Applied Probability and Mathematical Finance Theory 1 Introduction Can you answer the following questions and explain your answer logically? Example 1.1 (Flipping Coins [6]). Suppose

Applied Probabilityand Mathematical Finance Theory

Hiroshi Toyoizumi 1

January 16, 2006

1E-mail: [email protected]

Contents

1 Introduction 5

2 Basic Probability Theory 62.1 Why Probability? . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 Probability Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.3 Conditional Probability and Independence . . . . . . . . . . . . . . . 82.4 Random Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.5 Expectation, Variance and Standard Deviation . . . . . . . . . . . . . 102.6 Covariance and Correlation . . . . . . . . . . . . . . . . . . . . . . . 122.7 How to Make a Random Variable . . . . . . . . . . . . . . . . . . . . 132.8 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3 Normal Random Variables 163.1 What is Normal Random Variable? . . . . . . . . . . . . . . . . . . . 163.2 Lognormal Random Variables . . . . . . . . . . . . . . . . . . . . . 173.3 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

4 Useful Probability Theorems 224.1 The Law of Large Numbers . . . . . . . . . . . . . . . . . . . . . . . 224.2 Poisson Random Variables and the Law of Small Numbers . . . . . . 234.3 The Central Limit Theorem . . . . . . . . . . . . . . . . . . . . . . . 234.4 Useful Estimations . . . . . . . . . . . . . . . . . . . . . . . . . . . 244.5 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

5 Brownian Motions and Poisson Processes 285.1 Geometric Brownian Motions . . . . . . . . . . . . . . . . . . . . . 285.2 Discrete Time Tree Binomial Process . . . . . . . . . . . . . . . . . 305.3 Brownian Motions . . . . . . . . . . . . . . . . . . . . . . . . . . . 335.4 Poisson Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365.5 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

2

CONTENTS 3

6 Simulating Brownian Motions 386.1 Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

6.1.1 As an Advanced Calculator . . . . . . . . . . . . . . . . . . 386.2 Generating Random Variables . . . . . . . . . . . . . . . . . . . . . 396.3 Two-Dimensional Brownian Motions . . . . . . . . . . . . . . . . . . 39

6.3.1 Random Variable on Unit Circle . . . . . . . . . . . . . . . . 396.3.2 Generating Brownian Motion . . . . . . . . . . . . . . . . . 39

6.4 Geometric Brownian Motions . . . . . . . . . . . . . . . . . . . . . 416.4.1 Generating Bernouilli Random Variables . . . . . . . . . . . 416.4.2 Generating geometric brownian motions . . . . . . . . . . . . 41

7 Present Value Analysis 427.1 Interest Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 427.2 Continuously Compounded Interest Rates . . . . . . . . . . . . . . . 437.3 Present Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447.4 Rate of Return . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 457.5 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 477.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

8 Risk Neutral Probability and Arbitrage 488.1 Option to Buy Stocks . . . . . . . . . . . . . . . . . . . . . . . . . . 48

8.1.1 Risk Neutralization . . . . . . . . . . . . . . . . . . . . . . . 488.1.2 Arbitrage and Price of Option . . . . . . . . . . . . . . . . . 49

8.2 Duplication and Law of One Price . . . . . . . . . . . . . . . . . . . 508.3 Arbitrage Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 518.4 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 548.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

9 Black-Scholes Formula 559.1 Risk-neutral Tree Binomial Model . . . . . . . . . . . . . . . . . . . 559.2 Option Price on the Discrete Time . . . . . . . . . . . . . . . . . . . 569.3 Black-Scholes Model . . . . . . . . . . . . . . . . . . . . . . . . . . 579.4 Examples of Option Price via Black-Scholes Formula . . . . . . . . . 619.5 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 629.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

10 Delta Hedging Strategy 6310.1 Binomial Hedging Model . . . . . . . . . . . . . . . . . . . . . . . . 6310.2 Hedging in Black-Scholes Model . . . . . . . . . . . . . . . . . . . . 6710.3 Partial Derivative ∆ . . . . . . . . . . . . . . . . . . . . . . . . . . . 6810.4 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7010.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

4 CONTENTS

11 More Vanilla Options in Detail 7111.1 American Call Options . . . . . . . . . . . . . . . . . . . . . . . . . 7111.2 Put Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7211.3 Pricing American Put Options . . . . . . . . . . . . . . . . . . . . . 7311.4 Stock with Continuous Dividend . . . . . . . . . . . . . . . . . . . . 7711.5 Stock with Fixed-time-Dividend . . . . . . . . . . . . . . . . . . . . 7911.6 Forward and Futures Contract . . . . . . . . . . . . . . . . . . . . . 8011.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

Chapter 1

Introduction

Can you answer the following questions and explain your answer logically?

Example 1.1 (Flipping Coins [6]). Suppose you are flipping the coin twice. Can youfind any differences between the following probabilities?

1. the probability that the both flips land on heads given that the first flip lands onheads.

2. the probability that the both flips land on heads given that at least one of flipslands on heads.

Example 1.2 (Wait or Not Wait). Assume you would like to enter a building, but yourealized that you forgot your ID card to enter the building. You have been waiting foryour colleagues with ID card 10 minutes so far. Is it reasonable to think that you wouldenter the building more likely next 1 minute?

Example 1.3 (Easy Money). Imagine you are betting on a gamble. Here’s the bestscheme [2] to get easy money on gambling. When you lose, you bet twice as muchnext time. In that way, even if you lose one time, you can get the loss next time. So,you can win eventually. Do you use this scheme?

These questions are typical for decision making under uncertainty. In order toanswer the above question, we need to study probability theory. The aim of this lectureis to learn

1. the basic of applied probability,

2. the basic of finance theory,

3. learn how to use them to answer those questions in above examples.

This handout is based on the Text:

• Sheldon M. Ross, An Elementary Introduction to Mathematical Finance: Op-tions and Other Topics[6].

Note that the items covered in [6] are not complete enough to cover all of thislecture. You can find other example of these questions in [2].

5

Chapter 2

Basic Probability Theory

2.1 Why Probability?Example 2.1. Here’s examples where we use probability:

• Lottery.

• Weathers forecast.

• Gamble.

• Baseball,

• Life insurance.

• Finance.

Since our intuition sometimes leads us mistake in those random phenomena, weneed to handle them using extreme care in rigorous mathematical framework, calledprobability theory. (See Exercise 2.1).

2.2 Probability SpaceBe patient to learn the basic terminology in probability theory. To determine the prob-abilistic structure, we need a probability space, which is consisted by a sample space,a probability measure and a family of (good) set of events.

Definition 2.1 (Sample Space). The set of all events is called sample space, and wewrite it as Ω. Each element ω ∈Ω is called an event.

Example 2.2 (Lottery). Here’s the example of Lottery.

• The sample space Ω is first prize, second prize,..., lose.

6

2.2. PROBABILITY SPACE 7

• An event ω can be first prize, second prize,..., lose, and so on.

Sometimes, it is easy to use sets of events in sample space Ω.

Example 2.3 (Sets in Lottery). The followings are examples in Ω of Example 2.2.

W = win= first prize, second prize,..., sixth prize (2.1)L = lose (2.2)

Thus, we can say that “what is the probability of win?”, instead of saying “what isthe probability that we have either first prize, second prize,..., or sixth prize?”.

Definition 2.1 (Probability measure). The probability of A, P(A), is defined for eachset of the sample space Ω, if the followings are satisfyed:

1. 0≤ P(A)≤ 1 for all A⊂Ω.

2. P(Ω) = 1.

3. For any sequence of mutually exclusive A1,A2...

P(∞⋃

i=1

Ai) =∞

∑i=1

P(Ai). (2.3)

In addition, P is said to be the probability measure on Ω.

Mathematically, all function f which satisfies Definition 2.1 can regarded as prob-ability. Thus, we need to be careful to select which function is suitable for probability.

Example 2.4 (Probability Measures in Lottery). Suppose we have a lottery such as 10first prizes, 20 second prizes · · · 60 sixth prizes out of total 1000 tickets, then we havea probability measure P defined by

P(n) = P(win n-th prize) =n

100(2.4)

P(0) = P(lose) =79100

. (2.5)

It is easy to see that P satisfies Definition 2.1. According to the definition P, we cancalculate the probability on a set of events:

P(W ) = the probability of win= P(1)+P(2)+ · · ·+P(6)

=21

100.

Of course, you can cheat your customer by saying you have 100 first prizes insteadof 10 first prizes. Then your customer might have a different P satisfying Definition2.1. Thus it is pretty important to select an appropriate probability measure. Selectingthe probability measure is a bridge between physical world and mathematical world.Don’t use wrong bridge!

8 CHAPTER 2. BASIC PROBABILITY THEORY

Remark 2.1. There is a more rigorous way to define the probability measure. Indeed,Definition 2.1 is NOT mathematically satisfactory in some cases. If you are familiarwith measure theory and advanced integral theory, you may proceed to read [3].

2.3 Conditional Probability and IndependenceNow we introduce the most uselful and probably most difficult concepts of probabilitytheory.

Definition 2.2 (Conditional Probability). Define the probability of B given A by

P(B | A) =P(B & A)

P(A)=

P(B∩A)P(A)

. (2.6)

We can use the conditional probability to calculate complex probability. It is ac-tually the only tool we can rely on. Be sure that the conditional probability P(B|A) isdifferent with the regular probability P(B).

Example 2.5 (Lottery). Let W = win and F = first prize in Example 2.4. Thenwe have the conditional probability that

P(F |W ) = the probability of winning 1st prize given you win the lottery

=P(F ∩W )

P(W )=

P(F)P(W )

=10/1000210/1000

=1

21

6= 101000

= P(F).

Remark 2.2. Sometimes, we may regard Definition 2.2 as a theorem and call Bayserule. But here we use this as a definition of conditional probability.

Definition 2.3 (Independence). Two sets of events A and B are said to be independentif

P(A&B) = P(A∩B) = P(A)P(B) (2.7)

Theorem 2.1 (Conditional of Probability of Independent Events). Suppose A and B areindependent, then the conditional probability of B given A is equal to the probability ofB.

Proof. By Definition 2.2, we have

P(B | A) =P(B∩A)

P(A)=

P(B)P(A)P(A)

= P(B),

where we used A and B are independent.

2.4. RANDOM VARIABLES 9

Example 2.6 (Independent two dices). Of course two dices are independent. So

P(The number on the first dice is even while the one on the second is odd)= P(The number on the first dice is even)P(The number on the second dice is odd)

=12· 1

2.

Example 2.7 (Dependent events on two dice). Even though the two dices are indepen-dent, you can find dependent events. For example,

P(The sum of two dice is even while the one on the second is odd)6= P(The sum of two dice is even)P(The number on the second dice is odd).

See Exercise 2.4 for the detail.

2.4 Random VariablesThe name random variable has a strange and stochastic history1. Although its fragilehistory, the invention of random variable certainly contribute a lot to the probabilitytheory.

Definition 2.4 (Random Variable). The random variable X = X(ω) is a real-valuedfunction on Ω, whose value is assigned to each outcome of the experiment (event).

Remark 2.3. Note that probability and random variables is NOT same! Random vari-ables are function of events while the probability is a number. To avoid the confusion,we usually use the capital letter to random variables.

Example 2.8 (Lottery). A random variable X can be designed to formulate a lottery.

• X = 1, when we get the first prize.

• X = 2, when we get the second prize.

Example 2.9 (Bernouilli random variable). Let X be a random variable with

X =

1 with probability p.0 with probability 1− p.

(2.8)

for some p ∈ [0,1]. The random variable X is said to be a Bernouilli random variable.

1J. Doob quoted in Statistical Science. (One of the great probabilists who established probability asa branch of mathematics.) While writing my book [Stochastic Processes] I had an argument with Feller.He asserted that everyone said “random variable” and I asserted that everyone said “chance variable.” Weobviously had to use the same name in our books, so we decided the issue by a stochastic procedure. Thatis, we tossed for it and he won.


Sometimes we use random variables to indicate the set of events. For example,instead of saying the set that we win first prize, ω ∈Ω : X(ω) = 1, or simply X = 1.

Definition 2.5 (Probability distribution). The probability distribution function F(x) isdefined by

F(x) = PX ≤ x. (2.9)

The probability distribution function fully-determines the probability structure ofa random variable X . Sometimes, it is convenient to consider the probability densityfunction instead of the probability distribution.

Definition 2.6 (probability density function). The probability density function f (t) isdefined by

f (x) =dF(x)

dx=

dPX ≤ xdx

. (2.10)

Sometimes we use dF(x) = dPX ≤ x = P(X ∈ (x,x + dx]) even when F(x) hasno derivative.

Lemma 2.1. For a (good) set A,

PX ∈ A=∫

AdPX ≤ x=

∫A

f (x)dx. (2.11)

2.5 Expectation, Variance and Standard DeviationLet X be a random variable. Then, we have some basic tools to evaluate random vari-able X . First we have the most important measure, the expectation or mean of X .

Definition 2.7 (Expectation).

E[X ] =∫

∞

−∞

xdPX ≤ x=∫

∞

−∞

x f (x)dx. (2.12)

Remark 2.4. For a discrete random variable, we can rewrite (2.12) as

E[X ] = ∑n

xnP[X = xn]. (2.13)

Lemma 2.2. Let (Xn)n=1,...,N be the sequence of random variables. Then we canchange the order of summation and the expectation.

E[X1 + · · ·+XN ] = E[X1]+ · · ·+E[XN ] (2.14)

2.5. EXPECTATION, VARIANCE AND STANDARD DEVIATION 11

Proof. See Exercise 2.6.

E[X ] gives you the expected value of X , but X is fluctuated around E[X ]. So weneed to measure the strength of this stochastic fluctuation. The natural choice may beX −E[X ]. Unfortunately, the expectation of X −E[X ] is always equal to zero. Thus,we need the variance of X , which is indeed the second moment around E[X ].

