 004 8

V Rd = e−rh (πuVud + πdVdd) = e−(0.05)1 (0.425 557× 19. 565 3 + 0.574 443 × 0) =

7. 920 1Vd = max

¡V Rd , EVd

¢= 7. 920 1

Period 0 1Vu = 49. 004 8

S = 100EV = 100− 95 = 5V =?

Vd = 7. 920 1

V R = e−rh (πuVu + πdVu) = e−(0.05)1 (0.425 557× 49. 004 8 + 0.574 443 × 7. 920 1) =24. 165 0

V = max¡V R, EV

¢= 24. 165 0

So the American call premium is $24. 165 0.

b. Calculate the American put premiumPeriod 2 3

Suuu = 100¡1. 377 1283

¢= 261. 169 8

Suu = 100¡1. 377 1282

¢= 189. 648 2 Vuuu = 0

EVuu = 0Vuu =?

Suud = 100¡1. 377 1282

¢(0.755 784) = 143. 333 0

Sud = 100 (1. 377 128) (0.755 784) = 104. 081 1 Vuud = 0EVud = 0Vud =?

Sudd = 100 (1. 377 128)¡0.755 7842

¢= 78. 662 9

Sudd = 100¡0.755 7842

¢= 57. 120 9 Vudd = 95− 78. 662 9 = 16. 337 1

EVdd = 95− 57. 120 9 = 37. 879 1Vdd =? Sddd = 100

¡0.755 7843

¢= 43. 171 1

Vddd = 95− 43. 171 1 = 51. 828 9

V Ruu = e−rh (πuVuuu + πdVuud) = e−(0.05)1 (0.425 557× 0 + 0.574 443 × 0) =

0Vuu = max

¡V Ruu, EVuu

¢= 0

V Rud = e−rh (πuVuud + πdVudd) = e−(0.05)1 (0.425 557× 0 + 0.574 443 × 16. 337 1) =

8. 927 0Vud = max

¡V Rud, EVud

¢= 8. 927 0

V Rdd = e−rh (πuVudd + πdVddd) = e−(0.05)1 (0.425 557× 16. 337 1 + 0.574 443 × 51. 828 9) =

34. 934 0Vud = max

¡V Rdd, EVdd

¢= 37. 879 1



Period 1 2Vuu = 0

Su = 100 (1. 377 128) = 137. 712 8EVu = 0Vu =?

Vud = 8. 927 0Sd = 100 (0.755 784) = 75. 578 4EVd = 85− 75. 578 4 = 9. 421 6Vd =?

Vdd = 37. 879 1

V Ru = e−rh (πuVuu + πdVud) = e−(0.05)1 (0.425 557× 0 + 0.574 443 × 8. 927 0) =

4. 8780Vu = max

¡V Ru , EVu

¢= 4. 8780

V Rd = e−rh (πuVud + πdVdd) = e−(0.05)1 (0.425 557× 8. 927 0 + 0.574 443 × 37. 879 1) =

24. 311 8Vd = max

¡V Rd , EVd

¢= 24. 311 8

Period 0 1Vu = 4. 8780

S = 100EV = 0V =?

Vd = 24. 311 8

V R = e−rh (πuVu + πdVu) = e−(0.05)1 (0.425 557× 4. 8780 + 0.574 443 × 24. 311 8) =15. 259 3

V = max¡V R, EV

¢= 15. 259 3

So the American put premium is $15. 259 3

c. If we switch S and K and switch r and δ and recalculate the option price,what happens?After the switch, we have:

• S = 95

• K = 100

• r = 3%

• δ = 5%

u = e(r−δ)h+σ√h = e(0.03−0.05)1+0.3

√1 = 1. 323 13

d = e(r−δ)h−σ√h = e(0.03−0.05)1−0.3

√1 = 0.726 149




u− d=

e(0.03−0.05)1 − 0.726 1491. 323 13− 0.726 149 = 0.425 557

πd = 1− πu = 1− 0.425 557 = 0.574 443

Calculate the price of the American call optionPeriod 2 3

Suuu = 95¡1. 323 133

¢= 220. 0550

Suu = 95¡1. 323 132

¢= 166. 313 9 Vuuu = 220. 0550− 100 = 120. 055

EVuu = 166. 313 9− 100 = 66. 313 9Vuu =?

Suud = 95¡1. 323 132

¢(0.726 149) = 120. 768 7

Sud = 95 (1. 323 13) (0.726 149) = 91. 275 0 Vuud = 120. 768 7− 100 = 20. 768 7EVud = 0Vud =?

Sudd = 95 (1. 323 13)¡0.726 1492

¢= 66. 279 3

Sudd = 95¡0.726 1492

¢= 50. 092 8 Vudd = 0


¡0.726 1493

¢= 36. 374 8

Vddd = 0

V Ruu = e−rh (πuVuuu + πdVuud) = e−(0.03)1 (0.425 557× 120. 055 + 0.574 443 × 20. 768 7) =

61. 158 1Vuu = max

¡V Ruu, EVuu

¢= 66. 313 9

V Rud = e−rh (πuVuud + πdVudd) = e−(0.03)1 (0.425 557× 20. 768 7 + 0.574 443 × 0) =

8. 577 1Vud = max

¡V Rud, EVud

¢= 8. 577 1

V Rdd = e−rh (πuVudd + πdVddd) = 0


¢= 0

Period 1 2Vuu = 66. 313 9

Su = 95 (1. 323 13) = 125. 697 4EVu = 125. 697 4− 100 = 25. 697 4Vu =?

Vud = 8. 577 1Sd = 95 (0.726 149) = 68. 984 2EVd = 0Vd =?

Vdd = 0

V Ru = e−rh (πuVuu + πdVud) = e−(0.03)1 (0.425 557× 66. 313 9 + 0.574 443 × 8. 577 1) =

32. 167 7Vu = max

¡V Ru , EVu

¢= 32. 167 7



V Rd = e−rh (πuVud + πdVdd) = e−(0.03)1 (0.425 557× 8. 577 1 + 0.574 443 × 0) =

3. 542 2


¢= 3. 542 2

Period 0 1Vu = 32. 167 7

S = 95EV = 0V =?

Vd = 3. 542 2

V R = e−rh (πuVu + πdVu) = e−(0.03)1 (0.425 557× 32. 167 7 + 0.574 443 × 3. 542 2) =15. 2593

V = max¡V R, EV

¢= 15. 2593

So the American call premium is $15. 2593, which is equal to the Americanput premium before the switch.

Calculate the American put premiumPeriod 2 3

Suuu = 95¡1. 323 133

¢= 220. 0550

Suu = 95¡1. 323 132

¢= 166. 313 9 Vuuu = 0

EVuu = 0Vuu =?

Suud = 95¡1. 323 132

¢(0.726 149) = 120. 768 7

Sud = 95 (1. 323 13) (0.726 149) = 91. 275 0 Vuud = 0EVud = 100− 91. 275 0 = 8. 725Vud =?

Sudd = 95 (1. 323 13)¡0.726 1492

¢= 66. 279 3

Sudd = 95¡0.726 1492

¢= 50. 092 8 Vudd = 100− 66. 279 3 = 33. 720 7

EVdd = 100− 50. 092 8 = 49. 907 2Vdd =? Sddd = 95

¡0.726 1493

¢= 36. 374 8

Vddd = 100− 36. 374 8 = 63. 625 2

V Ruu = e−rh (πuVuuu + πdVuud) = e−(0.03)1 (0.425 557× 0 + 0.574 443 × 0) =

0


¢= 0

V Rud = e−rh (πuVuud + πdVudd) = e−(0.03)1 (0.425 557× 0 + 0.574 443 × 33. 720 7) =

18. 798 1


¢= 18. 798 1

V Rdd = e−rh (πuVudd + πdVddd) = e−(0.03)1 (0.425 557× 33. 720 7 + 0.574 443 × 63. 625 2) =

49. 394 8

Vud = max¡V Rdd, EVdd

¢= 49. 907 2



Period 1 2Vuu = 0

Su = 95 (1. 323 13) = 125. 697 4EVu = 0Vu =?

Vud = 18. 798 1Sd = 95 (0.726 149) = 68. 984 2EVd = 100− 68. 984 2 = 31. 015 8Vd =?

Vdd = 49. 907 2

V Ru = e−rh (πuVuu + πdVud) = e−(0.03)1 (0.425 557× 0 + 0.574 443 × 18. 798 1) =

10. 479 3Vu = max

¡V Ru , EVu

¢= 10. 479 3

V Rd = e−rh (πuVud + πdVdd) = e−(0.03)1 (0.425 557× 18. 798 1 + 0.574 443 × 49. 907 2) =

35. 584 8Vd = max

¡V Rd , EVd

¢= 35. 584 8

Period 0 1Vu = 10. 479 3

S = 95EV = 100− 95 = 5V =?

Vd = 35. 584 8

V R = e−rh (πuVu + πdVu) = e−(0.03)1 (0.425 557× 10. 479 3 + 0.574 443 × 35. 584 8) =24. 165 04

V = max¡V R, EV

¢= 24. 165 0

So the American put premium is $24. 165 0, which is equal to the Americancall premium before the switch.

Problem 10.21.

Suppose u < e(r−δ)h. Since d < u, we have d < u < e(r−δ)h. This meansthat the savings account is always better than the stock. So at t = 0, we shortsell e−δh share of stock and investment the short sale proceeds Se−δh into thesavings account. Then at time h, we close our short position by buying onestock from the market. The stock price at time h is either uS or dS.

t = 0 t = h, u mode t = h, d modeshort e−δh stock Se−δh −uS −dSdeposit Se−δh in savings −Se−δh Se(r−δ)h Se−δh

Total 0 S¡e−δh − u

¢> 0 S

¡e−δh − d

¢> 0

So initial cost is zero yet we have positive payoff at time h. This is anarbitrage opportunity.



Suppose d > e(r−δ)h. Since d < u, we have u > d > e(r−δ)h. This means thatthe investing in stocks is always better off than investing in a savings account.So at t = 0, we buy e−δh share of stock and borrow money from a bank tofinance the purchase.

t = 0 t = h, u mode t = h, d modebuy e−δh stock −Se−δh uS dS

borrow Se−δh in savings Se−δh −Se(r−δ)h −Se−δhTotal 0 S

¡u− e−δh

¢> 0 S

¡d− e−δh

¢> 0

So initial cost is zero yet we have positive payoff at time h. This is anarbitrage opportunity.




Chapter 11

Binomial option pricing II

Problem 11.1.

u = e(r−δ)h+σ√h = e(0−0.08)1+0.3

√1 = 1. 246 077

d = e(r−δ)h−σ√h = e(0−0.08)1−0.3

√1 = 0.683 861


u− d=

e(0−0.08)1 − 0.683 8611. 246 077− 0.683 861 = 0.425 56

πd = 1− πu = 1− 0.425 56 = 0.574 44

• K = 70

t = 0 t = 1Su = 100 (1. 246 077) = 124. 607 7Vu = 124. 607 7− 70 = 54. 607 7

S = 100EV = 100− 70 = 30V Sd = 100 (0.683 861 ) = 68. 386 1

Vd = 0

V R = e−rh (πuVu + πdVu) = e−0(1) (0.425 56× 54. 607 7 + 68. 386 1× 0) =23. 24

V = max¡V R, EV

¢= max (23. 24, 30) = 30

• K = 80

t = 0 t = 1Su = 100 (1. 246 077) = 124. 607 7Vu = 124. 607 7− 80 = 44. 607 7

S = 100EV = 100− 80 = 20V Sd = 100 (0.683 861 ) = 68. 386 1

Vd = 0

91

CHAPTER 11. BINOMIAL OPTION PRICING II

V R = e−rh (πuVu + πdVu) = e−0(1) (0.425 56× 44. 607 7 + 68. 386 1× 0) =18. 98

V = max¡V R, EV

¢= max (18. 98, 20) = 20

• K = 90

t = 0 t = 1Su = 100 (1. 246 077) = 124. 607 7Vu = 124. 607 7− 90 = 34. 607 7

S = 100EV = 100− 90 = 10V Sd = 100 (0.683 861 ) = 68. 386 1

Vd = 0

V R = e−rh (πuVu + πdVu) = e−0(1) (0.425 56× 34. 607 7 + 68. 386 1× 0) =14. 73

V = max¡V R, EV

¢= max (14. 73, 10) = 14. 73

• K = 100

t = 0 t = 1Su = 100 (1. 246 077) = 124. 607 7Vu = 124. 607 7− 100 = 24. 607 7

S = 100EV = 100− 100 = 0V Sd = 100 (0.683 861 ) = 68. 386 1

Vd = 0

V R = e−rh (πuVu + πdVu) = e−0(1) (0.425 56× 24. 607 7 + 68. 386 1× 0) =10. 47

V = max¡V R, EV

¢= max (10. 47, 0) = 10. 47

a. Early exercise occurs at t = 0 with K = 70, 80

b. The European options satisfy the put-call parity:CEur +Ke−rT = PEur + Se−δT

CEur = PEur + Se−δT −Ke−rT = PEur + 100e−0.08 −K = PEur −K +92.

311 6Since K = 70, 80, 90, 100 ≤ 100, we haveEV = max (0, S −K) = max (0, 100−K) = 100−KTo have EV > CEur, we need to have:100−K > PEur −K + 92. 311 6→ PEur < 100− 100e−0.08 = 7. 69To early exercise the American call at t = 0, the European put premium

needs to be less than 7. 69.



Clearly, the smaller the European put premium is, the more likely the earlyexercise of the American call is optimal at t = 0.

Everything else equal, the higher the strike price, the higher the price of aEuropean put. As a result, it can be optimal to early exercise the Americancall with K = 70, 80, while it’s not optimal to early exercise the American withK = 90, 100.Use the European put premium using the Black-Scholes option formula (the

textbook Equation 12.3):

P = Ke−rTN (−d2)− Se−δTN (−d1)

We find the following put price (S = 100, T = 1, r = 0, δ = 0.08, σ = 0.3):K PEur70 $2.308780 $5.253790 $9.7517100 $15.7113

Clearly, the condition PEur < 7. 69 is met when K = 70, 80 and violatedwhen K = 90, 100.

By the way, the price calculated using the Black-Scholes option formula won’tmatch the price calculated under the binomial option formula under h = 1.This is because the Black-Scholes option formula is the binomial option pricingmethod where n→∞ and h = T/n→ 0 .

c.To early exercise the American call, we need to have PEur < 7. 69. Thiscondition is met if K = 70, 80 and violated when K = 90, 100.

Problem 11.2.

Now r = 0.08 instead of r = 0. This increases the cost of early exercisingan American call. By early exercising an American call, you lost interest onthe strike asset K. We expect that it’s still not optimal to early exercise theAmerican call at K = 90, 100 (it’s not optimal to early exercise the Americancall at these strike prices even when r = 0, let alone when r = 0.08). However,we are not clear whether it’s optimal to early exercise the American call whenK = 70, 80. We have to check.

u = e(r−δ)h+σ√h = e(0.08−0.08)1+0.3

√1 = 1. 349 859

d = e(r−δ)h−σ√h = e(0.08−0.08)1−0.3

√1 = 0.740 818


u− d=

e(0.08−0.08)1 − 0.740 8181. 349 859− 0.740 818 = 0.425 558

πd = 1− πu = 1− 0.425 558 = 0.574 442



• K = 70

t = 0 t = 1Su = 100 (1. 349 859) = 134. 985 9Vu = 134. 985 9− 70 = 64. 985 9

S = 100EV = 100− 70 = 30V Sd = 100 (0.740 818 ) = 74. 081 8

Vd = 74. 081 8− 70 = 4. 081 8

V R = e−rh (πuVu + πdVu) = e−0.08(1) (0.425 558× 64. 985 9 + 0.574 442× 4. 081 8) =27. 69

V = max¡V R, EV

¢= max (27. 69, 25) = 27. 69

• K = 80

t = 0 t = 1Su = 100 (1. 349 859) = 134. 985 9Vu = 134. 985 9− 80 = 54. 985 9

S = 100EV = 100− 80 = 20V Sd = 100 (0.740 818 ) = 74. 081 8

Vd = 0

V R = e−rh (πuVu + πdVu) = e−0.08(1) (0.425 558× 54. 985 9 + 0.574 442× 0) =21. 60

V = max¡V R, EV

¢= max (21. 60, 20) = 21. 60

• K = 90

t = 0 t = 1Su = 100 (1. 349 859) = 134. 985 9Vu = 134. 985 9− 90 = 44. 985 9

S = 100EV = 100− 90 = 10V Sd = 100 (0.740 818 ) = 74. 081 8

Vd = 0

V R = e−rh (πuVu + πdVu) = e−0.08(1) (0.425 558× 44. 985 9 + 0.574 442× 0) =17. 67

V = max¡V R, EV

¢= max (17. 67, 10) = 17. 67

• K = 100



t = 0 t = 1Su = 100 (1. 349 859) = 134. 985 9Vu = 134. 985 9− 100 = 34. 985 9

S = 100EV = 0V Sd = 100 (0.740 818 ) = 74. 081 8

Vd = 0

V R = e−rh (πuVu + πdVu) = e−0.08(1) (0.425 558× 34. 985 9 + 0.574 442× 0) =13. 74

V = max¡V R, EV

¢= max (13. 74, 0) = 13. 74

a. Early exercise occurs at t = 0 with K = 70. Now the risk free rate is nolonger zero and we’ll lose interest on the strike price by exercising the Americancall early. The higher the strike price, the more interest we lose. In contrast,the previous problem has r = 0 and we don’t lose any interest by early exercise.It makes sense that if r is not zero (i.e. positive) then fewer strike prices willlead to optimal exercise than if r = 0.

b, c. The European options satisfy the put-call parity:CEur +Ke−rT = PEur + Se−δT

CEur = PEur + Se−δT −Ke−rT = PEur + 100e−0.08 −Ke−0.08

Since K = 70, 80, 90, 100 ≤ 100, we haveEV = max (0, S −K) = max (0, 100−K) = 100−KTo have EV > CEur, we need to have:100−K > PEur + 100e

−0.08 −Ke−0.08

→ PEur < 100− 100e−0.08 −K¡1− e−0.08

¢= 7. 69− 0.0769K

Using the Black-Scholes option pricing formula, we find the following putprice (S = 100, T = 1, r = 0.08, δ = 0, σ = 0.3):

K PEur 7. 69− 0.0769K70 0.7752 7. 69− 0.0769 (70) = 2. 30780 2.0904 7. 69− 0.0769 (80) = 1. 53890 4.4524 7. 69− 0.0769 (90) = 0.769100 8.0229 7. 69− 0.0769 (100) = 0

Among the 4 strike prices, only K = 70 satisfies the condition PEur < 7.69− 0.0769K. Hence of the 4 strike prices given, only K = 70 leads to optimalearly exercise.

Problem 11.3.

If δ = 0, then the stock doesn’t pay any dividend. It’s never optimal to earlyexercise an American call on a non-dividend paying stock. So early exercise willnever occur.



Problem 11.4.

u = e(r−δ)h+σ√h = e(0.08−0)1+0.3

√1 = 1. 462 285

d = e(r−δ)h−σ√h = e(0.08−0)1−0.3

√1 = 0.802 519


u− d=

e(0.08−0)1 − 0.802 5191. 462 285− 0.802 519 = 0.425 557

πd = 1− πu = 1− 0.425 557 = 0.574 443

• K = 100

t = 0 t = 1Su = 100 (1. 462 285) = 146. 228 5Vu = 0

S = 100EV = 0V Sd = 100 (0.802 519 ) = 80. 251 9

Vd = 100− 80. 251 9 = 19. 748 1

V R = e−rh (πuVu + πdVu) = e−0.08(1) (0.425 557× 0 + 0.574 443× 19. 748 1) =10. 4720

V = max¡V R, EV

¢= max (10. 4720, 0) = 10. 4720

• K = 110

t = 0 t = 1Su = 100 (1. 462 285) = 146. 228 5Vu = 0

S = 100EV = 110− 100 = 10V Sd = 100 (0.802 519 ) = 80. 251 9

Vd = 110− 80. 251 9 = 29. 748 1

V R = e−rh (πuVu + πdVu) = e−0.08(1) (0.425 557× 0 + 0.574 443× 29. 748 1) =15. 774 8

V = max¡V R, EV

¢= max (15. 774 8, 10) = 15. 774 8

• K = 120

t = 0 t = 1Su = 100 (1. 462 285) = 146. 228 5Vu = 0

S = 100EV = 120− 100 = 20V Sd = 100 (0.802 519 ) = 80. 251 9

Vd = 120− 80. 251 9 = 39. 748 1



V R = e−rh (πuVu + πdVu) = e−0.08(1) (0.425 557× 0 + 0.574 443× 39. 748 1) =21. 077 5

V = max¡V R, EV

¢= max (21. 077 5, 20) = 21. 077 5

• K = 130

t = 0 t = 1Su = 100 (1. 462 285) = 146. 228 5Vu = 0

S = 100EV = 130− 100 = 30V Sd = 100 (0.802 519 ) = 80. 251 9

Vd = 130− 80. 251 9 = 49. 748 1

V R = e−rh (πuVu + πdVu) = e−0.08(1) (0.425 557× 0 + 0.574 443× 49. 748 1) =26. 380 3

V = max¡V R, EV

¢= max (26. 380 3, 30) = 30

a. Early exercise occurs at t = 0 with K = 130.

b, c. The European options satisfy the put-call parity:CEur +Ke−rT = PEur + Se−δT

PEur = CEur +Ke−rT − Se−δT = CEur +Ke−0.08(1) − 100e−0(1) = CEur +Ke−0.08 − 100

The early exercise value isEV = max (0,K − S)Since K = 100, 100, 120, 130 ≥ 100, we haveEV = max (0,K − S) = K − S = K − 100To have EV > CEur, we need to have:K − 100 > CEur +Ke−0.08 − 100→ CEur < K

¡1− e−0.08

¢= 0.07 69K

Using the Black-Scholes option pricing formula, we find the following putprice (S = 100, T = 1, r = 0.08, δ = 0, σ = 0.3):

K CEur 0.07 69K100 15.7113 0.07 69 (100) = 7. 69110 11.2596 0.07 69 (110) = 8. 459120 7.8966 0.07 69 (120) = 9. 228130 5.4394 0.07 69 (130) = 9. 997

Among the 4 strike prices, only K = 130 satisfies the condition CEur < 0.0769K. Hence of the 4 strike prices given, only K = 130 leads to optimal earlyexercise.



The main reason for early exercising an American put is to earn interest onthe strike price. Hence higher the strike price, everything else equal, the morelikely an American put may be early exercised.

Problem 11.5.

Now δ = 0.08 as opposed to δ = 0 in the previous problem. This increasesthe cost of early exercising an American put. If you early exercise an Americanput, you can earn interest on the strike asset, but you also lose the opportunityof earning dividend on the stock.With δ = 0.08, we expect that it’s still not optimal to early exercise the

American put whenK = 100, 110, 120. Because it’s not optimal to early exercisethe American put with these strike prices even when δ = 0, let alone δ = 0.08.However, we are not clear whether it’s optimal to early exercise the Americanput when K = 130. We have to check.The put-call parity is:CEur +Ke−rT = PEur + Se−δT

CEur +Ke−0.08 = PEur + 100e−0.08

→ PEur = CEur +Ke−0.08 − 100e−0.08EV = max (0,K − S) = K − S = K − 100Early exercise if:K − 100 > CEur +Ke−0.08 − 100e−0.08→ CEur < K −Ke−0.08 − 100 + 100e−0.08 = 0.07 69K − 7. 69

Using the Black-Scholes option pricing formula, we find the following putprice (S = 100, T = 1, r = 0.08, δ = 0.08, σ = 0.3):

K CEur 0.07 69K 0.07 69K − 7. 69100 15.6584 0.07 69 (100)− 7. 69 = 0110 11.2165 0.07 69 (110)− 7. 69 = 0.769120 7.8627 0.07 69 (120)− 7. 69 = 1. 538130 5.4136 0.07 69 (130)− 7. 69 = 2. 307The condition CEur < 0.07 69K − 7. 69 is met only when K = 130. So it’s

optimal to early exercise the American put only when K = 130.

Problem 11.6.

The only reason to early exercise an American put is to earn interest on thestrike asset. If r = 0 we’ll never earn any interest on the strike price. So it’snever optimal to early exercise an American put if r = 0.We can also use the put-call parity to verify that it’s never optimal to early

exercise the American put when r = 0.CEur +Ke−rT = PEur + Se−δT

CEur +K = PEur + 100e−0.08



→ PEur = CEur +K − 100e−0.08EV = max (0,K − S) = K − S = K − 100Early exercise if:K − 100 > CEur +K − 100e−0.08→ CEur < −100 + 100e−0.08 = −7. 688Since CEur ≥ 0, it’s never optimal to early exercise the American put.

Problem 11.7.

Skip. This is a spreadsheet problem. Most likely SOA and CAS won’t askyou to build a 10 period binomial model in the exam.

Problem 11.8.


Problem 11.9.


Problem 11.10.


Problem 11.11.


Problem 11.12.


Problem 11.13.


Problem 11.14.



The textbook Section 10 Equation 10.7 says that the undiscounted risk-neutral stock price is the forward price:

time t time t+ hSu

SSd

πuSu + πdSd = Se(r−δ)h = Ft,t+h (Textbook Equation 10.7)

We are reusing the data in the textbook Figure 11.4 except that we setS = 100. We have:

S = 100 σ = 0.3 T = 1 h = T/n = 1/3r = 0.08 δ = 0

Then:u = e(r−δ)h+σ

√h = e(0.08−0)1/3+0.3

√1/3 = 1. 221 246

d = e(r−δ)h−σ√h = e(0.08−0)1/3−0.3

√1/3 = 0.863 693


u− d=

e(0.08−0)1/3 − 0.863 6931. 221 246 − 0.863 693 = 0.456 806

πd = 1− πu = 1− 0.456 806 = 0.543 194

The stock price tree is:Period 0 1

Su = 100 (1. 221 246) = 122. 124 6S = 100

Sd = 100 (0.863 693) = 86. 369 3


¡1. 221 2462

¢= 182. 141 7

Suu = 100¡1. 221 2462

¢= 149. 144 2

Suud = 100¡1. 221 2462

¢(0.863 693) = 128. 814 8

Sud = 100 (1. 221 246) (0.863 693) = 105. 478 2Sudd = 100 (1. 221 246)

¡0.863 6932

¢= 91. 100 8

Sdd = 100¡0.863 6932

¢= 74. 596 6

Sddd = 100¡0.863 6933

¢= 64. 428 5

a.The forward price is calculated using the textbook Equation 5.7:F0,T = S0e

(r−δ)T

The 4-month forward price is:F0,1/3 = 100e

(0.08−0)1/3 = 102. 702 5

The 8-month forward price is:F0,2/3 = 100e

(0.08−0)2/3 = 105. 478 1



The 1-year forward price is:F0,1 = 100e

(0.08−0)1 = 108. 328 7

b.u = e(r−δ)h+σ

√h = e(0.08−0)1/3+0.3

√1/3 = 1. 221 246

d = e(r−δ)h−σ√h = e(0.08−0)1/3−0.3

√1/3 = 0.863 693


u− d=

e(0.08−0)1/3 − 0.863 6931. 221 246 − 0.863 693 = 0.456 806

πd = 1− πu = 1− 0.456 806 = 0.543 194

t = 1/3 Stock Price Risk neutral probSu = 100 (1. 221 246) = 122. 124 6 πuSd = 100 (0.863 693) = 86. 369 3 πdTotal 1The undiscounted risk-neutral expected stock price at t = 1/3 is:πuSu + πdSd = 0.456 806 (122. 124 6) + 0.543 194 (86. 369 3) = 102. 702 53 =

F0,1/3

t = 2/3 Stock Price Risk neutral probSuu = 100

¡1. 221 2462

¢= 149. 144 2 π2u

Sud = 100 (1. 221 246) (0.863 693) = 105. 478 2 2πuπdSdd = 100

¡0.863 6932

¢= 74. 596 6 π2d

Total 1The undiscounted risk-neutral expected stock price at t = 2/3 is:π2uSuu+π

2dSdd+2πuπdS

ud = 0.456 8062 (149. 144 2)+2 (0.456 806) (0.543 194) (105. 478 2)+0.543 1942 (74. 596 6) = 105. 478 1 = F0,2/3

t = 1 Stock Price Risk neutral probSuuu = 100

¡1. 221 2462

¢= 182. 141 7 π3u

Suud = 100¡1. 221 2462

¢(0.863 693) = 128. 814 8 3π2uπd

Sudd = 100 (1. 221 246)¡0.863 6932

¢= 91. 100 8 3πuπ

2d

Sddd = 100¡0.863 6933

¢= 64. 428 5 π3d

Total 1

The undiscounted risk-neutral expected stock price at t = 2/3 is:π3uSuuu + 3π

2uπdS

uud + 3πuπ2dS

udd + π3dSddd

= 0.456 8063 (182. 141 7)+3¡0.456 8062

¢(0.543 194) 128. 814 8+3 (0.456 806)

¡0.543 1942

¢91.

100 8 +¡0.543 1943

¢64. 428 5 = 108. 328 7 = F0,1

Problem 11.15.


Problem 11.16.




Problem 11.17.


Problem 11.18.

Skip. If you understand the volatility calculation examples in my studyguide, you are fine.

Problem 11.19.

Skip. If you understand the volatility calculation examples in my studyguide, you are fine.

Problem 11.20.

This is a labor intensive problem. However, it’s a good practice problemfor using the Schroder method. The solution is similar to the textbook Figure11.11.

First, we’ll build a prepaid forward price tree.

