mathematical finance - waseda university

Mathematical Finance

Hiroshi Toyoizumi 1

September 26, 2012

1E-mail: [email protected]

Contents

1 Introduction 6

2 Normal Random Variables 72.1 What is Normal Random Variable? . . . . . . . . . . . . . . . . . 72.2 Lognormal Random Variables . . . . . . . . . . . . . . . . . . . 92.3 Lognormal Random Variables as the Stock Price Distribution . . . 102.4 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3 Large Sum of Random Variables 143.1 The Law of Large Numbers . . . . . . . . . . . . . . . . . . . . . 143.2 Poisson Random Variables and the Law of Small Numbers . . . . 163.3 The Central Limit Theorem . . . . . . . . . . . . . . . . . . . . . 173.4 Examples of Sum of Random Variables . . . . . . . . . . . . . . 173.5 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

4 Useful Probability Theorems 214.1 Useful Estimations . . . . . . . . . . . . . . . . . . . . . . . . . 214.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

5 Easy Money 235.1 Virtual Market . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

5.1.1 Risk Neutralization . . . . . . . . . . . . . . . . . . . . . 245.1.2 No Easy Money? . . . . . . . . . . . . . . . . . . . . . . 25

5.2 Ideal Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . 265.3 Duplication and Law of One Price . . . . . . . . . . . . . . . . . 285.4 Arbitrage Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 30

2

CONTENTS 3

5.5 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

6 Tools for Modeling Dynamics of Stock Prices 346.1 Geometric Brownian Motions . . . . . . . . . . . . . . . . . . . 346.2 Discrete Time Tree Binomial Process . . . . . . . . . . . . . . . 366.3 Brownian Motions . . . . . . . . . . . . . . . . . . . . . . . . . 416.4 Poisson Processes . . . . . . . . . . . . . . . . . . . . . . . . . . 426.5 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 446.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

7 On a Way to Black-Scholes Formula 457.1 The Only Possible Process to Avoid Arbitrage . . . . . . . . . . . 457.2 Option Price on the Discrete Time . . . . . . . . . . . . . . . . . 477.3 Black-Scholes Model . . . . . . . . . . . . . . . . . . . . . . . . 477.4 Examples of Option Price via Black-Scholes Formula . . . . . . . 527.5 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 537.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

8 Delta Hedging Strategy 558.1 Bernouilli Hedging Model . . . . . . . . . . . . . . . . . . . . . 558.2 Binomial Hedging Model . . . . . . . . . . . . . . . . . . . . . . 578.3 Hedging in Black-Scholes Model . . . . . . . . . . . . . . . . . . 608.4 Partial Derivative ∆ . . . . . . . . . . . . . . . . . . . . . . . . . 628.5 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 638.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

9 Stochastic Integral 659.1 Diffusion Process . . . . . . . . . . . . . . . . . . . . . . . . . . 659.2 Information . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 679.3 Definition of Stochastic Integral . . . . . . . . . . . . . . . . . . 689.4 Martingale . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 709.5 Calculus of Stochastic Integral . . . . . . . . . . . . . . . . . . . 70

10 Examples of Stochastic Integral 7410.1 Evaluation of E[W (t)4] . . . . . . . . . . . . . . . . . . . . . . . 7410.2 Evaluation of E[eαW (t)] . . . . . . . . . . . . . . . . . . . . . . . 76

4 CONTENTS

11 Differential Equations 7711.1 Ordinary Differential Equation . . . . . . . . . . . . . . . . . . . 7711.2 Geometric Brownian Motion . . . . . . . . . . . . . . . . . . . . 7811.3 Stochastic Process and Partial Differential Equation . . . . . . . . 79

12 Portfolio Dynamics 8112.1 Portfolio Model . . . . . . . . . . . . . . . . . . . . . . . . . . . 8112.2 Rate of Return of Stock and Risk-free Asset . . . . . . . . . . . . 8312.3 Arbitrage and Portfolio . . . . . . . . . . . . . . . . . . . . . . . 83

13 Pricing via Arbitrage 8513.1 Way to Black-Scholes Model . . . . . . . . . . . . . . . . . . . . 85

Bibliography

T. Bjork. Arbitrage Theory in Continuous Time. Oxford Finance. Oxford Univ Pr,2nd edition, 2004.

R. Durrett. Probability: Theory and Examples. Thomson Learning, 1991.

K. Ito. Probability Theory. Iwanami, 1991.

B. Oksendal. Stochastic Differential Equations: An Introduction With Applica-tions. Springer-Verlag, 2003.

S. M. Ross. An Elementary Introduction to Mathematical Finance: Options andOther Topics. Cambridge Univ Pr, 2002.

H. Toyoizumi. Statistics and probability for business.http://www.f.waseda.jp/toyoizumi/classes/accounting/stat/2007/business.pdf,2007a.

