Options and Other Derivatives The One-Period Model The previous

Copyright © 2006 by Krzysztof Ostaszewski

Options and Other Derivatives

The One-Period ModelThe previous chapter introduced the following two methods:• Replicate the option payoffs with known securities, and calculate the

price of the replicating portfolio,• Use state price vectors to discount the option’s cash flows.

Rather than using the state-price vector to discount the cash flows, now wedo what’s referred in the book to as normalizing the cash flows, and thendiscount them back using a risk-neutral probability. Normalizing the cashflows means dividing the cash flows of one security by the cash flows ofanother security. The risk-neutral probabilities are the probabilities thatequate the discounted normalized cash flows of the security to its normalizedprice.

The following are known securities:

S1:

S2:

When S2 is normalized, it becomes:

S2:

100

105

105

100

110

95

100/100

110/105

95/105


We solve for the risk-neutral probability of an “up” move, q, by discountingthe normalized cash flows of S

2.

100100

= q ⋅110105

+ 1− q( ) ⋅95

105

, thus q =

23

.

The system is arbitrage-free if and only if risk-neutral probabilities (betweenzero and one) exist. Knowing the risk-neutral probabilities allows us tocalculate, for example, the price of a call option on S

2 with strike price 104.

We note that the option pays 6 in the “up” scenario and 0 in the “down”scenario. Its normalized cash flows are:

6

105c

100

0

105where c is the call price. We discount its cash flows with interest and therisk-neutral probabilities to get its price:

c100

=23

6105

+ 1−

23

0( )

and c = 3.81.


The Multiperiod ModelThis is the example in the book illustrating the use of arbitrage-free pricingin a multiperiod model. Here are the payoffs of two securities in all fourpossible states, in a two-period model:

S1

1,( ) S2

1,( ) S1

2,( ) S2

2,( )1

105 110 110 120

21105 110 100 100

3100 95 95 95

4100 95 100 90

We have the following evolution of S1:

110 105

100 100 95

100

100

Let us normalize S2 by dividing its cash flows by those of S

1 at each node:

120/110 110/105

100/100100100

95/95

95/100 90/100

We can use the normalized cash flows to determine the risk-neutral “up”probabilities at each node (they need not the same at each node). For theupper node at time 1 (denote this probability by q(1,1)):

110105

= q 1,1( )120110

+ 1− q 1,1( )( )100100

⇒ q 1,1( ) =1121


For the lower node at time 1, q(1,0):95100

= q 1,0( ) 9595

+ 1− q 1,0( )( ) 90100

⇒ q 1,0( ) =12

For the node at time 0, q(0,0):100100

= q 0,0( )110105

+ 1− q 0,0( )( ) 95100

⇒ q 0,0( ) =2141

We obtain unique solutions, and all of the risk-neutral probabilities arebetween zero and one, indicating that this is an arbitrage-free model. Now,we can use the risk-neutral probabilities we just found to calculate theArrow-Debreu securities prices. This is useful, because all securities arelinear combinations of Arrow-Debreu securities, and price is a linearoperator. For

1, we must go up at the first node, and then up again at the

second node. So the risk-neutral probability of getting to state 1 at time 2, inour risk-neutral notation, is:

Q1

( ) = q 0,0( )q 1,1( ) =2141

⋅1121

=1141

The remaining probabilities are:

Q2( ) = q 0,0( ) 1− q( ) 1,1( )( ) =

21

41⋅ 1−

11

21

=

10

41

Q3

( ) = 1− q 0,0( )( )q 1,0( ) = 1−2141

⋅

12

=1041

Q4( ) = 1− q 0,0( )( ) 1− q 1,1( )( ) = 1− 21

41

⋅ 1− 1

2

=

1041

To calculate the prices of the Arrow-Debreu securities we use thenormalized cash flows and discount utilizing the risk-neutral probabilities.For the Arrow-Debreu security for state 1 at time 2, its price is denoted by

2,1

( ) . It will have normalized payoffs:1/110

1/1050/100

2,1

( ) 0/95 0/100

0/100


Therefore:

2,1

( )100

= Q1

( ) 1110

⇒ 2,1

( ) =1141

100110

=

1041

.