σF = σS ×StFPt,T

FPt,T = St − PVt (D) = 50− 4e−0.08(0.25) = 46. 079 21

σF = σS ×StFPt,T

= 0.3× 50

46. 079 2= 0.325 5265

u = erh+σF√h = e0.08(0.25)+0.325 5265

√0.25 = 1. 200 53

d = erh−σF√h = e0.08(0.25)−0.325 5265

√0.25 = 0.866 959


Time 0 0.25 0.5 0.75 195.71879

79.7304466.41269 69.12305

55.31947 57.5771046.07921 47.95973 49.91701

39.94880 41.5791434.63398 36.04742

30.0262526.03154



For example, the prepaid forward price at t = 0.25 is calculated as follows:³FPt+h,T

ú= FP

t,Tu = 46. 079 21× 1. 200 53 = 55. 319 47³FPt+h,T

´d= FP

t,Td = 46. 079 21× 0.866 959 = 39. 948 8

The prepaid forward price at t = 0.5 is calculated as follows:³FPt+2h,T

úu= FP

t,Tu2 = 46. 079 21× 1. 200 532 = 66. 412 7³

FPt+2h,T

úd= FP

t,Tud = 46. 079 21× 1. 200 53× 0.866 959 = 47. 959 73³FPt+2h,T

´dd= FP

t,Td2 = 46. 079 21× 0.866 9592 = 34. 633 98

(My numbers are calculated using Excel so you might not be able to fullymatch mine.)Next, convert the prepaid forward price tree into a stock price tree. The

one-to-one mapping between the prepaid forward price and the stock price is

St+∆t = FPt+∆t,T+PV (Div) = FP

t+∆t,T+

½De−r(TD−t−∆t) if TD ≥ t+∆t

0 if TD < t+∆t

The PV of the dividend at each interval is:Time 0 0.25 0.5 0.75 1Dividend time 0.25Dividend amount $4PV (Div) 3. 920 79 4 0 0 0

PVt=0 (Div) = 4e−0.08(0.25−0) = 3. 920 79We discount $4 from t = 0.25 to time zero

PVt=0.25 (Div) = 4e−0.08(0.25−0.25) = 4We discount $4 from t = 0.25 to t = 0.25

For t > 0.25, we have PVt (Div) = 0

Next, we add the PV (Div) to the prepaid forward price:Time 0 0.25 0.5 0.75 1

95.7187979.73044

66.41269 69.1230555.31947 + 4 57.57710

46.07921 + 3. 920 79 47.95973 49.9170139.94880 + 4 41.57914

34.63398 36.0474230.02625

26.03154

Now the stock price tree is:



Time 0 0.25 0.5 0.75 195.718794

79.73044066.412694 69.123048

59.319474 57.57710450 47.959733 49.917007

43.948797 41.57913834.633981 36.047421

30.02625326.031540

The risk neutral probabilities are:

πu =erh − d

u− d=

e0.08(0.25) − 0.866 9591. 200 53− 0.866 959 = 0.459 399

πd = 1− 0.459 399 = 0.540 601

From this point on, we can just use the standard binomial tree formula.First, we calculate the European call premium. We start from right to left,calculating the roll-back value.

Time 0 0.25 0.5 0.75 150.71879

35.6215023.17717 24.12305

14.27212 13.468168.43381 7.23801 4.91701

3.78761 2.214140.99703 0

00

In the above table, the final column is the call payoff. For example,50.71879 = max (0, 95.718794− 45) = 50. 718 794

The first 4 columns are the roll-back values. For example,35.62150 = e−0.08(0.25) (50.71879× 0.459 399 + 24.12305× 0.540 601) = 35.

621 5048.43381 = e−0.08(0.25) (14.27212× 0.459 399 + 3.78761× 0.540 601) = 8. 433 809The European call premium is 8.43381.

To calculate the American call premium, we still work from right to left.However, we’ll need to compare the roll back value and the exercise value andtake the greater of the two.

The American call premium is calculated as follows:



Time 0 0.25 0.5 0.75 150.71879

35.6215023.17717 24.12305

max (14.27212, 14.31947) 13.468168.45513 7.23801 4.91701

3.78761 2.214140.99703 0

00

It’s optimal to exercise the American call at the upper node at t = 0.25. Theroll back value is 14.27212. The exercise value is 59.319474− 45 = 14. 319 474,which is greater than the roll back value. The premium of 8.45513 is calculatedas follows:The roll back value is: e−0.08(0.25) (14. 319 474× 0.459 399 + 3.78761× 0.540 601) =

8. 455 132 8

The early exercise value at t = 0 is 50− 45 = 5.We take the greater of the two. So the American call premium is 8.45513.

By the way, if you bother to calculate the European put and the Americanput premium with strike price K = 45, here are the results:The prepaid forward price tree and the stock price tree won’t change whether

the option is a call or put.

The European put premium:Time 0 0.25 0.5 0.75 1

00

0 01.33205 0

3.89484 2.51380 06.21822 4.74394

9.59857 8.9525814.08269

18.96846

So the European put premium is 3.89484.Check the put-call parity: C + PV (K) = P + S − PV (Div)C + PV (K) = 8.43381 + 45e−0.08(1) = 49. 974 05P + S − PV (Div) = 3.89484 + 50− 3. 920 79 = 49. 974 05So C + PV (K) = P + S − PV (Div) holds.

The American put premium:



Time 0 0.25 0.5 0.75 10

00 0

1.33205 04.11033 2.51380 0

6.62489 4.7439410.36602 8.95258

14.9737518.96846

The two bold numbers indicate that early exercise is optimal.The American put premium is 4.11033.


Chapter 12

Black-Scholes formula

Problem 12.1.

Skip this spreadsheet problem.

Problem 12.2.

Skip this problem but remember the following key point. As n gets bigger,the price calculated using the discrete binomial tree method approach the pricecalculated using the Black-Scholes formula.

Problem 12.3.

a. r = 8% δ = 0T European call price1 7.896610 56.2377100 99.96311, 000 1001, 0000 1001, 0000 100

As T →∞, the European call premium approaches the current stock price.

b. r = 8% δ = 0.1%T European call price1 7.854210 55.3733100 90.44711, 000 36.78791, 0000 0.004510, 0000 0

107

CHAPTER 12. BLACK-SCHOLES FORMULA

Dividend reduces the value of the stock. Over a long period of time, thevalue of the underlying stock is reduced to zero; the value of the call is reducedto zero.

Problem 12.4.

a. r = 0% δ = 8%T European call price1 18.670510 10.1571100 0.00341, 000 0.0000

Dividend reduces the value of the stock. Over a long period of time, thevalue of the underlying stock is reduced to zero; the value of the call is reducedto zero.

b. r = 0.1% δ = 8%T European call price1 18.728110 10.2878100 0.00361, 000 0.0000

Dividend reduces the value of the stock, but you can earn interest on thestrike asset with r = 0.1% (vs. r = 0 in the previous problem).Consequently, the call is more valuable when r = 0.1% than when r = 0.

However, even with r = 0.1%, over a long period of time, the value of theunderlying stock is reduced to zero due to the dividend paid out; the call valueapproaches zero.

Problem 12.5.

a. We need to find the 90-strike yen-denominated euro put.

• The underlying asset is 1 euro.

• The strike asset is expressed in yen.

• The current price of the underlying is 95 yen. S = 95

• The strike asset is 90 yen. K = 90

• The strike asset 90 yen earns 1.5%. So the risk free rate is r = 1.5%(always remember that r is the earning rate of the strike asset)



• The underlying (i.e. 1 euro) earns 3.5%. So the dividend rate is δ = 3.5%(always remember that δ is the earning rate of the underlying asset)

• T = 0.5

d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

=

ln95

90+

µ0.015− 0.035 + 1

2× 0.12

¶0.5

0.1√0.5

=

0.658 560

N (d1) = NormalDist (0.658 560) = 0.744 910 8N (−d1) = 1−N (d1) = 1− 0.744 910 8 = 0.255 089d2 = d1 − σ

√T = 0.658 560− 0.1

√0.5 = 0.587 849

N (d2) = 0.721 683N (−d2) = 1−N (d2) = 1− 0.721 683 = 0.278 317

P = −Se−δTN (−d1) +Ke−rTN (−d2)= −95e−0.035(0.5)0.255 089 + 90e−0.015(0.5)0.278 317 = 1. 048 3 (yen)

b. Now we need to find the price of a1

90strike euro-denominated yen call

with 6 months to expiration.

• The underlying asset is 1 yen. The earning rate of the underlying asset is1.5%. So δ = 1.5%

• The current price of the underlying asset is 195euro. So S =

1

95

• The strike asset is 1

90euro. So K =

1

90. The earning rate of the strike

asset is 3.5%. So r = 3.5%

• Volatility is still 10%

C = Se−δTN (d1)−Ke−rTN (d2)

d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

=

ln1/95

1/90+

µ0.035− 0.015 + 1

2× 0.12

¶0.5

0.1√0.5

=

−0.587 849N (d1) = 0.278 316 8

d2 = d1 − σ√T = −0.587 849 − 0.1

√0.5 = −0.658 559 7

N (d2) = 0.255 089

C = Se−δTN (d1)−Ke−rTN (d2) =1

95e−0.015(0.5)0.278 316 8− 1

90e−0.035(0.5)

0.255 089 = 0.0001 226 1 (euro)

c. A 90-strike yen-denominated euro put is worth 1. 048 3 (yen). This put is"give 1 euro and get 90 yen."



• The privilege of "give 1 euro and get 90 yen" at T = 0.5 is worth 1. 048 3(yen) at t = 0 (statement a)

• This can be expressed at (1€→ 90Y ) = 1. 048 3Y

A1

90strike euro-denominated yen call is worth 0.0001 226 1 (euro). This

call is "give1

90euro and get 1 yen."

• The privilege of "give 1

90euro and get 1 yen" at T = 0.5 is worth 0.0001

226 1 (euro) at t = 0 (statement b)

• This can be expressed atµ1

90€→ 1Y

¶= 0.0001 226 1€

Statement a and b are essentially the same statement. Using Statement a,we can derive Statement b; using Statement b, we can derive Statement a.First, we derive Statement b from Statement a.(1€→ 90Y ) = 1. 048 3Y (statement a)

⇒µ1

90€→ 1Y

¶=1

90(1€→ 90Y ) =

1. 048 3

90Y

The above equation means this: If you can sell 1€ for a guaranteed price

90Y , then you must be able to sell1

90€ for a guaranteed price of 1Y . This

should make intuitive sense.

We convert1. 048 3

90Y into euros. Since at t = 0, 1€=95Y or 1Y =

1

95€,

then1. 048 3

90Y =

1. 048 3

90× 1

95€= 0.0001226 1€

⇒µ1

90€→ 1Y

¶=1

90(1€→ 90Y ) =

1. 048 3

90Y = 0.0001226 1€

This is exactly Statement b

Next, we derive Statement b from Statement a.µ1

90€→ 1Y

¶= 0.0001 226 1€ (statement b)

⇒ (1€→ 90Y ) = 90

µ1

90€→ 1Y

¶= 90 (0.0001 226 1€) = 90 (0.0001 226 1) 95Y =

1. 048 3Y

This is exactly Statement a.

Problem 12.6.



a.C = Se−δTN (d1)−Ke−rTN (d2)

d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

=

ln100

105+

µ0.06− 0 + 1

2× 0.42

¶1

0.4√1

= 0.228 024 589 ≈0.228 025

N (d1) = 0.590 186 6

d2 = d1 − σ√T = 0.228 025 − 0.4

√1 = −0.171 975

N (d2) = 0.431 729C = Se−δTN (d1) − Ke−rTN (d2) = 100e−0(1)0.590 186 6 − 105e−0.06(1)

0.431 729 = 16. 327 019

b.F0,T = Se(r−δ)T = 100e(0.06−0)1 = 106. 183 65

cAccording to the textbook Equation 12.7, if the underlying asset is futures

instead of stocks, we can use the general Black-Scholes formula except

• set F = S (replace the stock price with futures price)

• set δ = r

d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

=ln

F

K+1

2σ2T

σ√T

=ln100e(0.06−0)1

105+1

2× 0.42 (1)

0.4√1

=

ln100

105+

µ0.06− 0 + 1

2× 0.42

¶1

0.4√1

= 0.228 024 589 ≈ 0.228 025

N (d1) = 0.590 186 6

d2 = d1 − σ√T = 0.228 025 − 0.4

√1 = −0.171 975

N (d2) = 0.431 729C = Se−rTN (d1) − Ke−rTN (d2) = 100e−0(1)0.590 186 6 − 105e−0.06(1)

0.431 729 = 16. 327 019

The call premium in c is equal to the call premium in a. Why? We canprove this mathematically. Replace S with F = Se(r−δ)T and set δ = r , weget:

d1 =

lnSe(r−δ)T

K+

µr − r +

1

2σ2¶T

σ√T

=

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

d2 = d1 − σ√T

C =£Se(r−δ)T

¤e−rTN (d1)−Ke−rTN (d2) = Se−δTN (d1)−Ke−rTN (d2)



This is the same call formula when the underlying asset is a stock.P = −

£Se(r−δ)T

¤e−rTN (−d1) +

¡Ke−rT

¢N (−d2) = −Se−δTN (−d1) +

Ke−rTN (−d2)This is the same put formula when the underlying asset is a stock.

We can also prove that the futures option premium is equal to the underlyingstock option premium intuitively. On the maturity date T , the futures price isequal to the stock price. So if we stand at T , the payoff of a futures option andthe payoff of an otherwise identical stock option are identical. Consequently, thepremium of a futures option is equal to the premium of an otherwise identicalstock option.

Problem 12.7.

a.

d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

=

ln100

95+

µ0.08− 0.03 + 1

2× 0.32

¶0.75

0.3√0.75

=

0.471 669d2 = d1 − σ

√T = 0.471 669− 0.3

√0.75 = 0.211 861

N (d1) = 0.681 418N (d2) = 0.583 892C = Se−δTN (d1)−Ke−rTN (d2) = 100e

−0.03(0.75)0.681 418− 95e−0.08(0.75)0.583 892 = 14. 386 295

b.

d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

=

ln100e−0.03(0.75)

95e−0.08(0.75)+

µ0− 0 + 1

2× 0.32

¶0.75

0.3√0.75

=

0.471 669100d2 = d1 − σ

√T = 0.471 669− 0.3

√0.75 = 0.211 861


−0.03(0.75)0.681 418− 95e−0.08(0.75)0.583 892 = 14. 386 295

The call premium is b is the same as the call premium in a. This is becausethe call and put premium formula can be rewritten as:

C =¡Se−δT

¢N (d1)−

¡Ke−rT

¢N (d2)

P = −¡Se−δT

¢N (−d1) +

¡Ke−rT

¢N (−d2)

d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

=

lnSe−δT

Ke−rT+

µ0− 0 + 1

2σ2¶T

σ√T

d2 = d1 − σ√T



From the above equations, you see that instead of using r and δ, we canreplace S with Se−δT , replace K with Ke−rT , and set r = δ = 0. This will alsogive us the correct option premium.

Problem 12.8.

a. F0,T=0.75 = Se(r−δ)T = 100e(0.08−0.03)0.75 = 103. 821 20

b. To find the futures option premium, we replace S with F0,T and replaceδ with r in the standard Black-Scholes formula:

d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

=

ln103. 821 20

95+

µ0.08− 0.08 + 1

2× 0.32

¶0.75

0.3√0.75

=

0.471 669d2 = d1 − σ

√T = 0.471 669− 0.3

√0.75 = 0.211 861

N (d1) = 0.681 418N (d2) = 0.583 892C = Se−δTN (d1)−Ke−rTN (d2) = 103. 821 20e

−0.08(0.75)0.681 418−95e−0.08(0.75)0.583 892 = 14. 386 295

Here is another method. Since future option premium is equal to the stockoption premium, we can just calculate the stock option premium. Actually, thestock call option premium is calculated in 12.7 a.

c. 12.7 a and 12.8 b have the same premium.

Problem 12.9.

a.When a stock pays discrete dividend, we’ll use the textbook Equation 12.5.C = FP

0,TN (d1)− PV (K)N (d2)

FP0,T = S0 − PV (Div) = 50− PV (Div)

PV (Div) = 2e−0.08(1/360) = 1. 999 555 6 = 2FP0,T (S) = 50− 2 = 48

d1 =

lnFP0,T

PV (K)+1

2σ2T

σ√T

=ln

48

40e−0.08(0.5)+1

2× 0.32 × 0.5

0.3√0.5

= 1. 154 1

d2 = d1 − σ√T = 1. 154 1− 0.3

√0.5 = 0.9420

N (d1) = 0.875 77N (d2) = 0.826 90

C = 48 (0.875 77)− 40e−0.08(0.5) (0.826 90) = 10. 258

So the European call premium is 10. 258



If it’s ever optimal to early exercise an American call, the best time to earlyexercise is immediately before the dividend payment. Since the dividend ispaid tomorrow, the best time to early exercise is today. The exercise value isEV = 50− 40 = 10, which is less than the European call premium 10. 258. Soit’s not optimal to early exercise the American call.Some of you might think that it’s optimal to early exercise the American

call. If we early exercise today, we get a stock, which will pay us a dividend 2.So the total exercise value is 12.This reasoning is flawed. If you exercise the call today and take ownership

of a stock, you’ll get $2 dividend tomorrow. However, after the dividend is paid,the price of your stock is reduced by the amount of the dividend $2 (so you alsolose $2). So your exercise value is 10, not 12. And it’s not optimal to earlyexercise the American call.

b. C = FP0,TN (d1)− PV (K)N (d2)

FP0,T = S0 − PV (Div) = 50− PV (Div)

PV (Div) = 2e−0.08(1/360) = 1. 999 555 6 = 2FP0,T (S) = 50− 2 = 48

d1 =

lnFP0,T

PV (K)+1

2σ2T

σ√T

=ln

48

40e−0.08(0.5)+1

2× 0.32 × 0.5

0.3√0.5

= 1. 154 1

d2 = d1 − σ√T = 1. 154 1− 0.3

√0.5 = 0.9420

N (d1) = 0.875 77N (d2) = 0.826 90

C = 48 (0.875 77)− 40e−0.08(0.5) (0.826 90) = 10. 258

The exercise value is EV = 60− 40 = 20 > CHence it’s optimal to early exercise the American put at t = 0.

c. It’s optimal to early exercise the American call at t = 0 if the exercisevalue is greater than the European call premium. It’s not optimal to exercisethe American call at t = 0 if the exercise value is equal to or less than the heEuropean call premium. However, keep in mind that if the stock doesn’t paydividend, then it’s never optimal to early exercise an American call.

Problem 12.10.

The statement means that the absolute value |θ (t) | = |∂V∂t | reaches its max-imum value when t→ T . In other words, the closer to the expiration date, thehigher the |θ (t) |. This statement is not correct. I don’t know of any intuitiveway to explain why this statement is not correct. That’s why the textbook asksyou to test this statement using a spreadsheet.



If you want to test the statement using a spreadsheet, you can use thespreadsheet titled "optbasic2." Then you can make up a case and validate thestatement.However, for the purpose of passing the exam, you can ignore this problem.

Problem 12.11.

a. If you look at the Appendix 12.A (which is excluded from both CAS andSOA exam), you can see that Vega is the derivative of an option price regardingvolatility:

V ega = ∂V∂σ

The formula V ega = V (σ+ )−V (σ− )2 is an approximation. For this approx-

imation to work, needs to be small. Since Appendix 12.A is excluded fromboth CAS and SOA exam, you can ignore this part.

b. If you want to solve this problem, you can set up some test cases andcompare the approximated Vega with the actual Vega (using the spreadsheet).For the purpose of passing the exam, you can ignore this part.

Problem 12.12.

Since Appendix 12.A is excluded from both CAS and SOA exam, you canignore this problem.

Problem 12.13.

Let’s not worry about drawing a diagram and focus on how to calculate theprofit. I’ll do some sample calculations.

Let’s calculate the profit after 6 months (i.e. at expiration) assuming thestock price after 6 months is $60.Using the Black-Scholes formula, we can find:

• The 40-strike call premium is 4.1553. This call premium is calculatedusing the Black-Scholes formula by setting S = 40,K = 40, T = 0.5, r =0.08, σ = 0.3, δ = 0

• The 45-strike call premium is 2.1304. This call premium is calculatedusing the Black-Scholes formula by setting S = 40,K = 45, T = 0.5, r =0.08, σ = 0.3, δ = 0

• The net cost of the bull spread at t = 0 is 4.1553− 2.1304 = 2. 024 9

• Its future value at expiration is 2. 024 9e0.08(0.5) = 2. 107 5

The stock price at expiration is 60.



• The 40-strike call payoff is 60− 40 = 20

• The 45-strike call payoff is − (60− 45) = −15

• The total payoff is 20− 15 = 5

So our profit is: −2. 107 5 + 5 = 2. 892 5Let’s calculate the profit after 3 months assuming the stock price after 3

months is $60.Using the Black-Scholes formula, we can find:

• The 40-strike call premium is 4.1553



• Its future value after 3 months is 2. 024 9e0.08(0.25) = 2. 065 8

3 months later (i.e. at t = 0.25), we close our position. Right now, ourpurchased 40-strike call and sold 45-strike call both have 3 months to expiration.To close our position (i.e. to cancel out our position), at t = 0.25, we sell a 40-strike call and buy a 45-strike call. After this, our net position is zero. Att = 0.25, the stock price is 60.

• At t = 0.25, the 40-strike call with 3 months to expiration is worth 20.7969(so we receive 20.7969). This call premium is calculated using the Black-Scholes formula by setting S = 60,K = 40 T = 0.25, r = 0.08, σ = 0.3, δ =0.

• At t = 0.25, the 45-strike call with 3 months to expiration is worth 15.9480(so we pay 15.9480). This call premium is calculated using the Black-Scholes formula by setting S = 60,K = 45 T = 0.25, r = 0.08, σ =0.3, δ = 0

• The net receipt is 20.7969− 15.9480 = 4. 848 9

So our profit is: −2. 065 8 + 4. 848 9 = 2. 783 1Let’s calculate the profit after 1 day assuming the stock price after 1day is

$60.Using the Black-Scholes formula, we can find:




• Its future value after 1 day is 2. 024 9e0.08(1/365) = 2. 025 34



1 day later (i.e. at t = 364/365), we close our position. Right now, ourpurchased 40-strike call and sold 45-strike call both have 364 days to expiration.To close our position (i.e. to cancel out our position), at t = 364/365 we sell a40-strike call and buy a 45-strike call. After this, our net position is zero. Att = 364/365, the stock price is 60.

• At t = 364/365, the 40-strike call with 364 days to expiration is worth23.3775 (so we receive 23.3775). This call premium is calculated usingthe Black-Scholes formula by setting S = 60,K = 40 T = 364/365, r =0.08, σ = 0.3, δ = 0.

• At t = 364/365, the 45-strike call with 364 days to expiration is worth19.2391 (so we pay 19.2391). This call premium is calculated using theBlack-Scholes formula by setting S = 60,K = 45 T = 364/365, r =0.08, σ = 0.3, δ = 0

• The net receipt is 23.3775− 19.2391 = 4. 138 4

So our profit is: −2. 038 44 + 4. 138 4 = 2. 099 96

Problem 12.14.

Appendix 12.A is excluded from both CAS and SOA exam. So SOA andCAS can’t ask you to calculate option Greeks using formulas in Appendix 12.A.(SOA and CAS can ask you to calculate delta ∆ using the textbook formula10.1).However, SOA and CAS can ask you to calculate the Greeks for a portfolio

using the formula presented in Page 389. Page 389 is on the syllabus.So you need to learn how to calculate portfolio’s Greeks using the following

formula (in the textbook Page 389):

Greekoption =Pn

i=1 ωiGreeki

a. S = 40ω1 Greek1 ω2 Greek2 Greekoption =

Pni=1 ωiGreeki

bought 40K call sold 45K call Portfolio Price and GreeksPrice 1 4.1553 1 −2.1304 1 (4.1553) + 1 (−2.1304) = 2. 024 9Delta 1 0.6159 1 −0.3972 1 (0.6159) + 1 (−0.3972) = 0.218 7Gamma 1 0.0450 1 −0.0454 1 (0.0450) + 1 (−0.0454) = −0.000 4Vega 1 0.1080 1 −0.1091 1 (0.1080) + 1 (−0.1091) = −0.001 1Theta 1 −0.0134 1 0.0120 1 (−0.0134) + 1 (0.0120) = −0.001 4Rho 1 0.1024 1 −0.0688 1 (0.1024) + 1 (−0.0688) = 0.033 6

Column 3 and 5 are calculated using the "optbasic2" spreadsheet. Youdon’t need to worry about these two columns. You just need to focus on thelast column and practice the formula Greekoption =

Pni=1 ωiGreeki.



b. S = 45ω1 Greek1 ω2 Greek2 Greekoption =

Pni=1 ωiGreeki

bought 40K call sold 45K call Portfolio Price and GreeksPrice 1 7.7342 1 −4.6747 1 (7.7342) + 1 (−4.6747) = 3. 059 5Delta 1 0.8023 1 −0.6159 1 (0.8023) + 1 (−0.6159) = 0.186 4Gamma 1 0.0291 1 −0.0400 1 (0.0291) + 1 (−0.0400) = −0.010 9Vega 1 0.0885 1 −0.1216 1 (0.0885) + 1 (−0.1216) = −0.033 1Theta 1 −0.0135 1 0.0150 1 (−0.0135) + 1 (0.0150) = 0.001 5Rho 1 0.1418 1 −0.1152 1 (0.1418) + 1 (−0.1152) = 0.026 6

c. Ignore this part. Not sure what this problem wants to accomplish.

Problem 12.15.

a. S = 40ω1 Greek1 ω2 Greek2 Greekoption =

Pni=1 ωiGreeki

bought 40K call sold 45K call Portfolio Price and GreeksPrice 1 2.5868 1 −5.3659 1 (2.5868) + 1 (−5.3659) = −2. 779 1Delta 1 −0.3841 1 0.6028 1 (−0.3841) + 1 (0.6028) = 0.218 7Gamma 1 0.0450 1 −0.0454 1 (0.0450) + 1 (−0.0454) = −0.000 4Vega 1 0.1080 1 −0.1091 1 (0.1080) + 1 (−0.1091) = −0.001 1Theta 1 −0.0049 1 0.0025 1 (−0.0049) + 1 (0.0025) = −0.002 4Rho 1 −0.0898 1 0.1474 1 (−0.0898) + 1 (0.1474) = 0.057 6

b. S = 45ω1 Greek1 ω2 Greek2 Greekoption =

Pni=1 ωiGreeki

bought 40K call sold 45K call Portfolio Price and GreeksPrice 1 1.1658 1 −2.9102 1 (1.1658) + 1 (−2.9102) = −1. 744 4Delta 1 −0.1977 1 0.3841 1 (−0.1977) + 1 (0.3841) = 0.186 4Gamma 1 0.0291 1 −0.0400 1 (0.0291) + 1 (−0.0400) = −0.010 9Vega 1 0.0885 1 −0.1216 1 (0.0885) + 1 (−0.1216) = −0.033 1Theta 1 −0.0051 1 0.0056 1 (−0.0051) + 1 (0.0056) = 0.000 5Rho 1 −0.0503 1 0.1010 1 (−0.0503) + 1 (0.1010) = 0.050 7

c. Ignore

Problem 12.16.

There’s no easy way to solve this problem manually. This type of problemsshouldn’t show up in the exam.

Problem 12.17.

There’s no easy way to solve this problem manually. This type of problemsshouldn’t show up in the exam.



Problem 12.18.

S = 50 K = 60 r = 0.06 σ = 0.4 δ = 0.03

a. Solve for h.1

2σ2h (h− 1) + (r − δ)h− r = 0

1

2× 0.42h (h− 1) + (0.06− 0.03)h− 0.06 = 0

h = −0.608 182 5 h = 1. 233 182 5Use the bigger h for call and the smaller h for put.

H∗Call =hcall

hcall − 1K =

1. 233 182 5

1. 233 182 5− 1 × 60 = 317. 309 19


µS

H∗Call

¶hcall= (317. 309 19− 60)

µ50

317. 309 19

¶1. 233 182 5=

26. 351 83

The call should be exercised when the stock price reaches 317. 309 19. Thisprice is called the barrier.The call premium is 26. 351 83.

b. Now δ = 0.04 (instead of δ = 0.03). Everything else is the same as in a.Higher dividend yield means that the stock price will decrease more quickly.

Recall that the stock price drops by the dividend amount immediately afterthe dividend payment time. We expect that under δ = 0.04 the optimal stockprice (i.e. barrier) is lower than the barrier when δ = 0.03. In addition, weexpect that the option price under δ = 0.04 is lower than the option price underδ = 0.03.Solve for h.1

2σ2h (h− 1) + (r − δ)h− r = 0

1

2× 0.42h (h− 1) + (0.06− 0.04)h− 0.06 = 0


H∗Call =hcall

hcall − 1K =

1. 318 729 3

1. 318 729 3− 1 × 60 = 248. 247 52


µS

H∗Call

¶hcall= (248. 247 52− 60)

µ50

248. 247 52

¶1. 318 729 3=

22. 751 3

The call should be exercised when the stock price reaches 248. 247 52.The call premium is 22. 751 28

c.Now r = 0.07 (instead of r = 0.06). Everything else is the same as in a.



When the risk free interest rate goes up, stocks are expected to generate highreturns. Consequently, the value of a stock goes up; the call option premiumgoes up (while the put option premium goes down). We expect that both thebarrier and the call premium go up when r goes up from 0.06 to 0.07.Solve for h.1

2σ2h (h− 1) + (r − δ)h− r = 0

1

2× 0.42h (h− 1) + (0.07− 0.03)h− 0.07 = 0

h = 1. 218 245 8 h = −0.718 245 8Use the bigger h for call and the smaller h for put.