H. Toyoizumi. Mathematics for management.http://www.f.waseda.jp/toyoizumi/classes/accounting/math/2007/matht.pdf,2007b.

5

Chapter 1

Introduction

Using the knowledge of probabilities and statistics studied in

• Mathematics for Management (Toyoizumi [2007b]) and

• Statistics and Probability for Business (Toyoizumi [2007a]),

we will learn the advanced probability models and its evaluation related to Math-ematical Finance.

Especially, we pick up the risk management, portfolio, and evaluation of deriva-tives. Also, we will even learn the basics of the advanced mathematics such asgeometric Brownian motion, Black-Scholes formula and Ito-calculus.

6

Chapter 2

Normal Random Variables

Normal random variable is important tool to analyze a series of independent ran-dom variables.

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

Definition 2.1 (Normal random variable). Let X be a random variable with itsprobability density function

dP{X ≤ x}= f (x)dx =1

(2π)1/2σe−(x−µ)2/2σ2

dx, (2.1)

for some µ and σ . This is called the normal random variable with the parametersµ and σ , or written by N[µ,σ2].

Problem 2.1. Draw the graph of the probability density function f (x) of normalrandom variable N[µ,σ ].

Theorem 2.1 (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, (2.2)

and the variance

Var[X ] = σ2. (2.3)

7

8 CHAPTER 2. NORMAL RANDOM VARIABLES

Proof. omit.

Definition 2.2 (Standard normal random variable). Let X be a normal randomvariable with µ = 0 and σ = 1. The random variable is called the standard normalrandom variable, which is often denoted by N[0,1].

Lemma 2.1. Let X be a normal random variable with its mean µ and standarddeviation σ . Set Z = (X−µ)/σ . Then, Z is the standard normal random variable.

Proof. See Exercise 2.5.

Theorem 2.2. Let (Xi)i=1,2,...,n are independent normal random variables withits mean µi and standard deviation σi. Then the sum of these random variablesX = ∑

ni=1 Xi is again a normal random variable with

µ =n

∑i=1

µi, (2.4)

σ2 =

n

∑i=1

σ2i . (2.5)

Proof. We just prove X satisfies (2.4) and (2.5). We have

µ = E[X ] = E

[n

∑i=1

Xi

]=

n

∑i=1

E [Xi] =n

∑i=1

µi.

Also, 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 charac-teristic function method (or Fourier transform).

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

2.2. LOGNORMAL RANDOM VARIABLES 9

2.2 Lognormal Random VariablesNow it is the turn of Lognormal random variable. Lognormal random variablesare used to model the stock price distribution in Mathematical Finance.

Definition 2.3 (lognormal random variable). The random variable Y is said to belognormal if log(Y ) is a normal random variable.

Thus, a lognormal random variable can be expressed as

Y = eX , (2.6)

where X is a normal random variable. Lognormal random variables plays a mea-sure role in finance theory!

Theorem 2.3 (lognormal). If X is a normal random variable having the mean µ

and the standard deviation σ2, the lognormal random variable Y = eX has themean and the variance as

E[Y ] = eµ+σ2/2, (2.7)

Var[Y ] = e2µ+2σ2− e2µ+σ2

. (2.8)

Remark 2.1. It is important to see that although the mean of the lognormal randomvariable is subjected to not only the mean of the original normal random variablebut also the standard deviation.

Problem 2.2. Explain why the mean E[Y ] is larger than the “expected” meaneE[X ].

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

m(t) = E[etX ]. (2.9)

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 ]. (2.10)

Further, we have

m′′(0) = E[X2].