The remaining Arrow-Debreu security values are:

2,2( )

100= Q

2( ) 1100

⇒ 2,2( ) = 10

41

100100

=

1041

,

2,3

( )100

= Q3

( ) 195

⇒ 2,3

( ) =1041

10095

=

200779

,

2,4

( )100

= Q4

( ) 1100

⇒ 2,4

( ) =1041

100100

=

1041

.

Knowing these values we can establish the price of any European option.The book asks us to price the value of a European call option on Asset 2with a strike price of 100 and expiring at time 2. Such an option will have apayoff of 20 in state 1 at time 2 and zero in all other states. Therefore theprice of such an option is

20 2,1

( ) = 201041

=

20041

= 4.878049

Another example in the book prices a derivative with a payoff equal to

S2

2( )( )2

. Squaring the time 2 values for S2, we obtain the following payoffs

for this derivative:

114,400

210,000

3 9,025

4 8,100

The price of this derivative is:14,400 2,

1( ) +10,000 2,

2( ) + 9,025 2,

3( ) + 8,100 2,

4( ) =

= 10,243.90


The Binomial Option Pricing ModelIn this model, in each time period, the asset price either goes up by a factorof u or down by a factor of d. Each time period is h years (if a year is a unitof time, which is common, as interest rates are quoted as annual rates) long,and there are N total time periods (T = Nh). The risk-free force of interest isr. To make this model arbitrage-free, we must assume: u > e rh > d . The

probability of an “up” move in the asset price is q =e rh − du − d

(this is exactly

the same formula as q =1+ i − d

u − d proved previously, except for the use of

the force of interest and possibly fractional time t).

European Call and Put OptionsLet m be the minimum number of upward (downward) moves such that thecall (put, respectively) option is in the money. Then m is the smallestinteger such that

m >ln

K

d NS 0( )

lnud

European call/put option prices:

c 0( ) = S 0( )Φ m;N ,q*( ) − Ke− rNhΦ m;N ,q( ) , and

p 0( ) = KerNh 1− Φ m; N,q( )( ) − S 0( ) 1− Φ m;N ,q *( )( ),where

Φ m;N ,q( ) =N!

j! N − j( )!j =m

N

∑ q j 1− q( )N − j

and q* = que rh .

Spring 2003 Notes for SoA Course 6 exam, Copyright © 2003 by Krzysztof Ostaszewski - 248 -

The continuous-time limit of the binomial model is the Black-Scholesmodel. If we select u = e h , then the distribution of stock price changes islognormal with mean r − 2 / 2( )T and variance 2T . As N gets large,under these parameter selections, the binomial model’s valuation willapproach the Black-Scholes and can be taken to be the same for sufficientlylarge N.

Binomial model parametersIn practical applications of the binomial model, it is worth noting that arecombining tree dramatically reduces the number of computations required.Also, one has to be careful in practice, as if parameter values are choseninappropriately, the model will no longer be arbitrage-free.

Recursive valuationOne can work the valuation tree backwards (right-to-left) to establish risk-neutral probabilities and values of all securities. In general:

Value at current node =

= (Value in up node ⋅ Pr up( ) + +Value in down node⋅ Pr down( ))

or

V i, j( ) = e−rh qV i +1, j +1( ) + 1− q( )V i + 1, j( )( ) .

In order to value any security, one uses trading strategies that replicate thosesecurities with those with available prices. We generally work recursively,solving for the replicating portfolio at each node

1i, j( ) = e−r i+1( )h

uV i + 1, j( ) − dV i + 1, j + 1( )u − d

2i, j( ) =

V i +1, J +1( ) − V i + 1, j( )u − d( )S i, j( )

2, a solution of the above, is often referred to as the delta of the derivative;

you will see why this is when you get to the section on delta hedging.


Dividends and other incomeRemember that the price of a stock is reduced by the amount of the dividendat the moment the dividend is paid. To incorporate the dividend paymentsinto the recursive model, the following formulas are used:

For dollar dividends

S i −1, j( ) = e− rh q S i, j +1( ) + D i, j + 1( )( ) + 1− q( ) S i, j( ) + D i, j( )( )( )S i, j +1( ) + D i, j + 1( ) = S i −1, j( )uS i, j( ) + D i, j( ) = S i −1, j( )d

Thus, when working recursively, you must add the dividend to the stockprice before discounting. For dividends, which are proportions of stockprice,

S i −1, j( ) = e− r−( ) h qS i −1, j( )u1+ 1− q( )S i −1, j( )d

1( ),

where: e h = 1+ h( ) , h( ) =D i, j( )S i, j( ) , u

1=

u1+

, d1

=d

1+,

q =e rh − du − d

. This is intuitively similar to assuming that the stock price

grows at a rate of r − . Note that the stock price tree will not recombinewhen a dividend is paid as a constant dollar amount; it will, however, it willrecombine when the dividend is a constant proportion of the stock value.