H∗Call =hcall

hcall − 1K =

1. 218 245 8

1. 218 245 8− 1 × 60 = 334. 919 4


µS

H∗Call

¶hcall= (334. 919 4− 60)

µ50

334. 919 4

¶1. 218 245 8=

27. 100 1

The call should be exercised when the stock price reaches 334. 919 4The call premium is 27. 1001

d. The higher the volatility, the more valuable an option is. As σ goesup from 0.4 to 0.5, we expect both the barrier and the call option premium willgo up.

1

2σ2h (h− 1) + (r − δ)h− r = 0

1

2× 0.52h (h− 1) + (0.06− 0.03)h− 0.06 = 0


H∗Call =hcall

hcall − 1K =

1. 170 189 9

1. 170 189 9− 1 × 60 = 412. 547 36


µS

H∗Call

¶hcall= (412. 547 36− 60)

µ50

412. 547 36

¶1. 170 189 9=

29. 835 5

The call should be exercised when the stock price reaches 412. 547 36The call premium is 29. 835 5

Problem 12.19.

S = 50 K = 60 r = 0.06 σ = 0.4 δ = 0.03

a.Solve for h.1

2σ2h (h− 1) + (r − δ)h− r = 0



1

2× 0.42h (h− 1) + (0.06− 0.03)h− 0.06 = 0


H∗Put =hput

hcall − 1K =

−0.608 182 5−0.608 182 5− 1 × 60 = 22. 690 80

Pperpetual = (K −H∗Put)

µS

H∗Put

¶hput= (60− 22. 690 80)

µ50

22. 690 80

¶−0.608 182 5=

23. 074 7

The put should be exercised when the stock price reaches 22. 690 80.The put premium is 23. 074 7

b. Now δ = 0.04 (instead of δ = 0.03). Everything else is the same as in a.Higher dividend yield means that the stock price will decrease more quickly.

Since the value of a put goes up if the stock price goes down, we expect that thebarrier price will go down (i.e. exercise occurs later because we want to exercisewhen the stock price is low) and the put premium will go up.

Solve for h.1

2σ2h (h− 1) + (r − δ)h− r = 0

1

2× 0.42h (h− 1) + (0.06− 0.04)h− 0.06 = 0

h = −0.568 729 3 h = 1. 318 729 3

Use the bigger h for call and the smaller h for put.

H∗Put =hput

hcall − 1K =

−0.568 729 3−0.568 729 3− 1 × 60 = 21. 752 5


µS

H∗Put

¶hput= (60− 21. 752 5)

µ50

21. 752 5

¶−0.568 729 3=

23. 824 8

The put should be exercised when the stock price reaches 21. 752 5The put premium is 23. 824 8

c. As r goes up, people expect to get increased return from the stock. Thestock price goes up and the put option premium goes down.We expect the barrier will go up (compared with a, meaning that exercise

occurs sooner).1

2σ2h (h− 1) + (r − δ)h− r = 0

1

2× 0.42h (h− 1) + (0.07− 0.03)h− 0.07 = 0


H∗Put =hput

hcall − 1K =

−0.718 245 8−0.718 245 8− 1 × 60 = 25. 080 67




µS

H∗Put

¶hput= (60− 25. 080 67)

µ50

25. 080 67

¶−0.568 729 3=

23. 824 8


d. The higher the volatility, the more valuable an option is. As σ goesup from 0.4 to 0.5, we expect both the barrier will go down (i.e. exercise occurslater) and the put option premium will go up.

1

2σ2h (h− 1) + (r − δ)h− r = 0

1

2× 0.52h (h− 1) + (0.06− 0.03)h− 0.06 = 0


H∗Put =hput

hcall − 1K =

−0.410 189 9−0.410 189 9− 1 × 60 = 17. 452 5


µS

H∗Put

¶hput= (60− 17. 452 5)

µ50

17. 452 5

¶−0.410 189 9=

27. 629 4


Problem 12.20.

For a and b, if you use the Black-Scholes formula, you’ll find that call andput are both worth 17.6988.For part c. After we switch S and K and switch r and δ, the put after

the switch and the call before the switch have the same value. This is not acoincidence. It’s explained in my solution to Problem 10.19.

Problem 12.21.

a.1

2σ2h (h− 1) + (r − δ)h− r = 0

1

2× 0.32h (h− 1) + (0.08− 0.05)h− 0.08 = 0

h = −1. 177 043 h = 1. 510 376Use the bigger h for call and the smaller h for put.

H∗Call =hcall

hcall − 1K =

1. 510 376

1. 510 376− 1 × 90 = 266. 340 6




µS

H∗Call

¶hcall= (266. 340 6− 90)

µ100

266. 340 6

¶1. 510 376=

40. 158 9The call premium is 40. 158 9

b.1

2σ2h (h− 1) + (r − δ)h− r = 0

1

2× 0.32h (h− 1) + (0.05− 0.08)h− 0.05 = 0

h = −0.510 38 h = 2. 177 04

Use the bigger h for call and the smaller h for put.

H∗Put =hput

hcall − 1K =

−0.510 38−0.510 38− 1 × 100 = 33. 791 5


µSt

H∗Put

¶hput= (100− 33. 791 5)

µ90

33. 791 5

¶−0.510 38=

40. 158 9The put premium is 40. 158 9

c. After we switch S and K and switch r and δ, the perpetual Americanput after the switch and the perpetual American call before the switch have thesame value. This is a minor fact probably not worth memorizing. We are notgoing to worry about the proof.




Chapter 13

Market making and deltahedging

Problem 13.1.

d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

=

ln40

45+

µ0.08− 0 + 1

2× 0.32

¶91

365

0.3

r91

365

=

−0.578 25

d2 = d1 − σ√T = −0.578 25− 0.3

r91

365= −0.728 04


−0(91/365)0.281 55− 45e−0.08(91/365)0.233 29 = 0.971 3

A call on 100 stocks is worth 100 (0.971 3) = 97. 13Use the formula in Appendix 12.8 to find ∆∆ = e−δTN (d1) = e−0×91/365N (d1) = 0.281 55 (positive delta means buy-

ing stocks)

Suppose a trader sells a call option on 100 stocks. To hedge his risk, thetrader should at t = 0

• sell the call and receive 100 (0.971 3) = 97. 13

• buy 0.281 55 (100) = 28. 155 stocks costing 40 (28. 155) = 1126. 2

• borrow 1126. 2− 97. 13 = 1029. 07 from the bank

125

CHAPTER 13. MARKET MAKING AND DELTA HEDGING

The trader’s net position is zero at t = 0

If the stock price is 39 on Day 1.To close his short call position, the trader can buy, from the open market,

a 45-strike call expiring in 90 days. This new (purchased) call and the original(sold) call have the same underlying, same expiration date, same strike price.They will cancel out each other and the trader doesn’t have any liabilities asso-ciated either call.The cost of the new call is 70.55. It’s calculated as follows:

d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

=

ln39

45+

µ0.08− 0 + 1

2× 0.32

¶90

365

0.3

r90

365

=

−0.753 706d2 = d1 − σ

√T = −0.753 706 346− 0.3

r90

365= −0.902 675


−0(90/365)0.225 513−45e−0.08(90/365)0.183 349 = 0.705 5

A call on 100 stocks is worth 100 (0.705 5) = 70. 55So the trader needs to pay 70. 55 to buy a new call to cancel out the original

call he sold.At t = 0, the trader owns 28. 155 stocks. Now one day later, the stocks

are worth 28. 155 (39) = 1098. 045. So the trader sells out his stocks, receiving1098. 05.In addition, the trader needs to pay back the loan borrowed from the bank.

The future value of the loan one day later is:1029. 07e0.08(1/365) = 1029. 30So the trader’s net wealth at t = 1/365 is:1098. 05− (1029. 30 + 70. 55) = −1. 8So the trader lost $1.8.

If you want to following the textbook calculation on Page 417 "Day 1:Marking-to-market," here it is:Day 1Gain on 28. 155 shares 28. 155 (39− 40) = −28. 155Gain on the written call 97. 13− 70. 55 = 26. 58Interest 1029. 07

¡e0.08(1/365) − 1

¢= 0.226

Overnight profit −28. 155 + 26. 58− 0.226 = −1. 8

If the stock price is 40.5 on Day 1.To close his short call position, the trader can buy, from the open market,

a 45-strike call expiring in 90 days. This new (purchased) call and the original(sold) call have the same underlying, same expiration date, same strike price.



They will cancel out each other and the trader doesn’t have any liabilities asso-ciated either call.The cost of the new call is 110.46. It’s calculated as follows:

d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

=

ln40.5

45+

µ0.08− 0 + 1

2× 0.32

¶90

365

0.3

r90

365

=

−0.500 363d2 = d1 − σ

√T = −0.500 363− 0.3

r90

365= −0.649 332

N (d1) = 0.308 41N (d2) = 0.258 062C = Se−δTN (d1)−Ke−rTN (d2) = 40.5e

−0(90/365)0.308 41−45e−0.08(90/365)0.258 062 = 1. 104 65

A call on 100 stocks is worth 100 (1. 104 65) = 110. 465

One Day 1, the trader

• pays 110. 46 and buy a new call to cancel out the original call he sold

• sells out his 28. 155 stocks, receiving 28. 155 (40.5) = 1140. 28

• pays 1029. 07e0.08(1/365) = 1029. 30 to the bank to payoff the loan

So the trader’s net wealth on Day 1 is:1140. 28− (1029. 30 + 110. 46) = 0.52

If you want to use the textbook notation, here you go:Day 1Gain on 28. 155 shares 28. 155 (40.5− 40) = 14. 077 5Gain on the written call 97. 13− 110. 465 = −13. 335Interest 1029. 07

¡e0.08(1/365) − 1

¢= 0.226

Overnight profit 14. 077 5− 13. 335− 0.226 = 0.52So the trader gained $0.52.

Problem 13.2.

d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

=

ln40

40+

µ0.08− 0 + 1

2× 0.32

¶91

365

0.3

r91

365

=

0.208 05

d2 = d1 − σ√T = 0.208 05− 0.3

r91

365= 0.05 825

N (−d1) = 0.417 59 N (−d2) = 0.476 77P = 40e−0.08(91/365)0.476 77− 40e−0(91/365)0.417 59 = 1. 990 6



A put on 100 stocks is worth 100 (1. 990 6) = 199. 06Use the formula in Appendix 12.8 to find ∆∆ = −e−δTN (−d1) = −e−0×91/3650.417 59 = −0.417 6 (negative delta

means short selling stocks)

Suppose a trader sells a put option on 100 stocks. To hedge his risk, thetrader should at t = 0

• sell the put and receive 100 (1. 990 6) = 199. 06

• buy −0.417 6 (100) = −41. 76 stocks (i.e. short sell 41. 76 stocks) receiving41. 76 (40) = 1670. 4

• lend 199. 06 + 1670. 4 = 1869. 46 to a bank

The trader’s net position is zero at t = 0

If the stock price is 39.5 on Day 1.To close his short put position, the trader can buy, from the open market,

a 40-strike put expiring in 90 days. This new (purchased) put and the original(sold) put have the same underlying, same expiration date, same strike price.They will cancel out each other and the trader doesn’t have any liabilities asso-ciated either put.The cost of the new put is as follows:

d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

=

ln39

40+

µ0.08− 0 + 1

2× 0.32

¶90

365

0.3

r90

365

= 0.0

3695

d2 = d1 − σ√T = 0.03695− 0.3

r90

365= −0.112 02

N (−d1) = 0.485 26 N (−d2) = 0.544 60P = Ke−rTN (−d2)−Se−δTN (−d1) = 40e−0.08(90/365)0.544 60−39e−0(90/365)

0.485 26 = 2. 433 4

On Day 1 a put on 100 stocks is worth 100 (2. 433 4) = 243. 34


• pays 243. 34, buying a new put to cancel out the original put he sold

• buys back 41. 76 stocks to close the short sale position, paying 41. 76 (39) =1628. 64

• receives 1869. 46e0.08(1/365) = 1869. 87 from the bank



So the trader’s net wealth on Day 1 is:−243. 34− 1628. 64 + 1869. 87 = −2. 11So the trader lost 2. 11.If you want to use the textbook notation, here you go:Day 1Gain on 41. 76 shares 41. 76 (40− 39) = 41. 76Gain on the written put 199. 06− 243. 34 = −44. 28Interest income 1869. 46

¡e0.08(1/365) − 1

¢= 0.41

Overnight profit 41. 76− 44. 28 + 0.41 = −2. 11


a 40-strike call expiring in 90 days. This new (purchased) call and the original(sold) call have the same underlying, same expiration date, same strike price.They will cancel out each other and the trader doesn’t have any liabilities asso-ciated either call.The cost of the new call is calculated as follows:

d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

=

ln40.5

40+

µ0.08− 0 + 1

2× 0.32

¶90

365

0.3

r90

365

=

0.290 29

d2 = d1 − σ√T = 0.290 29− 0.3

r90

365= 0.141 32

N (−d1) = 0.385 80 N (−d2) = 0.443 81P = Ke−rTN (−d2)−Se−δTN (−d1) = 40e−0.08(90/365)0.443 81−40.5e−0(90/365)

0.385 80 = 1. 780 8




• buys back 41. 76 stocks to close the short sale position, paying 41. 76 (40.5) =1691. 28


So the trader’s net wealth on Day 1 is:−178. 08− 1691. 28 + 1869. 87 = 0.51So the trader gains 0.51.If you want to use the textbook notation, here you go:Day 1Gain on 41. 76 shares 41. 76 (40− 40.5) = −20. 88Gain on the written put 199. 06− 178. 08 = 20. 98Interest income 1869. 46

¡e0.08(1/365) − 1

¢= 0.41

Overnight profit −20. 88 + 20. 98 + 0.41 = 0.51



Next, I’m going to redo Problem 13.2 using the strike price K = 45. Thisway, Problem 13.2 is similar to Problem 13.1 except that in Problem 13.1 there’sa call and in Problem 13.2 there’s a put.

d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

=

ln40

45+

µ0.08− 0 + 1

2× 0.32

¶91

365

0.3

r91

365

=

−0.578 25d2 = d1 − σ

√T = −0.578 25− 0.3

r91

365= −0.728 04

N (−d1) = 0.718 45N (−d2) = 0.766 71P = Ke−rTN (−d2)−Se−δTN (−d1) = 45e−0.08(91/365)0.766 71−40e−0(91/365)

0.718 45 = 5. 082 6

A put on 100 stocks is worth 100 (5. 082 6) = 508. 26Use the formula in Appendix 12.8 to find ∆∆ = −Se−δTN (−d1) = −e−0×91/3650.718 45 = −0.718 45 (negative delta

means short selling stocks)

Suppose a trader sells a put option on 100 stocks. To hedge his risk, thetrader should at t = 0

• sell the call and receive 100 (5. 082 6) = 508. 26

• buy −0.718 45 (100) = −71. 845 stocks (i.e. short sell 71. 845 stocks) re-ceiving 71. 845 (40) = 2873. 8

• lend 2873. 8 + 508. 26 = 3382. 06 to a bank

The trader’s net position is zero at t = 0If the stock price is 39.5 on Day 1.To close his short put position, the trader can buy, from the open market,

a 45-strike put expiring in 90 days. This new (purchased) put and the original(sold) put have the same underlying, same expiration date, same strike price.They will cancel out each other and the trader doesn’t have any liabilities asso-ciated either put.The cost of the new call is 110.46. It’s calculated as follows:

d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

=

ln39

45+

µ0.08− 0 + 1

2× 0.32

¶90

365

0.3

r90

365

=

−0.753 706d2 = d1 − σ

√T = −0.753 706 346− 0.3

r90

365= −0.902 675

N (−d1) = 0.774 49



N (−d2) = 0.816 65P = Ke−rTN (−d2)−Se−δTN (−d1) = 45e−0.08(90/365)0.816 65−39e−0(90/365)

0.774 49 = 5. 826 3




• buys back 71. 845 stocks to close the short sale position, paying 71. 845 (39) =2801. 955


So the trader’s net wealth on Day 1 is:−582. 63− 2801. 955 + 3382. 801 = −1. 784So the trader lost 1.78.

If you want to use the textbook notation, here you go:Day 1Gain on 71. 845 shares 71. 845 (40− 39) = 71. 845Gain on the written put 508. 26− 582. 63 = −74. 37Interest income 3382. 06

¡e0.08(1/365) − 1

¢= 0.741

Overnight profit 71. 845− 74. 37 + 0.741 = −1. 784


a 45-strike call expiring in 90 days. This new (purchased) call and the original(sold) call have the same underlying, same expiration date, same strike price.They will cancel out each other and the trader doesn’t have any liabilities asso-ciated either call.The cost of the new call is 110.46. It’s calculated as follows:

d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

=

ln40.5

45+

µ0.08− 0 + 1

2× 0.32

¶90

365

0.3

r90

365

=

−0.500 363d2 = d1 − σ

√T = −0.500 363− 0.3

r90

365= −0.649 332

N (−d1) = 0.691 59N (−d2) = 0.741 94P = Ke−rTN (−d2)−Se−δTN (−d1) = 45e−0.08(90/365)0.741 94−40.5e−0(90/365)

0.691 59 = 4. 725 76

On Day 1 a put on 100 stocks is worth 100 (4. 725 76) = 472. 576One Day 1, the trader




• buys back 71. 845 stocks to close the short sale position, paying 71. 845 (40.5) =2909. 72


So the trader’s net wealth on Day 1 is:−472. 576− 2909. 72 + 3382. 801 = 0.505So the trader gains 0.505.If you want to use the textbook notation, here you go:Day 1Gain on 71. 845 shares 71. 845 (40− 40.5) = −35. 92Gain on the written put 508. 26− 472. 576 = 35. 684Interest income 3382. 06

¡e0.08(1/365) − 1

¢= 0.741

Overnight profit −35. 92 + 35. 684 + 0.741 = 0.505

Problem 13.3.

buy a 40-45 bull spread=buy a 40-strike call and sell a 45-strike putI used a spreadsheet to calculate the following so you might not be able to

fully match my result

Day 0: Expiration T = 91/365 Stock price 40Position Strike Call premium Deltabuy 40 −2.7804 −0.58240sell 45 0.9710 0.28155Net (one stock) −2.7804 + 0.9710 = −1. 809 4 −0.28155 + 0.58240 = −0.300 85Net (100 stocks) 100 (−1. 809 4) = −180. 94 100 (−0.300 85) = −30. 085

In the above table

• positive premium means cash inflow (i.e. receiving premium)

• negative premium means cash outgo (i.e. paying premium)

• positive delta means buying stocks

• negative delta means short selling stocks

On Day 0, the trader:

• Buy a 40-strike call and sell a 45-strike call, paying 180. 94

• Short sell 30. 085 stocks, paying 30. 085 (40) = 1203. 4. The negative deltameans short selling.



• Deposit 1203. 4− 180. 94 = 1022. 46 to a savings bank to earn a risk freerate

The trader’s net position is zero.Day 1: Expiration T = 90/365 Stock price 39Position Strike price Call premiumSell 40 2.2144Buy 45 −0.7054Net (one stock) 2.2144− 0.7054 = 1. 509Net (100 stocks) 100 (1. 509) = 150. 9

Day 1, the trader closes out his position

• Sell a 40-strike call to cancel out the original 40-strike call he bought; buya 45-strike call to cancel out the original 45-strike call he sold. The traderreceives 150. 9

• Buy 30. 085 stocks from the market to close the short sale, paying 30.085 (39) = 1173. 315

• Receive 1022. 46e0.08(1/365) = 1022. 684 from the savings account

The trader’s net profit: 150. 9− 1173. 315 + 1022. 684 = 0.269If the stock price is 39 on Day 1, the trader gets 0.269 profit.

Day 1: Expiration T = 90/365 Stock price 40.5Position Strike price Call premiumSell 40 3.0621Buy 45 −1.1046Net (one stock) 3.0621− 1.1046 = 1. 957 5Net (100 stocks) 100 (1. 957 5) = 195. 75


• Sell a 40-strike call to cancel out the original 40-strike call he bought; buya 45-strike call to cancel out the original 45-strike call he sold. The traderreceives 195. 75

• Buy 30. 085 stocks from the market to close the short sale, paying 30.085 (40.5) = 1218. 442 5

• Receive 1022. 46e0.08(1/365) = 1022. 684 from the savings account

The trader’s net profit: 195. 75−1218. 442 5+1022. 684 = −0.008 5 = −0.01If the stock price is 39 on Day 1, the trader losses 0.01.

Problem 13.4.



Day 0: Expiration T = 91/365 Stock price 40Position Strike Put premium Deltabuy 45 −5.0824 − (−0.71845) = 0.718 451sell 40 1.9905 −0.41760Net −5.0824 + 1.9905× 2 = −1. 101 4 0.71845− 0.41760× 2 = −0.116 75Net 100 (−1. 101 4) = −110. 14 100 (−0.116 75) = −11. 675

On Day 0, the trader:

• Buy a 45-strike put and sell two 45-strike put, paying 110. 14

• Short sell 11. 675 stocks, receiving 11. 675 (40) = 467

• Deposit 467− 110. 14 = 356. 86 to a savings bank to earn a risk free rate

The trader’s net position is zero.Day 1: Expiration T = 90/365 Stock price 39Position Strike price Call premiumSell 45 5.8265Buy 40 −2.4331Net 5.8265− 2.4331× 2 = 0.960 3Net 100 (0.960 3) = 96. 03


• Sell a 45-strike put to cancel out the original 40-strike put he bought; buytwo 40-strike puts to cancel out the original two 40-strike puts he sold.The trader receives 96. 03

• Buy 11. 675 stocks from the market to close the short sale, paying 11.675 (39) = 455. 325

• Receive 356. 86e0.08(1/365) = 356. 938 2 from the savings account

The trader’s net profit: 96. 03− 455. 325 + 356. 938 2 = −2. 36If the stock price is 39 on Day 1, the trader loses 2.36.

Day 1: Expiration T = 90/365 Stock price 40.5Position Strike price Call premiumSell 45 4.7257Buy 40 −1.7808Net 4.7257− 1.7808× 2 = 1. 164 1Net 100 (1. 164 1) = 116. 41

Day 1, the trader closes out his position1The delta formula in Appendix 12.B is the delta for the market maker if he sells an option.

If the market maker buys an option, add a negative sign to the formula. In this case, the deltais -0.71845 if the trader sells the 45-strike put. However, since the trader buys a 45-strike put,the delta is − (−0.91845) = 0.918 45



• Sell a 45-strike put to cancel out the original 40-strike put he bought; buytwo 40-strike puts to cancel out the original two 40-strike puts he sold.The trader receives 116. 41

• Buy 11. 675 stocks from the market to close the short sale, paying 11.675 (40.5) = 472. 84

• Receive 356. 86e0.08(1/365) = 356. 938 2 from the savings account

The trader’s net profit: 116. 41− 472. 84 + 356. 938 2 = 0.51If the stock price is 39 on Day 1, the trader gains 0.51.

Problem 13.5.

T = 91/365 r = 8% δ = 0σ = 30% K = 40

Let’s walk through Day 0 and Day 1 calculations.Day 0 1stock $40.00 $40.50put $199.05 $178.08delta −0.417596 −0.385797Investment −1, 869.43 −$1, 740.56Interest credited (end of the day) $0.41 $0.381 5Capital gain (end of the day) $0.09 −$4. 245 4Daily profit (end of the day) $0.50 −$3. 864

Day 0:Using the Black-Scholes formula, you can verify the put premium is C0 =

$199.05 and delta per stock is −0.417596. Negative delta means buying negativenumber of stocks (i.e. the trader needs to short sell stocks). The trader buys∆0 = 100 (−0.417596) = −41. 759 6 stocks (i.e. short-sell 41. 759 6 stocks),receiving 41. 759 6×40 = 1670. 384. In addition, the trader receives put premiumC0 =$199.05. So at t = 0 the trader’s asset (i.e. investment)

MV (0) = ∆0S0 − C0 = (−41. 759 6) 40− 199.05 = −1869. 434The negative amount means that the trader receives 1869. 434The trader lends out $1869. 434 (i.e. depositing $1869. 434) in a savings

account. Now his net position is zero.

At the end of Day 0, the stock price goes up from S0 = 40 to S1 = 40.5.The put liability goes down from C0 = $199.05 to C1 = $178.08. The trader’sasset before he rebalances his portfolio is (BR stands for before rebalance):

MV BR (1) = ∆0S1 − C1 = (−41. 759 6) 40.5− 178.08 = −1869. 343 8The trader’s profit at the end of Day 0 is:MV BR (1)−MV (0) erh = −1869. 343 8−(−1869. 434) e0.08(1/365) = 0.499 98 =

0.5



To find the capital gain and the interest earned at the end of Day 0, we justneed to break down the profit at the end of Day 0 into two parts:


+−MV (0)¡erh − 1


Capital gain at the end of Day 0: MV BR (1) −MV (0) = −1869. 343 8 −(−1869. 434) = 0.090 2

Interest earned at the end of Day 0: −MV (0)¡erh − 1

¢= − (−1869. 434)

¡e0.08/365 − 1

¢=

0.409 78 = 0.41The investment at the beginning of Day 0 is: MV (0) = 1869. 434Day 1:In the beginning of Day 1, the trader starts from a clean slate. He buys sells

∆1 = 100× (−0.385797) = −38. 579 7 stocks (i.e. short sell stocks) and receivesthe put premium C1 = $178.08.His asset (or investment) is MV (1) = ∆1S1 − C1 = (−38. 579 7) 40.5 −

178.08 = −1740. 557 85The trader’s asset at the end of Day 1 before he rebalances the portfolio (i.e.

before he starts over from a clean slate the next day) is:MV BR (2) = ∆1S2 − C2 = (−38. 579 7) 39.25− 230.55 = −1744. 803 225The trader’s profit at the end of Day 1 is:MV BR (2) − MV (1) erh = −1744. 803 225 − (−1740. 557 85) e0.08(1/365) =

−3. 864

To find the capital gain and the interest earned at the end of Day 1, we justneed to break down the profit at the end of Day 1 into two parts:


+−MV (1)¡erh − 1


The capital gain credited at the end of Day 1 is:

MV BR (2)−MV (1) = −1744. 803 225− (−1740. 557 85) = −4. 245 4The interest earned at the end of Day 1 is:−MV (1)

¡erh − 1

¢= − (−1740. 557 85)

¡e0.08(1/365) − 1

¢= 0.381 5

I’ll omit the calculations for the other days. Here is the result for all days:Day 0 1 2stock $40.00 $40.50 $39.25put $199.05 $178.08 $230.55delta −0.417596 −0.385797 −0.46892Investment −1, 869.43 −$1, 740.56 −$2, 071.07Interest credited (end of the day) $0.41 $0.381 5 $0.45Capital gain (end of the day) $0.09 −$4. 245 4 −$0.05Daily profit (end of the day) $0.50 −$3. 864 $0.40



Day 3 4 5stock $38.75 $40.00 $40.00put $254.05 $195.49 $194.58delta −0.50436 −0.41940 −0.41986Investment −$2, 208.46 −$1, 873.10 −$1, 874.02Interest credited (end of the day) $0.48 $0.41Capital gain (end of the day) −$4.48 $0.91Daily profit (end of the day) −$4.00 $1.32

Problem 13.6.

Day 0 1 2stock $40.00 $40.642 40.018put $199.05 172.6644 196.5319delta −0.41760 −0.37684 −0.41731Investment −$1, 869.43 −$1, 704.22 −$1, 866.51Interest credited (end of the day) $0.41 $0.37 $0.41Capital gain (end of the day) −$0.42 −$0.35 −$0.40Daily profit (end of the day) −$0.01 $0.02 $0.01

Day 3 4 5stock 39.403 $38.80 39.420put 222.5962 250.8701 220.0727delta −0.45918 −0.50202 −0.45940Investment −$2, 031.89 −$2, 198.56 −$2, 031.02Interest credited (end of the day) $0.45 $0.48Capital gain (end of the day) −$0.45 −$0.48Daily profit (end of the day) $0.00 $0.00

Problem 13.7.

If SOA tests this type of problems in the exam, they’ll need to give you atleast Γ and θ. You can calculate the option premium V and delta ∆. Once youhave Greeks, just use the formula:

V (St+h, T − t− h) ≈ V (St, T − t) +∆t + θh+1

2Γt

2 (Textbook 13.6)

I’m not going to do all the parts in the problem. I’m just going to show yousome examples.

First, you’ll need to get Greeks. You can use the Black-Scholes formula andfind the option premium V and delta ∆. To find Γ and θ, you’ll need to use thespreadsheet attached to the textbook.

σ = 0.3 r = 8% δ = 0 K = 40



Day 0 1time t 0 h = 1/365Expiration T − t 180/365 = 0.493 150 684 9 179/365 = 0.490 410 958 9stock price St 40 44Call price Vt 4.1217 6.8991Delta ∆t 0.6151 0.7720Gamma Γt 0.0454 0.0327Theta θt −0.0134 −0.0137

V (St, T − t) = V (40, 180/365− 0) = 4.1217V (St+h, T − t− h) = V

¡S0+1/365, 180/365− 0− 1/365

¢= 6.8991

= St+h − St = 44− 40 = 4∆t = 0.6151θt = −0.0134 × 365 (The spreadsheet gives the per-day theta; we need to

annualize it.)Γt = 0.0454

V (St, T − t) +∆t + θth+1

2Γt

2

= 4.1217 + 0.6151× 4− 0.0134× 365× 1

365+1

2× 0.0454× 42

= 6. 931 9The true value is V (St+h, T − t− h) = 6.8991The error percentage is:6. 931 9− 6.8991

6.8991= 0.4754%

σ = 0.3 r = 8% δ = 0 K = 40

Day 0 5time t 0 h = 5/365Expiration T − t 180/365 = 0.493 150 684 9 175/365 = 0.479 452 054 8stock price St 40 44Call price Vt 4.1217 6.8440Delta ∆t 0.6151 0.7726Gamma Γt 0.0454 0.0330Theta θt −0.0134 −0.0138

V (St, T − t) = V (40, 180/365− 0) = 4.1217V (St+h, T − t− h) = V

¡S0+1/365, 180/365− 0− 5/365

¢= 6.8440

= St+h − St = 44− 40 = 4∆t = 0.6151 θt = −0.0134× 365 Γt = 0.0454

V (St, T − t) +∆t + θth+1

2Γt

2

= 4.1217 + 0.6151× 4− 0.0134× 365× 5

365+1

2× 0.0454× 42

= 6. 878 3The true value is V (St+h, T − t− h) = 6.8440The error percentage is:



6. 878 3− 6.84406.8440

= 0.5 01%

Problem 13.8.