Problem 2.3. Derive 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. (2.11)

(See exercise 2.6.) Since Y = eX , we have

E[Y ] = m(1) = eµ+σ2/2, (2.12)

and

E[Y 2] = m(2) = e2µ+2σ2. (2.13)

Thus,

Var[Y ] = E[Y 2]−E[Y ]2 = e2µ+2σ2− e2µ+σ2

. (2.14)

2.3 Lognormal Random Variables as the Stock PriceDistribution

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)(2.15)

2.3. LOGNORMAL RANDOM VARIABLES AS THE STOCK PRICE DISTRIBUTION11

In mathematical finance, it is commonly assumed that Y (n) is independent andidentically distributed as the lognormal random variable. Taking log on both sideof (2.15), then we have

logS(n) = logS(n−1)+X(n), (2.16)

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

Example 2.1 (Stock price rises in two weeks in a row). Suppose Y (n) is thegrowth rate of a stock at the n-th week, which is independent and lognormallydistributed with the parameters µ and σ . We will find the probability that thestock price rises in two weeks in a row.

First, we will estimate P{the stock rises}. Since it is equivalent that y > 1 andlogy > 0, we have

P{the stock rises}= P{S(1) > S(0)}

= P{

S(1)S(0)

> 1}

= P{

log(

S(1)S(0)

)> 0

}= P{X > 0}

where X = logY (1) is N(µ,σ2), and we can define

Z =X −µ

σ, (2.17)

as the standard normal distribution. Hence, we have

P{S(1) > S(0)}= P{

X −µ

σ>

0−µ

σ

}= P{Z >−µ/σ}= P{Z < µ/σ},

where we used the symmetry of the normal distribution (see exercise 2.1).Now we consider the probability of two consecutive stock price rise. Since

Y (n) is assumed to be independent, we have


P{the stock rises two week in a row}= P{Y (1) > 1,Y (2) > 1}= P{Y (1) > 0}P{Y (2) > 0}= P{Z < µ/σ}2

Problem 2.4. 1. Derive the probability that stock price down two days in arow.

2. Derive the probability that stock price up at the first and down at the secondday.

2.4 References

2.5 ExercisesExercise 2.1. 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,we have

P{X >−x}= P{X < x}. (2.18)

Exercise 2.2 (moment generating function). Let X be a random variable. Thefunction m(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 2.3. Let X be a normal random variable with the parameters µ = 0 andσ = 1.

1. Find the moment generating function of X .

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

2.5. EXERCISES 13

Exercise 2.4. Use Microsoft Excel to draw the graph of probability distributionf (x) = dP{X ≤ 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.

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

Exercise 2.5. Let X be a normal random variable with its mean µ and standarddeviation σ . Set Z = (X−µ)/σ . Prove Z is the standard normal random variable.

Exercise 2.6. Verify (2.11) by using Lemma 2.1

Exercise 2.7. Let Y = eX be a lognormal random variable where X is N(µ,σ).Find E[Y ] and Var[Y ].

Chapter 3

Large Sum of Random Variables

Problem 3.1. Do you think these figures are similar?

3.1 The Law of Large NumbersIn many cases, we need to evaluate the average of large number of independentand identically-distributed random variables. Perhaps, you can intuitively assumethe expectation of a random variable X will be approximated by the sample aver-age of the data, as

E[X ]≈ 1n

n

∑i=1

Xi. (3.1)

This intuition can be validated by the following theorems.

Theorem 3.1 (Weak law of large numbers). Let X1,X2, .... be an i.i.d. sequenceof random variables with

µ = E[Xn]. (3.2)

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

P{|Sn/n−µ|> ε}→ 0 as n→ ∞. (3.3)

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

with high probability.

14

3.1. THE LAW OF LARGE NUMBERS 15

20 40 60 80 100

80

90

100

110

120

130

Figure 3.1: An example of Tree binomial process with d = 0.970446,u =1.03045, p = 0.516667 and starting from S0 = 100.

Figure 3.2: Exchange rate of yen at 2005/10/19. Adopted fromhttp://markets.nikkei.co.jp/kawase/index.cfm

16 CHAPTER 3. LARGE SUM OF RANDOM VARIABLES

You can say, “OK. I understand, we can expect the average will be close to themean most of the times. So, it might be possible to have some exceptions...”

Problem 3.2. Take the case of coin flipping. Give an example of this kind ofexception.

Actually, the exceptions should follow the rule, they will vanish...

Theorem 3.2 (Strong law of large numbers). Let X1,X2, .... be an i.i.d. sequenceof random variables with

µ = E[Xn]. (3.4)

With probability 1, we haveSn

n→ µ as n→ ∞. (3.5)

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

Problem 3.3. What is the difference between “most of the time” and “almost all”?