Definition 2.8 (Variance).

Var[X ] = E[(X −E[X ])2]. (2.15)

Lemma 2.3. We have an alternative to calculate Var[X ],

Var[X ] = E[X2]−E[X ]2. (2.16)


Unfortunately, the variance Var[X ] has the dimension of X2. So, in some cases, itis inappropriate to use the variance. Thus, we need the standard deviation σ [X ] whichhas the order of X .

Definition 2.9 (Standard deviation).

σ [X ] = (Var[X ])1/2. (2.17)

Example 2.10 (Bernouilli random variable). Let X be a Bernouilli random variablewith P[X = 1] = p and P[X = 0] = 1− p. Then we have

E[X ] = 1p+0(1− p) = p. (2.18)

Var[X ] = E[X2]−E[X ]2 = E[X ]−E[X ]2 = p(1− p), (2.19)

where we used the fact X2 = X for Bernouille random variables.

In many cases, we need to deal with two or more random variables. When theserandom variables are independent, we are very lucky and we can get many useful result.Otherwise...

Definition 2.2. We say that two random variables X and Y are independent when thesets X ≤ x and Y ≤ y are independent for all x and y. In other words, when X andY are independent,

P(X ≤ x,Y ≤ y) = P(X ≤ x)P(Y ≤ y) (2.20)

Lemma 2.4. For any pair of independent random variables X and Y , we have


• E[XY ] = E[X ]E[Y ].

• Var[X +Y ] = Var[X ]+Var[Y ].

Proof. Extending the definition of the expectation, we have a double integral,

E[XY ] =∫

xydP(X ≤ x,Y ≤ y).

Since X and Y are independent, we have P(X ≤ x,Y ≤ y) = P(X ≤ x)P(Y ≤ y). Thus,

E[XY ] =∫

xydP(X ≤ x)dP(Y ≤ y)

=∫

xdP(X ≤ x)∫

ydP(X ≤ y)

= E[X ]E[Y ].

Using the first part, it is easy to check the second part (see Exercise 2.9.)

Example 2.11 (Binomial random variable). Let X be a random variable with

X =n

∑i=1

Xi, (2.21)

where Xi are independent Bernouilli random variables with the mean p. The randomvariable X is said to be a Binomial random variable. The mean and variance of X canbe obtained easily by using Lemma 2.4 as

E[X ] = np, (2.22)Var[X ] = np(1− p). (2.23)

2.6 Covariance and CorrelationWhen we have two or more random variables, it is natural to consider the relation ofthese random variables. But how? The answer is the following:

Definition 2.10 (Covariance). Let X and Y be two (possibly not independent) randomvariables. Define the covariance of X and Y by

Cov[X ,Y ] = E[(X −E[X ])(Y −E[Y ])]. (2.24)

Thus, the covariance measures the multiplication of the fluctuations around theirmean. If the fluctuations are tends to be the same direction, we have larger covariance.

2.7. HOW TO MAKE A RANDOM VARIABLE 13

Example 2.12 (The covariance of a pair of Binomial random variables). Let X1 and X2be the independent Binomial random variables with the same parameter n and p. Thecovariance of X1 and X2 is

Cov[X1,X2] = E[X1X2]−E[X1]E[X2] = 0,

since X1 and X2 are independent. Thus, more generally, if the two random variablesare independent, their covariance is zero. (The converse is not always true. Give someexample!)

Now, let Y = X1 +X2. How about the covariance of X1 and Y ?

Cov[X1,Y ] = E[X1Y ]−E[X1]E[Y ]= E[X1(X1 +X2)]−E[X1]E[X1 +X2]

= E[X21 ]−E[X1]2

= Var[X1] = np(1− p) > 0.

Thus, the covariance of X1 and Y is positive as can be expected.

It is easy to see that we have

Cov[X ,Y ] = E[XY ]−E[X ]E[Y ], (2.25)

which is sometimes useful for calculation. Unfortunately, the covariance has the orderof XY , which is not convenience to compare the strength among different pair of ran-dom variables. Don’t worry, we have the correlation function, which is normalized bystandard deviations.

Definition 2.11 (Correlation). Let X and Y be two (possibly not independent) randomvariables. Define the correlation of X and Y by

ρ[X ,Y ] =Cov[X ,Y ]σ [X ]σ [Y ]

. (2.26)

Lemma 2.5. For any pair of random variables, we have

−1≤ ρ[X ,Y ]≤ 1. (2.27)

Proof. See Exercise 2.11

2.7 How to Make a Random VariableSuppose we would like to simulate a random variable X which has a distribution F(x).The following theorem will help us.


Theorem 2.2. Let U be a random variable which has a uniform distribution on [0,1],i.e

P[U ≤ u] = u. (2.28)

Then, the random variable X = F−1(U) has the distribution F(x).

Proof.

P[X ≤ x] = P[F−1(U)≤ x] = P[U ≤ F(x)] = F(x). (2.29)

2.8 ReferencesThere are many good books which useful to learn basic theory of probability. Thebook [5] is one of the most cost-effective book who wants to learn the basic appliedprobability featuring Markov chains. It has a quite good style of writing. Those whowant more rigorous mathematical frame work can select [3] for their starting point. Ifyou want directly dive into the topic like stochatic integral, your choice is maybe [4].

2.9 ExercisesExercise 2.1. Find an example that our intuition leads to mistake in random phenom-ena.

Exercise 2.2. Define a probability space according to the following steps.

1. Take one random phenomena, and describe its sample space, events and proba-bility measure

2. Define a random variable of above phenomena

3. Derive the probability function and the probability density.

4. Give a couple of examples of set of events.

Exercise 2.3. Explain the meaning of (2.3) using Example 2.2

Exercise 2.4. Check P defined in Example 2.4 satisfies Definition 2.1.

Exercise 2.5. Calculate the both side of Example 2.7. Check that these events aredependent and explain why.

Exercise 2.6. Prove Lemma 2.2 and 2.3 using Definition 2.7.

2.9. EXERCISES 15

Exercise 2.7. Prove Lemma 2.4.

Exercise 2.8. Let X be the Bernouilli random variable with its parameter p. Draw thegraph of E[X ], Var[X ], σ [X ] against p. How can you evaluate X?

Exercise 2.9. Prove Var[X +Y ] =Var[X ]+Var[Y ] for any pair of independent randomvariables X and Y .

Exercise 2.10 (Binomial random variable). Let X be a random variable with

X =n

∑i=1

Xi, (2.30)

where Xi are independent Bernouilli random variables with the mean p. The randomvariable X is said to be a Binomial random variable. Find the mean and variance of X .

Exercise 2.11. Prove for any pair of random variables, we have

−1≤ ρ[X ,Y ]≤ 1. (2.31)

Chapter 3

Normal Random Variables

Normal random variables are important tool to analyze a series of independent randomvariables.

3.1 What is Normal Random Variable?Let’s begin with the definition of normal random variables.

Definition 3.1 (Normal random variable). Let X be a random variable with its proba-bility density function

dPX ≤ x= f (x)dx =1

(2π)1/2σe−(x−µ)2/2σ2

dx, (3.1)

for some µ and σ . The random variable is called the normal random variable with theparameters µ and σ .

Theorem 3.1 (Mean and variance of normal random variables). Let X be a normalrandom variable with the parameters µ and σ . Then, we have the mean

E[X ] = µ, (3.2)

and the variance

Var[X ] = σ2. (3.3)

Proof.

Definition 3.2 (Standard normal random variable). Let X be a normal random variablewith µ = 0 and σ = 1. The random variable is called the standard normal randomvariable.

16

3.2. LOGNORMAL RANDOM VARIABLES 17

Lemma 3.1. Let X be a normal random variable with its mean µ and standard devia-tion σ . Set Z = (X −µ)/σ . Then, Z is the standard normal random variable.


Theorem 3.2. Let (Xi)i=1,2,...,n are independent normal random variables with its meanµi and standard deviation σi. Then the sum of these random variables X = ∑

ni=1 Xi is

again a normal random variable with

µ =n

∑i=1

µi, (3.4)

σ2 =

n

∑i=1

σ2i . (3.5)

Proof. We just prove X satisfies (3.4) and (3.5). By Lemma 2.2,

µ = E[X ] = E

[n

∑i=1

Xi

]=

n

∑i=1

E [Xi] =n

∑i=1

µi.

Also, by Lemma 2.4, we have

σ2 = E[X ] = Var

[n

∑i=1

Xi

]=

n

∑i=1

Var [Xi] =n

∑i=1

σ2i .

We can prove that X is a normal random variable by using so-called characteristicfunction method (or Fourier transform).

So, it is very comfortable to be in the world of normal random variables. Veryclosed!

3.2 Lognormal Random VariablesDefinition 3.3 (lognormal random variable). The random variable Y is said to be log-normal if log(Y ) is a normal random variable.

Thus, a lognormal random variable can be expressed as

Y = eX , (3.6)

where X is a normal random variable. Lognormal random variables plays a measurerole in finance theory!

18 CHAPTER 3. NORMAL RANDOM VARIABLES

Theorem 3.3 (lognormal). If X is a normal random variable having the mean µ andthe standard deviation σ2, the lognormal random variable Y = eX has the mean andthe variance as

E[Y ] = eµ+σ2/2, (3.7)

Var[Y ] = e2µ+2σ2 − e2µ+σ2. (3.8)

It is important to see that although the mean of the lognormal random variable issubjected to not only the mean of the original normal random variable but also thestandard deviation.

Proof. Let us assume X is the standard normal random variable, for a while. Let m(t)be the moment generation function of X , i.e.,

m(t) = E[etX ]. (3.9)

Then, by differentiate the left hand side and setting t = 0, we have

m′(0) =ddt

m(t)|t=0 = E[XetX ]|t=0 = E[X ]. (3.10)

Further, we have

m′′(0) = E[X2].

On the other hand, since X is the standard normal random variable, we have

m(t) = E[etX ] =1√2π

∫∞

−∞

etxe−x2/2dx.

Since tx− x2/2 = t2− (x− t)2/2, we have

m(t) =1√2π

et2/2∫

∞

−∞

e−(x−t)2/2dx,

where the integrand of the right hand side is nothing but the density of the normalrandom variable N(t,1). Thus,

m(t) = et2/2.

More generally, when X is a normal random variable with N(µ,σ), we can obtain

m(t) = E[etX ] = eµt+σ2t2/2. (3.11)

(See exercise 3.6.) Since Y = eX , we have

E[Y ] = m(1) = eµ+σ2/2, (3.12)

3.2. LOGNORMAL RANDOM VARIABLES 19

and

E[Y 2] = m(2) = e2µ+2σ2. (3.13)

Thus,

Var[Y ] = E[Y 2]−E[Y ]2 = e2µ+2σ2 − e2µ+σ2. (3.14)

Let S(n) be the price of a stock at time n. Let Y (n) be the growth rate of the stock,i.e.,

Y (n) =S(n)

S(n−1)(3.15)

In mathematical finance, it is commonly assumed that Y (n) is independent and identi-cally distributed as the lognormal random variable. Taking log on both side of (3.15),then we have

logS(n) = logS(n−1)+X(n), (3.16)

where if X(n) = logY (n) is regarded as the error term and normally distributed, theabove assumption is validated.

Example 3.1 (Stock price rises in two weeks in a row). Suppose Y (n) is the growthrate of a stock at the n-th week, which is independent and lognormally distributed withthe parameters µ and σ . We will find the probability that the stock price rises in twoweeks in a row.

First, we will estimate Pthe stock rises. Since it is equivalent that y > 1 andlogy > 0, we have

Pthe stock rises= PS(1) > S(0)

= P

S(1)S(0)

> 1

= P

log(

S(1)S(0)

)> 0

= PX > 0

where X = logY (1) is N(µ,σ), and we can define

Z =X −µ

σ, (3.17)

as the standard normal distribution. Hence, we have

PS(1) > S(0)= P

X −µ

σ>

0−µ

σ

= PZ >−µ/σ= PZ < µ/σ,

20 CHAPTER 3. NORMAL RANDOM VARIABLES

where we used the symmetry of the normal distribution (see exercise 3.1).Now we consider the probability of two consecutive stock price rise. Since Y (n) is

assumed to be independent, we have

Pthe stock rises two week in a row= PY (1) > 0,Y (2) > 0= PY (1) > 0PY (2) > 0= PZ < µ/σ2

3.3 References

3.4 ExercisesExercise 3.1. Prove the symmetry of the normal distribution. That is, let X be thenormal distribution with the mean µ and the standard deviation σ , then for any x, wehave

PX >−x= PX < x. (3.18)

Exercise 3.2 (moment generating function). Let X be a random variable. The functionm(t) = E[etX ] is said to be the moment generating function of X .

1. Prove E[X ] = m′(0).

2. Prove E[Xn] = dn

dtn m(t)|t=0.

3. Rewrite the variance of X using m(t).

Exercise 3.3. Let X be a normal random variable with the parameters µ = 0 and σ = 1.

1. Find the moment generating function of X .

2. By differentiation, find the mean and variance of X .

Exercise 3.4. Use Microsoft Excel to draw the graph of probability distribution f (x) =dPX ≤ x/dx of normal random variable X with

1. µ = 5 and σ = 1,

2. µ = 5 and σ = 2,

3. µ = 4 and σ = 3.

3.4. EXERCISES 21

What can you say about these graphs, especially large x? Click help in your Excel tofind the appropriate function. Of course, it won’t help you to find the answer though...

Exercise 3.5. Let X be a normal random variable with its mean µ and standard devia-tion σ . Set Z = (X −µ)/σ . Prove Z is the standard normal random variable.

Exercise 3.6. Verify (3.11) by using Lemma 3.1

Exercise 3.7. Let Y = eX be a lognormal random variable where X is N(µ,σ). FindE[Y ] and Var[Y ].