I’m not going to do all the parts in the problem. I’m just going to show yousome examples.

σ = 0.3 r = 8% δ = 0 K = 40Day 0 1time t 0 h = 1/365Expiration T − t 180/365 = 0.493 150 684 9 179/365 = 0.490 410 958 9Stock price St 40 44Put price Vt 2.5744 1.3602Delta ∆t −0.3849 −0.2280Gamma Γt 0.0454 0.0327Theta θt −0.0050 −0.0053

V (St, T − t) = V (40, 180/365− 0) = 2.5744V (St+h, T − t− h) = V

¡S0+1/365, 180/365− 0− 1/365

¢= 1.3602

= St+h − St = 44− 40 = 4∆t = −0.3849 θt = −0.0050× 365 Γt = 0.0454

V (St, T − t) +∆t + θth+1

2Γt

2

= 2.5744− 0.3849× 4− 0.0050× 365× 1

365+1

2× 0.0454× 42

= 1. 393The true value is V (St+h, T − t− h) = 1.3602The error percentage is:1. 393− 1.3602

1.3602= 0.2 41%

σ = 0.3 r = 8% δ = 0 K = 40

Day 0 5time t 0 h = 5/365Expiration T − t 180/365 = 0.493 150 684 9 175/365 = 0.479 452 054 8Stock price St 40 44Put price Vt 2.5744 1.3388Delta ∆t −0.3849 −0.2274Gamma Γt 0.0454 0.0330Theta θt −0.0050 −0.0054

V (St, T − t) = V (40, 180/365− 0) = 2.5744V (St+h, T − t− h) = V

¡S0+1/365, 180/365− 0− 1/365

¢= 1.3602

= St+h − St = 44− 40 = 4∆t = −0.3849 θt = −0.0050× 365 Γt = 0.0454

V (St, T − t) +∆t + θth+1

2Γt

2



= 2.5744− 0.3849× 4− 0.0050× 365× 5

365+1

2× 0.0454× 42

= 1. 373The true value is V (St+h, T − t− h) = 1.3602

The error percentage is: 1.33881. 373− 1.3388

1.3388= 2. 55%

Problem 13.9.

I’m going to do one set of calculation assuming the stock price one day lateris $30. However, I’m not going to produce a graph.

σ = 0.3 S = 40 r = 8% δ = 0 K = 40

Day 0 1time t 0 h = 1/365Expiration T − t 91/365 90/365Stock price St 40 30Call price Vt 2.7804 0.0730Delta ∆t 0.5824 0.0423Gamma Γt 0.0652 0.0202Theta θt −0.0173 −0.9134a. The price of a 40-strike, 90 day to expiration call is worth 0.0730b. Use delta approximation:= 30− 40 = −10

V0= V0 +∆t = 2.7804 + 0.5824 (−10) = −3. 043 6

We get a nonsense value of −3. 043 6. This is because the delta approxima-tion is good when is small. Here we have a large change of = −10.c. Use delta-gamma approximation:

V0= V0 +∆t +

1

2Γt

2

= 2.7804 + 0.5824 (−10) + 12× 0.0652× (−10)2 = 0.216 4

d. Use delta-gamma-theta approximation:

V0= V0 +∆t +

1

2Γt

2 + θth

= 2.7804 + 0.5824 (−10) + 12× 0.0652× (−10)2 − 0.0173× 365× 1

365= 0.199 1We see in a, b, c, and d, the approximation is not good. This is because the

approximation is good when is small. Here we have a large change of = −10.

Problem 13.10.



I’m not going to produce a graph. I’ll do one set of calculation assuming thestock price one day later is $41.

σ = 0.3 S = 40 r = 8% δ = 0 K = 40

Day 0 1time t 0 h = 1/365Expiration T − t 1 364/365Stock price St 40 30Call price Vt 6.2845 6.9504Delta ∆t 0.6615 0.6909Gamma Γt 0.0305 0.0287Theta θt −0.0104 −0.0106a. The price of a 40-strike, 364 day to expiration call is worth 6.9504b. Use delta approximation:= 41− 40 = 1

V0= V0 +∆t = 6.2845 + 0.6615 (1) = 6. 946

c. Use delta-gamma approximation:

V0= V0 +∆t +

1

2Γt

2

= 6.2845 + 0.6615 (1) +1

2× 0.0305× (1)2 = 6. 961 3


V0= V0 +∆t +

1

2Γt

2 + θth

= 6.2845 + 0.6615 (1) +1

2× 0.0305× (1)2 − 0.0104× 365× 1

365= 6. 950 9

We see in a, b, c, and d, the approximation is good. This is because issmall.

Problem 13.11.

I’m going to do one set of calculation assuming the stock price one day lateris $30.

σ = 0.3 S = 40 r = 8% δ = 0 K = 40

Day 0 1time t 0 h = 1/365Expiration T − t 91/365 90/365Stock price St 40 30Put price Vt 1.9905 9.2917Delta ∆t −0.4176 −0.9577Gamma Γt 0.0652 0.0202Theta θt −0.0088 0.0061a. The price of a 40-strike, 90 day to expiration put is worth 9.2917



b. Use delta approximation:= 30− 40 = −10

V0= V0 +∆t = 1.9905− 0.4176 (−10) = 6. 166 5


V0= V0 +∆t +

1

2Γt

2

= 1.9905− 0.4176 (−10) + 12× 0.0652× (−10)2 = 9. 426 5


V0= V0 +∆t +

1

2Γt

2 + θth

= 1.9905− 0.4176 (−10) + 12× 0.0652× (−10)2 − 0.0088× 365× 1

365= 9. 417 7

Problem 13.12.

I’m going to do one set of calculation assuming the stock price one day lateris $41.

σ = 0.3 S = 40 r = 8% δ = 0 K = 40

Day 0 1time t 0 h = 1/365Expiration T − t 1 364/365Stock price St 40 41Put price Vt 3.2092 2.8831Delta ∆t −0.3385 −0.3091Gamma Γt 0.0305 0.0287Theta θt −0.0023 −0.0025a. The price of a 40-strike, 364 day to expiration put is worth 2.8831b. Use delta approximation:= 41− 40 = 1

V0= V0 +∆t = 3.2092− 0.3385 (1) = 2. 870 7


V0= V0 +∆t +

1

2Γt

2

= 3.2092− 0.3385 (1) + 12× 0.0305× (1)2 = 2. 885 95


V0= V0 +∆t +

1

2Γt

2 + θth

= 3.2092− 0.3385 (1) + 12× 0.0305× (1)2 − 0.0023× 365× 1

365= 2. 883 65

Problem 13.13.



Please note that in DM 13.9, θ is annualized. However, in Table 13.1, θ isa per-day theta. (DM page 411 says that if the expiry is measured in years,then theta is the annualized change in option value; to get the per-day theta,divide by 365. The θ in Table 13.1 and in the author’s Excel spreadsheet isdaily theta.) So we need to annualize θ before using DM 13.9.

−µ1

2× 0.32 × 402 × 0.0652− 0.0173× 365 + 0.08 (0.5824× 40− 2.7804)

¶×

1

365= −5. 79 4× 10−5 ≈ 0

Problem 13.14.

We’ll use the Excel spreadsheet attached in the DM textbook and calculatethe following:Inputs:Stock Price 40Exercise Price 45Volatility 30%Risk-free interest rate 8%Time to Expiration (years) 0.5Dividend Yield 0%

Outputs:Black-Scholes (European)

Call PutPrice 2.1304 5.3659Delta 0.3972 −0.6028Gamma 0.0454 0.0454Vega 0.1091 0.1091Theta −0.0120 −0.0025Rho 0.0688 −0.1474Psi −0.0794 0.1206Elasticity 7.4578 −4.4936

DM 13.9 is:

−µ1

2× 0.32 × 402 × 0.0454− 0.0025× 365 + 0.08 (−0.6028× 40− 5.3659)

¶×

1

365= 5. 29× 10−6 ≈ 0

Problem 13.15.

Use DM’s spreadsheet. The inputs are:



Stock Price 40Exercise Price 45Volatility 30%Risk-free interest rate 8%Time to Expiration (years) = 180/365Dividend Yield 0%

We’ll get the following outputs:

Black-Scholes (European)40-strike call 45-strike call

Price 4.1217 2.1004Delta 0.6151 0.3949Gamma 0.0454 0.0457

We bought one 45-strike call. Suppose we need to sell (i.e. write) X unitof 40-strike call. The Gamma of the bought 45-strike call is −0.0457 (negativebecause we bought a call). The Gamma of the sold X unit of 40-strike call is0.0454X. The total Gamma of our portfolio is 0.0454X − 0.0457. To Gammahedge, set 0.0454X − 0.0457 = 0. This gives us X = 1. 006 6. The total Deltaof our portfolio is 0.6151× 1. 006 6− 0.3949 = 0.224 3. To delta hedge, we needto buy 0.224 3 share of the underlying stock.This is our final portfolio at time zero:Transactions at time zero CostBuy a 45-strike call 2.1004Sell 1. 006 6 unit of 40-strike call (to Gamma hedge) −1. 006 6× 4.1217 = −4. 148 9Buy 0.224 3 share of the stock (to Delta hedge) 0.224 3× 40 = 8. 972Borrow 2.1004− 4. 148 9 + 8. 972 = 6. 923 5 −6. 923 5Total 2.1004− 4. 148 9 + 8. 972− 6. 923 5 = 0

One day later, you close out your position:Transactions (one day later) RevenueSell a 45-strike call C

0

1

Buy 1. 006 6 unit of 40-strike call −1. 006 6C 0

2

Sell 0.224 3 share of the stock 0.224 3S0

Repay the borrowed amount 6. 923 5e0.08/365

Total C0

1 − 1. 006 6C0

2 + 0.224 3S0 − 6. 923 5e0.08/365

C0

1 is the price of a 45-strike call with 179 day to expiration.C

0

2 is the price of a 40-strike call with 179 day to expiration.S0is the price of the stock one day later.

If you want to draw the diagram of the overnight profit C0

1 − 1. 006 6C0

2 +0.224 3S

0 − 6. 923 5e0.08/365, you can get different profit by changing S0. For

example,



assume S0= 50. Also assume that the stock volatility and risk free interest

rate are not changed. Using the Black-Scholes formula (use T = 179/365), weget:

C0

1 = 8.1511 C0

2 = 12.0043The overnight profit is:8.1511− 1. 006 6× 12.0043 + 0.224 3× 50− 6. 923 5e0.08/365 = 0.36

If S0= 50, then the overnight profit is 0 − 1. 006 6 × 0 + 0.224 3 × 0 − 6.

923 5e0.08/365 = −6. 93.

By changing S0, you’ll get different profits. Then you draw a graph on how

the overnight profit varies by S0. Since it’s time-consuming to calculate the

overnight profit by changing S0, I’m not going to do it.

Problem 13.16.

Use DM’s spreadsheet. The inputs are:

Stock Price 40Exercise Price 45Volatility 30%Risk-free interest rate 8%Time to Expiration (years) = 180/365Dividend Yield 0%

We’ll get the following outputs:

Black-Scholes (European)40-strike call 45-strike put

Price 4.1217 5.3596Delta 0.6151 −0.6051Gamma 0.0454 0.0457We sell one 45-strike put and buy X unit of 40-strike call. The total Gamma

is 0.0457 − 0.0454X. To Gamma hedge, set 0.0457 − 0.0454X = 0. This givesus X = 1. 006 6.

The delta of 1. 006 6 unit of 40-strike call is −0.6151× 1. 006 6.The total Delta is −0.6151× 1. 006 6− 0.6051 = −1. 224 3, meaning that we

need to short sell 1. 224 3 unit of the underlying stock.

Final portfolio at time zero:Transactions at time zero RevenueSell a 45-strike put 5.3596Buy 1. 006 6 unit of 40-strike call (to Gamma hedge) −1. 006 6× 4.1217 = −4. 148 9Short sell 1. 224 3 share of the stock (to Delta hedge) 1. 224 3× 40 = 48. 972Lend 5.3596− 4. 148 9 + 48. 972 = 50. 182 7 −50. 182 7Total 5.3596− 4. 148 9 + 48. 972− 50. 182 7 = 0



One day later, you close out your position:Transactions (one day later) RevenueBuy a 45-strike put −P 0

1

Sell 1. 006 6 unit of 40-strike call 1. 006 6C0

2

Buy 1. 224 3 share of the stock 1. 224 3S0

Collect loan 50. 182 7e0.08/365

Total −P 0

1 + 1. 006 6C0

2 − 1. 224 3S0+ 50. 182 7e0.08/365

P0

1 is the price of a 45-strike put with 179 day to expiration.C

0

2 is the price of a 40-strike call with 179 day to expiration.S0is the price of the stock one day later.

Problem 13.17.

Assume the butterfly spread is asymmetric.

λ =K3 −K2

K3 −K1=45− 4045− 35 = 0.5

Written butterfly spread consists of selling one 35-strike call, buying two40-strike calls, and selling one 45-strike call.Inputs: σ = 0.3 S = 40 r = 8% δ = 0 T = 91/365strike K 35 40 45 butterfly spreadcall price 6.1315 2.7804 0.9710 6.1315− 2 (2.7804) + 0.9710 = 1. 541 7Delta 0.8642 0.5824 0.2815 0.8642− 2 (0.5824) + 0.2815 = −0.019 1Gamma 0.0364 0.0652 0.0563 0.0364− 2 (0.0652) + 0.0563 = −0.037 7Theta −0.0134 −0.0173 −0.0134 −0.0134− 2 (−0.0173)− 0.0134 = 0.007 8

Inputs: σ = 0.3 S = 40 r = 8% δ = 0 T = 180/365strike K 49call price 4.1217Delta 0.6151Gamma 0.0454Theta −0.0134

Suppose we sell X units of 180 day to expiration call. The total Gamma ofthe written butterfly spread and written X units of 180 day to expiration callis −0.037 7 + 0.0454X.To Gamma hedge, set −0.037 7 + 0.0454X = 0, X = 0.830 4. The total

Delta of the written butterfly and the written X units of 180 day to expirationcall is −0.019 1+ 0.830 4× 0.6151 = 0.491 7. So we need to buy 0.491 7 share ofthe underlying stock.

Transactions at time zero Cost (negative cost means receiving money)Sell the spread and buy 0.491 7 stock −1. 541 7 + 0.491 7× 40 = 18. 126 3Borrow 18. 126 3 from a bank −18. 126 3Total 0



One day later, we close out our position:Transactions (one day later) RevenueBuy a 35-strike 90 day to expiry call −C0

1

Sell two 40-strike 90 day to expiry calls 1. 006 6 2C0

2

Buy a 45-strike 90 day to expiry call −C0

3

Sell 0.491 7 stock 0.491 7S0

Repay the borrowed amount −18. 126 3e0.08/365Total −C0

1 + 2C0

2 − C0

3 + 0.491 7S0 − 18. 126 3e0.08/365

So the total overnight profit is −C0

1+2C0

2−C0

3+0.491 7S0−18. 126 3e0.08/365.

Problem 13.18.

inputs: σ = 0.3 S = 40 r = 8% δ = 0 T = 91/365strike K 40 45 ratio spreadput price 1.9905 5.0824 2× 1.9905− 5.0824 = −1. 101 4Delta −0.4176 −0.7185 2× (−0.4176)− (−0.7185) = −0.116 7Gamma 0.0652 0.0563 2× 0.0652− 0.0563 = 0.074 1Theta −0.0088 −0.0037 2× (−0.0088)− (−0.0037) = −0.013 9

in puts: σ = 0.3 S = 40 r = 8% δ = 0 T = 180/365strike K 40call price 4.1217Delta 0.6151Gamma 0.0454Theta −0.0134

We buy X unit of 40-strike 180-day to expiration call. The total Gammaof the ratio spread and bought X unit of 40-strike 180-day to expiration callis 0.074 1 + 0.0454X. To Gamma hedge, set 0.074 1 + 0.0454X = 0, X = −1.632 2. So we need to short sell 1. 632 2 shares of the underlying stock.

Transactions at time zero RevenueEnter the ratio spread and short 1. 632 2 stock 1. 101 4 + 1. 632 2× 40 = 66. 389 4Deposit 66. 389 4 in a savings account −66. 389 4Total 0

One day later, we close out our position:Transactions (one day later) RevenueBuy two 40-strike 90 day to expiry puts −2P 0

1

Sell a 45-strike 90 day to expiry put P0

2

Buy 1. 632 2 stock −1. 632 2S0

Receive money from the bank 66. 389 4e0.08/365

Total −2P 0

1 + P0

2 − 1. 632 2S0+ 66. 389 4e0.08/365



So the total overnight profit is −2P 0

1 + P0

2 − 1. 632 2S0+ 66. 389 4e0.08/365.

Skip Problem 13.19 and 13.20. Vega hedge and rho hedge are far outsidethe scope of the Exam MFE syllabus.


Chapter 14

Exotic options: I

Problem 14.1.

For n non-negative numbers (such as stock prices) S1, S2, ..., Sn, the arith-metic mean can never be less than the geometric mean:

S1 + S2 + ...+ Snn

≥ (S1S2...Sn)1/n

If and only if S1 = S2 = ... = Sn, we have:S1 + S2 + ...+ Sn

n= (S1S2...Sn)

1/n

The proof can be found at Wikipedia:http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_

meansFor example, this is the proof for n = 2.(S1 − S2)

2 ≥ 0→ S21 + S22 ≥ 2S1S2(S1 + S2)

2= S21 + S22 + 2S1S2 ≥ 4S1S2µ

S1 + S22

¶2≥ S1S2

S1 + S22

≥√S1S2

Problem 14.2.

Arithmetic average:5 + 4 + 5 + 6 + 5

5= 5.0

Geometric average:(5× 4× 5× 6× 5)1/5 = 4. 959

Ignore the question "What happens to the difference between the two mea-sures of the averages as the standard deviation of the observations increase?"This question is vague. I’m not sure what the author is after.

149

CHAPTER 14. EXOTIC OPTIONS: I

Problem 14.3.

u = e(r−δ)h+σ√h = e(0.08−0)0.5+0.3

√0.5 = 1. 286 765 9

u = e(r−δ)h−σ√h = e(0.08−0)0.5−0.3

√0.5 = 0.841 868 0


u− d=

e(0.08−0)0.5 − 0.841 8681. 286 766 − 0.841 868 = 0.447 165

πd = 1− πu = 1− 0.447 165 = 0.552 835

t = 0 t = 0.5 t = 1Suu = 165.57665

Su = 128.67659S = 100 Sud = 108.32871

Sd = 84.18680Sdd = 70.87417

The four arithmetic averages are:128.67659 + 165.57665

2= 147. 126 62 (The path is uu or up up)

128.67659 + 108.32871

2= 118. 502 65 (ud)

84.18680 + 108.32871

2= 96. 257 8 (du)

84.18680 + 70.87417

2= 77. 530 485 (dd)

Path Average asset Call Payoff Risk Neutral Probuu 147. 126 62 47. 126 62 π2uud 118. 502 65 18. 502 65 πuπddu 96. 257 8 0 πuπddd 77. 530 485 0 π2d

The price of the Asian arithmetic average price call:e−0.08(47. 126 62× 0.447 1652 + 18. 502 65× 0.447 165× 0.552 835+0× 0.447 165× 0.552 835 + 0× 0.552 8352) = 12. 921

If we were to calculate the price of the Asian arithmetic average asset put,then

Path Average asset Put Payoff Risk Neutral Probuu 147. 126 62 0 π2uud 118. 502 65 0 πuπddu 96. 257 8 3. 742 2 πuπddd 77. 530 5 22. 469 5 π2d

The price of the Asian arithmetic average price put:



e−0.08(0× 0.447 1652 + 0× 0.447 165× 0.552 835+3. 742 2× 0.447 165× 0.552 835 + 22. 469 5× 0.552 8352 = 7. 193

The four geometric averages are:√128.67659× 165.57665 = 145. 965 2 (uu)

√128.67659× 108.32871 = 118. 065 1 (ud)

√84.18680× 108.32871 = 95. 497 9 (du)

√84.18680× 70.87417 = 77. 244 2 (dd)

Path Average asset Call Payoff Risk Neutral Probuu 145. 965 2 45. 965 2 π2uud 118. 065 1 18. 065 1 πuπddu 95. 497 9 0 πuπddd 77. 244 2 0 π2d

The price of the Asian geometric average price call:e−0.08(45. 965 2× 0.447 1652 + 18. 065 1× 0.447 165× 0.552 835+0× 0.447 165× 0.552 835 + 0× 0.552 8352) = 12. 607

If we were asked to calculate the price of the Asian average price put, then:Path Average asset Call Payoff Risk Neutral Probuu 145. 965 2 0 π2uud 118. 065 1 0 πuπddu 95. 497 9 4. 502 1 πuπddd 77. 244 2 22. 755 8 π2d

The price of the Asian geometric average price put:e−0.08(0× 0.447 1652 + 0× 0.447 165× 0.552 835+4. 502 1× 0.447 165× 0.552 835 + 22. 755 8× 0.552 8352) = 7. 447

Problem 14.4.

a. Asian arithmetic average strike call:Path ST K Call Payoff Risk Neutral Probuu 165.57665 147. 126 62 18. 450 03 π2uud 108.32871 118. 502 65 0 πuπddu 108.32871 96. 257 8 12. 070 91 πuπddd 70.87417 77. 530 485 0 π2d

The price of the Asian arithmetic average strike call:e−0.08(18. 450 03× 0.447 1652 + 0× 0.447 165× 0.552 835+12. 070 91× 0.447 165× 0.552 835 + 0× 0.552 8352) = 6. 160



If we were to calculate the price of the Asian arithmetic average strike put,then

Path ST K Put Payoff Risk Neutral Probuu 165.57665 147. 126 62 0 π2uud 108.32871 118. 502 65 10. 173 94 πuπddu 108.32871 96. 257 8 0 πuπddd 70.87417 77. 530 485 76. 656 315 π2d

The price of the Asian arithmetic average strike put:e−0.08(0× 0.447 1652 + 10. 173 94× 0.447 165× 0.552 835+0× 0.447 165× 0.552 835 + 6. 656 315× 0.552 8352) = 4. 20

b. The price of the Asian geometric average strike call:Path ST K Call Payoff Risk Neutral Probuu 165.57665 145. 965 2 19. 611 45 π2uud 108.32871 118. 065 1 0 πuπddu 108.32871 95. 497 9 12. 830 81 πuπddd 70.87417 77. 244 2 0 π2d

The price of the Asian geometric average strike call:e−0.08(19. 611 45× 0.447 1652 + 0× 0.447 165× 0.552 835+12. 830 81× 0.447 165× 0.552 835 + 0× 0.552 8352) = 6. 548

If we were to calculate the price of the Asian geometric average strike put,then

Path ST K Put Payoff Risk Neutral Probuu 165.57665 145. 965 2 0 π2uud 108.32871 118. 065 1 9. 736 39 πuπddu 108.32871 95. 497 9 0 πuπddd 70.87417 77. 244 2 6. 370 03 π2d

The price of the Asian geometric average strike put:e−0.08(0× 0.447 1652 + 9. 736 39× 0.447 165× 0.552 835+0× 0.447 165× 0.552 835 + 6. 370 03× 0.552 8352) = 4. 019

Problem 14.5.

Stock price tree:182.14179

165.57665128.67659 128.814749

100 108.3287184.18680 91.10067

70.8741764.42843



Calculate the price of the arithmetic average asset call.Path Arithmetic Average Payoff Risk Neutral Probuuu 151.1369 51.1369 π3u = 0.0953uud 133.3612 33.3612 π2uπd = 0.1133udu 118.8058 18.8058 π2uπd = 0.1133udd 106.2345 6.2345 πuπ

2d = 0.1348

duu 106.8874 6.8874 π2uπd = 0.1133dud 94.3160 0 πuπ

2d = 0.1348

ddu 84.0221 0 πuπ2d = 0.1348

ddd 75.1314 0 π3d = 0.1603Total 1

The price of the arithmetic average asset call is:e−0.08(51.1369× 0.0953 + 33.3612× 0.1133+18.8058× 0.1133 + 6.2345× 0.1348 + 6.8874× 0.1133) = 11. 45

Calculate the price of the geometric average asset call.Path Arithmetic Average Payoff Risk Neutral Probuuu 149.1442 49.1442 π3u = 0.0953uud 132.8796 32.8796 π2uπd = 0.1133udu 118.3887 18.3887 π2uπd = 0.1133udd 105.4781 5.4781 πuπ

2d = 0.1348

duu 105.4781 5.4781 π2uπd = 0.1133dud 93.9754 0 πuπ

2d = 0.1348

ddu 83.7272 0 πuπ2d = 0.1348

ddd 74.5965 0 π3d = 0.1603Total 1

The price of the geometric average asset call is:e−0.08(49.1442× 0.0953 + 32.8796× 0.1133+18.3887× 0.1133 + 5.4781× 0.1348 + 5.4781× 0.1133) = 10. 94

Problem 14.6.

a. Using the Black-Scholes formula, we find that the call price is $4.1293.

b and c.knock-in call + knock-out call = ordinary callHowever, the knock-out call is worthless because the barrier 44 is lower than

the strike price 45. For the call to be worth something, the stock price mustexceed the strike price 45. However, as soon as the stock reaches the barrier 44,the call is knocked out dead (i.e. the call doesn’t exist any more). Hence thisknock-out call will never have a positive payoff.Hence the knock-in call becomes the ordinary call. The premium of the

knock-in call is also $4.1293.Key point to remember:



1. A knock-out call is worthless if the barrier is less than or equal to thestrike price.

2. A knock-in call is just an ordinary call if the barrier is less than or equalto the strike price.

Problem 14.7.

DM Chapter 14 doesn’t have any formula on the price of a barrier option. Tocalculate the barrier option price, you have to use the textbook’s spreadsheet.So this problem is out of the scope of Exam MFE. I used the textbook’s

Excel spreadsheet and calculated the following:

T BS DO BS/DO0.25 0.9744 0.7323 1.33060.5 2.1304 1.2482 1.70671 4.1293 1.8217 2.26672 7.4398 2.4505 3.03603 10.2365 2.8529 3.58814 12.6969 3.1559 4.02325 14.9010 3.4003 4.3823100 39.9861 5.3112 7.5286

BS: the price of a standard option using the Black-Scholes formula.DO: Down-and-out barrier option. Since the current stock price 40 exceeds

the barrier 38, the knock-out call with 38 barrier is really a down-and-out option(if the stock price drops to 38, the call is dead).The greater the expiration, the greater the ratio of BS/DO. As the expira-

tion increases, both the standard option and the down-and-out option becomemore valuable. However, as T increases, the value of a standard option goes upmuch faster than a down-and-out option.The value of a down-and-out option has a moderate increase. As T increase,

the stock price may be more volatile, increasing the value of the barrier option.However, there’s also more chance that the stock price may hit the barrier.

Problem 14.8.

Out of the scope of MFE. However, I used the textbook spreadsheet andfound the following:



T BS UO BS/UO0.25 5.0833 3.8661 1.31490.5 5.3659 3.4062 1.57531 5.6696 2.8626 1.98062 5.7862 2.2233 2.60263 5.6347 1.8109 3.11154 5.3736 1.5094 3.56015 5.0654 1.2761 3.9695100 0.0012 0.0001 24.4960

As T increases, the BS/UO increases too (just like in Problem 14.7). How-ever, if T is too big, the value of a standard put and the value of an up-and-output both approach zero. If T is too big, we lose the time value of money (themoney we spent in buying a put could have been invested elsewhere and earnedlot of interest).

Problem 14.9.

I used the spreadsheet and calculated the following (T is expressed in months):T BS UO BS/UO1 0.1727 0.1727 1.00032 0.5641 0.5479 1.02963 0.9744 0.8546 1.14014 1.3741 1.0384 1.32345 1.7593 1.1243 1.56496 2.1304 1.1468 1.85777 2.4886 1.1316 2.19918 2.8353 1.0954 2.58859 3.1718 1.0482 3.026010 3.4991 0.9962 3.512411 3.8180 0.9430 4.048812 4.1293 0.8906 4.6365

As T goes up, the BS/UO ratio goes up too.

Problem 14.10.

With K = 0.9, the standard put is worth 0.0188. With barrier 1 or 1.05, theup-and-out barrier is also worth 0.0188. Why? DM page 452 and 453 have anexplanation. Here is the main point. If the exchange rate never hits 1, then thestandard put and the up-and-out put with 1 or 1.5 barrier have the same value.The only way that the standard put is more valuable than the up-and-out putis when the exchange rate goes up from 0.9 to 1 or to 1.05 and then falls below0.9, in which case the up-and-out is dead yet the standard put has a positivepayoff of K − ST = 0.9− ST . However, such a scenario is rare and 0.9− ST is



small. Hence the standard put and the up-and-out put have roughly the samevalue.In contrast, when the strike price K = 1, the payoff of the standard put at

the above mentioned scenario is 1− ST , which is greater than 0.9− ST . Hencethe standard put is slightly more valuable than the up-and-out put.