3.2 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. Firstof all, we define Poisson random variable.

Definition 3.1 (Poisson random variable). A random variable N is said to be aPoisson random variable, when

P{N = n}=λ n

n!e−λ , (3.6)

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

Theorem 3.3. Let N be a Poisson random variable.

Var[N] = λ . (3.7)

Theorem 3.4 (The law of Poisson small number). The number of many indepen-dent rare events can be approximated by a Poisson random variable.

Example 3.1. Let N be the number of customers arriving to a shop. If eachcustomer visits this shop independently and its frequency to visit there is relativelyrare, then N can be approximated by a Poisson random variable.

3.3. THE CENTRAL LIMIT THEOREM 17

3.3 The Central Limit TheoremAccording to Section 3.1,

E[X ]≈ 1n

n

∑i=1

Xi. (3.8)

for a large n. However, there is an “error” even though n is large. Even this errorshould follow some rule!

Let X1,X2, .... be an i.i.d. sequence of random variables with

µ = E[Xn], (3.9)

σ2 = Var[Xn]. (3.10)

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

Sn−nµ =n

∑i=1

Xi−nµ. (3.11)

Theorem 3.5 (Central limit theorem). For a large n, we have

P{

Sn−nµ

σ√

n≤ x

}≈Φ(x), (3.12)

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

Sn ∼ N(nµ,nσ2). (3.13)

The central limit theorem indicates that no matter what the random variable Xiis like, the sum Sn can be regarded as a normal random variable.

Instead of stating lengthy and technically advanced proof of laws of large num-bers and central limit theorem, we give some examples of how Sn converges to aNormal random variable.

3.4 Examples of Sum of Random VariablesExample 3.2 (Average of Bernouilli Random Variables). Let Xi be i.i.d Bernoullirandom variables with p = 1/2. Or more simply, Xi is the result of ith coin flip-


ping. Suppose we are going to evaluate the sample average A of Xi’s;

A =1n

n

∑i=1

Xi. (3.14)

Of course, we expect that A is close to the mean E[Xi] = 1/2 but how close for agiven n?

Figure 3.3 shows the histogram of 1000 different runs of A with n = 10. Sincen = 10 is relatively small samples, we have large deviation from the expectedmean 1/2. On the other hand, when we have more samples, the sample average Atends to be 1/2 as we see in Figure 3.4 and 3.5, which can be a demonstration ofWeak Law of Large Numbers (Theorem 3.1).

As we see in in Figure 3.3 to 3.5, we may still have occasional high and lowaverages. How about individual A for large sample n. Figure 3.6 shows the sam-ple path level convergence of the sample average A to E[X ] = 1/2, as it can beexpected from Strong Law of Large Numbers (Theorem 3.2).

Now we know that A converges to 1/2. Further, due to Central Limit Theorem(Theorem 3.5), magnifying the histogram, the distribution can be approximateby Normal distribution. Figure 3.7 show the detail of the histogram of A. Thehistogram is quite similar to the corresponding Normal distribution.

Remark 3.1. In this example, Sn/n→ 1/2 but not saying your wealth will be even-tually balancing if you bet on coin-tossings. Sn will diverge instead of converge,as n getting large, which means that you have great chance to be bankrupting.

3.5 ReferencesThe proofs omitted in this chapter can be found in those books such as [Ito, 1991]and [Durrett, 1991].

3.6 ExercisesExercise 3.1. Something here!

3.6. EXERCISES 19

0.2 0.4 0.6 0.8 1

2

4

6

8

Figure 3.3: 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 3.4: 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 3.5: The sample average A when n = 1000.


100 200 300 400 500

0.2

0.4

0.6

0.8

1

Figure 3.6: 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 3.7: 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 isthe corresponding Normal distribution.

Chapter 4

Useful Probability Theorems

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

Theorem 4.1 (Markov’s Inequality). Let X be a nonnegative random variable,then we have

P{X ≥ a} ≤ E[X ]/a, (4.1)

for all a > 0.

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

aI{X≥a} ≤ X . (4.2)

Taking expectation on the both side, we have

aP{X ≥ a} ≤ E[X ]. (4.3)

Theorem 4.2 (Chernov Bounds). Let X be the random variable with moment gen-erating function M(t) = E[etX ]. Then, we have

P{X ≥ a} ≤ e−taM(t) for all t > 0, (4.4)

P{X ≤ a} ≥ e−taM(t) for all t < 0. (4.5)

21

22 CHAPTER 4. USEFUL PROBABILITY THEOREMS

Proof. For t > 0 we have

P{X ≥ a}= P{etX ≥ eta}. (4.6)

Using Theorem 4.1,

P{etX ≥ eta} ≤ e−taE[etX ]. (4.7)

Thus, we have the result.