Chapter 4

Useful Probability Theorems

4.1 The Law of Large NumbersIn many cases, we need to evaluate the average of large number of independent andidentically-distributed random variables. Perhaps, you can intuitively assume the aver-age will be approximated by its mean. This intuition can be validated by the followingtheorems.

Theorem 4.1 (Weak law of large numbers). Let X1,X2, .... be an i.i.d. sequence ofrandom variables with

µ = E[Xn]. (4.1)

Let Sn = ∑ni=1 Xi. Then, for all ε > 0,

P|Sn/n−µ|> ε→ 0 as n→ ∞. (4.2)

Or, if we take sufficiently large number of samples, the average will be close to µ withhigh probability.

You can say, “OK. I understand, for the large sample, we can expect the averagewill be close to the mean most of the times. So, it might be possible to have someexceptions...” In that case, we have another answer.

Theorem 4.2 (Strong law of large numbers). Let X1,X2, .... be an i.i.d. sequence ofrandom variables with

µ = E[Xn]. (4.3)

With probability 1, we have

Sn

n→ µ as n→ ∞. (4.4)

So, now we cay say that almost all trials, the average will be converges to µ .

Example 4.1. Gambler example

22

4.2. POISSON RANDOM VARIABLES AND THE LAW OF SMALL NUMBERS23

4.2 Poisson Random Variables and the Law of SmallNumbers

Compare to the the theorems of large numbers, the law of small numbers are lessfamous. But sometimes, it gives us great tool to analyze stochastic events. First of all,we define Poisson random variable.

Definition 4.1 (Poisson random variable). A random variable N is said to be a Poissonrandom variable, when

PN = n=(λ )n

n!e−λ , (4.5)

where λ is a constant equal to its mean E[N].

Theorem 4.3. Let N be a Poisson random variable.

Var[N] = λ . (4.6)

Theorem 4.4 (The law of Poisson small number). The number of many independentrare events can be approximated by a Poisson random variable.

Example 4.2. Let N be the number of customers arriving to a shop. If each customervisits this shop independently and its frequency to visit there is relatively rare, then Ncan be approximated by a Poisson random variable.

4.3 The Central Limit TheoremLet X1,X2, .... be an i.i.d. sequence of random variables with

µ = E[Xn], (4.7)

σ2 = Var[Xn]. (4.8)

Now we would like to estimate the sum of this random variables, i.e.

Sn =n

∑i=1

Xi. (4.9)

Theorem 4.5 (Central limit theorem). For a large n, we have

P

Sn−nµ

σ√

n≤ x

≈Φ(x), (4.10)

where Φ(x) is the distribution function of the standard normal distribution N(0,1). Or

Sn ∼ N(nµ,σ√

n). (4.11)

24 CHAPTER 4. USEFUL PROBABILITY THEOREMS

The central limit theorem indicate that no matter what the random variable Xi islike, the sum Sn can be regarded

Instead of stating lengthy and technically advanced proof of laws of large numbersand ccentral limit theorem, we give some examples of how Sn converges to a Normalrandom variable.

Example 4.3 (Average of Bernouilli Random Variables). Let Xi be i.i.d Bernoulli ran-dom variables with p = 1/2. Suppose we are going to evaluate the sample average Aof Xi’s;

A =1n

n

∑i=1

Xi. (4.12)

Of course, we expect that A is close to the mean E[Xi] = 1/2 but how close?Figure 4.1 shows the histogram of 1000 different runs of A with n = 10. Since

n = 10 is relatively small samples, we have large deviation from the expected mean1/2. On the other hand, when we have more samples, the sample average A tends tobe 1/2 as we see in Figure 4.2 and 4.3, which can be a demonstration of Weak Law ofLarge Numbers (Theorem 4.1).

As we see in in Figure 4.1 to 4.3, we may still have occasional high and low av-erages. How about individual A for large sample n. Figure 4.4 shows the sample pathlevel convergence of the sample average A to E[X ] = 1/2, as it can be expected fromStrong Law of Large Numbers (Theorem 4.2).

Now we know that A converges to 1/2. Further, due to Central Limit Theorem(Theorem 4.5), magnifying the histogram, the distribution can be approximate by Nor-mal distribution. Figure 4.5 show the detail of the histogram of A. The histogram isquite similar to the corresponding Normal distribution.

4.4 Useful EstimationsWe have the following two estimation of the distribution, which is sometimes veryuseful.

Theorem 4.6 (Markov’s Inequality). Let X be a nonnegative random variable, then wehave

PX ≥ a ≤ E[X ]/a, (4.13)

for all a > 0.

Proof. Let IA be the indicator function of the set A. It is easy to see

aIX≥a ≤ X . (4.14)

Taking expectation on the both side, we have

aPX ≥ a ≤ E[X ]. (4.15)

4.4. USEFUL ESTIMATIONS 25

0.2 0.4 0.6 0.8 1

2

4

6

8

Figure 4.1: Histogram of the sample average A = 1n ∑

ni=1 Xi when n = 10, where Xi is a

Bernouilli random variable with E[Xi] = 1/2.

0.2 0.4 0.6 0.8 1

2

4

6

8

Figure 4.2: The sample average A when n = 100.

0.2 0.4 0.6 0.8 1

2.5

5

7.5

10

12.5

15

17.5

Figure 4.3: The sample average A when n = 1000.

26 CHAPTER 4. USEFUL PROBABILITY THEOREMS

20 40 60 80 100

0.2

0.4

0.6

0.8

1

Figure 4.4: The sample path of the sample average A = 1n ∑

ni=1 Xi up to n = 100, where

Xi is a Bernouilli random variable with E[Xi] = 1/2.

0.46 0.48 0.5 0.52 0.54

5

10

15

20

25

30

Figure 4.5: The detailed histgram of the sample average A = 1n ∑

ni=1 Xi when n = 10,

where Xi is a Bernouilli random variable with E[Xi] = 1/2. The solid line is the corre-sponding Normal distribution.

4.5. REFERENCES 27

Theorem 4.7 (Chernov Bounds). Let X be the random variable with moment generat-ing function M(t) = E[etX ]. Then, we have

PX ≥ a ≤ e−taM(t) for all t > 0, (4.16)

PX ≤ a ≥ e−taM(t) for all t < 0. (4.17)

Proof. For t > 0 we have

PX ≥ a= PetX ≥ eta. (4.18)

Using Theorem 4.6,

PetX ≥ eta ≤ e−taE[etX ]. (4.19)

Thus, we have the result.

4.5 ReferencesThe proofs omitted in this chapter can be found in those books like ITO KIYOSHI andDurret [3].

4.6 ExercisesExercise 4.1. Something here!

Chapter 5

Brownian Motions and PoissonProcesses

Brownian Motions and Poisson Processes are the most useful and practical tools tounderstand stochastic processes appeared in the real world.

5.1 Geometric Brownian MotionsDefinition 5.1 (Geometric Brownian motion). We say S(t) is a geometric Brownianmotion with the drift parameter µ and the volatility parameter σ if for any y≥ 0,

1. the growth rate S(t + y)/S(t) is independent of all history of S up to t, and

2. log(S(t +y)/S(t)) is a normal random varialble with its mean µt and its varianceσ2t.

Let S(t) be the price of a stock at time t. In mathematical finance theory, often,we assume S(t) to be a geometric Brownian motion. If the price of stock S(t) is ageometric Brownian motion, we can say that

1. the future price growth rate is independent of the past price, and

2. the distribution of the growth rate is distributed as the lognormal with the param-eter µt and σ2t..

The future price is probabilistically decided by the present price. Sometimes, thiskind of stochastic processes are refereed to a Markov process.

Lemma 5.1. If S(t) is a geometric Brownian motion, we have

E[S(t)|S(0) = s0] = s0et(µ+σ2/2). (5.1)

28

5.1. GEOMETRIC BROWNIAN MOTIONS 29

Proof. Set

Y =S(t)S(0)

=S(t)s0

.

Then, Y is lognormal with (tµ, tσ2). Thus by using Theorem 3.3 we have

E[Y ] = et(µ+σ2/2).

On the other hand, we have

E[Y ] =E[S(t)]

s0.

Hence, we have (5.1).

Remark 5.1. Here the mean of S(t) increase exponentially with the rate µ + σ2/2,not the rate of µ . The parameter σ represents the fluctuation, but since the lognormaldistribution has some bias, σ affects the average exponential growth rate.

Theorem 5.1. Let S(t) be a geometric Brownian motion with its drift µ and volatilityσ , then for a fixed t ≥ 0, S(t) can be represented as

S(t) = S(0)eW , (5.2)

where W is the normal random variable N(µt,σ2t).

Proof. We need to check that (5.2) satisfies the second part of Definition 5.1, which iseasy by taking log on both sides in (5.2) and by seeing

log(

S(t)S(0)

)= W. (5.3)

Sometimes instead of Definition 5.1, we use the following stochastic differentialequation to define geometric Brownian motions,

dS(t) = µS(t)dt +σS(t)dB(t), (5.4)

where B(t) is the standard Brownian motion and dB(t) is define by so-called Ito calcu-lus.

Note that the standard Brownian motion is continuous function but nowhere differ-entiable, so the terms like dB(t)/dt should be treated appropriately. But these treatmentincludes the knowledge beyond this book. Thus, the following parts is not mathemati-cally rigorous. The solution to this equation is

S(t) = S(0)eµt+σB(t). (5.5)

Formally, if we differentiate (5.5), we can verify this satisfy (5.4). It easy to see thatS(t) satisfies Definition 5.1.

30 CHAPTER 5. BROWNIAN MOTIONS AND POISSON PROCESSES

Theorem 5.2. The geometric Brownian motion S(t) satisfies the stochastic differentialequation,

dS(t) = µS(t)dt +σS(t)dB(t), (5.6)

with the initial condition S(0).

5.2 Discrete Time Tree Binomial ProcessIn this section, we imitate a stock price dynamic on the discrete time, which is subjectto geometric Brownian motion.

Definition 5.2 (Tree binomial process). The process Sn is said to be a Tree Binomialprocess, if the dynamics is express as

Sn =

uSn−1 with probability p,

dSn−1 with probability 1− p,(5.7)

for some constant factors u≥ d ≥ 0 and some probability p.

Figure 5.1 depicts a sample path of Tree Binomial Process. Do you think it isreasonably similar to Figure 5.2, which is a exchange rate of yen?

Theorem 5.3 (Approximation of geometric brownian motion). A geometric Brownianmotion S(t) with the drift µ and the volatillity σ can be regardede as the limit of a treebinomial process with the parameters:

d = e−σ√

∆, (5.8)

u = eσ√

∆, (5.9)

p =12

(1+

µ

σ

√∆

). (5.10)

Proof. Let ∆ be a small interval and n = t/∆. Define a tree binomial process on thediscrete time i∆ such as

S(i∆) =

uS((i−1)∆) with probability p,dS((i−1)∆) with probability 1− p,

(5.11)

for all n. The multiplication factors d and u as well as the probability p have somespecific value, as described in the following, to imitate the dynamics of the Brownianmotion. Let Yi be the indicator of “up”s (Bernouilli random variable), i.e.,

Yi =

1 up at time i∆,0 down at time i∆.

(5.12)

5.2. DISCRETE TIME TREE BINOMIAL PROCESS 31

20 40 60 80 100

80

90

100

110

120

130

Figure 5.1: An example of Tree binomial process with d = 0.970446,u = 1.03045, p =0.516667 and starting from S0 = 100.

Figure 5.2: Exchange rate of yen at 2005/10/19. Adopted fromhttp://markets.nikkei.co.jp/kawase/index.cfm


Thus, the number of “up”s up to time n∆ is ∑ni=1 Yi, and the number of “down”s is

n−∑ni=1 Yi. Hence, the stock price at time n∆ given that the initial price S(0) is

S(n∆) = S(0)u∑Yidn−∑Yi

= S(0)dn( u

d

)∑Yi

.

Now set

S(t) = S(0)dt/∆(u/d)∑t/∆

i=1 Yi , (5.13)

and show that the limit S(t) is geometric Brownian motion with the drift µ and thevolatility σ as ∆→ 0. Taking log on the both side, we have

log(

S(t)S(0)

)=

t∆

logd + log( u

d

) t/∆

∑i=1

Yi. (5.14)

To imitate the dynamics of geometric Brownian motion, here we artificially set

d = e−σ√

∆,

u = eσ√

∆,

p =12

(1+

µ

σ

√∆

).

Note that logd and logu are symmetric, while d and u are asymmetric, and p→ 1/2 as∆→ 0. Now using these, we have

log(

S(t)S(0)

)=−tσ√

∆+2σ

√∆

t/∆

∑i=1

Yi. (5.15)

Consider taking the limit ∆→ 0, then the number of terms in the sum increases. UsingCentral Limit Theorem (Theorem 4.5), we have the approximation:

t/∆

∑i=1

Yi ∼ Normal distribution. (5.16)

Thus, log(S(t)/S(0)) has also the normal distribution with its mean and variance:

E[log(S(t)/S(0)] =−tσ√

∆+2σ

√∆

t/∆

∑i=1

E[Yi]

=−tσ√

∆+2σ

√∆

t∆

p,

=−tσ√

∆+σ

√∆

t∆

(1+

µ

σ

√∆

)= µt,

5.3. BROWNIAN MOTIONS 33

and

Var[log(S(t)/S(0)] = 4σ2∆

t/∆

∑i=1

Var[Yi]

= 4σ2t p(1− p)

→ σ2t, as ∆→ 0,

since p→ 1/2. Thus,

log(S(t)/S(0))∼ N[µt,σ2t], (5.17)

and moreover S(t + y)/S(t) is independent with the past because of the constructionof S(t). Hence the limiting distribution of S(t) with ∆ → 0 is a geometric Brownianmotion, which means all the geometric Brownian motion S(t) with the drift µ and thevolatility σ can be approximated by an appropriate tree Binomial process.