Problem 14.11.

a.Using the Black-Scholes formula, we find the call premium is $9.6099.

b. If we buy the compound option, then we can, at t1 = 1, buy, for aguaranteed price of x = 2, a call that expires at T = 2. We’ll exercise thecompound option at t1 = 1 only if the stock price at t = 1 is such that the callwe are entitled to buy is equal to or greater than the premium 2.We’ll use the textbook Equation 14.11 to find S∗, the minimum price such

that exercising the compound option is worthwhile.C (S∗,K, T − t1) = xC (S∗, 40, 2− 1) = 2C (S∗, 40, 1) = 2To solve for S∗, we have use the trial-and-error approach. I found that

S∗ = 31.723.In other words, if the stock price at t1 = 1 is 31.723, then a call option

written at t1 = 1 and expires at T = 2 is exactly worth $2.In this case, we’ll notexercise the compound option. We’ll let the compound option expire worthless.If the stock price at t1 = 1 is less than 31.723, then the call written at t1 = 1

and expires at T = 2 is worth less than $2 and we’ll not exercise the compoundoption. We’ll let the compound option expire worthless.Only if the stock price at t1 = 1 is greater than 31.723 should we exercise

the compound option.c. To find the price of the compound call (i.e. call on call), we’ll use worksheet

called "Compound" provided by the DM textbook.The inputs are:Stock Price 40Exercise Price to buy asset 40Exercise Price to buy option 2Volatility 30%Risk-free interest rate 8%Expiration for Option on Option (years) 1Expiration for Underlying Option (years) 2Dividend Yield 0%

Compound Option PricesCall on Call 7.9482Put on Call 0.1845Call on Put 2.2978



Put on Put 0.4484Critical S for compound call 31.723Critical S for compound put 44.3494So the compound call premium is 7.9482. (From the output, we see that the

critical stock price for the compound call is 31.723, which matches our answerin part b).

d. From Part c, we see that the price of the put on call is 0.1845.Alternatively, we can use DM 14.12 to calculate the put on the call, given

we know the price of the call on call.CallOnCall − PutOnCall + xe−rt1 = BSCall

7.9482− PutOnCall + 2e−0.08×1 = 9.6099PutOnCall = 0.184 5

Problem 14.12.

a. Use the Black-Scholes formula, we find the put price is 3.6956.b. Use the "Compound" worksheet.InputsStock Price 40Exercise Price to buy asset 40Exercise Price to buy option 2Volatility 30%Risk-free interest rate 8%Expiration for Option on Option (years) 1Expiration for Underlying Option (years) 2Dividend Yield 0%

Outputs:Call on Call 7.9482Put on Call 0.1845Call on Put 2.2978Put on Put 0.4484Critical S for compound call 31.723Critical S for compound put 44.3494

So if S1 < 44.3494, we’ll exercise the put at t = 1.

c. The put on put price is 0.4484.

Problem 14.13.

a. DM 14.15 gives you the price formula for a gap call. A gap put formulais

P (K1,K2) = K1e−rTN (−d2)− Se−δTN (−d1)



d1 =

lnS

K2+

µr − δ +

1

2σ2¶T

σ√T

d2 = d1 − σ√T

For foreign currency, δ = r€, r = r$, and S = x0.

b. The gap put payoff is 0.8− x if x < 1. If σ = 0, then x = x0 = 0.9. Thegap out payoff is −0.1. The gap put premium is 0.8e−0.06×0.5 − 0.9e−0.03×0.5 =−0.110 2.As the volatility increases, the value of the gap put increases.

Problem 14.14.

Skip.

Problem 14.15.

Skip.

Problem 14.16.

Price of a standard 40-strike call on S. The inputs to the Black Scholesformulas are:

S = K = 40 σ = 0.3 r = 0.08 T = 1 δ = 0The call price is 6.2845.

The price of an exchange option with S as underlying and 0.667Q as thestrike asset is 7. 577.

S = 40 K = 0.667× 60 = 40σ =√0.32 + 0.52 − 2× 0.5× 0.3× 0.5 = 0.435 89

d1 =ln

40e−0×1

40e−0.04×1+1

2× 0.435 892 × 1

0.435 89√1

= 0.309 7

d2 = 0.309 7− 0.435 89√1 = −0.126 2

N (d1) = 0.621 6 N (d2) = 0.449 8C = 40e−0×1 × 0.621 6− 40e−0.04×1 × 0.449 8 = 7. 577

Problem 14.17.

a. Inputs to the exchange call formula:S = 40 δS = 0 σS = 0.3K = 60 δK = 0 σK = 0.5ρ = 0.5 T = 1



You should get: C = 2. If you change δS = 0.1, then the call price isC = 1.2178

In the replicating portfolio, the number of underlying stocks to hold at timezero is e−δST . If we increase δS , then we hold fewer shares of the underlyingasset and the call price drops.

b. inputs to the exchange call formula:S = 40 δS = 0 σS = 0.3K = 60 δK = 0 σK = 0.5ρ = 0.5 T = 1Once again, C = 2.If δK = 0.1, then C = 2.86.If δK goes up, then the present value of the strike price goes down, making

the call more valuable.

c. inputs to the exchange call formula:S = 40 δS = 0 σS = 0.3K = 60 δK = 0 σK = 0.5ρ = 0.5 T = 1

Once again, C = 2.If ρ = −0.5, then C = 5.79.If ρ is negative, then σ =

pσ2S + σ2K − 2ρσSσK is higher then if ρ is positive.

So a negative ρ increases σ, making an option more valuable.

Problem 14.18.

ρ = 1σ =

pσ2S + σ2K − 2ρσSσK =

√0.32 + 0.32 − 2× 1× 0.3× 0.3 = 0

a. The call price is zero. Because V ar [ln (S/K)] = σ2S + σ2K − 2ρσSσK = 0,S/K is a constant during [0, T ]. Since at time zero S = K, then S = K during[0, T ]. So the payoff of the call is zero and the call is worthless.

b. Now σ =pσ2S + σ2K − 2ρσSσK =

√0.42 + 0.32 − 2× 1× 0.4× 0.3 =

0.1. Using DM 14.16, we find that the exchange call premium is 1.60.Skip the remaining problems (Problem 14.19 through 14.22).




Chapter 18

Lognormal distribution

Problem 18.1.

Using DM 18.4 z =x− μ

σ, we find the equivalent draws from the a standard

normal distribution are:−7− (−8)√

15= 0.258 2

−11− (−8)√15

= −0.774 6 −3− (−8)√15

= 1. 2910

2− (−8)√15

= 2. 5820−15− (−8)√

15= −1. 807 4

Problem 18.2.

Using DM 18.4 z =x− μ

σ, we get x = μ+ zσ. The equivalent draws are:

0.8 + (−1.7)√25 = −7. 7 0.8 + (0.55)

√25 = 3. 55

0.8 + (−0.3)√25 = −0.7 0.8 + (−0.02)

√25 = 0.7

0.8 + (0.85)√25 = 5. 05

Problem 18.3.

Linear combination of normal random variables is also normal.x1 + x2 is normal. Its mean is E (x1 + x2) = E (x1) +E (x2) = 1− 2 = −1.

Its variance isV ar (x1 + x2) = V ar (x1)+V ar (x2) + 2Cov (x1, x2) = 5+ 2+2 (1.3) = 9. 6

x1−x2 is normal. Its mean is E (x1 − x2) = E (x1)−E (x2) = 1− (−2) = 3.Its variance is

V ar (x1 − x2) = V ar (x1)+V ar (x2)− 2Cov (x1, x2) = 5+ 2− 2 (1.3) = 4. 4

Problem 18.4.

161

CHAPTER 18. LOGNORMAL DISTRIBUTION

x1 + x2 is normal with mean E (x1 + x2) = E (x1) + E (x2) = 2 + 8 = 10and variance

V ar (x1 + x2) = V ar (x1) + V ar (x2) + 2Cov (x1, x2)Cov (x1, x2) = ρσx1σx2

⇒ V ar (x1 + x2) = 0.5 + 14 + 2 (−0.3)√0.5√14 = 12. 91

x1 − x2 is normal with mean E (x1 − x2) = 2− 8 = −6 and varianceV ar (x1 − x2) = V ar (x1) + V ar (x2)− 2Cov (x1, x2)V ar (x1 − x2) = 0.5 + 14− 2 (−0.3)

¡√0.5¢ ¡√

14¢= 16. 09

Problem 18.5.

Use DM 18.10.x1 + x2 + x3 is normal with mean

E (x1 + x2 + x3) = E (x1) +E (x2) +E (x3) = 1 + 2 + 2.5 = 5. 5varianceV ar (x1 + x2 + x3) = V ar (x1)+V ar (x2)+V ar (x3)+2Cov (x1, x2)+2Cov (x1, x3)+

2Cov (x2, x3)= 5 + 3 + 7 + 2 (0.3)

¡√5¢ ¡√

3¢+ 2 (0.1)

¡√5¢ ¡√

7¢+ 2 (0.4)

¡√3¢ ¡√

7¢=

22. 17

x1 + 3x2 + x3 is normal with mean

E (x1 + 3x2 + x3) = E (x1) + 3E (x2) +E (x3) = 1 + 3 (2) + 2.5 = 9. 5varianceV ar (x1 + 3x2 + x3) = V ar (x1)+3

2V ar (x2)+V ar (x3)+2×3Cov (x1, x2)+2Cov (x1, x3) + 2× 3Cov (x2, x3)= 5+32×3+7+2×3 (0.3)

¡√5¢ ¡√

3¢+2 (0.1)

¡√5¢ ¡√

7¢+2×3 (0.4)

¡√3¢ ¡√

7¢=

58. 15

x1 + x2 + 0.5x3 is normal with mean

E (x1 + x2 + 0.5x3) = E (x1)+E (x2)+ 0.5E (x3) = 1+2+0.5 (2.5) = 4. 25varianceV ar (x1 + x2 + 0.5x3) = V ar (x1)+V ar (x2)+0.5

2V ar (x3)+2Cov (x1, x2)+2× 0.5Cov (x1, x3) + 2× 0.5Cov (x2, x3)= 5 + 3 + 0.52 × 7 + 2 (0.3)

¡√5¢ ¡√

3¢+ 2 × 0.5 (0.1)

¡√5¢ ¡√

7¢+ 2 ×

0.5 (0.4)¡√3¢ ¡√

7¢= 14. 50

Problem 18.6.



Use DM 18.13.E (ex) = e2+0.5(5) = 90. 017 1

Let a represent the median of y = ex. Then P (y ≤ a) = 0.5.

y = ex is an increasing function. Hence of the median of ex corresponds tothe median of x.Let b represent the median of x.P (x ≤ b) = 0.5. Hence b = 2 (the median of a normal random variable is

its mean).=⇒ a = eb = e2 = 7. 389 1

Problem 18.7.

There’s a typo in the problem. The table should say "Month" instead of"Day" (i.e. should be Month 0, Month 1, ...,Day 4, not Day 0, Day 1, ...,Day4).This is similar to DM Table 11.1.a. Stock A:

Month Stock price rt=lnStSt−1

³rt −

∧r´2

0 100

1 105 ln105

100= 0.04879 016 0.00238048

2 102 ln102

105= −0.02898 75 0.00084028

3 97 ln97

102= −0.05026 183 0.00252625

4 100 ln100

97= 0.03 045 921 0.00092776

Total 0.00000000 0.00667477

mean monthly continuously compounded return:∧r =

4Pt=1

rt

4=0

4= 0

monthly standard deviation:

vuuut 4Pt=1

³rt −

∧r´2

4− 1 =

r0.00667477

3= 4. 716 909 3×

10−2

annual standard deviation:√12¡4. 716 909 3× 10−2

¢= 16.34%

b. Stock B



Month Stock price rt=lnStSt−1

³rt −

∧r´2

0 1001 105 0.04879016 0.002380482 102 0.35667494 0.127217023 97 −0.43592432 0.190030014 100 0.00000000 0.00092776

Total 0 0.32055527

mean monthly continuously compounded return:∧r =

4Pt=1

rt

4=0

4= 0

monthly standard deviation:

vuuut 4Pt=1

³rt −

∧r´2

4− 1 =

r0.32055527

3= 0.326 881 87

annual standard deviation:√12 (0.326 881 87) = 1. 132 4 = 113.24%

c. The statement is correct.For example, the mean monthly continuously compounded return for Stock

A is

∧r =

4Pt=1

rt

4=

4Pt=1ln

StSt−14

=

4Pt=1(lnSt − lnSt−1)

4=lnS4 − lnS0

4=ln

S4S04

=

ln100

1004

= 0

So the mean monthly continuously compounded return for Stock A dependsonly on S4 and S0; S1, S2, and S3 are irrelevant.However, S1, S2, and S3 matters when we are calculating the monthly stan-

dard deviation in the following formula:

vuuut 4Pt=1

³rt −

∧r´2

4− 1 .

Problem 18.8.

lnSt is normally distributed with mean lnS0+¡α− δ − 0.5σ2

¢t and variance

σ2t.lnSt ∼ N

£lnS0 +

¡α− δ − 0.5σ2

¢t, σ2t

¤Then for a given stock price c

P (St ≤ c) = P (lnSt ≤ ln c) = ΦÃln c− lnS0 −

¡α− δ − 0.5σ2

¢t

σ√t

!



One subtle point. Since St is continuous, the probability is zero that Sttakes on a single value. Hence P (St = c) = 0, P (St ≤ c) = P (St < c), andP (St > c) = P (St > c).

In this problem, S0 = 100, α = 0.08, δ = 0, σ = 0.3

lnSt ∼ N£ln 100 +

¡0.08− 0− 0.5× 0.32

¢t, 0.32t

¤lnSt ∼ N

£4. 605 17 + 0.035 t, 0.32t

¤P (St ≤ c) = Φ

µln c− 4. 605 17− 0.035 t

0.3√t

¶For t = 1, we have:lnS1 ∼ N

£4. 640 17, 0.32

¤P (S1 ≤ c) = P (lnS1 ≤ ln c) = Φ

µln c− 4. 640 17

0.3

¶

P (S1 ≤ 105) = Φµln 105− 4. 640 17

0.3

¶= Φ (0.0460) = NormalDist (0.0460) =

0.518 3

P (S1 > 105) = 1− 0.518 3 = 0.481 7

Next, we consider how P (St > c) changes if we change t or σ.

We know that P (St ≤ c) = Φ

Ãln c− lnS0 −

¡α− δ − 0.5σ2

¢t

σ√t

!

Set z =ln c− lnS0 −

¡α− δ − 0.5σ2

¢t

σ√t

, where z is a normal random vari-

able.

d

dtz =

d

dt

ln c− lnS0 −¡α− δ − 0.5σ2

¢t

σ√t

= −α− δ − 0.5σ2σ

d

dt

√t = −α− δ − 0.5σ2

2σ√t

For this problem, α = 0.08, δ = 0, σ = 0.3d

dtz = −0.08− 0− 0.5× 0.3

2

2× 0.3√t

= −0.005 833√t

So z is a decreasing function of t. Remember that the accumulative normaldistribution Φ (z) is an increasing function of z, we conclude:

The higher the t, the lower the z, the lower the P (St ≤ c) = Φ (z), and thehigher the P (St > c) = 1− Φ (z).

Similarly,



d

dσz =

d

dσ

ln c− lnS0 −¡α− δ − 0.5σ2

¢t

σ√t

=d

dσ

−¡α− δ − 0.5σ2

¢t

σ√t

= −√td

dσ

α− δ − 0.5σ2σ

= −√td

dσ

µα− δ

σ− 0.5σ

¶=√t

µ0.5 +

α− δ

σ2

¶

For this problem, α = 0.08, δ = 0 sod

dσz =√t

µ0.5 +

0.08

σ2

¶> 0.

So z is an increasing function of σ.The higher the σ, the higher the z, the higher the P (St ≤ c) = Φ (z), and

the lower the P (St > c) = 1− Φ (z).

Problem 18.9.

Use DM 18.30.

E (St|St > K) = Se(α−δ)tN

µ∧d1

¶N

µ∧d2

¶

∧d1 =

lnS0K+¡α− δ + 0.5σ2

¢t

σ√t

∧d2 =

∧d1 − σ

√T

In this problem,

∧d1 =

ln100

105+¡0.08− 0 + 0.5× 0.32

¢1

0.3√1

= 0.254 0

N

µ∧d1

¶= NormalDist (0.2540) = 0.600 3

∧d2 = 0.2540− 0.3

√1 = −0.046

N

µ∧d2

¶= NormalDist (−0.046 ) = 0.481 7

E (S1|S1 > 105) = 100e(0.08−0)1 ×0.600 3

0.481 7= 135

Next, we wan to analyze how E (St|St > 105) changes if we change t (whilekeeping other parameters unchanged), σ (while keeping other parameters unchanged),and α (while keeping other parameters unchanged). We see that E (St|St > 105)increases if we increase t, σ, or α. We consider t = 0.25, 0.5, 0.75, 1, ..., 6.



t E (S1|S1 > 105) σ E (S1|S1 > 105) α E (S1|S1 > 105)0.25 117.19 0.1 115.15 0.01 131.630.5 123.98 0.2 124.65 0.02 132.080.75 129.74 0.3 135.00 0.03 132.541 135.00 0.4 146.25 0.04 133.01

1.25 139.97 0.5 158.48 0.05 133.491.5 144.76 0.6 171.78 0.06 133.981.75 149.42 0.7 186.26 0.07 134.492 154.01 0.8 202.04 0.08 135.00

2.25 158.54 0.9 219.26 0.09 135.532.5 163.05 1 238.06 0.1 136.072.75 167.54 1.1 258.61 0.11 136.633 172.04 1.2 281.11 0.12 137.19

3.25 176.54 1.3 305.77 0.13 137.783.5 181.06 1.4 332.82 0.14 138.373.75 185.61 1.5 362.55 0.15 138.984 190.19 1.6 395.25 0.16 139.60

4.25 194.81 1.7 431.27 0.17 140.244.5 199.48 1.8 471.02 0.18 140.904.75 204.19 1.9 514.93 0.19 141.575 208.95 2 563.51 0.2 142.25

5.25 213.76 2.1 617.34 0.21 142.955.5 218.64 2.2 677.07 0.22 143.675.75 223.58 2.3 743.46 0.23 144.416 228.58 2.4 817.37 0.24 145.16

Problem 18.10.

Use DM 18.23. P (St < K) = N (−d2)

∧d1 =

lnS0K+¡α− δ + 0.5× σ2

¢t

σ√t

=ln100

98+¡0.08− 0 + 0.5× 0.32

¢1

0.3√1

=

0.484 0

∧d2 =

∧d1 − σ

√t = 0.484 0− 0.3

√1 = 0.184

N

µ−∧d2

¶= NormalDist (−0.184 ) = 0.427 0

P (St < 98) = 0.427 0

Next, we analyze how P (St < 98) changes if we change t. We find thatP (St < 98) increases with t initially and then decreases with t.



t d2 P (St < 98) = N (−d2)0.25 0.1930 0.42350.5 0.1777 0.42950.75 0.1788 0.42901 0.1840 0.4270

1.25 0.1907 0.42441.5 0.1979 0.42161.75 0.2052 0.41872 0.2126 0.4158

2.25 0.2199 0.41302.5 0.2271 0.41022.75 0.2341 0.40753 0.2410 0.4048

3.25 0.2477 0.40223.5 0.2543 0.39963.75 0.2607 0.39724 0.2670 0.3947

4.25 0.2732 0.39244.5 0.2792 0.39004.75 0.2852 0.38785 0.2910 0.3855

5.25 0.2967 0.38335.5 0.3023 0.38125.75 0.3078 0.37916 0.3133 0.3770

Problem 18.11.

Use DM 18.28.

E (St|St < K) = Se(α−δ)tN

µ−∧d1

¶N

µ−∧d2

¶If K = 98:

∧d1 =

lnS0K+¡α− δ + 0.5× σ2

¢t

σ√t

=ln100

98+¡0.08− 0 + 0.5× 0.32

¢1

0.3√1

=

0.484 0

∧d2 =

∧d1 − σ

√t = 0.484 0− 0.3

√1 = 0.184

N

µ−∧d1

¶= NormalDist (−0.484 0 ) = 0.3142

N

µ−∧d2

¶= NormalDist (−0.184 ) = 0.427 0



E (S1|S1 < 98) = 100e(0.08−0)1 ×0.3142

0.427 0= 79. 71

If K = 120:

∧d1 =

ln100

120+¡0.08− 0 + 0.5× 0.32

¢1

0.3√1

= −0.191 1∧d2 =

∧d1 − σ

√t = −0.191 1− 0.3

√1 = −0.491 1

N

µ−∧d1

¶= NormalDist (0.191 1 ) = 0.575 8

N

µ−∧d2

¶= NormalDist (0.491 1 ) = 0.688 3

E (S1|S1 < 120) = 100e(0.08−0)1 ×0.575 8

0.688 3= 90. 62

If K = 98. We find that E (S1|S1 < K) decreases with t.t d1 d2 E (S1|S1 < K)

0.25 0.4235 0.1930 88.120.5 0.4295 0.1777 84.410.75 0.4290 0.1788 81.791 0.4270 0.1840 79.71

1.25 0.4244 0.1907 77.971.5 0.4216 0.1979 76.471.75 0.4187 0.2052 75.152 0.4158 0.2126 73.97

2.25 0.4130 0.2199 72.902.5 0.4102 0.2271 71.922.75 0.4075 0.2341 71.013 0.4048 0.2410 70.18

3.25 0.4022 0.2477 69.393.5 0.3996 0.2543 68.663.75 0.3972 0.2607 67.974 0.3947 0.2670 67.32

4.25 0.3924 0.2732 66.714.5 0.3900 0.2792 66.124.75 0.3878 0.2852 65.575 0.3855 0.2910 65.04

5.25 0.3833 0.2967 64.535.5 0.3812 0.3023 64.045.75 0.3791 0.3078 63.586 0.3770 0.3133 63.13

If K = 120. We find that E (S1|S1 < K) decreases with t.



t d1 d2 E (S1|S1 < K)0.25 0.8764 −1.1571 98.140.5 0.7814 −0.7770 95.090.75 0.7260 −0.6007 92.641 0.6883 −0.4911 90.62

1.25 0.6602 −0.4131 88.881.5 0.6381 −0.3533 87.361.75 0.6198 −0.3051 86.002 0.6044 −0.2647 84.78

2.25 0.5910 −0.2302 83.662.5 0.5792 −0.1999 82.632.75 0.5687 −0.1730 81.683 0.5591 −0.1488 80.80

3.25 0.5504 −0.1268 79.973.5 0.5424 −0.1066 79.203.75 0.5350 −0.0879 78.464 0.5281 −0.0705 77.77

4.25 0.5216 −0.0543 77.114.5 0.5156 −0.0390 76.494.75 0.5098 −0.0246 75.895 0.5044 −0.0109 75.32

5.25 0.4992 0.0021 74.785.5 0.4942 0.0145 74.265.75 0.4895 0.0263 73.766 0.4850 0.0377 73.28

The following table lists the value of E (St|St < K)



K = 98 K = 120 K = 98 K = 120t σ = 0.3 σ = 0.3 σ = 0.1 σ = 0.1

0.25 88.12 98.14 95.26 102.010.5 84.41 95.09 94.22 103.690.75 81.79 92.64 93.52 104.721 79.71 90.62 92.99 105.35

1.25 77.97 88.88 92.56 105.741.5 76.47 87.36 92.21 105.991.75 75.15 86.00 91.90 106.162 73.97 84.78 91.64 106.28

2.25 72.90 83.66 91.40 106.372.5 71.92 82.63 91.19 106.432.75 71.01 81.68 91.00 106.483 70.18 80.80 90.83 106.51

3.25 69.39 79.97 90.68 106.533.5 68.66 79.20 90.53 106.553.75 67.97 78.46 90.40 106.564 67.32 77.77 90.28 106.57

4.25 66.71 77.11 90.16 106.574.5 66.12 76.49 90.06 106.574.75 65.57 75.89 89.96 106.575 65.04 75.32 89.86 106.57

5.25 64.53 74.78 89.78 106.575.5 64.04 74.26 89.69 106.575.75 63.58 73.76 89.61 106.566 63.13 73.28 89.54 106.56

The diagram:



Problem 18.12.

This problem states that KT = S0erT , but it doesn’t specify r .To solve the

problem, we set r = α.

As T increases, P (ST < KT ) increases, P (ST > KT ) decreases, and thecall/put price increases.



t d2 P (ST < KT ) P (ST > KT ) Call price Put price0.25 −0.0750 0.5299 0.4701 $5.98 $5.980.5 −0.1061 0.5422 0.4578 $8.45 $8.450.75 −0.1299 0.5517 0.4483 $10.34 $10.341 −0.1500 0.5596 0.4404 $11.92 $11.92

1.25 −0.1677 0.5666 0.4334 $13.32 $13.321.5 −0.1837 0.5729 0.4271 $14.58 $14.581.75 −0.1984 0.5786 0.4214 $15.73 $15.732 −0.2121 0.5840 0.4160 $16.80 $16.80

2.25 −0.2250 0.5890 0.4110 $17.80 $17.802.5 −0.2372 0.5937 0.4063 $18.75 $18.752.75 −0.2487 0.5982 0.4018 $19.64 $19.643 −0.2598 0.6025 0.3975 $20.50 $20.50

3.25 −0.2704 0.6066 0.3934 $21.32 $21.323.5 −0.2806 0.6105 0.3895 $22.10 $22.103.75 −0.2905 0.6143 0.3857 $22.85 $22.854 −0.3000 0.6179 0.3821 $23.58 $23.58

4.25 −0.3092 0.6214 0.3786 $24.29 $24.294.5 −0.3182 0.6248 0.3752 $24.97 $24.974.75 −0.3269 0.6281 0.3719 $25.63 $25.635 −0.3354 0.6313 0.3687 $26.27 $26.27

5.25 −0.3437 0.6345 0.3655 $26.89 $26.895.5 −0.3518 0.6375 0.3625 $27.50 $27.505.75 −0.3597 0.6405 0.3595 $28.09 $28.096 −0.3674 0.6433 0.3567 $28.67 $28.67

Why do the European call and the European put have the same price? Usethe call-put parity:

Ke−rT

C = S0e−δTN (d1)−Ke−rTN (d2)

Since K = S0erT and δ = 0, we have:

C = S0N (d1)− S0N (d2)

d1 =

lnS0

S0erT+

µr − δ +

1

2σ2¶T

σ√T

=

−rT +µr +

1

2σ2¶T

σ√T

= 0.5σ√T

d2 = d2 − σ√T = −0.5σ

√T

=⇒ C = S0N³0.5σ√T´−S0N

³−0.5σ

√T´= S0N

³0.5σ√T´−S0

³1−N

³0.5σ√T´´=

S0

h2N

³0.5σ√T´− 1i

P = −S0e−δTN (−d1)+Ke−rTN (−d2) = −S0N³−0.5σ

√T´+S0N

³−0.5σ

√T´=

−S0³1−N

³0.5σ√T´´+ S0N

³−0.5σ

√T´= S0

h2N

³0.5σ√T´− 1i



For example, for T = 0.25

C = P = 100 (2N (0.075)− 1) = 100 (2× 0.529 9− 1) = 5. 98

By the way, we can use the put-call parity to see why C = P .

C +Ke−rT = P + S0e−δT

Since K = S0erT and δ = 0, we have: C + S0 = P + S0 and C = P .

Typically, the strike price K is a fixed amount (as opposed to the increasingamount K = S0e

rT ); the call price and put price move in opposite directions.In this problem, K = S0e

rT increases with T ; the call and the put options bothbecome more valuable as T increases. This is a pure mathematical coincidence.

Intuitively, how to reconcile the fact that as T increases, P (ST < KT ) in-creases, P (ST > KT ) decreases, and the call and the put prices both increase?Please note that the call/put price not only depends on the probability of theoption being in the money, not also depends on the payoff. For example, eventhough P (ST > KT ) decreases with T , if ST −KT increases at a faster speed,the call price will go up.

Problem 18.13.

Parameters:

S = 100 K = 90 σ = 0.3 α = 4.5% δ = 0

As T increases, P (St < K) increases and then decreases. So the impact ofT on P (St < K) is ambiguous. However, E(St|St < K) decreases over time.



T d2 P (St < K) = N(−d2) E(St|St < K)0.5 0.5438 0.2933 $79.201 0.4179 0.3380 $74.64

1.5 0.3684 0.3563 $71.452 0.3426 0.3659 $68.96

2.5 0.3275 0.3716 $66.893 0.3182 0.3752 $65.13

3.5 0.3124 0.3774 $63.594 0.3089 0.3787 $62.23

4.5 0.3070 0.3794 $61.005 0.3061 0.3798 $59.88

5.5 0.3061 0.3798 $58.866 0.3067 0.3795 $57.91

6.5 0.3077 0.3791 $57.047 0.3091 0.3786 $56.22

7.5 0.3108 0.3780 $55.468 0.3127 0.3772 $54.74

8.5 0.3148 0.3764 $54.069 0.3171 0.3756 $53.42

9.5 0.3194 0.3747 $52.8210 0.3219 0.3738 $52.24

10.5 0.3244 0.3728 $51.7011 0.3270 0.3718 $51.17

11.5 0.3296 0.3708 $50.6812 0.3323 0.3698 $50.20




Chapter 19

Monte Carlo simulation

Please note that due to rounding, you may not be able to fully reproduce myresult. For example, when you see my z = 0.1234, the actual z used in mycalculation could be 0.123436. If you plug in z = 0.1234, you may not be ableto reproduce my result.

Problem 19.1.

Here is the snapshot of the simulation done in Excel:

A B C1 i u u2

2 1 0.104689 0.0109603 2 0.491579 0.2416504 3 0.085629 0.0073325 4 0.878402 0.7715906 5 0.199163 0.039666... ... ... ...