Chapter 5

Easy Money

Problem 5.1. Are there any easy money? If so, what will happen?

5.1 Virtual MarketConsider a simple virtual market of options based on stocks. Let r be the interestrate. We consider to give the price of an option to purchase a stock at a future timeat a fixed price. Let Xn be the price of stock at time n. Suppose, given X0 = 100,the price of the stock can be either $200 or $50 at time 1 in this virtual market.

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

Problem 5.2. If you were a dealer, how are you going to buy and sell options?

At time 0, how can we determine the price of this option? Or, how much canwe pay for this option? Intuitively, the probability that the stock rises is large, thevalue of the option will decrease.

Problem 5.3. Explain the reason intuitively.

Here’s the solution. Surprisingly, regardless of the probability of the stock’sup and down, the price of the option c should be like this.

c =100−50(1+ r)−1

3. (5.1)

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

23

24 CHAPTER 5. EASY MONEY

5.1.1 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 stockis short, or we are in the position to selling the stocks without actually having thestocks. Likewise, when y < 0, we are actually selling the option.

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

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 thestock at $150. On the other hand, if X1 = 200, you can use the option to buy thestocks at $150. Moreover, you can sell the stocks at the market price and obtainthe difference $50 immediately. Thus, the value of the investment V1 at time 1depends on the value of X1 and is

V1 =

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

(5.2)

So, if the equation

200x+50y = 50x, (5.3)

holds regardless of the stock price X1, the value of the investment V1 can be fixed.By solving (5.3), we have

y =−3x, (5.4)

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.You can neutralize the uncertainty of stock and option by using the above

procedure.

Problem 5.4. Can you consider other methods to neutralize the uncertainty of thestock price?

5.1. VIRTUAL MARKET 25

5.1.2 No Easy Money?Now we evaluate the overall investment performance. Suppose we borrow themoney needed for this investment. At time 0, we need to invest the money

100x+ cy. (5.5)

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

G1 = V1− (100x+ cy)(1+ r). (5.6)

Now from the argument in Section 5.1.1, if we adopt the investment strategy suchas in (5.4), we know that we can avoid the risk and V1 = 50x. Thus,

G1 = 50x− (100x+3xc)(1+ r)

= (1+ r)x{3c−100+50(1+ r)−1}. (5.7)

Now you are ready to learn how to get easy money. Consider the followingsituations:

1. Suppose 3c > 100− 50(1 + r)−1. Set some positive x, then G1 > 0. Thatmeans our investment guaranteed to be profitable. Actually, in this case, theoption price c satisfies

c >100−50(1+ r)−1

3. (5.8)

Compared to (5.1), the option price is expensive. So, we can exploit thesituation by selling the options in short.

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

c <100−50(1+ r)−1

3, (5.9)

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


3. Suppose 3c = 100− 50(1 + r)−1. The option price satisfies (5.1). We’resorry but you cannot obtain easy money.

Of course, in some temporary cases, we may experience case 1 and 2. We callthese sure-win situation arbitrage. However, for example, if the option price isexpensive compared to (5.1), as soon as many investors and the dealer of optionsrealized that they can exploit it by selling the option, everybody starts to sell theoptions. Thus, in the end the price of the options will drop. If the option priceis lower than (5.1), everybody wants to buy them, and eventually the option pricewill soar. Hence the price of option will converge to (5.1).

Example 5.1. Consider the following thought experiment. Option price is largerthan c = {100−50(1+ r)−1}/3.

Problem 5.5. What are you going to do, when you find the option price is largerthan c = {100−50(1+ r)−1}/3?

Of course, you sell options in short and obtain easy money. However, if otherplayers read this text, they also start selling options. (They may realize the situa-tion independent of this text...) Eventually, the price of option falls. But how faryou can continue selling? It should be just before the option price hit c = {100−50(1+ r)−1}/3. Thus, the option price should be c = {100−50(1+ r)−1}/3.

5.2 Ideal DynamicsThis is another way to establish the price of call option. Unlike Section 5.1, herewe will not use the portfolio but buy (or sell) whichever stock or option.