Remark 5.2. In the following, to show the properties of geometric Brownian motions,we sometimes check the properties holds for the corresponding tree Binomial Processand then taking appropriate limit to show the validity of the properties of geometricBrownian motions.

Example 5.1 (Trajectories of geometric brownian motions). This is an example of dif-ferent trajectories of Geometric Brownian Motions. Here we set the drift µ = 0.1, thevolatility σ = 0.3 and the interval ∆ = 0.01 in the geometric brownian motion. Aspointed out in Remark 5.2, we use tree binomial process as an approximation of geo-metric brownian motion. Thus, the corresponding parameters are d = 0.970446,u =1.03045 and p = 0.516667.

It is important to see that the path from geometric brownian motion will differ timeby time, even we have the same parameters. It can rise and down. Figure 5.4 gathersthose 10 different trajectories. Since we set the drift µ = 0.1 > 0, we can see up-wardtrend while each trajectories are radically different.

On the other hand, Figure 5.5 shows the geometric brownian motion with the neg-ative drift µ =−0.1, and the probability of the upward change p = 0.483333. Can yousee the difference between Figure 5.5 and Figure 5.4?

5.3 Brownian MotionsDefinition 5.3 (Brownian motion). A stochastic process S(t) is said to be a Brownianmotion, if S(t +y)−S(y) is a normal random variable N[µt,σ2t], which is independentwith the past history up to the time y.

Note that for a Brownian motion, we have

E[S(t)] = µt, (5.18)

Var[S(t)] = σ2t. (5.19)


20 40 60 80 100

80

90

100

110

120

130

Figure 5.3: Another example of geometric brownian motion with the drift µ = 0.1, thevolatility σ = 0.3 and the interval ∆ = 0.01 and starting from S0 = 100.

20 40 60 80 100

80

100

120

140

160

180

200

Figure 5.4: Many trajectories of geometric brownian motion with the upward driftµ = 0.1, the volatility σ = 0.3 and the interval ∆ = 0.01 and starting from S0 = 100.

5.3. BROWNIAN MOTIONS 35

20 40 60 80 100

100

120

140

Figure 5.5: Many trajectories of geometric brownian motion with the downward driftµ =−0.1.

Thus, both quantities are increasing as t increases.Brownian motions is more widely used than geometric Brownian motions. How-

ever, to apply them in finance theory, Brownian motion has two major draw backs.First, Brownian motions allows to have negative value. Second, in Brownian motion,the price differences on the same time intervals with same length is stochastically same,no matter the initial value, which is not realistic in the real finance world.

An example of trajectroy of two-dimensional brownian motion can be found inFigure 5.6.

-20 20 40 60

-80

-60

-40

-20

Figure 5.6: An example of trajectory of two-dimensional brownian motion.


5.4 Poisson ProcessesDefinition 5.4 (Counting process). Let N(t) be the number of event during [0, t). Theprocess N(t) is called a counting process.

Definition 5.5 (Independent increments). A stochastic process N(t) is said to haveindependent increment if

N(t1)−N(t0),N(t2)−N(t1), ...,N(tn)−N(tn−1), (5.20)

are independent for all choice of the time instants t0 < t1,< .... < tn.

Thus, intuitively the future direction of the process which has independent incre-ment are independent with its past history.

Definition 5.6 (Poisson Processes). Poisson process N(t) is a counting process ofevents, which has the following features:

1. N(0) = 0,

2. N(t) has independent increments,

3. PN(t + s)−N(s) = n= e−λ t (λ t)n

n! , where λ > 0 is the rate of the events.

Theorem 5.4. Let N(t) be a Poisson process with its rate λ , then we have

E[N(t)] = λ t, (5.21)Var[N(t)] = λ t. (5.22)

Or rewriting this, we have

λ =E[N(t)]

t, (5.23)

which validates that we call λ the rate of the process.

Proof. By Definition 5.6, we have

E[N(t)] =∞

∑n=0

nPN(t) = n

=∞

∑n=0

ne−λ t (λ t)n

n!

= λ te−λ t∞

∑n=0

(λ t)n−1

(n−1)!

= λ te−λ teλ t

= λ t.

The other is left for readers to prove (see Exercise 5.2).

5.5. REFERENCES 37

5.5 References

5.6 ExercisesExercise 5.1. Find E[Sn] and Var[Sn] for tree binomial process.

Exercise 5.2. Let N(t) be a Poisson process with its rate λ , and prove

Var[N(t)] = λ t. (5.24)

Chapter 6

Simulating Brownian Motions

We simulate Brownian motions using Mathematica, following the steps below.

6.1 Mathematica

6.1.1 As an Advanced CalculatorMathematica can handle “mathematical formulas” and derive the result as much preci-sion as possible. You can write an appropriate formula after the prompt “In[ ]:=”, andhit “shift + return” to obtain the result.

Example 6.1 (Calculations). These are some examples of calculations in Mathematica.

In[2]:=1/3 + 1/4

Out[2]= 712

In[3]:=Expand[(a+b)2]

Out[3]= a2 +2ab+b2

In[4]:=D[ax2 +bx+ c,x]

Out[4]= b+2ax

Those mathematical symbols are built-in.

Example 6.2 (Symbols). Here’s examples of symbols in Mathematica.

Pi = π (6.1)I = i imaginary number (6.2)Log[x] = loge x (6.3)Sin[θ ] = sin(θ). (6.4)

(6.5)

38

6.2. GENERATING RANDOM VARIABLES 39

6.2 Generating Random VariablesGenerating random variables is a core of simulation. The function Random[ ] willcreate a uniform random variable on the interval [0,1]. Each time the function called,

Example 6.3 (Random variable sequence). Here’s a simple example how to generatea sequence of random variables.

In[3]:= SeedRandom[128]

In[4]:= Random[],Random[],Random[]

Out[4]= 0.355565,0.486779,0.00573919

In[5]:= SeedRandom[128]

In[6]:= Random[],Random[],Random[]

Out[6]= 0.355565,0.486779,0.00573919

The function SeedRandom set the seed of the random number. If no SeedRandomis set, Mathematica set the seed from the current time, so as to generate the differentrandom sequence each time.

Example 6.4. You can create table by using the function ’Table’.

In[1]:=Table[Random[Integer],10]

Out[1]= 1, 1, 0, 0, 0, 1, 0, 1, 1, 0

Here’s one more example to generate a kind of random walk.

In[1]:=S = 0; Table[ t = Random[Integer]; S = S + t , 10]

Out[1]= 0, 1, 2, 2, 2, 2, 3, 4, 5, 5

6.3 Two-Dimensional Brownian Motions

6.3.1 Random Variable on Unit CircleGenerate a pair of random variable (X ,Y ) on unit circle as shown in Figure 6.1. Thepair should have the relation X2 +Y 2 = 1.

Check your pairs of random variables drawing a graph like Figure 6.1. You can usethe function ListPlot to display the results.

6.3.2 Generating Brownian MotionUsing the pairs of random variables as the movement, simulate a brownian motion asshown in Figure 6.2.

40 CHAPTER 6. SIMULATING BROWNIAN MOTIONS

-1 -0.5 0.5 1

-1

-0.5

0.5

1

Figure 6.1: An example of Random Variables on Unit Circle.

-20 20 40 60

-80

-60

-40

-20

Figure 6.2: An example of trajectroy of two-dimensional brownian motion.

6.4. GEOMETRIC BROWNIAN MOTIONS 41

6.4 Geometric Brownian MotionsSimulate geometric brownian motions with the drift µ and the volatility σ . Use anapproximation of tree-binomial process.

6.4.1 Generating Bernouilli Random VariablesGenerate a Bernouilli random variable T with its mean E[T ] = p. You may use thepackage Statistics‘DiscreteDistributions‘, if needed.

Make a sequence of Bernouilli random variables as

Out[xx] = 0,1,1,0,1,0,0,0,1,0,0,1,1,0,0,0,0,1,0,0,1,0,0,1,1,1,1,0,1,0,1,0,0,1,0,1,1,1.

Then, use this sequence to generate the multiplication factors of tree-binomial pro-cess as shown below.

Out[xx] = 1.03045,0.970446,0.970446,0.970446,1.03045,1.03045,1.03045,0.970446.

6.4.2 Generating geometric brownian motionsTake ∆ = 0.01 and set appropriate u,d and p for the tree-binomial process. Generatean approximation of geometric brownian motion with the drift µ and the volatility σ

as in Figure 6.3.

20 40 60 80 100

80

90

100

110

120

130

Figure 6.3: Another example of geometric brownian motion with the drift µ = 0.1, thevolatility σ = 0.3 and the interval ∆ = 0.01 and starting from S0 = 100.

Chapter 7

Present Value Analysis

7.1 Interest RatesLet r be the interest rate on some term, and P be the amount of money you saved in thebank. Then, at the end of the term you get

P+ rP = (1+ r)P. (7.1)

Example 7.1 (Compounding). If you have the interest rate r, which is compoundedsemi-annually, you can get

P(1+ r/2)(1+ r/2) = P(1+ r/2)2, (7.2)

at the end of the year. Sometimes, we call r the nominal interest rate or yearly interestrate.

Definition 7.1 (Effective interest rate). Let P0 be the amount of money you investedinitially, P1 be the amount of money you obtained at the end of the year. Then, thevalue re is said to be the effective interest rate, where

re =P1−P0

P0. (7.3)

In other words, given you invested P0 initially, the amount you get in the end of ayear can be expressed as

P1 = P0(1+ re). (7.4)

Example 7.2 (Compound interest rate). Assume the same situation as in Example 7.1.The effective interest rate re in this case is

re =P1−P0

P0

=P(1+ r/2)2−P

P= (1+ r/2)2−1.

42

7.2. CONTINUOUSLY COMPOUNDED INTEREST RATES 43

Remark 7.1. In economics, the word “nominal interest rate” is used as the interest ratenot including the effect of inflation, while the word “effective interest rate” is used asindicated to be adjusted to the inflation factor.

7.2 Continuously Compounded Interest RatesFirst, let us recall the definition of exponential, or Euler constant.

Definition 7.2 (exponential). We define e by the following:

e = limn→∞

(1+1/n)n. (7.5)

The following is a useful expression of the so-called exponential function Exp(x) =ex.

Lemma 7.1 (exponential).

ex = limn→∞

(1+ x/n)n. (7.6)

Proof. By the Definition 7.2, we have

ex =

limn→∞

(1+1/n)nx

= limn→∞

(1+1/n)xn

= limm→∞

(1+ x/m)m,

where we set m = xn.

Suppose we divide one year into n equal intervals, and consider n-compoundedyearly interest rate. Let r be the nominal interest rate and assume we initially invest P,then at the end of the year, we obtain

P(1+ r/n)n. (7.7)

Now, consider we have large n and take the limit n → ∞, which is turn out to be thecontinuous compounding.

limn→∞

P(1+ r/n)n = Per, (7.8)

where we used Lemma 7.1.

Theorem 7.1 (Continuous compounding). Let r be the nominal interest rate. Theeffective interest rate of continuous compounding on 1 year is

re = er−1. (7.9)

44 CHAPTER 7. PRESENT VALUE ANALYSIS

Proof. By Definition 7.1,

re =P1−P0

P0=

P0er−P0

P0= er−1.

Corollary 7.1. The effective interest rate is always larger than the nominal interestrate.

Proof. By Theorem 7.1, we have

re = er−1 > r, (7.10)

for r ≥ 0.

7.3 Present ValueSuppose we can get be the amount of money v at the end of the period i, when r is thenominal interest rate of the period. What can you say about the “value” of this moneynow? Or more precisely, when someone is offered to us the above investment, howmuch is the rational cost of this? The answer is this:

Suppose we lend the amount X at time 0, and we need to pay its interest at the timei. We need to pay X(1+ r)i at the end of period i. If we have the relation;

X(1+ r)i = v, (7.11)

then we can pay back the money. So, the present value of v of time i is equal tov(1+ r)−i. The term (1+ r)−i bring back the future value to the present value.

More generally, we have the following theorem for cash streams.

Theorem 7.2 (Present value of cash stream). Let r be the interest rate. Given the cashstream;

a = (a1,a2, . . . ,an), (7.12)

where ai is the return at the end of year i. Then, the cash stream a has the present valuePV (a) such as

PV (a) =n

∑i=1

ai

(1+ r)i . (7.13)

Or, if we deposit the amount PV (a) at time 0 at a bank, we can obtain the equivalentcash stream as a.

7.4. RATE OF RETURN 45

Proof. At the end of the first year, we withdraw a1 from our bank account. Thus, takinginto the interest rate r, we have as the remainder,

(1+ r)PV (a)−a1 =a2

1+ r+ · · ·+ an

(1+ r)n−1 . (7.14)

Continue this procedure to the end of the year i, and we have

ai+1

1+ r+ · · ·+ an

(1+ r)n−i , (7.15)

at our bank. At the end of year n−1, we have only an/(1 + r) at our bank. Thus, wepay an at the end of the year n, and all cleared.

We can even consider the perpetual payment.

Corollary 7.2 (Present value of perpetual payment). Suppose we have a perpetual cashstream of c, i.e.,

a = (c,c,c, . . .). (7.16)

Then the present value of this cash stream PV (a) = c/r, given the interest rate r.

Proof. By extending Theorem 7.2, we have

PV (a) =∞

∑i=1

c(1+ r)i

=c

1+ r

∞

∑i=0

1(1+ r)i

=c

1+ r1

1−1/(1+ r)

=cr.

Remark 7.2. The result of Corollary 7.2 is not surprising, if we consider to deposit c/rin a bank, and we can get c as annual interest without affecting the original principal.

7.4 Rate of ReturnDefinition 7.3 (Rate of return). Suppose we invest a as the initial payment of an in-vestment, and let b be its return after 1 period. Then we can define the rate of returnas

r =ba−1, (7.17)

which is the interest rate that makes the present value of the return equal to the initialpayment.