1000 999 0.132422 0.0175361001 1000 0.869963 0.7568351002 Total 498.294032 334.277115

Sample formulasCell B2 = rand() B3 = rand()

C2 = B2ˆ2 C3 = B3ˆ2

B1002 = sum(B2 : B1001) C1002 = sum(C2 : C1001)

E (u) =

Puin

=498.294032

1000= 0.498 294

177

CHAPTER 19. MONTE CARLO SIMULATION

V ar (u) =n

n− 1

Ã1

n

Pu2i −

µPuin

¶2!=1000

999

µ334.277115

1000− 0.498 2942

¶=

0.08 606 6

The correct mean of u ∼ (0, 1) is: E (u) = 0.5The correction variance: V ar (u) =

1

12= 0.08333 3

Next, we graph the histogram.D E F

1 bin range frequency2 0.1 0 < u < 0.1 1043 0.2 0.1 ≤ u < 0.2 1064 0.3 0.2 ≤ u < 0.3 945 0.4 0.3 ≤ u < 0.4 1006 0.5 0.4 ≤ u < 0.5 907 0.6 0.5 ≤ u < 0.6 1088 0.7 0.6 ≤ u < 0.7 919 0.8 0.7 ≤ u < 0.8 9310 0.9 0.8 ≤ u < 0.9 11411 1 0.9 ≤ u < 1 10012 Total 1000

Sample formulas for Column F.F2 = countif(B2 : B1001, ” < ”&D2)F3 = countif(B2 : B1001, ” < ”&D3)− F2F4 = countif(B2 : B1001, ” < ”&D4)− (F2 + F3)F5 = countif(B2 : B1001, ” < ”&D5)− (F2 + F3 + F4)...F11 = countif(B2 : B1001, ” < ”&D11)− (F2 + F3 + F4 + ...+ F10)

Problem 19.2.

u1 u2 u3 ... u12 x =P

ui − 6 x2

1 0.6922 0.1621 0.0013 ... 0.1888 −0.8461 0.7158422 0.0553 0.1343 0.5132 ... 0.4550 −0.2583 0.0667423 0.6137 0.5859 0.1285 ... 0.6732 0.5420 0.2938174 0.2610 0.5720 0.5668 ... 0.3890 −1.1239 1.2632035 0.7009 0.5851 0.4098 ... 0.1757 −0.6006 0.3606816 0.1991 0.8738 0.1330 ... 0.0924 0.2029 0.0411737 0.8552 0.3089 0.2594 ... 0.6550 0.4612 0.212701... ... ... ... ... ... ... ...

1000 0.4822 0.9639 0.9245 ... 0.7599 0.9899 0.979991Total 45.381885 949.311003



Figure 19.1:



Figure 19.2:

E (x) =45.381885

1000= 0.045 382 (close to zero)

V ar (x) =1000

999

µ1

1000× 949.311003− 0.045 3822

¶= 0.948 200 (close to 1)

The histogram looks like a bell curve.

Problem 19.3.

Snapshot of Excel for the simulation:A B C D E G H

1 i x1 x2 ex1 ex2 (ex1)2 (ex2)2

2 1 −0.88028 0.535024 0.414668 1.70749 0.171949 2.9155223 2 −0.71431 0.209528 0.489528 1.233096 0.239637 1.5205264 3 0.114864 −0.1811 1.121721 0.834356 1.258259 0.6961505 4 −0.8346 0.156575 0.434047 1.169498 0.188397 1.3677266 5 0.646959 −1.65712 1.909724 0.190687 3.647045 0.0363617 6 0.628469 0.48616 1.874738 1.626059 3.514643 2.644069

... ... ... ... ... ... ...2001 2000 0.779131 0.509077 2.179578 1.663754 4.750559 2.7680782002 sum 3, 210.5069 17, 452.8493 13, 501.1984 1, 753, 120.4685

Sample formulas.



CellB2 = NormInv (Rand(), 0, 1) C2 = NormInv (Rand(), 0.7, 0.3ˆ0.5)D2 = exp (B2) E2 = exp (C2)

E (ex1) =3210.5069

2000= 1. 605 3

The correct mean (using DM 18.13): E (ex1) = e0.5 = 1. 648 7

The estimated variance:

V ar (ex1) =2000

1999

µ1

2000× 13501.1984− 1. 605 3 2

¶= 4. 1757

The correct variance (DM18.14) : V ar (ex1) = e (e− 1) = 4. 670 8

E (ex2) =17452.8493

2000= 8. 726 4

The correct mean: E (ex2) = e0.7+0.5×3 = 9. 025 0

The estimated variance:

V ar (ex2) =2000

1999

µ1

2000× 1753120.4685− 9. 025 0 2

¶= 795. 507 4

The correct variance is:V ar (ex2) = e2×0.7+3

¡e3 − 1

¢= 1554. 533 6

Our estimate of E (ex2) and V ar (ex2) are way off. To improve our estimate,we need to increase the number of simulations.

Problem 19.4.

I performed 5,000 simulations.i z ST put payoff P P 2

1 −0.2273 38.9990 1.0010 0.9812 0.96272 −0.0408 40.1055 0.0000 0.0000 0.00003 1.7371 52.3623 0.0000 0.0000 0.00004 −0.8544 35.4978 4.5022 4.4131 19.47545 −0.5836 36.9693 3.0307 2.9707 8.8248... ... ... ... ... ...

5000 0.4792 43.3588 0.0000 0.0000 0.0000sum 10, 197.3969 9, 995.4749 61, 151.4337

The z column in the above table is generated using Excel’s formulaNormInv (Rand () , 0, 1)

Sample calculations for Row 1.Excel’s formula NormInv (Rand () , 0, 1) happens to return z = −0.2273.ST = S0e

(r−δ−0.5σ2)T+σ√Tz = 40e(0.08−0−0.5×0.3

2)0.25+0.3√0.25(−0.2273) =

38. 9990



Figure 19.3:



Figure 19.4:

The put payoff: 40− 38. 9990 = 1. 001The put price: P = e−0.08×0.25 × 1. 001 = 0.9812P 2 = 0.98122 = 0.962 7

The estimate put price is:9995.4749

5000= 1. 999 1

You can verify that the put price based on the Black-Scholes formula is1.9927.The estimated variance of the put price per simulation is:5000

4999

µ1

5000× 61151.4337− 1. 999 12

¶= 8. 235 5

Suppose we want to perform n simulations and take the average put priceof these n simulations as an estimate of the put price.

Then P =P1 + P2 + ...+ Pm

nhere Pi is the put price calculated from the i-th simulation

=⇒ V ar¡P¢=

nV ar (P per simulation)n2

=V ar (P per simulation)

n

0.012 =8. 235 5

nn =

8. 235 5

0.012= 82355

So we need to perform roughly 82, 400 simulations.



Problem 19.5.

i z S1 1/S1 F = e−rT (1/S1) F 2

1 −0.03768 40.9592 0.0244 0.0225 0.0005082 0.531987 48.5928 0.0206 0.0190 0.0003613 2.062351 76.9061 0.0130 0.0120 0.0001444 −2.46722 19.7610 0.0506 0.0467 0.0021825 2.579892 89.8235 0.0111 0.0103 0.0001066 −1.23246 28.6210 0.0349 0.0323 0.0010407 0.477051 47.7985 0.0209 0.0193 0.0003738 −1.37135 27.4530 0.0364 0.0336 0.001131... ... ... ... ... ...

5000 1.421245 63.4500 0.0158 0.0145 0.000212sum 116.9386 2.9935

For example, if z = −0.03768, thenS1 = S0e

(r−δ−0.5σ2)T+σ√Tz = 40e(0.08−0−0.5×0.3

2)1+0.3√1(−0.03768) = 40.

959 2

The forward price is:

F = e−rT (1/S1) = e−0.08×11

40.9592= 0.0225

The estimated forward price is:116.9386

5000= 0.023 39

The estimated variance of the forward price per simulation is:5000

4999

µ1

5000× 2.9935− 0.023 392

¶= 0.00005162

We can calculate the true forward price using DM 20.30:

FP0,T [S

a (T )] = e−rTSa (0) e[a(r−δ)+0.5a(a−1)σ2]T

Set a = −1. The true forward price is:FP0,T

£S−1 (T )

¤= e−0.08

¡40−1

¢e(−1(0.08−0)+0.5×(−1)(−1−1)0.3

2)1 = 0.02331

Problem 19.6.

a.



i z S1 S21 F = e−rT¡S21¢

F 2

1 0.5992 49.5825 2, 458.4195 2, 269.4072 5, 150, 209.11922 −1.3882 27.3147 746.0939 688.7315 474, 351.01583 −1.2717 28.2864 800.1177 738.6017 545, 532.53114 0.3870 46.5239 2, 164.4688 1, 998.0565 3, 992, 229.84435 −0.7793 32.7893 1, 075.1384 992.4778 985, 012.25926 0.4702 47.7004 2, 275.3243 2, 100.3891 4, 411, 634.26567 1.1824 59.0623 3, 488.3595 3, 220.1616 10, 369, 440.99628 0.4278 47.0974 2, 218.1649 2, 047.6243 4, 192, 765.3642... ... ... ... ... ...

5000 −0.1042 40.1502 1, 612.0360 1, 488.0968 2, 214, 432.0185sum 9, 517, 170.0248 26, 459, 271, 681.8572

The estimate forward price at time zero for a claim paying S21 at T = 1 is:9517170.0248

5000= 1903. 43

You can verify that the true forward price (using DM 20.30) is 1, 896.49


4999

µ1

5000× 26, 459, 271, 681.8572− 1903. 432

¶= 166, 9142. 40

b.i z S1 S0.51 F = e−rT

¡S0.51

¢F 2

1 0.112 42.8403 6.5453 6.0420 36.5060962 −0.8019 32.5673 5.7068 5.2680 27.7520223 0.0557 42.1228 6.4902 5.9912 35.8946824 −0.9828 30.8470 5.5540 5.1270 26.2860795 2.0755 77.2100 8.7869 8.1114 65.7940226 −1.0444 30.2822 5.5029 5.0798 25.8047897 1.0135 56.1445 7.4930 6.9169 47.843187... ... ... ... ... ...

5000 −1.2727 28.2776 5.3177 4.9088 24.096581sum 30, 059.267012 184, 880.552079

The estimate forward price at time zero for a claim paying S0.51 at T = 1 is:30059.267012

5000= 6. 01185

You can verify that true forward price (using DM 20.30) is 6.0086

The estimated variance of the forward price per simulation is:

5000

4999

µ1

5000× 184880.552079− 6.011852

¶= 0.833 9



c.i z S1 S−21 F = e−rT

¡S−21

¢F 2

1 0.0111 41.5630 0.0006 0.0005 0.000000292 −0.1893 39.1378 0.0007 0.0006 0.000000363 −0.6352 34.2374 0.0009 0.0008 0.000000624 0.4972 48.0883 0.0004 0.0004 0.000000165 −1.2449 28.5144 0.0012 0.0011 0.000001296 0.1526 43.3653 0.0005 0.0005 0.000000247 −0.5234 35.4052 0.0008 0.0007 0.00000054

... ... ... ... ... ...8 0.2497 44.6471 0.0005 0.0005 0.00000021

sum 3.216862 0.002918

The estimated forward price is:3.216862

5000= 0.000 643

You can verify that the true forward price is 0.000644


4999

µ1

5000× 0.002918− 0.0006442

¶= 1. 69× 10−7

Skip the remaining problems.


Chapter 20

Brownian motion and Ito’slemma

Problem 20.1.

I don’t want you to memorize Ito’s lemma. For problems related to It’slemma, all you need to know is two things:

1. Unlike a deterministic random variable X where (dX)2 = 0, for a sto-chastic random variable Z, the term (dZ)2 is not zero and hence can’t beignored. In fact, the textbook and my study guide have explained that(dZ)

2= dt. However, you can ignore higher order such as (dZ)3.

2. Ito’s lemma is just the stochastic counterpart of the Taylor series. Toderive Ito’s lemma, first write the Taylor series. For a stochastic randomvariable X and a function y = f (t,X), first write the Taylor expansion:

dy = d f (t,X) =∂f

∂tdt+

∂f

∂XdX+

1

2

∂2f

∂t2(dt)2+

1

2

∂2f

∂X2(dX)2+....Here we

ignored the higher order terms (dt)3, (dX)3, and above. Next, throw awaythe term (dt)2 since t is a deterministic random variable and (dt)2 → 0.

Now we have dy = d f (t,X) =∂f

∂tdt +

∂f

∂XdX +

1

2

∂2f

∂X2(dX)2. This is

Ito’s lemma.

With this point in mind, let’s solve the problem.

a. If the stock price S follows the textbook Equation 20.8, then:

dS (t) = αdt+ σdZ (t) (DM 20.8)

dS (t) is a linear function of dZ (t). Since [dZ (t)]2 = dt, we should keep(dS)

2 in the Taylor expansion but throw away all other higher order terms:

187

CHAPTER 20. BROWNIAN MOTION AND ITO’S LEMMA

→ d lnS =∂ lnS

∂tdt+

∂ lnS

∂SdS +

1

2

∂2 lnS

∂S2(dS)2

Since lnS doesn’t contain t,∂ lnS

∂t= 0. In addition,

∂ lnS

∂S=

1

Sand

∂2 lnS

∂S2=

∂

∂S

1

S= − 1

S2

→ d lnS =∂ lnS

∂SdS +

1

2

∂2 lnS

∂S2(dS)

2=1

SdS − 1

2

1

S2(dS)

2

(dS)2 = (αdt+ σdZ)2

= α2 (dt)2 + 2ασ (dZ) (dt) + σ2 (dZ)2

Next, use the multiplication rule:

dZ × dt = 0 (DM 20.17a)

(dt)2= 0 (DM 20.17b)

(dZ)2= dt (DM 20.17c)

→ (dS)2= (αdt+ σdZ)

2= σ2 (dZ)

2= σ2dt

→ d lnS =1

S(αdt+ σdZ)− 1

2

1

S2σ2dt

=

µ1

Sα− 1

2

1

S2σ2¶dt+

σ

SdZ

b.If the stock price S follows the textbook Equation 20.9, then:

dS (t) = λ (α− S) dt+ σdZ (DM 20.9)

→ d lnS =∂ lnS

∂SdS +

1

2

∂2 lnS

∂S2(dS)

2=1

SdS − 1

2

1

S2(dS)

2

(dS)2 = [λ (α− S) dt+ σdZ]2

= λ2 (α− S)2 (dt)2 + 2λ (α− S) dt× σdZ + σ2 (dZ)2 = σ2dt

→ d lnS =1

SdS − 1

2

1

S2(dS)

2

=λ (α− S) dt+ σdZ

S− 12

σ2dt

S2

=

∙λ (α− S)

S− 12

σ2

S2

¸dt+

σ

SdZ

c.If the stock price S follows the textbook Equation 20.27, then:

dS (t) =

∙∧α (S, t)−

∧δ (S, t)

¸dt+

∧σ (S, t) dZ (t) (DM 20.8)

When S (t) follows a geometric Brownian motion,



∧α (S, t) = αS (t)

∧δ (S, t) = δS (t)

∧σ (S, t) = σS (t)

Hence DM Equation 20.8 becomes:

dS (t) = S (α− δ) dt+ σSdZ (t) (20.1)

d lnS =∂ lnS

∂SdS +

1

2

∂2 lnS

∂S2(dS)

2=1

SdS − 1

2

1

S2(dS)

2

(dS)2 = [S (α− δ) dt+ σSdZ (t)]2 = σ2S2 (dZ)2 = σ2S2dt

→ d lnS =1

SdS − 1

2

1

S2(dS)2

=S (α− δ) dt+ σSdZ (t)

S− 12

σ2S2dt

S2

=

µα− δ − 1

2σ2¶dt+ σdZ

Problem 20.2.

dS2 =∂S2

∂tdt+

∂S2

∂SdS +

1

2

∂2S2

∂S2(dS)

2

=∂S2

∂SdS +

1

2

∂2S2

∂S2(dS)2 = 2SdS + (dS)2

a. Under DM Equation 20.8:dS (t) = αdt+ σdZ (t)

(dS)2 = σ2dt

→ dS2 = 2SdS + (dS)2

= 2S (αdt+ σdZ) + σ2dt =¡2αS + σ2

¢dt+ 2σSdZ

b. Under DM Equation 20.9:dS (t) = λ (α− S) dt+ σdZ

(dS)2 = σ2dt

→ dS2 = 2SdS + (dS)2

= 2S [λ (α− S) dt+ σdZ] + σ2dt=£2Sλ (α− S) + σ2

¤dt+ 2SσdZ

c. Under Geometric Brownian motiondS (t) = S (α− δ) dt+ σSdZ (t)

(dS)2 = σ2S2dt

→ dS2 = 2SdS + (dS)2

= 2S [S (α− δ) dt+ σSdZ] + σ2S2dt= 2S2 [(α− δ) dt+ σdZ] + σ2S2dt= S2

£2 (α− δ) + σ2

¤dt+ 2σS2dZ

Problem 20.3.



dS−1 =∂S−1

∂tdt+

∂S−1

∂SdS +

1

2

∂2S−1

∂S2(dS)2

∂S−1

∂t= 0

∂S−1

∂S= −S−2 ∂2S−1

∂S2= 2S−3

→ dS−1 = −S−2dS + S−3 (dS)2

a. Under DM Equation 20.8:

dS (t) = αdt+ σdZ (t)

(dS)2 = σ2dt

→ dS−1 = −S−2dS + S−3 (dS)2

= −S−2 (αdt+ σdZ) + S−3σ2dt=¡−S−2α+ S−3σ2

¢dt− S−2σdZ


(dS)2 = σ2dt

→ dS−1 = −S−2dS + S−3 (dS)2

= −S−2 (λ (α− S) dt+ σdZ) + S−3σ2dt=¡−S−2λ (α− S) + S−3σ2

¢dt− S−2σdZ


(dS)2 = σ2S2dt

→ dS−1 = −S−2dS + S−3 (dS)2

= −S−2 [S (α− δ) dt+ σSdZ (t)] + S−3σ2S2dt= −S−1 [(α− δ) dt+ σdZ (t)] + S−1σ2dt= S−1

£− (α− δ) + σ2

¤dt− S−1σdZ (t)

Problem 20.4.

dS0.5 =∂S0.5

∂tdt+

∂S0.5

∂SdS +

1

2

∂2S0.5

∂S2(dS)

2

∂S0.5

∂t= 0

∂S0.5

∂S= 0.5S−0.5

∂2S0.5

∂S2= −0.25S−1.5

→ dS0.5 = 0.5S−0.5dS − 0.125S−1.5 (dS)2a. Under DM Equation 20.8:

dS (t) = αdt+ σdZ (t)

(dS)2 = σ2dt→ dS0.5 = 0.5S−0.5 [αdt+ σdZ (t)]− 0.125S−1.5σ2dt


(dS)2= σ2dt

→ dS0.5 = 0.5S−0.5 [λ (α− S) dt+ σdZ]− 0.125S−1.5σ2dt




(dS)2 = σ2S2dt→ dS0.5 = 0.5S−0.5 [S (α− δ) dt+ σSdZ (t)]− 0.125S−1.5σ2S2dt

= 0.5S0.5 [(α− δ) dt+ σdZ (t)]− 0.125S0.5σ2dt

Problem 20.5.

dS (t) = S (αs − δs) dt+ σsSdZS (DM 20.37)

dQ (t) = Q (αQ − δQ) dt+ σQQdZQ (DM 20.38)

dSdQ = ρdt (DM 20.17d)

d¡S2Q0.5

¢=

∂¡S2Q0.5

¢∂S

dS+∂¡S2Q0.5

¢∂Q

dQ+∂2¡S2Q0.5

¢∂S∂Q

dSdQ+1

2

∂2¡S2Q0.5

¢∂S2

(dS)2+

1

2

∂2¡S2Q0.5

¢∂Q2

(∂Q)2

∂¡S2Q0.5

¢∂S

= 2SQ0.5∂¡S2Q0.5

¢∂Q

= 0.5Q−0.5S2∂2¡S2Q0.5

¢∂S∂Q

=

SQ−0.5

∂2¡S2Q0.5

¢∂S2

= 2Q0.5∂2¡S2Q0.5

¢∂Q2

= −0.25S2Q−1. 5

→ d¡S2Q0.5

¢=

2SQ0.5 [S (αs − δs) dt+ σsSdZS ]+0.5Q−0.5S2 [Q (αQ − δQ) dt+ σQQdZQ]+SQ−0.5ρdt+Q0.5 (σQQ)

2 dt

−0.125S2Q−1. 5 [Q (αQ − δQ) dt+ σQQdZQ]2

Problem 20.6.

From Problem 20.1.c, we have:

d lnS =

µαS − δS −

1

2σ2S

¶dt+ σSdZS

d lnQ =

µαQ − δQ −

1

2σ2Q

¶dt+ σQdZQ

→ d ln (SQ) = d lnS+d lnQ+

µαS − δS −

1

2σ2S

¶dt+σSdZS+

µαQ − δQ −

1

2σ2Q

¶dt+

σQdZQ



Problem 20.7.

dS (t)

S (t)= (α− δ) dt+ σdZ

FP0,T

£SA (T )

¤= e−rTSA (0) e[A(r−δ)+0.5A(A−1)σ

2]T

Assume T = 1If A = 2FP0,1

£S2 (1)

¤= e−0.06×11002e(2(0.06−0)+0.5×2(2−1)0.4

2)1 = 12460. 77

If A = 0.5FP0,1

£S0.5 (1)

¤= e−0.06×11000.5e(0.5(0.06−0)+0.5×0.5(0.5−1)0.4

2)1 = 9. 51

If A = −2FP0,1

£S−2 (1)

¤= e−0.06×1100−2e(−2(0.06−0)+0.5(−2)(−2−1)0.4

2)1 = 1. 35×10−4

The textbook asks you to compare your solution to the solution to Problem19.7. However, Problem 19.7 is out of the scope of Exam MFE. So you don’t todo such a comparison.Skip the remaining problems (Problem 20.8 and beyond); they are out of

the scope of Exam MFE.


Chapter 21

The Black-Scholes equation

Problem 21.1.

Equation 21.12 is V (t, T ) = e−r(T−t)

BS PDE is Equation 21.11: Vt +1

2σ2S2VSS + (r − δ)SVS − rV = 0

We want to prove that V (t, T ) = e−r(T−t) satisfies Equation 21.11.

V (t, T ) = e−r(T−t)

=⇒ VS =∂V

∂S= 0 VSS =

∂2V

∂S2= 0 Vt =

∂V

∂t= re−r(T−t) = rV

=⇒ Vt +1

2σ2S2VSS + (r − δ)SVS − rV = rV − rV = 0

The boundary condition is that V (T, T ) = 1. Clearly, V (t, T ) = e−r(T−t)

satisfies this boundary condition.

Problem 21.2.

V (t, T ) = ASaeγt

VS = a¡ASa−1eγt

¢=

aV

SVSS = a (a− 1)

¡ASa−2eγt

¢=

a (a− 1)VS2

Vt = γ (ASaeγt) = γV

Vt+1

2σ2S2VSS+(r − δ)SVS−rV = γV +

1

2σ2a (a− 1)V +(r − δ) aV −rV

To satisfy BS PDE, we just need to choose the a such that

γV +1

2σ2a (a− 1)V +(r − δ) aV −rV = 0 or γ+

1

2σ2a (a− 1)+(r − δ) a−r =

0

1

2σ2a2 +

µr − δ − 1

2σ2¶a+ (γ − r) = 0

Solving this equation, we get:

193

CHAPTER 21. THE BLACK-SCHOLES EQUATION

a =

−µr − δ − 1

2σ2¶±

sµr − δ − 1

2σ2¶2− 2σ2 (γ − r)

σ2=

µ1

2− r − δ

σ2

¶±sµ

r − δ

σ− 12

¶2− 2 (γ − r)

σ2

If we set a above, then V (t, T ) = ASaeγt satisfies the BS PDE.

Problem 21.3.

According to Proposition 20.30, the prepaid forward price at time zero valueof a claim paying Sa (T ) is

F0,T [Sa (T )] = e−rTSa (0) e[a(r−δ)+0.5a(a−1)σ

2]T

In the above formula, if we replace T with (T − t)ASaeγt and S (0) withS (t), we’ll get the prepaid forward price at time t value of a claim payingSa (T ):

V (t, T ) = Ft,T [Sa (T )] = e−r(T−t)Sa (t) e[a(r−δ)+0.5a(a−1)σ

2](T−t)

We need to prove that Ft,T [Sa (T )] satisfies the BS PDE.

Notice that V (t, T ) is in the form of ASaeγt where γ = r − a (r − δ) −0.5a (a− 1)σAccording to Problem 21.2, V (t, T ) satisfies the BS PDE if we set

a =

µ1

2− r − δ

σ2

¶±

sµr − δ

σ− 12

¶2− 2 (γ − r)

σ2

Problem 21.4.

First, let’s prove that if V 1 (S, t, T ) and V 2 (S, t, T ) each satisfy the BS PDE

Vt +1

2σ2S2VSS + (r − δ)SVS − rV = 0, then for any constants k1 and k2 the

linear combination V (S, t, T ) = k1V1 (S, t, T ) + k2V

2 (S, t, T ) also satisfies theBS PDE.

Proof.Vt = k1V

1t + k2V

2t VS = k1V

1S + k2V

2S VSS = k1V

1SS + k2V

2SS

Vt +1

2σ2S2VSS + (r − δ)SVS − rV

= k1

∙V 1t +

1

2σ2S2V 1

SS + (r − δ)SV 1S − rV 1

¸+k2

∙V 2t +

1

2σ2S2V 2

SS + (r − δ)SV 2S − rV 2

¸= k1 × 0 + k2 × 0 = 0

In this problem,Ke−r(T−t) and S (t) e−δ(T−t) are each in the form of ASaeγt.According to Problem 21.2, Ke−r(T−t) and S (t) e−δ(T−t) each satisfy the BSPDE. Hence the linear combination Ke−r(T−t)+ S (t) e−δ(T−t) satisfy the BSPDE.The boundary condition is that V (S, T, T ) = K + S.



Problem 21.5.

Some basics.N (x) =

1√2π

R x−∞ e−s

2/2ds

N0(x) =

1√2π

e−x2/2 Generally

d

dx

R xaf (s) ds = f (x)

N00(x) =

1√2π

d

dxe−x

2/2 = − 1√2π

xe−1

2x2

= −xN 0(x)

Now let’s prove.

V = S (t) e−δ(T−t)N (d1) d1 =

lnS (t)

K+

µr − δ +

1

2σ2¶(T − t)

σ√T − t

∂d1∂S (t)

=1

σ√T − t

× ∂

∂S (t)ln

S (t)

K=

1

σ√T − t

× 1

S (t)

∂

∂SN (d1) = N

0(d1)

∂d1∂S

= N0(d1)×

1

σ√T − t

× 1

S (t)

∂

∂SN

0(d1) = N

00(d1)

∂d1∂S

= N00(d1)×

1

σ√T − t

× 1

S (t)

→ VS = e−δ(T−t)∙N (d1) + S (t)

∂N (d1)

∂S

¸= e−δ(T−t)

"N (d1) +

N0(d1)

σ√T − t

#

=⇒ (r − δ)SVS = (r − δ)Se−δ(T−t)

"N (d1) +

N0(d1)

σ√T − t

#= (r − δ)V +

(r − δ)

σ√T − t

Se−δ(T−t)N0(d1)

→ VSS = e−δ(T−t)∙∂

∂SN (d1) +

1

σ√T − t

∂

∂SN

0(d1)

¸= e−δ(T−t)

∙N

0(d1)×

1

σ√T − t

× 1

S (t)+

1

σ√T − t

×N00(d1)×

1

σ√T − t

× 1

S (t)

¸= e−δ(T−t)

∙N

0(d1)×

1

σ√T − t

× 1

S (t)− 1

σ√T − t

× d1N0(d1)×

1

σ√T − t

× 1

S (t)

¸=

e−δ(T−t)

S (t)σ√T − t

N0(d1)

µ1− d1

σ√T − t

¶

=⇒ 1

2σ2S2VSS =

1

2σ2S2

e−δ(T−t)

Sσ√T − t

N0(d1)

µ1− d1

σ√T − t

¶= σS

e−δ(T−t)

2√T − t

N0(d1)

µ1− d1

σ√T − t

¶www.actuary88.com c°Yufeng Guo 195


→ Vt = S (t)∂

∂te−δ(T−t)N (d1) = S (t)

∙N (d1)

∂

∂te−δ(T−t) + e−δ(T−t)

∂

∂tN (d1)

¸= S (t)

∙N (d1) δe

−δ(T−t) + e−δ(T−t)N0(d1)

∂

∂td1

¸Please note that S (t) is a fixed constant for a given time t.