Put-call parity for dividend-paying assetsWe already know that for non-dividend paying stocks

c 0( ) − p 0( ) = S 0( )e− t − Ke − rT

The modifications to Black-Scholes model in order to accommodate adividend-paying stock are exactly the same as the modifications to the put-call parity.

Exotic derivativesThese are derivatives that go beyond the standard European put and call.Many exotic options are path dependent and therefore difficult to value witha binomial method, especially if they are American.


Main types of exotic derivatives:- Digital (a.k.a. Binary) OptionIts payoff depends on whether the underlying asset value is above or below afixed amount on the expiration date. Cash-or-nothing call (put) options pay afixed amount, X, if the underlying asset value is above (below, respectively)the exercise price K, and nothing otherwise. All-or-nothing call (put) optionspay the asset value if the terminal asset price is above (below) the exerciseprice at expiration, and nothing otherwise. One-touch all-or-nothing optionspay off if the underlying asset goes above (call) or below (put) the exerciseprice during the life of the option.

- Gap optionPayoff depends on whether the underlying asset value on the expiration dateis above or below a fixed amount that is different from that used for thepayoff. The payoff for a call option is S(T) – K if S(T) > H or zero otherwise.

- Asian optionPayoff depends on the average price of the underlying asset during the life ofthe option. Commonly used in equity-indexed products (e.g., equity-linkedannuities), foreign currency options, and interest rate options for hedgingpurposes.Asian options pay the difference between the average price and theexercise price provided this is a positive quantity. The exercise price can befixed such that the payoffs are:

Average price call: SAVG

− X( )+

Average price put: X − SAVG

( )+

The exercise price can also be the average (called “floating strike options”):

Average strike call: S T( ) − SAVG

( )+

Average strike put: SAVG

− S T( )( )+

- Lookback optionFor such an option, payoff depends on the underlying asset price atexpiration and also the minimum or maximum asset value attained during

the life of the option. Therefore, a lookback call pays S T( ) − Smin

( )+, and a

lookback put pays Smax

− S T( )( )+. These options are path dependent, and

thus quite difficult to price binomially. A variation is a high-water mark call


option, which pays Smax

− K( )+. Such an option is commonly found in

equity-indexed annuities.

- Barrier optionsFor such an option, its payoff depends on whether or not the value of theunderlying asset reaches a certain price level (a barrier) before expiration.Knockout options become worthless or pay a fixed rebate if the asset valuereaches a barrier but otherwise have payoffs identical to a standard option.- Down-and-out options become worthless or pay a fixed amount if the

asset value falls below a barrier.- Down-and-in options only come into existence if the asset price falls

below a barrier.- Up-and-out options become worthless or pay a fixed acount if the asset

value reaches a barrier.- Up-and-in options only come into existence if the asset price reaches a

barrier.- Double knockout options become worthless or pay a fixed amount if the

asset price reaches either of the lower or upper barrier.- A standard call option is the sum of a down-and-out call option and a

down-and-in call option with the same exercise price and barrier.

Options on the minimum or maximum (a.k.a. rainbow options)- Their payoffs depend on the values of several assets.- Options pay the maximum or minimum of the several assets.- A European option on the maximum of two assets is equivalent to

holding one of the assets plus a European option to exchange this assetfor the other one.

CliquetsA cliquet option is a series of standard call options that pay the annualincrease in the underlying asset.

QuantosShort for “quantity adjusted option”. Guaranteed exchange-rate contracts inwhich the payoff on a foreign currency derivative is converted to thedomestic currency at a fixed exchange rate.


Other exotics- Compound options = options on other options.- Chooser options are options, which permit the choice between buying a

call or put option so that the option holder can select the option theyrequire at expiration depending on which is more valuable.

- Spread options = payoffs reflect the difference between the values of twoassets.

American optionsPermit the holder to exercise at any time. This early exercise privilegeallows us to establish some bounds for the price of an American option.