46 CHAPTER 7. PRESENT VALUE ANALYSIS

How can we compare different investments which give us different cash streams.We can measure the effectiveness of our investment with the rate of return extendingDefinition 7.3.

Definition 7.4 (Rate of return for cash stream). Suppose we invest a as the initialpayment of an investment, and let bi be the cash stream gained by this investment.Define a function P(r) as

P(r) =−a+n

∑i=1

bi(1+ r)−i, (7.18)

for r > −1. P(r) is representing the present value of cash stream (a,b1,b2, . . . ,bn).Then, we will define the rate of return r∗ with

P(r∗) = 0. (7.19)

Remark 7.3. Note that we allow a negative rate of return, especially when ∑bi < a.

Theorem 7.3 (Uniqueness and existence of rate of return). Suppose we have a cashstream (a,b1,b2, . . . ,bn) with a > 0, bi ≥ 0 and bn > 0. Then, there is a unique rate ofreturn r∗.

Proof. First, we will show that P(r) is strictly decreasing. Take r < r′, and we have

P(r)−P(r′) =n

∑i=1

bi(1+ r)−i− (1+ r′)−i> 0, (7.20)

since bi are nonnegative and bn is positive. Thus, P(r) is strictly decreasing. Further,it is easy to see P(r) → ∞ as r →−1, and P(r) →−a < 0 as r → ∞. Since P(r) iscontinuous, there exist an unique r >−1 satisfies P(r) = 0.

Example 7.3. Suppose we have an investment which will pay us back $60 for twoyear. We are suppose to obtain $60 at the end of each year. We will put $100 as ourinvestment. Then,

P(r) =−100+60

1+ r+

60(1+ r)2 . (7.21)

Thus, the rate of return r∗ is the solution of the equation:

100 = +60

1+ r+

60(1+ r)2 . (7.22)

Now take x = 1/(1+ r), then

3x2 +3x−5 = 0. (7.23)

Solving this, we have

x =−3±

√69

6≈ 0.8844. (7.24)

Hence r∗= 1/x−1 = 0.131.

7.5. REFERENCES 47

7.5 Referencesreference here!


Chapter 8

Risk Neutral Probability andArbitrage

8.1 Option to Buy StocksLet r be the interest rate. We consider to give the price of an option to purchase a stockat a future time at a fixed price. Let Xn be the price of stock at time n. Suppose, givenX0 = 100, the price of the stock can be either $200 or $50 at time 1 for simplicity.

We can consider an option to buy the stock at $150 at time 1. Note that this optionis worthless if the X1 = 50, since in that case, you can buy the stock at $50 instead.

At time 0, how can we determine the price of this option? Or, how much can wepay for this option?

Here’s the solution. Surpassingly, regardless of the probability of the stock’s upand down, the price of the option c should be like this.

c =100−50(1+ r)−1

3. (8.1)

We have a couple of ways to formulate the option price. However, we will followthe simplest way.

8.1.1 Risk NeutralizationWe will see how we can avoid risk of the up and down of stock price.

Let us suppose at time 0 we buy x unit of stocks and y unit of options. We allowboth x and y to be negative. For example, when x < 0, the position of this stock isshort, or we are in the position to selling the stocks without actually having the stocks.Likewise, when y <, we are actually selling the option. Wonderful world of free trade.

At time 0, the cost of this investment is 100x+cy, where c is the price of the option.Of course, the value of this investment will be varied according to the fluctuation of thestock price. However, we can avoid this risk.

48

8.1. OPTION TO BUY STOCKS 49

At time 1, the option is worthless if X1 = 50, since in that case, you can buy thestock at $50 which is far less expensive than exercising the option obtaining the stockat $ 150. On the other hand, if X1 = 200, you can use the option to buy the stocks at$150. Moreover, you can sell the stocks at the market price and obtain the difference $50 immediately. Thus, the value of the investment V1 at time 1 depends on the value ofX1 and is

V1 =

200x+50y, if X1 = 200,50x if X1 = 50.

(8.2)

So, if the equation

200x+50y = 50x, (8.3)

holds regardless of the stock price X1, the value of the investment V1 can be fixed. Bysolving (8.3), we have

y =−3x, (8.4)

which means that either

• by buying x stocks and selling y option at time 0, or

• by selling x stock and buying y option at time 0,

we can avoid the risk of stock value fluctuation, and obtain V1 = 50x at time 1.

8.1.2 Arbitrage and Price of Option

Now we evaluate the overall investment performance. Suppose we borrow the moneyneeded for this investment. At time 0, we need to invest the money

100x+ cy. (8.5)

So, we will borrow the money form a bank. At time 1 we need to pay its interest, butat the same time we earn the money V1 from our investment. Thus, overall gain G1 is

G1 = V1− (100x+ cy)(1+ r). (8.6)

Now from the argument in Section 8.1.1, if we adopt the investment strategy such as in(8.4), we know that we can avoid the risk and V1 = 50x. Thus,

G1 = 50x− (100x+3xc)(1+ r)

= (1+ r)x3c−100+50(1+ r)−1. (8.7)

Consider the following situations:

50 CHAPTER 8. RISK NEUTRAL PROBABILITY AND ARBITRAGE

1. Suppose 3c > 100−50(1+ r)−1. Set some positive x, then G1 > 0. That meansour investment guaranteed to be profitable. Actually, in this case, the option pricec satisfies

c >100−50(1+ r)−1

3. (8.8)

Compared to (8.1), the option price is expensive. So, we can exploit the situationby selling the options in short.

2. Suppose 3c < 100−50(1+ r)−1. In this case by setting negative x, we can alsoget positive gain G1 at time 1. In this case

c <100−50(1+ r)−1

3, (8.9)

which means the options is being selling at a bargain. Thus, buying options, youcan earn money, while hedging the risk of stock price drop by stock on spot.

3. Suppose 3c = 100− 50(1 + r)−1. The option price satisfies (8.1). We’re sorrybut you cannot obtain easy money.

Of course, in some temporary cases, we may experience case 1 and 2. We call thesesure-win situation arbitrage. However, for example, if the option price is expensivecompared to (8.1), as soon as many investors and the dealer of options realized thatthey can exploit it by selling the option, everybody starts to sell the options. Thus,in the end the price of the options will drop. If the option price is lower than (8.1),everybody wants to buy them, and eventually the option price will soar. Hence theprice of option will converge to (8.1).

8.2 Duplication and Law of One PriceThe other way to set the option price is following law of one price.

Theorem 8.1 (law of one price). Consider two investments and their costs c1 and c2respectively. If the present values of their payoff are same, then either

1. c1 = c2, or

2. there is an arbitrage opportunity.

Proof. Suppose c1 < c2, then we can buy c1 and selling c2, and obtain money.

Corollary 8.1. If there is no arbitrage, the price of the investments with the same returnshould be identical.

Now using Theorem 8.1, we will derive the option cost as in the same situation asin Section 8.1 by considering two different investments:

8.3. ARBITRAGE THEOREM 51

1. Buy a call option. The payoff P1 at time 1 is

P1 =

50 if the price of stock is $200,0 if the price of stock is $50.

(8.10)

2. Borrow x from bank, and buy y shares. The initial investment is 100y− x. Attime 1 you need to pay back (1+ r)x, while selling the y share. Thus, the payoffQ1 is

Q1 =

200y− x(1+ r) if the price of stock is $200,50y− x(1+ r) if the price of stock is $50.

(8.11)

By choosing appropriate x and y, we have the same pay off P1 = Q1 for these twoinvestment, i.e.,

200y− x(1+ r) = 50, (8.12)50y− x(1+ r) = 0, (8.13)

or

y =13, (8.14)

x =50

3(1+ r). (8.15)

In this case, two investments have the identical payoff. In other word, we could dupli-cate the option by buying stocks at spot and borrowing money from bank. Thus, usingTheorem 8.1, the initial costs of two investments are identical. Thus the option price cis

c = 100y− x =100−50(1+ r)−1

3, (8.16)

which is identical to (8.1).

8.3 Arbitrage TheoremHere’s the third way to decide the option price.

Let J be an experiment (of a gamble?). The outcome of J can be 1, ...,m. Let ri( j)be the return fuction, that is, when we bet a unit money on wager i, if the experimentresult is J = j, we can get ri( j). Note that the amount of bet can be negative.

We can bet xi on wager i. We call the vector x = (x1, ...,xn) the betting strategy. Ifwe bet on the strategy x, we can get the amount of money

n

∑i=1

xiri(J). (8.17)


Definition 8.1 (risk neutral probability). The probability measure is said to be riskneutral when the expectation of outcome on all bets is fair.

Theorem 8.2 (arbitrage theorem). Either there exists risk neutral probability, or thereis the sure-win strategy (arbitrage). More precisely, either one of the followings true.

1. There exists a probability vector p = (p1, ..., pm) for which

E[ri(J)] =m

∑j=1

p jri( j) = 0 for all wager i, (8.18)

where we regard p j = PJ = j.

2. There is a betting strategy x = (x1, ...,xn) for which

n

∑i=1

xiri( j) > 0 for all outcome j. (8.19)

Assume we need to estimate the price of a call option of a stock. Suppose the initialprice of the stock is s. After a week, the stock can be either a > s or b < s. Let C bethe price of call option buying the stock at strike price K. We can either buy or sell thestock option with C.

We need to find out appropriate value of C, which gives no sure-win. We have thefollowing two choices:

• buy (or sell) stock,

• buy (or sell) option.

Let p = PX1 = a and 1− p = PX1 = b. Suppose we buy one stock at spot. Let rbe the interest rate. Then, the present value of this investment R1 is

R1 = X1(1+ r)−1− s (8.20)

taking into account of the initial payment s. Thus, the expected return of this investmentis

E[R1] = pa(1+ r)−1− s+(1− p)b(1+ r)−1− s

= pa−b1+ r

+b

1+ r− s. (8.21)

By setting E[R1] = 0, we have

p =s(1+ r)−b

a−b. (8.22)

Thus, given that the expected return of this investment R1 equals to zero, p shouldshould have the value like in (8.22).

8.3. ARBITRAGE THEOREM 53

Next, suppose we buy one option. The present value of the return of this investmentQ1 is

Q1 =

(a−K)(1+ r)−1−C if X1 = a,−C if X1 = b.

(8.23)

Thus, we have

E[Q1] = p(a−K)(1+ r)−1−C (8.24)

Assuming (8.22), we have

E[Q1] =(a−K)s−b(1+ r)−1

a−b−C (8.25)

To guarantee to have no sure-win, E[Q1] also has to be zero. Thus,

C =(a−K)s−b(1+ r)−1

a−b(8.26)

Hence, by Theorem 8.2, we don’t have sure-win if the option price is (8.26).

Remark 8.1. The option price (8.26) coincides with (8.1).

Example 8.1 (Risk-neutral probabilities). Assume the same setting in Section 8.1.Given no arbitrage, the stock price must have

p = PX1 = 200

=100(1+ r)−50

150

=2r +1

3,

from (8.22). Thus,

E[X1] = 200p+50(1− p)= 50+150p

= 50+1502r +1

3= 100(1+ r).

This means that the only “rational” probability structure for the future stock priceshould has the same expected present value. Of course, given that this probabilitystructure, we have the option price,

c =100−50(1+ r)−1

3.




Chapter 9

Black-Scholes Formula

9.1 Risk-neutral Tree Binomial ModelWe show the tree binomial process introduced in Section 5.2 can be derived form arbi-trage theorem. This is the process we assume for the stock dynamics.

Theorem 9.1 (Risk-neutral tree binomial process). Consider the stock price on thediscrete time. Let r be the interest rate, and let Sn be the stock price at time n andsatisfies the following special dynamics:

Sn =

uSn−1,

dSn−1,(9.1)

where we assume d < (1+ r) < u. If there is no arbitrage opportunity, then the processSn should be a tree binomial process with the probability of up is p = (1+ r−d)/(u−d).

Proof. Let Xn be the indicator of “up”s of stock price at time n, i.e.,

Xn =

1 up at n,

0 down at n,(9.2)

Note that (X1,X2, . . . ,Xn) determines the stock price Sn. By Arbitrage Theorem (Theo-rem 8.2), in order to avoid arbitrage, the expected return of all bet is equal to zero.

First, let us consider an investment scheme. The stock price data up to time n− 1are used only for the observation. According to some rule, we decide to buy stock ornot. More precisely, if

(X1,X2, . . . ,Xn−1) = (x1,x2, . . . ,xn−1), (9.3)

we buy the stock at time n−1 by borrowing money from a bank, and sell it at time n.(Isn’t this fairly common rule?)

55

56 CHAPTER 9. BLACK-SCHOLES FORMULA

Let

α = P(X1,X2, . . . ,Xn−1) = (x1,x2, . . . ,xn−1), (9.4)p = PXn = 1|(X1,X2, . . . ,Xn−1) = (x1,x2, . . . ,xn−1). (9.5)

According to the stock price change, the gain of this investment differs. By condi-tioning the event (X1,X2, . . . ,Xn−1), we can get the expected gain at time n of thisinvestment E[Gn]:

E[Gn] = αE[profit from buying the stock]+ (1−α)0, (9.6)

since we avoid buying stock with probability 1−α . The probability of stock rise is p,so

E[profit from buying the stock] = puSn−1 +(1− p)dSn−1− (1+ r)Sn−1, (9.7)

since we need to pay the interest for the money borrowed from the bank. Thus,

E[Gn] = αpuSn−1 +(1− p)dSn−1− (1+ r)Sn−1. (9.8)

Thus, setting the expected gain to be zero, we have

0 = E[Gn] = αpuSn−1 +(1− p)dSn−1− (1+ r)Sn−1, (9.9)

or,

p = PXn = 1|(X1,X2, . . . ,Xn−1) = (x1,x2, . . . ,xn−1)=1+ r−d

u−d, (9.10)

which is independent of our specific rule (x1,x2, . . . ,xn−1). This means that Xn isindependent of (X1,X2, . . . ,Xn−1). Accordingly, using inductive arguments, we con-clude that (X1,X2, . . . ,Xn) are independent. Thus, the process Sn is a tree binomialprocess.