∂

∂td1 =

∂

∂t

lnS (t)

K+

µr − δ +

1

2σ2¶(T − t)

σ√T − t

=∂

∂t

lnS (t)

Kσ√T − t

+∂

∂t

µr − δ +

1

2σ2¶(T − t)

σ√T − t

=1

2σ (T − t)

3

2

lnS (t)

K+

µr − δ +

1

2σ2¶

∂

∂t

√T − t

σ

=1

2σ (T − t)

3

2

lnS (t)

K− 1

2σ√T − t

µ1

2σ2 + r − δ

¶

=1

2 (T − t)

⎡⎢⎢⎣ lnS (t)

K

σ (T − t)

1

2

−√T − t

σ

µ1

2σ2 + r − δ

¶⎤⎥⎥⎦

=1

2 (T − t)

⎡⎢⎢⎣ lnS (t)

K+

µr − δ +

1

2σ2¶(T − t)

σ (T − t)

1

2

⎤⎥⎥⎦

− 1

2 (T − t)

⎡⎢⎢⎣√T − t

σ

µ1

2σ2 + r − δ

¶+

µr − δ +

1

2σ2¶(T − t)

σ (T − t)

1

2

⎤⎥⎥⎦

=1

2 (T − t)

⎡⎢⎢⎣d1 − 2µr − δ +

1

2σ2¶

σ

√T − t

⎤⎥⎥⎦=⇒ Vt = S (t)

∙N (d1) δe

−δ(T−t) + e−δ(T−t)N0(d1)

∂

∂td1

¸

= δV + S (t) e−δ(T−t)N0(d1)

1

2 (T − t)

⎡⎢⎢⎣d1 − 2µr − δ +

1

2σ2¶

σ

√T − t

⎤⎥⎥⎦www.actuary88.com c°Yufeng Guo 196


Vt +1

2σ2S2VSS + (r − δ)SVS

= δV + S (t) e−δ(T−t)N0(d1)

1

2 (T − t)

⎡⎢⎢⎣d1 − 2µr − δ +

1

2σ2¶

σ

√T − t

⎤⎥⎥⎦

+σSe−δ(T−t)

2√T − t

N0(d1)

µ1− d1

σ√T − t

¶

+(r − δ)V +(r − δ)

σ√T − t

Se−δ(T−t)N0(d1)

= rV

End of the proof.

Problem 21.6.

V = e−r(T−t)N (d2)

d2 = d1 − σ√T − t

∂

∂td2 =

∂

∂td1 −

∂

∂t

¡σ√T − t

¢=

∂

∂td1 +

1

2

σ√T − t

∂

∂Sd2 =

∂

∂Sd1 =

1

σ√T − t

× 1

S (t)=

1

Sσ√T − t

From the previous problem, we know that

∂

∂td1 =

1

2 (T − t)

⎡⎢⎢⎣d1 − 2µr − δ +

1

2σ2¶

σ

√T − t

⎤⎥⎥⎦

=1

2 (T − t)

⎡⎢⎢⎣d2 + σ√T − t−

2

µr − δ +

1

2σ2¶

σ

√T − t

⎤⎥⎥⎦=

1

2 (T − t)

∙d2 −

2 (r − δ)

σ

√T − t

¸

→ ∂

∂td2 =

1

2 (T − t)

∙d2 −

2 (r − δ)

σ

√T − t

¸+1

2

σ√T − t



=1

2 (T − t)

⎡⎢⎢⎣d1 − 2µr − δ +

1

2σ2¶

σ

√T − t

⎤⎥⎥⎦

=1

2 (T − t)

⎡⎢⎢⎣d2 + σ√T − t−

2

µr − δ +

1

2σ2¶

σ

√T − t

⎤⎥⎥⎦

=1

2 (T − t)

⎡⎢⎢⎣d2 + σ√T − t−

2

µr − δ +

1

2σ2¶

σ

√T − t

⎤⎥⎥⎦

=1

2 (T − t)

⎡⎢⎢⎣d2 − 2µr − δ − 1

2σ2¶

σ

√T − t

⎤⎥⎥⎦

⇒ Vt = e−r(T−t)∙rN (d2) +

∂

∂tN (d2)

¸= rV + e−r(T−t)N

0(d2)

∂

∂td2

= rV + e−r(T−t)N0(d2)

1

2 (T − t)

⎡⎢⎢⎣d2 − 2µr − δ − 1

2σ2¶

σ

√T − t

⎤⎥⎥⎦

VS = e−r(T−t)N0(d2)

∂

∂Sd2 = e−r(T−t)

N0(d2)

Sσ√T − t

=⇒ (r − δ)SVS = (r − δ)Se−r(T−t)N0(d2)

1

σ√T − t

× 1

S

= (r − δ) e−r(T−t)N0(d2)

1

σ√T − t

= N0(d2)

(r − δ) e−r(T−t)

σ√T − t

VSS = e−r(T−t)∂

∂S

"N

0(d2)

Sσ√T − t

#=

e−r(T−t)

σ√T − t

∂

∂S

"N

0(d2)

S

#

=e−r(T−t)

σ√T − t

"N

00(d2)

S

∂

∂Sd2 −

N0(d2)

S2

#

=e−r(T−t)

σ√T − t

"−d2N

0(d2)

S

1

Sσ√T − t

− N0(d2)

S2

#

= −N 0(d2)

e−r(T−t)

S2σ√T − t

µd2

σ√T − t

+ 1

¶= −N 0

(d2)e−r(T−t)

S2σ2 (T − t)

¡d2 + σ

√T − t

¢www.actuary88.com c°Yufeng Guo 198


Vt +1

2σ2S2VSS + (r − δ)SVS

= rV + e−r(T−t)N0(d2)

1

2 (T − t)

⎡⎢⎢⎣d2 − 2µr − δ − 1

2σ2¶

σ

√T − t

⎤⎥⎥⎦−12σ2S2N

0(d2)

e−r(T−t)

S2σ2 (T − t)

¡d2 + σ

√T − t

¢+(r − δ)Se−r(T−t)

N0(d2)

Sσ√T − t

= rV

Problem 21.7.

S (t) e−δ(T−t)N (d1) and e−δ(T−t)N (d2) each satisfy BS PDE. Hence theirlinear combination S (t) e−δ(T−t)N (d1)−Ke−δ(T−t)N (d2) also satisfies BS PDE.You can verify on your own that S (t) e−δ(T−t)N (d1)−Ke−δ(T−t)N (d2) satisfiesthe boundary condition when t = T .

Problem 21.8.

The forward price is Se(r−δ)T . If K = Se(r−δ)T , then

d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

=

lnS

Se(r−δ)T+

µr − δ +

1

2σ2¶T

σ√T

=

− (r − δ)T +

µr − δ +

1

2σ2¶T

σ√T

=1

2

√Tσ

d2 = d1 − σ√T =

1

2

√Tσ − σ

√T = −1

2

√Tσ

a. Bet #1: Getting $1 if ST > K = Se(r−δ)T and zero otherwise.Bet #2: Getting $1 if ST < K = Se(r−δ)T and zero otherwise.

Why Bet #1 is always worth less than Bet #2?Bet #1 is worth e−r(T−t)N (d2). Bet #2 is worth e−r(T−t)N (−d2).Since d2 < 0, we have −d2 > 0 > d2. Since the normal cdf is increasing

function, N (−d2) > N (d2). Hence Bet #2 is greater than Bet #1.

b. To make the bet fair, we need to have P ∗ (ST > K) = N (d2) = 0.5so there’ll be equal risk-neutral chance of ST > K and ST < K. To haveN (d2) = 0.5, we need to have d2 = 0 and d1 = σ

√T .

=⇒ d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

= σ√T ln

S

K+

µr − δ +

1

2σ2¶T =

σ2T



lnS

K= −

µr − δ − 1

2σ2¶T ln

K

S=

µr − δ − 1

2σ2¶T K = Se

r−δ−1

2σ2 T

So x = K = Ser−δ−

1

2σ2 T

c. price of an asset call option is Se−r(T−t)N (d1).N (d1) is the expected fractional share of the stock *if* the call is exercised.

To make the bet fair, set N (d1) = 0.5. This gives us d1 = 0

d1 =

lnS

K+

µr − δ +

1

2σ2¶T

σ√T

= 0 lnS

K+

µr − δ +

1

2σ2¶T

lnS

K= −

µr − δ +

1

2σ2¶T ln

K

S=

µr − δ +

1

2σ2¶T

=⇒ K = Ser−δ+

1

2σ2 T

So we need to set x = K = Ser−δ+

1

2σ2 T

Problem 21.9.

SinceK = Se(r−δ)T is the forward price of the stock, d1 =ln

S

Se(r−δ)T+

µr − δ +

1

2σ2¶T

σ√T

=

1

2

√Tσ and d2 = d1 − σ

√T =

1

2

√Tσ − σ

√T = −1

2

√Tσ.

Bet #1 pays ST−K if ST > K. Its value is V1 = Se−rTN (d1)−Ke−rTN (d2) =

Se−δTN³0.5√Tσ´−Ke−rTN

³−0.5

√Tσ´

Bet #2 pays K − ST if K > ST . Its value is V2 = Ke−rTN (−d2) −Se−δTN (−d1) = Ke−rTN

³0.5√Tσ´− Se−δTN

³−0.5

√Tσ´

V1−V2 = Se−δTNh³0.5√Tσ´+N

³−0.5

√Tσí−Ke−rT

hN³−0.5

√Tσ´+N

³0.5√Tσí

Using the formula N (−x) = 1−N (x), we get: V1−V2 = Se−δT −Ke−rT =Se−δT − e−rTSe(r−δ)T = 0The two bets have the same value.

Problem 21.10.

Please note that the continuous payment rate Γ in this problem is not theoption Gamma. To avoid confusion, we’ll use β to represent the option’s con-tinuous payment rate.At time t



• We buy one option on the stock. We pay V

• We buy N shares of the stock (a negative N means short selling stocks).We pay NS .

• We deposit W means into a savings account. We pay W .

To have zero-financing, we set our total initial cost to zero

I = V +NS +W = 0 (DM 21.7)

Next, let’s consider the change of I during [t, t+ dt]:

dI = βdt+ dV +N (dS + δSdt) + dW (DM 21.8)

dI is the interest we earned during [t, t+ dt]. SinceW is invested in a savingsaccount, we have dW = rWdt. This says that the interest earned on W during[t, t+ dt] is rWdt.

βdt is the payment we receive from the derivative during [t, t+ dt].Notice that the change of S is dS+δSdt (the sum of the change of the stock

price dS and the dividend received δSdt). Apply Ito’s lemma:dV = Vtdt+ VSdS + 0.5σ

2S2VSSdt→ dI = βdt+ Vtdt+ VSdS + 0.5σ

2S2VSSdt+N (dS + δSdt) + dW = βdt+Vtdt+ (VS +N) dS + 0.5σ2S2VSSdt+NδSdt+ rWdt

Set N = −VS . Then dS term becomes zero and W = − (V − VSS).Because our initial cost is zero, the interest we earned dI should be zero.βdt+ Vtdt+ 0.5σ

2S2VSSdt− VSδSdt− r (V − VSS) dt = 0

This gives us:β + Vt + 0.5σ

2S2VSS − VSδS − r (V − VSS) = 0⇒ Vt + 0.5σ

2S2VSS + (r − δ)VSS + β − rV = 0





Chapter 22

Exotic options: II

Skip all the problems.

203

CHAPTER 22. EXOTIC OPTIONS: II


Chapter 23

Volatility

Problem 23.1.

This problem can be solved using the approach used in DM Table 11.1 (DMpage 361). In other words, if you can reproduce DM Table 11.1, you should beable to solve this problem.

Problem 23.2.

This problem can be solved using the approach used in DM Table 11.1 (DMpage 361).

Problem 23.3.

This problem is out of the scope of the exam MFE. However, if you want tosolve it, you can use DM 23.6 to find the answer.

Problem 23.4.

Out of the scope of the exam MFE. See DM Example 23.2 if you want to

know how to solve it.

Problem 23.5.

Out of the scope of the exam MFE. See DM Example 23.2 if you want to

know how to solve it.

Problem 23.6.

205

CHAPTER 23. VOLATILITY

a. You can do this using the option price formula. Alternatively, you canfind the call price using the put-call parity C + PV (K) = P + S. Since thematurity is very short, P is close to zero; there’s little chance that the stock pricewill drop from $100 to below $50 during T = 0.01. Hence C = S − PV (K) =100− 50e−0.06×0.01 = 50. 03.

b. Using the spreadsheet in the CD that comes with the textbook, you shouldfind vega is almost zero. Vega measures the sensitivity of the option price tovolatility (see DM section 12.3). Vega is almost zero because the option maturityT = 0.01 is very short. It’s hard for the stock price to change significantly duringthis short amount of time (unless the volatility is very very big).

c. Under 5% or 100% volatility, the option price is still about 50.03. Thevolatility isn’t huge enough to move the stock price during T = 0.01.However, under 500% volatility, the option price is 51.3. The volatility is

huge; the stock price can change significantly during T = 0.01

d. It’s difficult to calculate the implied volatility for a call option that’s deepin the money and that has a short time to maturity.

Problem 23.7.

a. Using the put-call parity, we have C = P +S−PV (K) > S−PV (K) =100− 50e−0.06×0.01 = 50. 03. So the call is worth at east 50.03; it can never beworth less than 50. 03. Hence the bid price 50 is never possible. We can’t findthe implied volatility under this bid price.

b. Once again, we use the put-call parity. The ask price 50.1 is greater thanthe minimum value of the call 50.03. In order for the call to be worth morethan 50. 03, the put value must be greater than zero. To make this happen,there must be some change that the stock price will drop from 100 to below 50during T = 0.01. The only way this can happen is that the stock volatility isvery high.

c. Skip.

d. It’s difficult to calculate the implied volatility for a deep in-the-moneycall with a short maturity.

Problem 23.8.

a. For the call to have any value, the stock price must move up from $50and be greater than K = 100 at maturity. There’s little chance that the stockprice will move up by that much during the short maturity T = 0.01 and undera small volatility of 30%. Hence the call price is virtually zero.



b. The maturity is short. The option price is not very sensitive to thevolatility. Hence vega is close to zero.

c. The implied volatility is about 30% if the bid price is zero; the impliedvolatility must be huge if the ask price is 0.05 (there must be some chance thatthe stock price can move up by more than $50 at maturity in order for the callto be worth 0.05).

d. The market maker wants to buy the option for $0 price and sell the optionfor $0.05. The market-maker thinks that the option isn’t worth much. At thesame time, the market maker still wants to make a little profit.

e. It’s difficult to calculate and interpret the implied volatility for a deepout-of-the-money option that has a short maturity.





Chapter 24

Interest rate models

Problem 24.1.

a. We need to find the price of a 1-year bond issued at t = 1. The price ofthis bond is just the PV of $1 discounted from t = 2 to t = 1.

P (1, 2) =P (0, 2)

P (0, 1)To understand this formula, notice that P (0, 2) = P (1, 2)P (0, 1). This

equation means that to calculate the PV of $1 discounted from t = 2 to t = 0,we first discount $1 from t = 2 to t = 1 and next discount it from t = 1 to t = 0.

P (1, 2) =P (0, 2)

P (0, 1)=0.8495

0.9259= 0.917 485 7

b. C =Time zero cost of what you get at T ×N (d1)−Time zero cost ofwhat you give at T ×N (d2)



d2 = d1 − σ√T

Make sure you know the formula for σ. DM page 790 shows that

σ =pV ar (ln (Ft,T (P [T, T + s])))

If you buy this option, then at T = 1, you can pay K = 0.9009 and buy a1-year bond. This 1-year bond will give you $1 at time T + s = 2.

• The value of this bond at T = 0 is P (0, 2) = 0.8495, the PV of $1discounted from T + s = 2 to T = 1

• Time zero cost of the strike price K at T = 1 is just PV of K discountedfrom T = 1 to time zero. So PV (K) = 0.9009× 0.9259

C =Time zero cost of what you get at T ×N (d1)−Time zero cost of whatyou give at T ×N (d2)

209

CHAPTER 24. INTEREST RATE MODELS

→ d1 =ln 0.8495

0.9009×0.9259+0.5×0.12×1

0.1√1

= 0.232 43

→ d2 = d1 − σ√T = 0.232 43− 0.1

√1 = 0.132 43

N (d1) = 0.591 90 N (d2) = 0.552 68C = 0.8495× 0.591 90− 0.9009× 0.9259× 0.552 68 = 0.0418

c.P =Time zero cost of what you give at T ×N (−d2)−Time zero cost of what

you get at T ×N (−d1)P = 0.9009× 0.9259× (1− 0.552 68)− 0.8495× (1− 0.591 90) = 0.0264

Alternative calculation using the put-call parity.C + P (0, T )K = P + P (0, T + s)0.0418 + 0.9009× 0.9259 = P + 0.8495P = 0.0264d. Let’s walk through the notations and formula.Notation

• RT=1 (T = 1, T + s = 2). The (not annualized) interest rate agreed uponat time T = 1 that applies to the time interval [T = 1, T + s = 2].

• Caplet. A Caplet gives the buyer the right to buy the time-T = 1 mar-ket interest rate RT (T, T + s) = RT=1 (T = 1, T + s = 2) by paying afixed strike interest rate KR = 11%. If KR ≥ RT=1 (T = 1, T + s = 2),the caplet expires worthless. The payoff of the caplet at T + s = 2 ismax [0, RT=1 (T = 1, T + s = 2)− 0.11]. The payoff of the caplet at T ismax [0, RT=1 (T = 1, T + s = 2)− 0.11]

RT=1 (T = 1, T + s = 2)

Please note that the actual interest rate during [T = 1, T + s = 2] is a ran-dom variable. One might be tempted to think that we already know the Year 2interest rate as:

1

P (1, 2)− 1 = 8. 993 5%

However, this thinking is flawed. 8. 993 5% is the implied Yr 2 interest ratebased on the information available to us at t = 0. This rate can be differentfrom the Year 2 spot rate.

To calculate the price of the caplet, we first modify the payoff:max [0, RT=1 (T = 1, T + s = 2)− 0.11]

RT=1 (T = 1, T + s = 2)= 1.11max

∙0,RT=1 (T = 1, T + s = 2)− 1.11RT=1 (T = 1, T + s = 2) 1.11

¸=

1.11max

∙0,

1

1.11− 1

RT=1 (T = 1, T + s = 2)

¸www.actuary88.com c°Yufeng Guo 210


max

∙0,

1

1.11− 1

RT=1 (T = 1, T + s = 2)

¸= max

∙0, 0.900 9− 1

RT=1 (T = 1, T + s = 2)

¸is the payoff of a put on a bond. This put gives the buyer the right, at T = 1,

to sell a bond that matures at T + s = 2 for a guaranteed price1

1.11= 0.900 9.

From Part c, we already know that this put price is 0.0264. Hence the capletprice is:1.11× 0.0264 = 0.02 93

Problem 24.2.

a. P (2, 3) =P (0, 3)

P (0, 2)=0.7722

0.8495= 0.909 01

b. C = P (0, T + s)N (d1)− P (0, T )KN (d2)P (0, T + s) = P (0, 2 + 1) = P (0, 3) = 0.7722P (0, T ) = P (0, 2) = 0.8495

→ C = 0.7722N (d1)− 0.8495× 0.9N (d2)

d1=

lnP (0, T + s)

P (0, T )K+ 0.5σ2T

σ√T

=ln

0.7722

0.8495× 0.9 + 0.5× 0.1052 × 2

0.105×√2

= 0.141 29

d2 = d1 − σ√T = 0.141 29− 0.105×

√2 = −0.007 2

N (d1) = 0.556 18N (d2) = 0.497 13

→ C = 0.7722× 0.556 18− 0.8495× 0.9× 0.497 13 = 0.04 94

c. P = P (0, T )KN (−d2)− P (0, T + s)N (−d1)= 0.8495× 0.9× (1− 0.497 13)− 0.7722× (1− 0.556 18)= 0.04175 1

Alternative calculation using the put-call parity.C + P (0, T )K = P + P (0, T + s)0.04 94 + 0.8495× 0.9 = P + 0.7722P = 0.041 75d. To calculate the price of the caplet, we first modify the payoff:max [0, RT=2 (T = 2, T + s = 3)− 0.11]

RT=2 (T = 2, T + s = 3)= 1.11max

∙0,

1

1.11− 1

RT=2 (T = 2, T + s = 3)

¸max

∙0,

1

1.11− 1

RT=2 (T = 2, T + s = 3)

¸= max

∙0, 0.900 9− 1

RT=2 (T = 2, T + s = 3)

¸is the payoff of a put on a bond. This put gives the buyer the right, at T = 2,

to sell a bond that matures at T + s = 3 for a guaranteed price1

1.11= 0.900 9.



We can estimate the put price. Since1

1.11= 0.900 9 is close to 0.9, from Part

c we know that P = 0.041 75. Hence the price of the caplet is 1.1×0.041 75 = 0.0459Or we can calculate the put price.

P = P (0, T )KN (−d2)−P (0, T + s)N (−d1) = 0.8495× 0.900 9N (−d2)−0.7722N (−d1)

d1=

lnP (0, T + s)

P (0, T )K+ 0.5σ2T

σ√T

=ln

0.7722

0.8495× 0.900 9 + 0.5× 0.1052 × 2

0.105×√2

=

0.134 56d2 = d1 − σ

√T = 0.134 56− 0.105×

√2 = −1. 393 2× 10−2

N (−d1) = 0.446 48N (−d2) = 0.505 56P = 0.8495× 0.900 9× 0.505 56− 0.7722× 0.446 48 = 0.04214Hence the price of the caplet is 1.1× 0.04214 = 0.0464

Problem 24.3.

Make sure you understand that this problem is different from SOA May 2007#9. In SOA May 2007 #9, the first cash flow occurs at t = 1.The first yearmarket rate 6% is below the cap rate 7.5%. At the end of Year 1 (at t = 1), weget nothing from the cap .In this problem, the author wants us to repeatedly use using DM Equation

24.36 to calculate the cap price. So the author wants the first cash flow to occurat t = 2.The cap contract is signed at t = 0. During Yr 2, the cap rate 11.5% is com-

pared with the actual Yr 2 interest. The payoff at t = 2 is themax (0, 11.5%−Yr 2 rate).The PV of this payoff at t = 1 is:

max (0, 11.5%−Yr 2 rate)1 +Yr 2 rate

= 1.115max

µ0,

1

1.115− 1

1 +Yr 2 rate

¶

max

µ0,

1

1.115− 1

1 +Yr 2 rate

¶is the payoff a put. This put gives the buyer

the right at T = 1 to buy a one-year bond maturing at T + s = 2. You canverify that the time zero cost of this put is 0.0248.

P = P (0, 1)KN (−d2)− P (0, 2)N (−d1)

d1=

lnP (0, 2)

P (0, 1)K+ 0.5σ2T

σ√T

=

ln0.8495

0.9259× 1

1.115

+ 0.5× 0.12 × 1

0.1×√1

= 0.277 36

d2 = d1 − σ√T = 0.277 36− 0.1×

√1 = 0.177 36



P = 0.9259× 1

1.115N (−0.177 36)− 0.8495N (−0.277 36)

= 0.9259× 1

1.115× 0.429 61− 0.8495× 0.390 75 = 0.0248

So the time zero cost of the payoff at t = 2 is 1.115× 0.0248.

Similarly, the time zero cost of the payoff at t = 3 is 1.115× 0.0404

P = P (0, 2)KN (−d2)− P (0, 3)N (−d1)The put gives the buyer the right at T = 2 to buy a bond maturing at

T + s = 3

d1=

lnP (0, 3)

P (0, 2)K+ 0.5σ2T

σ√T

=

ln0.7722

0.8495× 1

1.115

+ 0.5× 0.1052 × 2

0.105×√2

= 0.164 82

d2 = d1 − σ√T = 0.164 82− 0.105×

√2 = 0.01633

P = 0.8495× 1

1.115N (−0.01633)− 0.7722N (−0.164 82)

= 0.8495× 1

1.115× 0.493 49− 0.7722× 0.434 54 = 0.040 4

The time zero cost of the payoff at t = 4 is 1.115× 0.0483

P = P (0, 2)KN (−d2)− P (0, 3)N (−d1)The put gives the buyer the right at T = 3 to buy a bond maturing at

T + s = 4

d1=

lnP (0, 4)

P (0, 3)K+ 0.5σ2T

σ√T

=

ln0.7020

0.7722× 1

1.115

+ 0.5× 0.112 × 3

0.11×√3

= 0.166 35

d2 = d1 − σ√T = 0.166 35− 0.11×

√3 = −0.02417

P = 0.7722× 1

1.115N (0.02417)− 0.7020N (−0.166 35)

= 0.7722× 1

1.115× 0.509 64− 0.7020× 0.433 94 = 0.04832 8

The time zero cost of the cap is:1.115× (0.0248 + 0.0404 + 0.0483) = 0.126 6

Problem 24.4.

At time zero, we

• But a 3-year bond. Cost: P (t, T2) = e−0.08×3 = 0.786 63

• To hedge, we buy N = −(T2 − t)P (t, T2)

(T1 − t)P (t, T1)= −1× e−0.08×3

2× e−0.08×6= −0.635 62

unit of 6-year bond (i.e. sell 0.635 62 unit of 6-year bond), receiving0.635 62e−0.08×6 = 0.393 31



• Borrow 0.786 63− 0.393 31 = 0.393 32 from a bank at 8% interest rate.

• Our net position is 0.786 63− (0.393 31 + 0.393 32) = 0

If at the end of the day the risk-free rate is 8.25%, we’ll close off our position.

• Sell the (3− 1/365)-year bond, receiving e−0.0825×(3−1/365) = 0.780 93

• Buy back the (6− 1/365)-year bond, paying 0.635 62e−0.0825×(6−1/365) =0.387 54

• Pay off the loan, paying 0.393 32e0.08×1/365 = 0.393 41 (we borrow theloan for only one day).

• Our net position: 0.780 93 − 0.387 54 − 0.393 41 = −0.000 02. So we lose0.000 02

Why do we lose money? Because of the convexity mismatch.We buy the 3-year bond. The convexity is (see DM 7.14 and DM Example

7.9):3× 41.082

= 10. 288 07

We sell 0.635 62 unit of the 6-year bond. The convexity is:

0.635 62× 6× 71.082

= 22. 887 55

Even though we hedged the duration, we didn’t hedge convexity. The twoportfolios have different convexities. As explained in my study guide, the high-convexity bond has a better value. To arbitrage, we should have bought thehigh-convexity bond and sold the low-convexity bond. So to arbitrage, we needto reverse our position (i.e. at t = 0, sell one unit of 3-year bond and buy0.635 62 unit of 6-year bond). Then we’ll earn 0.000 02 free money during Day1.How to reverse our position. At time zero, we

• Sell a 3-year bond, receiving P (t, T2) = e−0.08×3 = 0.786 63

• To hedge, we buy 0.635 62 unit of 6-year bond, paying 0.635 62e−0.08×6 =0.393 31

• Lend 0.786 63− 0.393 31 = 0.393 32 at 8% interest rate.

• Our net position is 0.786 63− (0.393 31 + 0.393 32) = 0

At the end of the day, the risk-free rate is 8.25%. We close off our position.

• Buy the (3− 1/365)-year bond, paying e−0.0825×(3−1/365) = 0.780 93

• Sell the (6− 1/365)-year bond, receiving 0.635 62e−0.0825×(6−1/365) = 0.387 54



• Close the loan, receiving 0.393 32e0.08×1/365 = 0.393 41.

• Our net position: 0.387 54 + 0.393 41 − 0.780 93 = 0.000 02. So we gain0.000 02

At time zero, our cost is zero. At the end of the day, we receive 0.000 02profit. This is arbitrage.

If at the end of the day the risk-free rate is 7.75%, we’ll close off our position.

• Sell the (3− 1/365)-year bond, receiving e−0.0775×(3−1/365) = 0.792 72

• Buy back the (6− 1/365)-year bond, paying 0.635 62e−0.0775×(6−1/365) =0.399 34

• Pay off the loan, paying 0.393 32e0.08×1/365 = 0.393 41 (we borrow theloan for only one day).

• Our net position: 0.792 72 − 0.399 34 − 0.393 41 = −0.000 03. So we lose0.000 03

To arbitrage, we reverse out position and gain 0.000 03 free money.This example shows that the assumption of the parallel shift of a flat yield

curve leads to arbitrage. To build a good model for bond price, we need tothrow away this bad assumption.

Problem 24.5.

a.4-year 5% annual coupon bond with yield 6% (Bond 1)Price: P1 = 0.05

¡e−0.06×1 + e−0.06×2 + e−0.06×3 + e−0.06×4

¢+ e−0.06×4 =

0.959 16Macaulay duration:

D1 =0.05

¡e−0.06×1 + 2e−0.06×2 + 3e−0.06×3 + 4e−0.06×4

¢+ 4e−0.06×4

0.959 16= 3.

716 7

8-year 7% annual coupon bond with yield 6% (Bond 2)Price:P2 = 0.07

¡e−0.06×1 + e−0.06×2 + e−0.06×3 + e−0.06×4 + ...+ e−0.06×8

¢+e−0.06×8 =

1. 050 3Macaulay duration:

D2 =0.07

¡e−0.06×1 + 2e−0.06×2 + 3e−0.06×3 + ...+ 8e−0.06×8

¢+ 8e−0.06×8

1. 050 3=

6. 433 2b. Use DM 7.13:

N = −D1B1 (y1) (1 + y1)

D2B2 (y2) (1 + y2)



Please also note that DM 24.7 is the special case of DM 7.13. If the twobonds each have one cash flow and two bonds have the same yield (i.e. y1 = y2),then DM 7.13 becomes DM 24.7.

In this problem, the yield is the same for the two bonds. So y1 = y2.

N = −D1B1 (y1) (1 + y1)

D2B2 (y2) (1 + y2)= −D1P1 (y)

D2P2 (y)= −3. 716 7× 0.959 16

6. 433 2× 1. 050 3 = −0.527 6

So we need to buy −0.527 6 (i.e. sell 0.527 6) unit of Bond 2. At time zero,we

• sell 0.527 6 unit of Bond 2, receiving 0.527 6× 1. 050 3 = 0.554 14

• buy one Bond 1, paying 0.959 16

• borrow the difference 0.959 16− 0.554 14 = 0.405 02 from a bank

0.959 16− 0.527 6× 1. 050 3 = 0.405 021 72Our net position is zero.

If at the end of the day, the yield is 6.25%:Back back 0.527 6 unit of Bond 2, which has (8− 1/365) year to maturity.