Consider a call option. We have S t( ) ≥ C t( ) ≥ S t( ) − K( )+. If the option

cost were greater than the asset price, then you could sell the option, buy theasset, and keep the difference, using the asset to fulfill the obligation on theoption you sold. If the option cost were less than the intrinsic value, then youcould buy the option and immediately exercise it for the intrinsic value,making a riskless profit. In fact, the call price will be strictly greater than theintrinsic value at all times, except possibly maturity or just immediatelyprior to a dividend payment. Thus a rational investor will not exercise anAmerican call option between dividend payment dates, since such an optioncan always be sold for more money than could be obtained by exercising it.Therefore if there are no dividend payments then the price of an Americancall option on a single asset is equal to the price of a European call option onthe same asset. It is rational to exercise an American option when theintrinsic value immediately prior to the dividend payment is greater than thevalue of the option after the dividend payment; so exercise whenS t( ) > S* t −( ), where S * t −( ) − K = D t *( ) (the underlying price at whichthe intrinsic value of the option immediately prior to the dividend payment isequal to the option value immediately after the dividend payment). ForAmerican put options, there is always a critical asset price below which it isoptimal to exercise the American put option early.


Numerical Valuation of American options using the Binomial MethodFor an American put option, the value at each node is

P = Max K − S,e rh q K − Su( )+ + 1− q( ) K − Sd( )+( )( )This means that you should calculate the value recursively from right-to-left,but replace the option value at any node with the intrinsic value at that nodeif the intrinsic value at that node is greater than the recursive valuecalculated. Note the three possibilities for the put option at each node thataffect the decision to exercise early:

- If the underlying price in both “up” and “down” states is less than thestrike price of the put option (i.e., the put is in the money in bothscenarios) then it is optimal to exercise early.

- If the underlying price in both “up” and “down” states is greater thanthe strike price of the put option (i.e., the put is out of the money inboth scenarios) then it is not optimal to exercise early.

- If the option is in the money in the “down” state but not in the “up”state, then you have to compute the critical price, S*; this is the priceat which the investor is indifferent between exercising early andretaining the option. It can be found by solving the equation:

K − S* = e− rh q ⋅0 + 1− q( ) ⋅ K − S *d( )( ), i.e., the current intrinsicvalue has to equal the current value if unexercised. This solves to:

S* = K1− e−rh 1− q( )

1− e− rh 1− q( )d . The investor should exercise the option if

the actual asset price, S, is below the critical price, S*.The procedure for an American call option is similar, except the comparisonbetween the option’s value and it’s intrinsic value for the purposes ofsubstituting the greater of the two is made only at nodes at which a dividendis paid.

Numerical MethodsThey have become very important in finance. Reasons for that are:

- Security models have become more complex,- Types of securities and derivatives are also complex,- New developments in risk management technology, and- Regulation and practice mandate more analysis (e.g., VaR).


Some comments on numerical methods:- Even closed form or analytic solutions still require numerical

techniques,- Prices of some derivatives are governed by partial differential

equations, which require numerical techniques to solve,- Lattice methods are numerical in nature,- Monte Carlo, i.e., simulation, is numerical in nature.

Lattice models- Lattice models approximate the distribution of the underlying with a

discrete one,- Trinomial models have been shown to be more stable and more

accurate with half as many time intervals as a standard binomiallattice. But they require more calculation.

The trinomial model can be described as follows:“Up” movement by a factor u with probability q

u

“Middle” movement by a factor m with probability 1− qu− q

d

“Down” movement by a factor d with probability qd

The probabilities are computed by equating the first three moments of thelognormal distribution:

quu + 1− q

u− q

d( )m + q

dd = e rh

quu2 + 1− q

u− q

d( )m 2 + q

dd 2 = e2rh+ 2h

quu3 + 1− q

u− q

d( )m 3 + qdd 3 = e3rh+ 3 2h

Here the parameters are u = e h ,m = 1,d =1u

and is chosen to make

the model arbitrage-free. This produces a recombining tree.

Monte Carlo simulationSteps:

- Generate N values from the distribution of the ending asset value,- Compute the option value for each underlying value,- Discount this value at the risk-free rate,- Compute the average option value for the sample.

As N gets large, the answer will converge to the true value.