9.2 Option Price on the Discrete TimeDefinition 9.1 (Strike price and expiration time). Let Sn be the price of a stock at timen. Consider an option that we can buy a stock at the price K at time n. The value K iscalled the strike price and n is called the expiration time of the option.

Theorem 9.2 (Option price on binomial process). If there is no arbitrage opportunity,the option price C should be

C = (1+ r)−nE[(Sn−K)+], (9.11)

where Sn is defined and proved to be the risk-neutral binomial tree process in Theorem9.1.

9.3. BLACK-SCHOLES MODEL 57

Proof. We need to find out appropriate value of C, which gives no sure-win. By Theo-rem 9.1, we know that Sn should be the risk-neutral binomial process when there is noarbitrage opportunities.

Suppose we buy one option at cost C. The option is worthless if the stock price Snis less than K. On the other hand, if Sn is larger than the strike price K, then the valueof this option is Sn−K, since we can sell the stock obtained by exercising the option atthe market price Sn. Thus, the payoff of the option at time n is

(Sn−K)+, (9.12)

where x+ = max(x,0). Thus, the present value of the expected return of this investmentR is

E[R] = (1+ r)−nE[(Sn−K)+]−C. (9.13)

By Arbitrage Theorem (Theorem 8.2), this should be zero, or we have an arbitrage.Thus,

C = (1+ r)−nE[(Sn−K)+]. (9.14)

9.3 Black-Scholes ModelNow we proceed to the continuous time model. Let S(t) be a stock price at time t.We assume the stock price S(t) is a geometric Brownian motion with the drift µ andthe volatility σ as defined in Section 5.1. As noted in Theorem 5.3, if we divide tinto n interval, all geometric Brownian motion can be approximated by discrete-timebinomial tree process with

Sn =

uSn−1 with probability p,

dSn−1 with probability 1− p,(9.15)

where

d = e−σ√

∆, (9.16)

u = eσ√

∆, (9.17)

p =12

(1+

µ

σ

√∆

). (9.18)

and ∆ = t/n is the time interval. However, by Theorem 9.1. if we assume there is noarbitrage, the binomial process should be risk-neutral with

p = Pup=1+ rt/n−d

u−d, (9.19)


since the nominal interest rate is rt/n. Using the Taylor expansion of the exponentialfunction, we have

d = e−σ√

t/n = 1−σ√

t/n+σ2t2n

+O((t/n)3/2), (9.20)

u = eσ√

t/n = 1+σ√

t/n+σ2t2n

+O((t/n)3/2), (9.21)

Using these in (9.19), we have

p≈ 12

(1+

r−σ2/2σ

√t/n

), (9.22)

for a large n. Comparing this to (9.18), we can conclude that the original geometricBrownian motion has to have the drift

µ = r−σ2/2, (9.23)

if we assume there is no arbitrage.Thus, we have the following theorem.

Theorem 9.3 (Risk neutral geometric Brownian motion). Let r be the nominal interestrate. If there is no arbitrage, the geometric Brownian motion representing stock pricedynamics should have the drift µ = r−σ2/2 and the volatility σ . This process is calledthe risk neutral geometric Brownian motion.

Further, we can prove the famous Black-Scholes formula in the simple form.

Theorem 9.4 (Black-Scholes formula). Consider a call option of the stock with strikeprice K and the expiration time t. Let C be the price of this option. If we assume thereis no arbitrage opportunity, then we have

C = e−rtE[(S(t)−K)+], (9.24)

where S(t) is the geometric Brownian motion with the drift µ = r− σ2/2 and thevolatility σ .

Proof. Since we assume no arbitrage, the expected gain of all bets, including purchaseof the option, should be zero. Thus, supposing to buy a option with cost C, we have

E[gain at time t] = E[(S(t)−K)+−Cert ] = 0, (9.25)

which results (9.24).

Remark 9.1. Since the term e−rt represents drawing back to present value, (9.24) isnothing but the present value of expectation of option payoff.

Although, (9.24) is simple for our eyes, it is hard to evaluate.

9.3. BLACK-SCHOLES MODEL 59

Corollary 9.1 (Original Black-Scholes formula). Given S(0) = s, the option price C isobtained by

C = e−rtE[(seW −K)+] (9.26)

= sΦ(ω)−Ke−rtΦ(ω−σ

√t), (9.27)

where W is a normal random variable with N((r−σ2/2)t,σ2t), Φ(x) is the distribu-tion function of the standard normal random variable and

ω =rt +σ2t/2− log(K/s)

σ√

t. (9.28)

To prove this Corollary, we need the following Lemmas. Note that S(t) can beexpressed in

S(t) = seW (9.29)

= se(r−σ2/2)t+σ√

tZ , (9.30)

where Z is the standard normal random variable.

Lemma 9.1. By using the representation in (9.29), the following two set of events areconsidered to be equivalent:

S(t) > K⇐⇒

Z > σ√

t−ω

. (9.31)

Proof. By (9.29), we have

S(t) > K⇐⇒

e(r−σ2/2)t+σ√

tZ >Ks

. (9.32)

Since log is increasing function, we havee(r−σ2/2)t+σ

√tZ >

Ks

⇐⇒

Z >

log(K/s)− (r−σ2/2)tσ√

t

⇐⇒

Z > σ

√t−ω

,

where we used the fact:

σ√

t−ω =log(K/s− (r−σ2/2)t

σ√

t. (9.33)


Lemma 9.2. Let I be the indicator such as

I =

1 if S(t) > K,

0 if S(t)≤ K.(9.34)

Then,

E[I] = PS(t) > K= Φ(σ√

t−ω). (9.35)

Proof. Using Lemma 9.1, we have

E[I] = PI = 1= PS(t) > K= PZ > σ

√t−ω

= PZ < ω−σ√

t.

Lemma 9.3.

e−rtE[IS(t)] = sΦ(ω). (9.36)

Proof. Set α = σ√

t−ω . Since I and S(t) can be regarded as a function of the randomvariable Z, we have

E[IS(t)] = E[1Z>αse(r−σ2/2)t+σ

√tZ

]=

∫∞

α

se(r−σ2/2)t+σx√

tdPZ ≤ x

=∫

∞

α

se(r−σ2/2)t+σx√

t 1√2π

e−x2/2dx

=1√2π

se(r−σ2/2)t∫

∞

α

e−(x2−2σx√

t)/2dx

Since x2−2σx√

t = (x−σ√

t)2−σ2t, we have

=1√2π

sert∫

∞

α

e−(x−σ√

t)2/2dx

=1√2π

sert∫

∞

−ω

e−y2/2dy,

where we put y = x−σ√

t. Since the integrant is indeed the density of standard normaldistribution, we have

= sertPZ >−ω= sert

Φ(ω),

by the symmetry of Φ.

9.4. EXAMPLES OF OPTION PRICE VIA BLACK-SCHOLES FORMULA 61

Proof of Corollary 9.1. Since (S(t)−K)+ = I(S(t)−K), by Theorem 9.4 and , wehave

C = e−rtE[(S(t)−K)+]

= e−rtE[I(S(t)−K)]

= e−rtE[I(S(t)]−Ke−rtE[I]

= sΦ(ω)−Ke−rtΦ(σ

√t−ω),

where we used Lemma 9.2 and 9.3.

9.4 Examples of Option Price via Black-Scholes For-mula

Example 9.1 (Option price via Black-Scholes). Suppose the current price of a stocks = S(0) = 30. The yearly interest rate r = 0.08 and the volatility σ = 0.20. Let C bethe price of call option with the strike price K = 36 and its expiration date is 3 monthfrom now. We will estimate C.

Then, t = 1/4 and ω in (9.28) is

ω =rt +σ2t/2− log(K/s)

σ√

t

=0.08 ·1/4+0.22 ·1/4 ·1/2− log(36/30)

0.2√

1/4

≈−1.57322.

Thus, by (9.27) in Corollary 9.1, we have

C = sΦ(ω)−Ke−rtΦ(ω−σ

√t),

= 30 ·Φ(−1.57322)−34e0.08·1/4Φ(−1.57322−0.2

√1/4)

= 0.0715115

Thus, the price of this option is 0.07. Table 9.1 shows the return of stock and optionpurchase. You can get more profit when the stock price rises.

Table 9.1: Return when the stock price risesvalue at 0 value at 1 return

Stock 30 40 4/3Option 0.07 4 4/0.07

However, by Lemma 9.2,

PS(t) > K= Φ(σ√

t−ω) = 0.0471424. (9.37)


With high probability you will lose your money with this option. Thus, you can under-stand that option is high risk and high return.

In Figure 9.1, we show the simulation of the stock price dynamics. Your opinion?

20 40 60 80 100

25

30

35

40

Figure 9.1: Example of risk-neutral geometric brownian motion with the volatilityσ = 0.2 starting from S0 = 30, when the interest rate r = 0.08. Here we used thetree-binomial approximation with the interval ∆ = 0.01.


9.6 ExercisesExercise 9.1. Estimate the value of following call options, given the current price ofa stock s = S(0) = 30, and compare the result with Example 9.1. What can you sayabout it.

1. The yearly interest rate r = 0.08 and the volatility σ = 0.40.

2. The yearly interest rate r = 0.04 and the volatility σ = 0.20.

Note that e0.1 = 1.10517 and e0.05 = 1.05127.

Exercise 9.2. Give two examples where buying options is more appropriate than buy-ing its stocks.

Chapter 10

Delta Hedging Strategy

10.1 Binomial Hedging ModelAs usual, we use the discrete-time model of stock price. Let S0 = s be the initial priceand

S1 =

us,ds.

(10.1)

Here’s our new assignment. Find the amount of money x at time 0 required to meet apayment P1 below varying according to stock price.

P1 =

a up,

b down.(10.2)

Actually, P1 is representing derivatives more general than option. Now assume we buyy shares at time 0, then we can deposit x− ys at a bank. Note that if x− ys is negative,we should borrow some money from the bank. The return of this investment R1 is

R1 =

ys ·u+(x− ys)(1+ r) up,

ys ·d +(x− ys)(1+ r) down,(10.3)

where r is the interest rate. Thus, if we can set (x,y) satisfying

ys ·u+(x− ys)(1+ r) = a, (10.4)ys ·d +(x− ys)(1+ r) = b, (10.5)

then our assignment is solved. Subtracting both sides, we have

y(us−ds) = a−b.

y =a−b

s(u−d).

63

64 CHAPTER 10. DELTA HEDGING STRATEGY

Substituting this y into (10.4), we have

a−bu−d

·u+(x− a−bu−d

)(1+ r) = a,

a−bu−d

· u− (1+ r)+ x(1+ r) = a,

Solving this with respect to x, we have

x =1

1+ r

[a

1+ r−du−d

+bu−1− r

u−d.

](10.6)

Here we use the risk-neutral probability p defined in Theorem 9.1 as

p =1+ r−d

u−d, (10.7)

1− p =u−1− r

u−d. (10.8)

Then,

x = pa

1+ r+(1− p)

b1+ r

=1

1+ rpa+(1− p)b, (10.9)

which means that if we prepare x at time 0, which is just the present value of theexpected payoff in the risk-neutral probability, we can cover the pay off at time 1. Notethat in this case, we should buy the stock as much as

y =a−b

s(u−d), (10.10)

which is the ratio of the difference of payoff to the stock price change.Now we proceed to the time 2. In this case, the possibilities of payoff are not only

two like a and b in previous discussion, but three cases (why?). Let xi,2 be the moneyrequired to cover the payoff at time 2, given that

S2 = uid2−is, (10.11)

where i = 0,1,2 is the indicator of the number of up’s up to time 2.First, suppose S1 = us. If S2 = u2s, we have payoff x2,2. On the other hand, if

S2 = uds, we have payoff x1,2. Thus, using previous argument at time 1 and time 2, weknow that the money required at the time 1, say x1,1, should be

x1,1 =1

1+ rpx2,2 +(1− p)x1,2. (10.12)

Note that p = (1+ r−d)/(u−d) depends on the stock price dynamics d,u but doesn’tdepend on the payoff. In order to achieve the payoff, we need to buy stock as much as

y1,1 =x2,2− x1,2

us(u−d), (10.13)

10.1. BINOMIAL HEDGING MODEL 65

where s, which is the initial stock price in the previous discussion, is replaced with us,and the rest will be put in the bank.

Second, assume S1 = ds. With similar argument, we need the money x0,1 at time 1as

x0,1 =1

1+ rpx1,2 +(1− p)x0,2, (10.14)

with

y0,1 =x1,2− x0,2

ds(u−d). (10.15)

Summarize the results (10.12) and (10.14), we have

xi,1 = E[ the time-1 present value of payoff at time n|S(1) = uis], (10.16)

where i = 0,1.Now we can re-evaluate our investment strategy at time 0. As shown above, we

need the payoff x0,1 and x0,1 with respect to the value of S1. Thus, to cover this payoff,we need the money as

x0,0 =1

1+ rpx1,1 +(1− p)x0,1. (10.17)

Using (10.12) and (10.14), we have

x0,0 =1

(1+ r)2 [ppx2,2 +(1− p)x1,2+(1− p)px1,2 +(1− p)x0,2]

=1

(1+ r)2

[p2x2,2 +2p(1− p)x1,2 +(1− p)2x0,2

]. (10.18)

The payoff at time 2 will be achieved by buying the stock

y0,0 =x1,1− x0,1

s(u−d). (10.19)

As it can be predicted, x0,0, the money required to cover the payoff at time 2, is thepresent value of the expected payoff at time 2, i.e.,

x0,0 = E[the present value of payoff at time n|S(0) = s]. (10.20)

Here’s an example how to meet the requirement by adjusting our stock position.