Now we are standing at t = 1/365. Bond 2’s price now is:P2 =

¡0.07

¡e−0.0625×1 + e−0.0575×2 + e−0.0625×3 + ...+ e−0.0625×8

¢+ e−0.0625×8

¢e0.0625/365

= 1. 034 39Sell Bond 1, which has (4− 1/365) year to maturityThe price isP1 =

¡0.05

¡e−0.0.0625×1 + e−0.0625×2 + e−0.0625×3 + e−0.0625×4

¢+ e−0.0625×4

¢e0.0625/365 =

0.953 48Repay the bank: 0.405 02e0.06/365 = 0.405 09The net profit at the end of Day 1 is: 0.953 48−0.527 6×1. 034 39−0.405 09 =

0.002 65If at the end of the day, the yield is 5.75%:Back back 0.527 6 unit of Bond 2, which has (8− 1/365) year to maturity.

Now we are standing at t = 1/365. Bond 2’s price now is:P2 = 0.07

£e−0.0575×(1−1/365) + e−0.0575×(2−1/365) + ...+ e−0.0575×(8−1/365)

¤+

e−0.0575×(8−1/365)

=¡0.07

¡e−0.0575×1 + e−0.0575×2 + e−0.0575×3 + ...+ e−0.0575×8

¢+ e−0.0575×8

¢e0.0575/365

= 1. 067 54

Sell Bond 1, which has (4− 1/365) year to maturityThe price isP1 =

¡0.05

¡e−0.0575×1 + e−0.0575×2 + e−0.0575×3 + e−0.0575×4

¢+ e−0.0575×4

¢e0.0575/365 =

0.968 27

Repay the bank: 0.405 02e0.06/365 = 0.405 09



The net profit at the end of Day 1 is: 0.968 27−0.527 6×1. 067 54−0.405 09 =−0.000 05

If we reserve our position, we can have 0.000 05 at the end of Day 1. Referto Problem 24.4 to see how to reverse our position.Summary: If the yield moves up to 6.25% in Day 1, we can make 0.002 65

profit; if the yield moves down to 5.75% in Day 1, we can make 0.000 05 profit.Once again, the assumption of the flat yield curve leads to arbitrage.

Problem 24.6.

Vasicek model:a. Zero risk premium means that α (r, t, T ) = r. Hence φ (r, t) =

α (r, t, T )− r

q (r, t, T )= 0

DM 24.26: r = b− 0.5σ2

α2= 0.1− 0.5× 0.1

2

0.22= −0.025

2-year bond:

B (T − t = 2) = aT−t|α =1− e−α(T−t)

α=1− e−0.2(2)

0.2= 1. 648 4

aT−t|α is a T − t year continuous annuity with the force of interest α.

A (T − t = 2) = er[B(2)−2]−B2(2)σ2/4α = e−0.025(1. 648 4−2)−1. 648 4

2×0.12/(4×0.2) =0.975 14The 2-year bond is worth:P (0, 2) = A (2) e−B(2)r = 0.975 14e−1. 648 4×0.05 = 0.897 99The delta is:d

drP (0, 2) =

d

drA (2) e−B(2)r = −B (2)A (2) e−B(2)r = −B (2)P (0, 2) =

−1. 648 4× 0.897 99 = −1. 480 2The gamma is:d2

dr2P (0, 2) = −B (2) d

drP (0, 2) = B2 (2)P (0, 2) = 1. 648 42 × 0.897 99 = 2.

44

10-year bond:

B (10) = a10|0.2 =1− e−0.2×10

0.2= 4. 323 3

A (10) = er[B(10)−10]−B2(10)σ2/4α = e−0.025(4. 323 3−10)−4. 323 3

2×0.12/(4×0.2) =0.912 36The 10-year bond is worth:P (0, 10) = A (10) e−B(10)r = 0.912 36e−4. 323 3×0.05 = 0.735The delta is:d

drP (0, 10) = −B (10)P (0, 10) = −4. 323 3× 0.735 = −3. 177 6

The gamma is:d2

dr2P (0, 10) = B2 (10)P (0, 10) = 4. 323 32 × 0.735 = 13. 738



The delta is:d

drP (0, 2) =

d

drA (2) e−B(2)r = −B (2)A (2) e−B(2)r = −B (2)P (0, 2) =

−1. 648 4× 0.897 99 = −1. 480 2The gamma is:d2

dr2P (0, 2) = −B (2) d

drP (0, 2) = B2 (2)P (0, 2) = 1. 648 42 × 0.897 99 = 2.

44

b. If we buy a 2-year bond, to duration hedge our risk, we need to buy

N = − 2P (0, 2)

10P (0, 10)= −2× 0.897 99

10× 0.735 = −0.244 35So we need to sell 0.244 35 unit of 10-year bond.

At t = 0, we

• buy a 2-year bond, paying P (0, 2) = 0.897 99

• sell 0.244 35 unit of 10-year bond, receiving 0.244 35× 0.735 = 0.179 60

• borrow 0.897 99− 0.179 60 = 0.718 39 at 5%

• Our net position is zero.

The one standard deviation of the interest rate under Vasicek is r±σp1/365 =

0.05± 0.1p1/365

ru = 0.05 + 0.1p1/365 = 0.055

rd = 0.05− 0.1p1/365 = 0.045

Under ru = 0.05 + 0.1p1/365 = 0.055

2− 1/365 year bond:

B (2− 1/365) = 1− e−α(T−t)

α=1− e−0.2(2−1/365)

0.2= 1. 647

A (2− 1/365) = e−0.025(1. 646 7−2+1/365)−1. 6472×0.12/(4×0.2) = 0.975 2

The 2− 1/365-year bond is worth:P (0, 2− 1/365) = 0.975 2e−1. 6467×0.055 = 0.890 8

10− 1/365-year bond:

B (10− 1/365) = 1− e−0.2×(10−1/365)

0.2= 4. 323 0

A (10− 1/365) = er[B(10−1/365)]−B2(10−1/365)σ2/4α = e−0.025(4. 323−10+1/365)−4. 323

2×0.12/(4×0.2) =0.912 34The 10− 1/365-year bond is worth:P (0, 10− 1/365) = A (10− 1/365) e−B(10−1/365)r = 0.912 34e−4. 323×0.055 =

0.719 3

At the end of the day, we close our position:



• sell a 2− 1/365-year bond, receiving 0.890 8

• buy 0.244 35 unit of 10−1/365-year bond, paying 0.244 35 × 0.719 3 =0.175 8

• pay back the loan, paying 0.718 39e0.05/365 = 0.718 5

• our net position is: 0.890 8− 0.175 8− 0.718 5 = −0.003 5

• So we lose 0.003 5

Under rd = 0.05− 0.1p1/365 = 0.045

2− 1/365 year bond:

B (2− 1/365) = 1− e−α(T−t)

α=1− e−0.2(2−1/365)

0.2= 1. 647

A (2− 1/365) = e−0.025(1. 646 7−2+1/365)−1. 6472×0.12/(4×0.2) = 0.975 2

The 2− 1/365-year bond is worth:P (0, 2− 1/365) = 0.975 2e−1. 6467×0.045 = 0.905 6

10− 1/365-year bond:

B (10− 1/365) = 1− e−0.2×(10−1/365)

0.2= 4. 323 0

A (10− 1/365) = er[B(10−1/365)]−B2(10−1/365)σ2/4α = e−0.025(4. 323−10+1/365)−4. 323

2×0.12/(4×0.2) =0.912 34The 10− 1/365-year bond is worth:P (0, 10− 1/365) = A (10− 1/365) e−B(10−1/365)r = 0.912 34e−4. 323×0.045 =

0.751 1


• sell a 2− 1/365 year bond, receiving 0.905 6

• buy 0.244 35 unit of 10 − 1/365 year bond, paying 0.244 35 × 0.751 1 =0.183 5


• our net position is: 0.905 6− 0.183 5− 0.718 5 = 0.003 6

• So we gain 0.003 6

Delta hedge. At t = 0, we buy a 2-year bond and buy N units of 10-yearbond. We already calculated the following:

• The delta of the 2-year bond is −1. 480 2

• The delta of the 10-year bond is −3. 177 6



The total delta of the portfolio is: −1. 480 2 +N (−3. 177 6).To delta hedge, set −1. 480 2 +N (−3. 177 6) = 0 → N = −0.465 8

So at t = 0, we


• sell 0.465 8 unit of 10-year bond, receiving 0.465 8× 0.735 = 0.342 36

• borrow 0.897 99− 0.342 36 = 0.555 63 at 5%


Under ru = 0.05 + 0.1p1/365 = 0.055

The 2− 1/365-year bond is worth: P (0, 2− 1/365) = 0.890 8The 10− 1/365-year bond is worth: P (0, 10− 1/365) = 0.719 3



• buy 0.465 8 unit of 10−1/365-year bond, paying 0.465 8×0.719 3 = 0.335 05



• So we gain 0.000 04

Under rd = 0.05− 0.1p1/365 = 0.045

The 2− 1/365-year bond is worth: P (0, 2− 1/365) = 0.905 6The 10− 1/365-year bond is worth: P (0, 10− 1/365) = 0.751 1


• sell a 2− 1/365 year bond, receiving 0.905 6

• buy 0.465 8 unit of 10−1/365 year bond, paying 0.465 8×0.751 1 = 0.349 86



• So we gain 0.000 03



CIR:

a. Zero risk premiummeans that α (r, t, T ) = r. Hence φ (r, t) =α (r, t, T )− r

q (r, t, T )=

0DM 24.28: φ (r, t) = φ

√r/σ. Hence φ = 0

DM 24.29:

γ =

q(a+ φ)

2+ 2σ2 =

q(0.2 + 0)

2+ 2× 0.447212 = 0.663 32

a+ φ+ γ = 0.2 + 0 + 0.663 32 = 0.863 32

2-year bond:

A (T − t = 2) =

"2γe(a+φ+γ)(T−t)/2¡

a+ φ+ γ¢ ¡eγ(T−t) − 1

¢+ 2γ

#2abσ2

=

∙2× 0.663 32e0.863 32×2/2

0.863 32 (e0.663 32×2 − 1) + 2× 0.663 32

¸2× 0.2× 0.10.447212

= 0.967 18

B (T − t = 2) =2¡eγ(T−t) − 1

¢¡a+ φ+ γ

¢ ¡eγ(T−t) − 1

¢+ 2γ

=2¡e0.663 32×2 − 1

¢0.863 32 (e0.663 32×2 − 1) + 2× 0.663 32 = 1. 489 72

P (0, 2) = A (2) e−B(2)r = 0.967 18e−1. 489 72×0.05 = 0.897 76The delta is:d

drP (0, 2) =

d

drA (2) e−B(2)r = −B (2)P (0, 2) = −1. 489 72 × 0.897 76 =

−1. 337 41The gamma is:d2

dr2P (0, 2) = −B (2) d

drP (0, 2) = B2 (2)P (0, 2) = 1. 489 722 × 0.897 76 =

1. 992 37

10-year bond:

A (T − t = 10) =

∙2× 0.663 32e0.863 32×10/2

0.863 32 (e0.663 32×10 − 1) + 2× 0.663 32

¸2× 0.2× 0.10.447212

=

0.685 54

B (T − t = 10) =2¡e0.663 32×10 − 1

¢0.863 32 (e0.663 32×10 − 1) + 2× 0.663 32 = 2. 311 96

P (0, 10) = A (10) e−B(10)r = 0.685 54e−2. 311 96×0.05 = 0.610 70



The delta is:d

drP (0, 10) = −B (10)P (0, 10) = −2. 311 96× 0.610 70 = −1. 411 91

The gamma is:d2

dr2P (0, 10) = B2 (10)P (0, 10) = 2. 311 962 × 0.610 70 = 3. 264 29

b. If we buy a 2-year bond, to duration hedge our risk, we need to buy Nunits of 10 year bond.

N = − 2P (0, 2)

10P (0, 10)= − 2× 0.897 76

10× 0.610 70 = −0.294 01So we need to sell 0.294 01 unit of 10-year bond.

At t = 0, we


• sell 0.294 01 unit of 10-year bond, receiving 0.294 01× 0.610 70 = 0.179 55

• borrow 0.897 76− 0.179 55 = 0.718 21 at 5%


Notice that under CIR, dr = a (b− r) dt+ σ√rdz. The one standard devia-

tion of the interest rate under CIR is r±σ√rp1/365 = 0.05± 0.44721

√0.05×p

1/365

0.44721√0.05×

p1/365 = 0.005 234 2

ru = 0.05 + 0.005 234 2 = 0.055 23rd = 0.05− 0.005 234 2 = 0.044 77

Under ru = 0.055 23The 2− 1/365-year bond is worth:

A (T − t = 2− 1/365) ="

2× 0.663 32e0.863 32×(2−1/365)/20.863 32

¡e0.663 32×(2−1/365) − 1

¢+ 2× 0.663 32

#2× 0.2× 0.10.447212

=

0.967 26

B (T − t = 2− 1/365) =2¡e0.663 32×(2−1/365) − 1

¢0.863 32

¡e0.663 32×(2−1/365) − 1

¢+ 2× 0.663 32

=

1. 488 40

→ P (0, 2− 1/365) = A (2− 1/365) e−B(2−1/365)r = 0.967 26e−1. 488 40×0.055 23 =0.890 93

The 10− 1/365-year bond is worth:



A (T − t = 10− 1/365) ="

2× 0.663 32e0.863 32×(10−1/365)/20.863 32

¡e0.663 32×(10−1/365) − 1

¢+ 2× 0.663 32

#2× 0.2× 0.10.447212

=

0.685 63

B (T − t = 10− 1/365) =2¡e0.663 32×(10−1/365) − 1

¢0.863 32

¡e0.663 32×(10−1/365) − 1

¢+ 2× 0.663 32

=

2. 311 95

→ P (0, 10− 1/365) = 0.685 63e−2. 311 95×0.055 23 = 0.603 44



• buy 0.294 01 unit of 10 − 1/365-year bond, paying 0.294 01 × 0.603 44 =0.177 42


• our net position is: 0.890 93− 0.177 42− 0.718 31 = −0.004 8

• So we lose 0.004 8

Under rd = 0.044 77

P (0, 2− 1/365) = A (2− 1/365) e−B(2−1/365)r = 0.967 26e−1. 488 40×0.044 77 =0.904 91

P (0, 10− 1/365) = 0.685 63e−2. 311 95×0.044 77 = 0.618 21






• So we gain 0.004 84

Delta hedge:

−1. 337 41 +N (−1. 411 91) = 0 N = −0.947 23So we need to sell 0.947 23 unit of the 10-year bond.

At t = 0, we




• sell 0.947 23 unit of 10-year bond, receiving 0.947 23× 0.610 70 = 0.578 47

• borrow 0.897 76− 0.578 47 = 0.319 29 at 5%


One day later, under ru = 0.055 23P (0, 2− 1/365) = A (2− 1/365) e−B(2−1/365)r = 0.967 26e−1. 488 40×0.055 23 =

0.890 93P (0, 10− 1/365) = 0.685 63e−2. 311 95×0.055 23 = 0.603 44





• our net position is: 0.890 93− 0.571 60− 0.319 33 = 0

• So we gain 0

Under rd = 0.044 77P (0, 2− 1/365) = 0.967 26e−1. 488 40×0.044 77 = 0.904 91

P (0, 10− 1/365) = 0.685 63e−2. 311 95×0.044 77 = 0.618 21





• our net position is: 0.904 91− 0.585 59− 0.319 33 ≈ 0

• So we gain 0

Problem 24.7.



t = 0 t = 1 t = 2 t = 30.16

0.040.12

0.120.12

0.100.08

0.100.12

0.100.08

0.080.08

0.060.04

Price of 1-Yr bond: P (0, 1) = e−0.1 = 0.904 84

The yield for 1-Yr bond is 0.1.

Price of 2-Yr bond:Path Prob Price0→u 0.5 e−0.12e−0.1

0→d 0.5 e−0.08e−0.1

P (0, 2) = 0.5e−0.12e−0.1 + 0.5e−0.08e−0.1 = 0.818 89

The yield for 2-Yr bond is y

e−2y = 0.818 89 y = − ln 0.818 892

= 9.990 3%

Price of 3-Yr bond:Path Prob Price0→ u→ uu 0.25 e−0.14e−0.12e−0.1

0→ u→ ud 0.25 e−0.1e−0.12e−0.1

0→ d→ du 0.25 e−0.1e−0.08e−0.1

0→ d→ dd 0.25 e−0.06e−0.08e−0.1

P (0, 5) = 0.25¡e−0.14e−0.12e−0.1 + e−0.1e−0.12e−0.1 + e−0.1e−0.08e−0.1 + e−0.06e−0.08e−0.1

¢=

0.741 56

Yield for the 3-Yr bond: −13ln 0.741 56 = 9. 966 6%

Price of 4-Yr bond:



Path Prob Price0→ u→ uu→ uuu 1/8 e−0.16e−0.14e−0.12e−0.1 = 0.594 520→ u→ uu→ uud 1/8 e−0.12e−0.14e−0.12e−0.1 = 0.618 780→ u→ ud→ udu 1/8 e−0.12e−0.10e−0.12e−0.1 = 0.644 040→ u→ ud→ udd 1/8 e−0.08e−0.10e−0.12e−0.1 = 0.670 320→ d→ du→ duu 1/8 e−0.12e−0.10e−0.08e−0.1 = 0.670 320→ d→ du→ dud 1/8 e−0.08e−0.10e−0.08e−0.1 = 0.697 680→ d→ dd→ ddu 1/8 e−0.08e−0.06e−0.08e−0.1 = 0.726 150→ d→ dd→ ddd 1/8 e−0.04e−0.06e−0.08e−0.1 = 0.755 78

P (0, 4) =0.594 52 + 0.618 78 + 0.644 04 + 0.670 32 + 0.670 32 + 0.697 68 + 0.726 15 + 0.755 78

8=

0.672 20

Yield for the 4-Yr bond: −14ln 0.672 20 = 9. 930 0%

Yields decline with maturity. This is explained in DM page 796. The averageof the exponential interest rates is less than the exponentiated average.

Problem 24.8.

Instead of working path by path, we can work backwards.

At t = 3, the bond price is:Vuuu = 0.8331 Vuud = 0.8644 Vddu = 0.8906 Vddd = 0.9123At t = 2, the bond price is:

Vuu = 0.8321×0.8331 + 0.8644

2= 0.706 24

Vud = Vdu = 0.8798×0.8644 + 0.8906

2= 0.772 02

Vdd = 0.9153×0.8906 + 0.9123

2= 0.825 10

At t = 1, the bond price is:

Vu = 0.8832×0.706 24 + 0.772 02

2= 0.652 80

Vd = 0.9023×0.772 02 + 0.825 10

2= 0.720 54

At t = 0, the bond price is:

Vu = 0.9091×0.652 80 + 0.720 54

2= 0.624 25 = 0.6243

Problem 24.9.



Next year, the bond price is either Vu = 0.652 80 with a yield of 0.652 80−1/3−1 = 0.152 76 or Vd = 0.720 54 with a yield of 0.720 54−1/3 − 1 = 0.115 44.Using DM 24.48, the yield volatility is 0.5 ln

0.152 76

0.115 44= 0.140 06 = 0.14

Problem 24.10.

See DM page 805.

Problem 24.11.

DM Page 806 and 807 explain that rA =P (0, 3)

P (0, 4)−1 = 0.7118

0.6243−1 = 0.140 16.

Since the interest rate cap applies during the interval [t = 3, t = 4], the fair caprate must be the implied forward rate rA = 0.140 16 during [t = 3, t = 4] . Wecan verify that rA = 0.140 16 using the binomial tree.Let r (3, 4) represent the actual interest rate during the interval [t = 3, t = 4].

The reference rate rA is the 1-year forward rate 3 years hence. So rA is alsothe rate during [t = 3, t = 4]. This is the difference between r (3, 4) and rA.r (3, 4) is the actual interest rate observed in the market during Year 3; rA israte agreed upon at t = 0 that applies to Year 3.Set the notional principal to $100. At t = 4, the payoff is 100 [r (3, 4)− rA]

= 100r (3, 4) − 100rA. We need to find rA such that the PV of the payoff iszero.First, we calculate PV of 100r (3, 4). At t = 4, 100r (3, 4) has 4 possible

values: 20.03,15.68,12.28, and 9.62. Discounting these 4 values to t = 3, we getthe 4 values:

100× 0.20031 + 0.2003

= 16. 687100× 0.15681 + 0.1568

= 13. 555

100× 0.12281 + 0.1228

= 10. 937100× 0.09621 + 0.0962

= 8. 775 8

We have 3 values at t = 2 (using 0.5 as the risk-neutral probability of up ordown)

Vuu = 0.5

µ16. 687

1 + 0.2017+

13. 555

1 + 0.2017

¶= 12. 583

Vud = 0.5

µ13. 555

1 + 0.1366+

10. 937

1 + 0.1366

¶= 10. 774

Vdd = 0.5

µ10. 937

1 + 0.0925+

8. 775 8

1 + 0.0925

¶= 9. 021 9

The value at t = 1:

Vu = 0.5

µ12. 583

1 + 0.1322+

10. 774

1 + 0.1322

¶= 10. 315



Vd = 0.5

µ10. 774

1 + 0.1082+

9. 021 9

1 + 0.1082

¶= 8. 931 6

The value at t = 0:

V = 0.5

µ10. 315

1 + 0.1+8. 931 6

1 + 0.1

¶= 8. 748 5

So PV of 100r (3, 4) is 8. 748 5.PV of 100rA is 100rA × P (0, 4) = 100rA × 0.6243P (0, 4) = 0.6243 is from DM Table 24.2.

100rA × 0.6243 = 8. 748 5 rA = 0.140 13

Problem 24.12.

I just solve for Tree #1. Once you understand the logic, you can do Tree#2.1-Yr bond price:

P (0, 1) =1

1.08= 0.925 93

2-Yr bond price:

Vd =1

1 + 0.07676Vu =

1

1 + 0.10363

V = 0.5× Vu + Vd1.08

= 0.5× 1

1.08

µ1

1 + 0.07676+

1

1 + 0.10363

¶= 0.849 45

3-Yr bond price:

Vdd =1

1 + 0.08170Vud =

1

1 + 0.10635Vuu =

1

1 + 0.13843

Vd = 0.5×1

1 + 0.07676

µ1

1 + 0.08170+

1

1 + 0.10635

¶= 0.849

Vu = 0.5×1

1 + 0.10363

µ1

1 + 0.10635+

1

1 + 0.13843

¶= 0.807 46

V = 0.5× 1

1.08(0.849 + 0.807 46) = 0.766 88

4-Yr bond price:Value at t = 3

1

1 + 0.07943

1

1 + 0.09953

1

1 + 0.12473

1

1 + 0.15630

Value at t = 2

Vdd = 0.5×1

1 + 0.08170

µ1

1 + 0.07943+

1

1 + 0.09953

¶= 0.848 62



Vud = 0.5×1

1 + 0.10635

µ1

1 + 0.09953+

1

1 + 0.12473

¶= 0.812 84

Vuu = 0.5×1

1 + 0.13843

µ1

1 + 0.12473+

1

1 + 0.15630

¶= 0.770 33

Value at t = 1

Vd = 0.5×1

1 + 0.07676(0.848 62 + 0.812 84) = 0.771 51

Vu = 0.5×1

1 + 0.10363(0.812 84 + 0.770 33) = 0.717 26

Value at t = 0

V = 0.5× 1

1.08(0.771 51 + 0.717 26) = 0.689 25

5-Yr bond price:value at t = 4

1

1 + 0.07552

1

1 + 0.09084

1

1 + 0.109271

1 + 013143

1

1 + 0.15809value at t = 3

0.5× 1

1 + 0.07943

µ1

1 + 0.07552+

1

1 + 0.09084

¶= 0.855 32

0.5× 1

1 + 0.09953

µ1

1 + 0.09084+

1

1 + 0.10927

¶= 0.826 82

0.5× 1

1 + 0.12473

µ1

1 + 0.10927+

1

1 + 0.13143

¶= 0.793 67

0.5× 1

1 + 0.15630

µ1

1 + 0.13143+

1

1 + 0.15809

¶= 0.755 57

value at t = 2

Vdd = 0.5×1

1 + 0.08170(0.855 32 + 0.826 82) = 0.777 54

Vud = 0.5×1

1 + 0.10635(0.826 82 + 0.793 67) = 0.732 36

Vuu = 0.5×1

1 + 0.13843(0.793 67 + 0.755 57) = 0.680 43

value at t = 1

Vd = 0.5×1

1 + 0.07676(0.777 54 + 0.732 36) = 0.701 13

Vu = 0.5×1

1 + 0.10363(0.732 36 + 0.680 43) = 0.640 07

Value at t = 0

V = 0.5× 1

1.08(0.701 13 + 0.640 07) = 0.620 93



Problem 24.13.

I’m going to solve for only Tree #1.

DM Page 799 states that the volatility in Year 1 is the standard deviationof the natural log of the yield for that bond 1 year hence. So the volatility inYr 1 for an n-year bond is the standard deviation of the natural log of the yieldof an (n− 1)-year bond issued at t = 1.Yr-1 yield for 1-yr bond is unknown.

Yr-1 yield for 2-yr bond . The up yield of an 2− 1 = 1 year bond issued att = 1 is ru = 0.10362; the down yield is 0.07676.Using DM 24.48, we get:

0.5 lnrurd= 0.5 ln

0.10362

0.07676= 0.150 02

Yr-1 yield for a 3-yr bond. We first calculate the price of a 2-yr bond issuedat t = 1.

From the previous problem regarding the price of the 3-year bond, we know

Vd = 0.5×1

1 + 0.07676

µ1

1 + 0.08170+

1

1 + 0.10635

¶= 0.849

Vu = 0.5×1

1 + 0.10363

µ1

1 + 0.10635+

1

1 + 0.13843

¶= 0.807 46

ru = 0.807 46−1/2 − 1 = 0.112 86

rd = 0.849−1/2 − 1 = 0.08529 1

Volatility: 0.5 lnrurd= 0.5 ln

0.112 85

0.08529 1= 0.140 00

Yr-1 yield for a 4-yr bond. We first calculate the price of a 3-yr bond issuedat t = 1.From the previous problem about the price of a 4-yr bond,

Vd = 0.5×1

1 + 0.07676(0.848 62 + 0.812 84) = 0.771 51

rd = 0.771 51−1/3 − 1 = 9. 031 7× 10−2

Vu = 0.5×1

1 + 0.10363(0.812 84 + 0.770 33) = 0.717 26

rd = 0.717 26−1/3 − 1 = 0.117 14


0.117 14

9. 031 7× 10−2 = 0.130 02Yr-1 yield for a 5-yr bond. We first calculate the price of a 4-yr bond issued

at t = 1.From the previous problem about the price of a 5-yr bond,

Vd = 0.5×1

1 + 0.07676(0.777 54 + 0.732 36) = 0.701 13

rd = 0.701 13−1/4 − 1 = 9. 282 4× 10−2



Vu = 0.5×1

1 + 0.10363(0.732 36 + 0.680 43) = 0.640 07

ru = 0.640 07−1/4 − 1 = 0.118


0.118

9. 282 4× 10−2 = 0.119 99 = 0.12Don’t worry about the question "Can you unambiguously say that rates in

one tree are more volatile than the other?"

Problem 24.14.

Skip. This problem is not worth your time.

Problem 24.15.

I’ll solve for only Tree #1.time 0 1 2 3 4

0.158090.15630

0.13843 0.131430.10362 0.12473

0.08 0.10635 0.109270.07676 0.09953

0.08170 0.090840.07943

0.07552

Set the notional amount to $1. The payoff at each node is1

1 + rmax (0, r − 0.105)

Payoff:time 0 1 2 3 4

0.15809− 0.1051 + 0.15809

0.15630− 0.1051 + 0.15630

0.13843− 0.1051 + 0.13843

0.13143− 0.1051 + 0.13143

00.12473− 0.1051 + 0.12473

00.10635− 0.1051 + 0.10635

0.10927− 0.1051 + 0.10927

0 00 0

00



Vuuu =0.15630− 0.1051 + 0.15630

+1

1 + 0.15630

µ0.5× 0.15809− 0.105

1 + 0.15809+ 0.5× 0.13143− 0.105

1 + 0.13143

¶=

7. 428 977 9× 10−2

Vuud =0.12473− 0.1051 + 0.12473

+1

1 + 0.12473

µ0.5× 0.13143− 0.105

1 + 0.13143+ 0.5× 0.10927− 0.105

1 + 0.10927

¶=

2. 963 786 7× 10−2

Vdud = 0 +1

1 + 0.09953

µ0.5× 0.10927− 0.105

1 + 0.10927+ 0.5× 0

¶= 1. 750 465 4 ×

10−3

Vuu =0.13843− 0.1051 + 0.13843

+1

1 + 0.13843

¡0.5× 7. 428 977 9× 10−2 + 0.5× 2. 963 786 7× 10−2

¢=

7. 501 016 6× 10−2

Vud =0.10635− 0.1051 + 0.10635

+1

1 + 0.10635

¡0.5× 2. 963 786 7× 10−2 + 0.5× 1. 750 465 4× 10−3

¢=

1. 540 576 3× 10−2

Vdd =1

1 + 0.08170

¡0.5× 1. 750 465 4× 10−3 + 0.5× 0

¢= 8. 091 270 2×10−4

Vu =1

1 + 0.10362

¡0.5× 7. 501 016 6× 10−2 + 0.5× 1. 540 576 3× 10−2

¢=

4. 096 334 3× 10−2

Vd =1

1 + 0.07676

¡0.5× 1. 540 576 3× 10−2 + 0.5× 8. 091 270 2× 10−4

¢=

7. 529 482 0× 10−3

V =1

1 + 0.08

¡0.5× 4. 096 334 3× 10−2 + 0.5× 7. 529 482 0× 10−3

¢= 2.

245 038 2× 10−2

If the notional amount is $1, the interest cap is worth 2. 245 038 2× 10−2 att = 0. Since the notional amount is 250 million, the interest cap is worth att = 0

250× 2. 245 038 2× 10−2 = 5. 612 595 5 (million)


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