Variance reduction techniquesThey reduce the standard error more efficiently than running thousands ofsimulations:

- Antithetic variable technique: Average a pair of unbiased estimatorsto produce a new unbiased estimator whose variance is less than eitherof the individual estimators assuming that the two individualestimators are negatively correlated.

- Control variate method: Replace the problem under consideration witha similar but simpler problem that has an analytic solution. The twomust be highly correlated.

- Stratified sampling: Break the sampling region into pieces (calledstrata), then sample from the strata. You may also sample from onlysignificant strata: For example, if pricing an option, which is deeplyout of the money, don’t simulate asset prices that will only result in azero value for the option.

- Low-discrepancy methods: Use deterministic points that are asuniformly distributed as possible rather than random simulations.

- Path dependency: One can combine simulated paths into “bundles” toapproximate the value of American-style derivatives.

HedgingDealers who write derivative contracts must manage the risk. To manage therisk, one generally uses hedging, i.e., assuming positions that balance therisk by offsetting it. The risk is measured with various sensitivity statistics.They can be estimated by the use of a binomial model:

- Delta is the sensitivity of an option’s price to changes in the price of theunderlying:

∆ i, j( ) =V i + 1, j +1( ) − V i +1, j( )

u − d( )S i, j( )The hedge of the derivative uses the number of units of the underlying assetin the replicating portfolio equal to the delta. If a portfolio of assets isconstructed so that the overall portfolio has a delta of zero, then the portfoliois said to be delta neutral. In the Black-Scholes model of a call option, Delta= N d

1( ) .


- Gamma is the sensitivity of delta to changes in the price of theunderlying asset

Γ =∆ i + 1, j +1( ) − ∆ i + 1, j( )

u − d( )S i, j( )In the Black-Scholes model for a call option, Γ =

Φ d1

( )S

0T

.

- Theta is the sensitivity of the value of a derivative with respect to time,time-decay measure:

Θ =V i + 2, j + 1( ) − V i, j( )

2h- Rho is the sensitivity of the value of the derivative to interest rates.- Vega is the sensitivity of the value of the derivative to volatility of the

underlying.

The sensitivities can be estimated using simulation. The hedge portfolio’ssensitivities will change over time, and one must rebalance the hedgerepeatedly. There is a tradeoff between the safety of frequently rebalancingthe hedge portfolio and the transaction costs required to do so. Path-dependent derivatives present the biggest challenges in hedging, since thehedging requirement can often change dramatically, especially when theoption is nearing expiration and the option is close to the money or someother meaningful boundary (such as the barrier in a knockout option). Astatic hedge is one that is established on the initial date and not rebalanced; itis constructed in an attempt to replicate the path-dependent option’s payoffsat critical price levels (such as near the boundary).

Economic roles of derivatives:- Provide efficient method to achieve payoffs that are not readily available

otherwise,- Path-dependent options can hedge stochastic volatility,- Can help manage tax and regulatory considerations,- Lower transaction costs than purchasing replicating securities

individually.


Insurance productsLife insurance products contain a variety of guarantees, including maturityguarantees and return of premium guarantees upon death or withdrawal, ofwhich minimum interest rate guarantees are most significant. Theseembedded options are quite a challenge to price. They require inputregarding the possible values of the asset portfolio backing the products, andregarding the behavior of the policyholders. Equity-indexed products arealso common in some insurance companies (equity-linked annuities are anew product since 1995).

Pension PlansEmbedded option: participant is entitled to the maximum of two alternativebenefit amounts (such as a defined benefit formula or the accumulated valueof a set of contributions). A trinomial valuation of this benefit based on twovariables (salary growth rate and implicit crediting rate) found that thestandard deterministic valuation typically employed by actuaries mayundervalue this option as much as 35%.


Exercise 1.A European derivative pays the excess of the underlying asset price over$23, plus a dollar (the dollar is paid even if the excess is zero), in twomonths’ time. The current price of the underlying is $22, and the volatility ofthe underlying asset’s continuously compounded rate of return is 10% peryear. The continuously compounded risk free interest rate in the risk-neutralworld is 0.25% per month. Us a two-period binomial model to estimate theprice of this European derivative.