1. Prepare the money equal to x0,0, and buy the stock as much as y0,0 at time 0. Therest is kept at the bank.

2. At time 1, we happen to have S1 = us (up). The value of our investment at time1 is x1,1, which is consisted by y0,0 stocks and (1 + r)(x0,0− y0,0) bank deposit.Then, adjust our position of stock holding to y1,1 by selling or buying, and putthe rest at the bank.


3. At the time 2, we have S2 = uds (up and down), then the value of our investmenthave x1,2. which is our requirement.

Now it’s time to generalize the argument. Let xi,n be the payoff at time n giventhat Sn = uidn−is. Also let xi,k be the money required at time k given that Sk = uidk−is.Then

xi,k = E[time-k present value of payoff at time n|Sk = uidk−is]

=1

1+ rpxi+1,k+1 +(1− p)xi,k+1 (10.21)

= (1+ r)−(n−k)n−k

∑j=0

(n− k

j

)p j(1− p)n−k− jxi+ j,n, (10.22)

for i = 0,1,2 . . . ,k and k = 0,1,2, . . . ,n. Also we need to adjust the position of stock as

yi,k =xi+1,k+1− xi,k+1

s(u−d), (10.23)

which is the ratio of the difference of next payoff to the stock price change.Suppose at the time n, we set the payoff having the form like

xi,n = (uidn−is−K)+. (10.24)

for i = 1,2, . . . ,n. Clearly, this is nothing but the payoff of the call option with strikeprice K and expiration time n. Thus, if we follow our investment strategy, we success-fully replicate the option payoff.

By the law of one price (Theorem 8.1), the initial cost of our investment strategyx0,0 should be equal to the option price C which was derived from Theorem 9.2).

Thus, we obtained the following theorem.

Theorem 10.1 (Delta hedge in discrete time binomial process). Let Sn be the risk-neutral binomial tree process. Set the initial investment money as

C = (1+ r)−nE[(Sn−K)+], (10.25)

which is equal to the call option price of strike price K and expiration time n. Letxi,n = (uidn−is−K)+, and recursively set xi,k

xi,k =1

1+ rpxi+1,k+1 +(1− p)xi,k+1 (10.26)

= (1+ r)−(n−k)n−k

∑j=0

(n− k

j

)p j(1− p)n−k− jxi+ j,n. (10.27)

We can hedge the option payoff by continuously adjusting the position of stock by

yi,k = (xi+1,k+1− xi,k+1)/(s(u−d)), (10.28)

when Sk = uidk−is for i = 0,1,2 . . . ,k and k = 0,1,2, . . . ,n.

10.2. HEDGING IN BLACK-SCHOLES MODEL 67

10.2 Hedging in Black-Scholes ModelHere’s a powerful tool to investigate limits.

Lemma 10.1 (l’Hospital’s Rule). Suppose f and g are differentiable on the interval(a,b). Assume f , g, f ′ and g′ are continuous on (a,b), and g′(c) for some fixed pointc ∈ (a,b).

If f (x)→ 0 and g(x)→ 0 as x→ c, and there exists limx→c f ′(x)/g′(x), then

limx→c

f (x)g(x)

=f ′(x)g′(x)

. (10.29)


Let us consider a call option for a stock. Assume the stock price dynamics isgoverned by a geometric brownian motion. Let K be the strike price of the option, t bethe expiration time and σ be the volatility of the geometric brownian motion.

As usual, we consider the discrete-time approximation of the geometric brownianmotion. Let h be the time interval and S(t) be the stock price at time t. Then,

S(h) =

seσ

√h,

se−σ√

h.(10.30)

Let C(s, t) be the cost of call option with the current price s and the expiration timet. After time h, if the stock price rises, we must prepare the money

C(seσ√

h, t−h), (10.31)

which is the Black-Schole’s option price when the stock price is seσ√

h and the expira-tion time t−h. Similarly, if stock price goes down, we must prepare

C(se−σ√

h, t−h). (10.32)

Thus, by the argument in Section 10.1 and (10.10), we know that buying the stock asmuch as D,

D(h) =C(seσ

√h, t−h)−C(se−σ

√h, t−h)

seσ√

h− se−σ√

h, (10.33)

we can cover the money required to buy the option at time h, given that we have C(s, t)at the time 0. Now take h→ 0 in (10.33), then

D(0) = limh→0

C(seσ√

h, t−h)−C(se−σ√

h, t−h)

seσ√

h− se−σ√

h

= lima→0

C(seσa, t−a2)−C(se−σa, t−a2)seσa− se−σa . (10.34)


where a =√

h. Since both the denominator and numerator are converges to 0 as h→ 0,we need to apply l’Hospital’s Rule (Lemma 10.1). Then,

D(0) = lima→0

dda

[C(seσa, t−a2)−C(se−σa, t−a2)

]d

da [seσa− se−σa](10.35)

Now

dda

C(seσa, t−a2) =∂C(y, t−a2)

∂a

∣∣∣∣y=seσa

+∂C(seσa,u)

∂a

∣∣∣∣u=t−a2

= lima→0

[Ct(seσa, t−a2)(−2a)+Cs(seσa, t−a2)sσeσa]

= sσCs(s.t),

where Ct(s, t) = ∂C(s, t)/∂ t and Cs(s, t) = ∂C(s, t)/∂ s. Similarly,

− dda

C(se−σa, t−a2) = lima→0

[−Ct(se−σa, t−a2)(−2a)−Cs(se−σa, t−a2)(−sσe−σa)

]= sσCs(s.t).

Thus, (10.35) can be rewritten as

D(0) = limh→0

D(h) =2sσCy(s.t)

2sσ

=∂C(s, t)

∂ s. (10.36)

The quantity D(0) = ∂C(s, t)/∂ s is the sensitivity of the option price with the initialstock price and is called ∆.

Thus, we have the following theorem:

Theorem 10.2 (Delta hedge in geometric brownian process). Let S(t) be the risk-neutral geometric brownian process. Set the initial investment money as

C(0, t) = e−rtE[(S(t)−K)+], (10.37)

which is equal to the call option price of strike price K and expiration time n. We canhedge the option payoff by continuously adjusting the position of stock by ∆

∆ =∂C(s, t)

∂ s, (10.38)

and the rest can be put (or borrowed) at a bank with the interest rate r.

10.3 Partial Derivative ∆

As in Section 10.2, the partial derivative called ∆ is important for hedging.

10.3. PARTIAL DERIVATIVE ∆ 69

Theorem 10.3 (Partial Derivative ∆). The partial derivative ∆ is obtained by

∆ =∂C(s, t)

∂ s= Ψ(ω). (10.39)

Proof. First, recall that

C(s, t) = E[e−rt(S(t)−K)+]. (10.40)

Set the indicator function I = 1S(t)>K as in Lemma 9.2, then

e−rt(S(t)−K)+ = e−rt(S(t)−K)I. (10.41)

Thus,

∆ =∂C(s, t)

∂ s

=∂

∂ sE[e−rt(S(t)−K)I]

= E[

∂

∂ se−rt(S(t)−K)I

].

Now, we have

∂

∂ se−rt(S(t)−K)I= I

∂e−rt(S(t)−K)∂ s

+e−rt(S(t)−K)∂ I∂ s

= I∂e−rt(S(t)−K)

∂ s, (10.42)

on S(t) 6= K. Since PS(t) = K= 0, we have

∆ = E[

∂


∣∣∣∣S(t) 6= K]

PS(t) 6= K

+E[

∂


∣∣∣∣S(t) = K]

PS(t) = K

= E[

I∂e−rt(S(t)−K)

∂ s

](10.43)

Since S(t) is the geometric brownian motion, S can be represented as

S(t) = se(r−σ2)t+σ√

tZ , (10.44)

for some standard normal random variable Z (see (9.29)). Thus,

∂

∂ se−rt(S(t)−K) = e−rt ∂S(t)

∂ s

=S(t)e−rt

s. (10.45)


Using (10.45) in (10.43), we have

∆ =e−rt

sE[IS(t)] (10.46)

In Lemma 9.3, we have already shown that

e−rtE[IS(t)] = sΦ(ω). (10.47)


10.5 ExercisesExercise 10.1. Use

f ′(x) = limx−c

f (x)− f (c)x− c

(10.48)

to prove l’Hospital’s Rule (Lemma 10.1).

Chapter 11

More Vanilla Options in Detail

Call options we have been studied are classified as in vanilla options, the basic typesof options. The options or derivatives more complicated are called exotic options.

11.1 American Call OptionsLet us re-define the call options in detail.

Definition 11.1 (Call option). A call option of a stock is defined to give you the optionof calling for the stock (or simply buy) at a strike price K with some specific expirationtime t.

Depending when the buyer of option can exercise the option, we have the followingtwo major types of options.

Definition 11.2 (European and American call option). When you buy an american typecall option, you can exercise your option at any time prior to the expiration time t. Oncontrary, in european type call options, you can exercise your option only when theexpiration time.

Remark 11.1. We have been implicitly assume only european call options up to thissection. However, it will be shown that this assumption is valid!

The freedom when we can exercise the option seems to make option more attrac-tive. The truth can be found in the following theorem.

Theorem 11.1 (No american call option). Never exercise an american option beforeits expiration time.

Proof. Let S be the present price of the stock, K be the strike price, and t be the expi-ration time. Let us assume we exercise the option now. Of course, it should be S > Kto obtain the fixed profit Pa:

Pa = S−K. (11.1)

71

72 CHAPTER 11. MORE VANILLA OPTIONS IN DETAIL

Consider that we sell the stock in short and wait exercising the option until the expira-tion time. Now you obtain S by selling the stock. Suppose at time t the stock price isabove the strike price:

S(t) > K. (11.2)

In this case, you can get the stock to sell by exercising the option with the price K.Thus, Pe, the profit we exercise at time t is given by

Pe = S−K = Pa. (11.3)

On the contrary, when the stock price declines at time t and

S(t)≤ K, (11.4)

Then, you don’t have to exercise the option to prepare the stock, but you can obtain thestock in the market, and

Pe = S−S(t)≥ S−K = Pa. (11.5)

Thus, you can obtain the profit more than or equal to the profit when you exercise youroption before the time t.

Remark 11.2. Suppose you have a call option. When the stock price rises beyond yourexpectation, you might have the temptation to exercise the option now and obtain thefixed profit. However, according to Theorem 11.1, what you should do is to sell short.Thus, american call option means nothing.

11.2 Put OptionsHere’s one other type of vanilla options.

Definition 11.3 (Put option). A put option of a stock is defined to give you the optionof putting the stock up for a sale (or simply sell) at a strike price K with some specificexpiration time t. As in the case of call options, when you buy an american put option,you can exercise your option at any time prior to the expiration time t. On contrary, ineuropean put options, you can exercise your option only when the expiration time.

Unlike call options, american put options have some advantage over european puts(see Exercise 11.1). Thus, the price of american put options can be more expensivethan the european one.

Theorem 11.2 (Put-call option parity). Let S be a stock price at time 0, and let C bethe price of the call option of the stock with its strike price K and the expiration time t.Also, let P be the price of european put option of strike price K and the expiration timet. Then, we have either

1. S +P = Ke−rt +C, or

11.3. PRICING AMERICAN PUT OPTIONS 73

2. there is an arbitrage opportunity,

where r is the interest rate.

Proof. Let S(t) be the stock price at time t. We consider two cases separately.

Case 1. S +P < Ke−rt +C. Then, we take the following strategy at time 0:

• buy one share,

• buy one put, and

• sell one call option.

Together with all these buying and selling, we need S + P−C > 0 at time 0,which we borrow from a bank with the interest rate r. In case when S(t) ≤ K,the call option we sold is worthless, but we can exercise the put option and sellthe share we bought, and obtain K. On the other hand, when S(t) ≥ K, the putoption is worthless, but the person who bought our call option will exercise it,and we need to sell the stock at the price K. In either cases, our position at timet is K. We assumed S +P < Ke−rt +C, or

K > ert(S +P−C), (11.6)

which means an arbitrage.

Case 2. S +P > Ke−rt +C. Then, we take the following inverse strategy at time 0:

• sell one share,

• sell one put, and

• buy one call option.

Together with all these buying and selling, we have S+P−C in our hand at time0, which we can deposit in a bank. At time t, we need K independent of thestock price (see Exercise 11.2). Meanwhile the money we deposited in our bankis worth (S +P−C)ert > K, by the assumption.

Thus, in either cases, there is an arbitrage.

11.3 Pricing American Put OptionsThere is no closed form solution of pricing american put options. You need solvedynamic programming.

Let S(t) be a stock price at time t which is the risk neutral geometric brownianmotion with the volatility sigma and the interest rate r. Let us consider the americanput option of the stock with expiration time t, which means the option of putting thestock up for a sale (or simply sell) at a strike price K with some specific expiration timet, and you can exercise your option any time up to time t.


First, as usual, we will use tree-binomial process to approximate the risk-neutralgeometric brownian motion. Let n be arbitrary large number, and we split the time tinto n intervals, and let tk = kt/n be the discrete time instances. Thus,

S(tk+1) =

uS(tk) with probability p,dS(tk) with probability 1− p.

(11.7)

where

u = eσ√

n/t ,

d = e−σ√

n/t ,

p =1+ rt/n−d

u−d.

With n → ∞, we have the above tree binomial process to be converged to the originalrisk-neutral geometric brownian motion.

Further, for approximation of the american put option, we assume we can exercisethe option only at time tk. Note that this may reduce the profit from the put option.However, when n→ ∞, this risk will be negligible.

Let us consider the stock price at time tk. In the case when we have i up’s and k− idowns, the stock price is

S(tk) = uidk−is, (11.8)

where s is the initial stock price, and i = 0,1, . . .k. Thus, potentially, according to thenumber of up’s, we have k + 1 different values of S(tk). Consider when we are at thetime tk without exercising the put option. Depending the value of S(tk), we shoulddetermine whether we should exercise the put option now or later.

Let Vk(i) be the conditional expected return of the put option at time tk, given that

1. the put option has never been exercise up to the time tk,

2. S(tk) = uidk−is,

3. after tk, we will use the optimal strategy to maximize our profit.

In other word, if we do not exercise the put option now, the average return is Vk(i).Assume that our investment objective is to maximize the expected return. Given

S(tk), we compare the return when we exercise the put option now with the expectedreturn when we won’t do it now. Only when K−S(tk) > Vk(i), we exercise the optionnow. Starting from time tn. we will derive all those conditional expected return back totime t0 = 0. These kind of decision making problem is called dynamic programing.

Theorem 11.3 (Dynamic programming for estimating american put options). Let S(tk)be the tree binomial process which imitates the dynamics of a stock price. The price ofan american put option can be approximately determined by the following procedure.

11.3. PRICING AMERICAN PUT OPTIONS 75

1. Start estimating Vn(i) as

Vn(i) = (K− suidn−i)+, (11.9)

for i = 0,1, . . . ,n.

2. Descend from n to estimate

Vk(i) = max(

K− suidk−i,e−rt/n pVk+1(i+1)+(1− p)Vk+1(i))

. (11.10)

3. We can use the above procedure recursively until k = 0, and the price of theamerican put option is approximated by V0(0).

Proof. We start from time tn. Suppose we haven’t exercise the put option yet. Theoptimal policy is to exercise the put option only when S(tn) < K and obtain the re-turn K−S(tn), otherwise the put option is worthless. Thus, when S(tn) = suidn−i, theconditional expected return Vn(i) is completely determined by

Vn(i) = (K− suidn−i)+. (11.11)

Now suppose that the stock price at time tk is S(tk) = suidk−i, and we haven’t exercisethe put option yet. Here’s two choices you can make:

1. Do not exercise the put option now. Then, the present value of the expectedreturn is

e−rt/n pVk+1(i+1)+(1− p)Vk+1(i) . (11.12)

2. Exercise the option now. Then, the return is fixed as

K−S(tk) = K− suidk−i, (11.13)

where of course K > suidk−i.

Thus, the optimal policy to maximize the present value of the expected return is tocompare the above two and make a decision. The resulted return from this optimalpolicy is

Vk(i) = max(K− suidk−i,e−rt/n pVk+1(i+1)+(1− p)Vk+1(i)), (11.14)

for i = 0,1,2, . . . i. We can use (11.14) recursively until k = 0 to obtain V0(0), which isthe price of the american put option.

Corollary 11.1 (When exercise your put option). You should exercise your put optionwhen you find your stock as

S(tk) < K− e−rt/n pVk+1(i+1)+(1− p)Vk+1(i) . (11.15)


Example 11.1 (American put option). Consider an american put option with the initialstock price s = 9, expiration time t = 0.25 (one quarter year), strike price K = 10. Letthe volatility of the stock be σ = 0.3 and the interest rate be r = 0.06.

Consider the approximation of put option with n = 5. Then

u = eσ√

n/t = 1.0694,

d = e−σ√

n/t = 0.9351,

p =1+ rt/n−d

u−d= 0.505.

Now at time k = 5, we have 5 different cases;

S(t5) =

9u5 > 10,

9u4d = 9u3 > 10,

9u3d2 = 9u = 9.635,

9u2d3 = 9d = 8.416,

9u1d4 = 9d2 = 7.359,

9d5 = 6.435.

Using (11.9) in Theorem 11.3, we have the conditional expected return:

V5(5) = 0,

V5(4) = 0,

V5(3) = (10−9.635)+ = 0.375,

V5(2) = (10−8.416)+ = 1.584,

V5(1) = (10−7.359)+ = 2.641,

V5(0) = (10−6.435)+ = 3.565.

Use (11.10) to have

V4(4) = max(10−9u4,e−rt/n pV5(5)+(1− p)V4(5)) = 0,

V4(3) = max(10−9u3d1,e−rt/n pV5(4)+(1− p)V5(3)) = e−rt/n pV5(4)+(1− p)V5(3)= 0.181,

V4(2) = max(10−9u2d2,e−rt/n pV5(3)+(1− p)V5(2)) = 10−9 = 1,

V4(1) = max(10−9ud3,e−rt/n pV5(2)+(1− p)V5(1)) = 10−9ud3 = 2.130,

V4(0) = max(10−9d4,e−rt/n pV5(1)+(1− p)V5(0)) = 10−9d4 = 3.119.

In other words, when you find that the stock price is less than 9 at time k = 4, we shouldexercise the put option . Continuing the above procedure, we have

V3(3) = e−rt/n pV4(4)+(1− p)V4(3)= 0.089,

V3(2) = max(0.375,e−rt/n pV4(3)+(1− p)V4(2)) = 0.375,

V3(1) = max(1.584,e−rt/n pV4(2)+(1− p)V4(1)) = 1.584,

V3(0) = max(2.641,e−rt/n pV4(1)+(1− p)V4(0)) = 2.641.

11.4. STOCK WITH CONTINUOUS DIVIDEND 77

V2(2) = e−rt/n pV3(3)+(1− p)V3(2)= 0.333,

V2(1) = max(1,e−rt/n pV3(2)+(1− p)V3(1)) = 1,

V2(0) = max(2.130,e−rt/n pV3(1)+(1− p)V3(0)) = 2.130.

V1(1) = e−rt/n pV2(2)+(1− p)V2(1)= 0.698,

V1(0) = max(1.592,e−rt/n pV2(1)+(1− p)V2(0)= 1.592.

Finally, we find the approximation of the price of the put option as

V0(0) = max(1,e−rt/n pV1(1)+(1− p)V1(0)= 1.137. (11.16)

We can set more large n to get more accurate price approximation.

11.4 Stock with Continuous DividendFirst, recall the solution of the following elementary ordinary differential equation;

Lemma 11.1. The differential equation ofdN(t)

dt = αN(t),N(0) = N0,

(11.17)

has an unique solution as

N(t) = N0eαt . (11.18)

Proof. Arranging and Integrating both side, we have∫ dN(t)N(t)

=∫

αdt. (11.19)

Since ∫ dxx

= logx+C, (11.20)

we have

logN(t) = αt +C, (11.21)

or

N(t) = Ceαt . (11.22)

Now taking t = 0, we can find the integral constant C = N0.


Now we consider the call option of a stock with dividend. According to the stockprice, we can expect its dividend.

First, we assume that the dividend is continuously paid according to S(t).

Theorem 11.4 (Stock price with continuously paying dividend). Let S(t) be the priceof a stock at time t whose dividend is continuously paid proportional to S(t). Then, thestock price is geometric brownian motion with

S(t) = S(0)e−αteW , (11.23)

where α is the ratio of dividend to the stock price and W is the normal random variableof N[(r−σ2/2)t, tσ2] and r is the interest rate.

Proof. Here we suppose we adopt the investment strategy that all the profit of dividendwill be use to purchase for the same stock. Let N(t) be the total number of stockwe have at time t, and let D(t) be the total dividend obtained up to time t. ThusdD(t) = D(t +dt)−D(t) be the dividend paid for a small interval dt, which is assumedto be proportional to the stock price S(t), i.e.,

dD(t) = αS(t)N(t)dt, (11.24)

We buy as much additional stock as possible when we get the dividend. Thus, thenumber of stock we can add at time t is

dN(t) =dD(t)S(t)

= αN(t)dt. (11.25)

Thus, we obtain a differential equation:

dN(t)dt

= αN(t), (11.26)

with its initial condition

N(0) = 1. (11.27)

Thus, setting initially we owned one stock, i.e., N0 = 1 in Lemma 11.1, we can find thenumber of stocks N(t) grows exponentially, and

N(t) = eαt . (11.28)

Let M(t) be the value of our investment at time t. Then,

M(t) = S(t)eαt . (11.29)

Note that we have initially,

M(0) = S(0). (11.30)

11.5. STOCK WITH FIXED-TIME-DIVIDEND 79

By using the same argument as before, it is reasonable to assume M(t) should bea risk-neutral geometric brownian motion. Otherwise, we have an arbitrage using thisinvestment strategy. Thus, like (9.29)M(t) can be expressed as

M(t) = M(0)eW , (11.31)

where W is the normal random variable of N[(r−σ2/2)t, tσ2]. Using the relationship(11.29), we have

S(t) = M(t)e−αt = M(0)eW e−αt = S(0)eW e−αt (11.32)

Using this Theorem 11.4, it is to estimate the value of call option of the stock withdividend.

Theorem 11.5 (Call option of the stock with dividend). Let S(t) be the price of astock at time t whose dividend is continuously paid proportional to S(t) as in Theorem11.4. Consider a call option of the stock with its expiration time t and the strike priceK. Then the price of this option C is evaluated by adjusting the initial stock price ass = S(0)e−αt in the Black-Scholes formula (Corollary 9.1) as

C = e−rtE[(S(0)e−αteW −K)+] (11.33)

= S(0)e−αtΦ(ω)−Ke−rt

Φ(ω−σ√

t). (11.34)

Proof. Since the price of option is estimated by the present value of the expected pay-off, we have

C = e−rtE[(S(t)−K)+]. (11.35)

By Theorem 11.4, we can rewrite the above as

C = e−rtE[(S(0)e−αteW −K)+]. (11.36)

Since e−αt is not a random variable, we can set s = S(0)e−αt .

11.5 Stock with Fixed-time-DividendThe assumption of continuous paying is rather artificial. Now let us assume the divi-dend is paying at some specified time.

Suppose at time t0 the dividend αS(t0) will be paid. Then, the stock price will bedropped as much as S(t0) at time t0. Otherwise, we can buy the stock just before timet0 and obtain the dividend, and then sell the stock. Thus, if we assume no arbitrage, thestock will be dropped at the time t0.


In this case, we cannot expect the stock price S(t) to be geometric brownian mo-tion, since it has a jump. However, we consider an investment strategy whose value isgeometric brownian motion. Let S(t0+) be the stock price just after t0, i.e.,

S(t0+) = limt↓t0

S(t). (11.37)

Since the stock price drops αS(t0) at time t0, we have

S(t0+) = S(t0)−αS(t0). (11.38)

Thus, at t0, after getting the dividend, we can buy the stock as much as

αS(t0)S(t0+)

=α

1−α. (11.39)

Also, the number of share we have after t0 is

1+α

1−α=

11−α

. (11.40)

Let M(t) be the market value of our investment at time t.

M(t) =

S(t) t < t0,

11−α

S(t) t ≥ t0.(11.41)

Now we assume the market value of our investment M(t) follows geometric brow-nian motion with drift r−σ2/2 and the volatility σ . Then, for t < t0, we have

S(t) = M(t), (11.42)

and we can use Black-Scholes formula to estimate the call option cost. On the otherhand, when t ≥ t0,

S(t) = (1−α)M(t) = (1−α)M(0)eW , (11.43)

where W is the normal random variable of N[(r−σ2/2)t, tσ2]. Since M(0) = S(0), wehave

S(t) = (1−α)S(0)eW . (11.44)

Thus, in this case, we can use Black-Scholes formula adjusting the initial stock priceto be s = S(0)(1−α).

11.6 Forward and Futures ContractNow consider pricing problem of forward contracts.

11.7. EXERCISES 81

Definition 11.4 (Forward contract). In a forward contract, you agree now to pay thefixed money at time t for one share of a stock. Note that the deal is fixed now and youcannot waive the purchase unlike the options, even when the stock price is descendedconsiderably.

Definition 11.5 (Futures contract). Futures contract is nothing but the standardizedformat of forward contracts so as to dealt within the market. Thus, you can sell or buythe futures contract before the expiration time.

Remark 11.3. The forward contract is mainly used among the bank and security com-pany. While, the futures contracts are used for anyone. See the detailed difference andexplanation of the forward and futures contracts for example in the book [1].

Theorem 11.6 (Pricing forward contract). Let S be the present market price of a stock.Given that the interest rate r, the price of forward contract buying the stock at time t is

F = Sert . (11.45)

Proof. Assume that F < Sert . Then, the forward contract is cheap, and we should buythe forward contract, along with sell the stock in short. Thus at time 0, we obtain S andput it in a bank. Then, at time t, we have Sert in cash. To cover the short-selling, we buya stock at the price F using the forward contract. At the end, we obtain Sert −F > 0.

On contrary, when F > Sert , we can sell the future contract, and at the same time,we will borrow the amount S to buy a stock. At time t, we need to return Sert to thebank, but we can sell the stock at the price F by the future contract. Thus, we can getthe amount F > Sert .

In either case, we have an arbitrage opportunity. Thus, by Arbitrage theorem (The-orem 8.2), the price of the forward contract should be F = Sert .

Remark 11.4. In the case of the futures contract for commodities like wheat, coffee,oil, gas etc., the price of the futures contracts will not satisfy Theorem 11.6, since weneed the extra storage cost for the commodities.

11.7 ExercisesExercise 11.1. Consider the case when american put options have some advantageover european calls unlike call options.

Exercise 11.2. In the proof of Theorem 11.2, show that we need K at time t in Case 2.

Bibliography

[1] Zvi Bodie and Robert C. Merton. Finance. Prentice–Hall, 1998.

[2] Colin Bruce. Conned Again, Watson!: Cautionary Tales of Logic, Math, and Prob-ability: Theory and Examples. Perseus Books Group, 2002.

[3] Richard Durrett. Probability: Theory and Examples. Thomson Learning, 1991.

[4] Bernt Oksendal. Stochastic Differential Equations: An Introduction With Applica-tions. Springer-Verlag, 2003.

[5] Sheldon M. Ross. Applied Probability Models With Optimization Applications.Dover Pubns, 1992.

[6] Sheldon M. Ross. An Elementary Introduction to Mathematical Finance: Optionsand Other Topics. Cambridge Univ Pr, 2002.

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