Solution. The basic formulae for the binomial model are (for each one-month period):

u = e h = e0.10

1

12 = 1.029288

d =1u

= 0.971545

q = e rh − du − d

= e0.0025 − 0.9715451.029288 − 0.971545

= 0.5361

e rh = e0.0025 =1.002503127

Using these parameters for the underlying we get this model:

$22.64 ⋅1.029288 = $23.31

$22 ⋅1.029288 = $22.64

$22 $22

$22 ⋅0.971545 = $21.37

$21.37 ⋅0.971545 = $20.76

The payoffs of the derivative are (note that 1 – 0.5361 = 0.4639):


$1.31

1

1.00253127$1.31⋅0.5361+ $1.00 ⋅0.4639( ) = $1.16918

$1.08 $1.00

1

1.00253127$1.00 ⋅0.5361+ $1.00 ⋅0.4639( ) = $0.99750

$1.00The price of this derivative is

$1.08 = 1

1.00253127$1.16918 ⋅0.5361+ $0.99750 ⋅0.4639( ).

Exercise 2.You are given the following binomial model for the value of the short-terminterest rate, risk-free over a one-year period:• One year from now this short-term interest rate is either:

r1u = r0 ⋅ 1+ γ( ) with probability 0.50, or

r1d =

r01+ γ

with probability 0.50,

where r0 in the initial short-term rate, and γ is a parameter. Theprobabilities given are risk-neutral probabilities.• Annualized volatility of this short-term interest rate is σ = 25%.• The current value of the short-term rate is 4%.A 2-year European (meaning that it pays only if the short-rate breaches thefloor at the end of two years) interest rate floor with a 3.5% strike level and anotional amount of 100 is issued. This derivative security will pay thedifference between 3.5% and then current short-term interest rate ascalculated for the 100 notional amount, if such difference is positive.Calculate the value of this interest rate floor.

Solution.Recall that in the binomial model u = eσ t and d = e−σ t . When t = 1, thisbecomes u = eσ and d = e−σ . In terms of the notation of this problem, youcan conclude that:

r1u

r1d = e2σ = 1+ γ( )2 .

We are given σ = 25%. This means that1+ γ = eσ = 1.28402542.

Given this, the short-term rate will be in two years:4% ⋅ 1.28402542( )2 = 6.59488508% with probability 0.25,4% with probability 0.50, and4% ⋅ 1.28402542( )−2 = 2.42612264% with probability 0.25.

Only the third outcome produces positive cash flows from the floor, which isthen worth

100 ⋅ 3.50% − 2.42612264%( ) = 100 ⋅1.07387736% = 1.07387736.This cash flow is paid with probability 0.25, and its expected present value isthe price of the floor. We use the risk-free rate over the next year as the ratefor discounting for that year. Therefore the value of the floor is:

1.07387736

1.04 ⋅ 1+ 0.041.28402542

⎛⎝⎜

⎞⎠⎟⋅0.25 ≈ 0.25034485.

Exercise 3.Now assume in the previous problem that the floor only pays the differencebetween 3.5% and the current short-term rate at the end of the first year, ifsuch difference is positive. What is the value of such one-year floor?

Solution.At the end of the first year, the short rate is

4% ⋅ e0.25 ≈ 5.136102% with probability 0.50, and4% ⋅ e−0.25 ≈ 3.115203% with probability 0.50.

Only the second outcome gives rise to a payment by the floor, with suchpayment being 0.38479687 on the 100 notional amount. Its expected presentvalue is

0.384796871.04

⋅0.50 ≈ 0.18499849.

If the floor were a two-year floor inclusive of both years, its total valuewould be

0.25034485 + 0.1849849 = 0.43534334.


Exercise 4.You are given the following securities:• 60-day Treasury Bill with face amount 1000.• 150-day Treasury Bill with face amount 1000 and current price 975.• Stock of ABCZ Corporation with current price of 24.7282442.• European call option on the ABCZ stock, with current price of 1, time toexercise of 60 days exactly, strike price of 30, for which the d1 parameter inthe Black-Scholes equation equals 0.70. The price of this option is equalexactly to the price given by the Black-Scholes equation.• European put option on the ABCZ stock, with time to exercise of 60 days,and strike price of 30. The price of this option is equal exactly to the pricegiven by the Black-Scholes equation. You can assume that put-call parityholds perfectly.• Futures contract on a 90-day Treasury Bill, time to delivery of 60 days,face amount of 1000, and current price of 984. You can assume that thefundamental relation betweenfutures and the underlying holds.You are also given the cumulative normal distribution values

z –1.4 –0.7 0 0.7 1.4N(z) 0.0808 0.242 0.5 0.758 0.9192

Calculate the current market price of the put option.

Solution. The fundamental relation between futures and underlying is:F = Se r T −t( )

.Here, the 90-day Treasury Bill futures go for 984, and the underlying, whichis the 150-day Treasury Bill, in 60 days deliverable as a 90-day Bill, goes for975. Therefore, the continuously-compounded risk-free rate, assuming, forsimplicity, 360 days in a year, must satisfy:

984 = 975e60

360r

,e60

360r

= 984975

= 328325

.

This means that to discount anything by 60 days, you multiply by 325/328,and to accumulate over 60 days, you multiply by 328/325.From put-call parity, we get:

1− P = 24.7282442 − 325328

30,

P = 1− 24.7282442 + 325328

30 =

= −23.7282442 + 29.725609756 == 5.997365556.


While the $30 call sells for $1, $30 put sells for $5.997365556. This shouldnot be in any way a surprise, the put has $5 intrinsic value.


Exercise 5. Using the same data as in problem No. 13, find the annualized volatility(standard deviation) of the rate of return of ABCZ stock. You may assumethat 1 year has 360 days, and the price of the call is determined by the Black-Scholes equation exactly, except that S, the price of the stock, is replaced bythe difference S – PV(Dividends paid on the stock during the life of theoption).

Solution.We have

d1

=ln

SPV X( )

+ 1

22t

t=

ln24.7282442

325328

30

+ 12

2 16

16

= 0.70

Therefore:

0.0833333 2 −0.2857738 - 0.1840629 = 0

=0.2857738 ± 0.28577382 − 4 0.0833333( ) -0.1840629( )

2 ⋅0.0833333= 3.98372953, or = −0.5544439 (can' t be).

Therefore, = 398.37%. Don’t think this answer is unreasonable, as youhave a call option deeply out of the money, and the said call still hassignificant value, $1. This can only happen if there is a reasonable chance ofstock exceeding $30 in price within sixty days, and for that to happen, thestock must move up by roughly 24% within 1/6 of a year.


Exercise 6.Using the same data as in the previous problem, demonstrate that a one-period market consisting of the 60-day T-bill, the put option, and the stock isarbitrage free over one period. You are additionally given that the period is60 days, and the stock price at the end of 60 days will be either 21, or 30, or36.

Solution.

$1 invested in the Treasury Bill will end up being $328325

no matter what

happens.$24.7282442 invested in stock of ABCZ ends up being $21, $30, or $36.

$5.997365556 invested in the put ends up being $9, $0, or $0, as the exerciseprice is $30.

The market matrices are:

S 0( )= 1 1 1[ ],S 1( ) =1.009231 0.849231 1.5006589

1.009231 1.213188 0

1.009231 1.455825 0

.

S(1) is an invertible matrix with determinant 0.4217175494. For the marketto be arbitrage-free, all you need is a state-price vector

1 2 3[ ]T,

i.e., solution to the system of equations:

1.0092311

+1.0092312+1.009231

3=1

0.8492311+1.213188

2+1.455825

3=1

1.50065891+ 0⋅

2+ 0⋅

3=1

Because S(1) is invertible, not only does a solution exist, it is unique, so thatthis market is complete as well, as long as the solution is positive. It is pretty

obvious that 1

= 11.5006589

= 0.66637395, and we get a new, simpler

system of equations by putting that value in the first two equations.

1.0092312

+1.0092313

= 0.32747491

1.2202012

+1.4602013

= 0.43409436


or1.220201

2+1.220201

3= 0.39365477

1.2202012

+1.4602013

= 0.43409436

2= 0.16666667,

3= 0.15781303. The market is arbitrage-free.


Exercise 7.Determine whether the market described in the previous problem iscomplete, and find its risk-neutral probabilities, if it is arbitrage-free.

Solution.The market is complete because it is arbitrage-free and the matrix S(1) isinvertible. Because we already solved for the state-price vector, the secondpart is almost a freebie.

1= 0.66637395,

2= 0.16666667,

3= 0.15781303

and therefore

p1

=1

1+ i( ) = 0.66637395328325

= 0.67252509

p2

=2

1+ i( ) = 0.16666667328325

= 0.16820514

p3

=3

1+ i( ) = 0.15781303328325

= 0.15926977

Documents

Options and Other Derivatives The One-Period Model The previous