Suppose the initial price of the stock is s. After a week, the stock can be eithera > s or b < s. Let C be the price of call option buying the stock at strike price K.We can either buy or sell the stock option with C.

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

• buy (or sell) stock,

• buy (or sell) option.

Let p = P{X1 = a} and 1− p = P{X1 = b}. Suppose we buy one stock at spot.Let r be the interest rate. Then, the present value of this investment R1 is

R1 = X1(1+ r)−1− s (5.10)

5.2. IDEAL DYNAMICS 27

taking into account of the initial payment s. Thus, the expected return of thisinvestment is

E[R1] = p{a(1+ r)−1− s}+(1− p){b(1+ r)−1− s}

= pa−b1+ r

+b

1+ r− s. (5.11)

To vanish the opportunity of easy money from a financial product, there shouldbe a particular situation where its expected profit should be equal to zero (SeeSection 5.4). Thus, by setting E[R1] = 0, we have

p =s(1+ r)−b

a−b. (5.12)

Thus, given that the expected return of this investment R1 equals to zero, p shouldshould have the value like in (5.12).

Next, suppose we buy one option. The present value of the return of thisinvestment Q1 is

Q1 =

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

(5.13)

Thus, we have

E[Q1] = p(a−K)(1+ r)−1−C (5.14)

Assuming (5.12), we have

E[Q1] =(a−K){s−b(1+ r)−1}

a−b−C (5.15)

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(5.16)

Hence, by Theorem 5.3, we don’t have sure-win if the option price is (5.16).

Remark 5.1. The option price (5.16) coincides with (5.1).


Example 5.2 (Risk-neutral probabilities). Go back to our virtual market. Assumethe same setting in Section 5.1. Given no arbitrage, the stock price must have

p = P{X1 = 200}

=100(1+ r)−50

150

=2r +1

3,

from (5.12). 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.

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

Theorem 5.1 (law of one price). Consider two investments and their costs c1 andc2 respectively. 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 5.1. If there is no arbitrage, the price of the investments with the samereturn should be identical.

5.3. DUPLICATION AND LAW OF ONE PRICE 29

Now using Theorem 5.1, we will derive the option cost as in the same situationas in Section 5.1 by considering two different investments:

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.

(5.17)

2. Borrow x from bank, and buy y shares. The initial investment is 100y− x.At time 1 you need to pay back (1+ r)x, while selling the y share. Thus, thepayoff Q1 is

Q1 =

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

(5.18)

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

200y− x(1+ r) = 50, (5.19)50y− x(1+ r) = 0, (5.20)

or

y =13, (5.21)

x =50

3(1+ r). (5.22)

In this case, two investments have the identical payoff. In other word, we couldduplicate the option by buying stocks at spot and borrowing money from bank.Thus, using Theorem 5.1, the initial costs of two investments are identical. Thusthe option price c is

c = 100y− x =100−50(1+ r)−1

3, (5.23)

which is identical to (5.1).


5.4 Arbitrage TheoremDefinition 5.1 (risk neutral probability). The probability measure is said to be riskneutral when the expectation of outcome on all bets is fair.

Consider two investment options whose gain is determined by the future priceof a stock X . Assume that the stock price X can be either a or b. Let R1 andR2 be the corresponding gain to the investments. Since the gain of investment isdetermined by the stock price, we have

Ri =

{ri(a) X = a,

ri(b) X = b.(5.24)

We can split our money x = x1 +x2 into the two investment, thus the total gainR is obtained by

R = x1R1 + x2R2. (5.25)

Theorem 5.2 (arbitrage theorem for two investments). Either one of the follow-ings holds.

1. There exist the risk-neutral probability. More precisely, there exists 0≤ p≤1 such that

pr1(a)+(1− p)r1(b) = 0, (5.26)pr2(a)+(1− p)r2(b) = 0 (5.27)

2. There is a sure-win strategy. That is there exist a betting strategy (x1,x2)for which

R = x1R1 + x2R2 > 0. (5.28)

Proof. First, suppose we have a risk-neutral probability as in (5.26) and we willshow that there is no sure win. Rewriting (5.26), we have

r1(a) =p−1

pr1(b), (5.29)

r2(a) =p−1

pr2(b) (5.30)

5.4. ARBITRAGE THEOREM 31

Suppose, we somehow find the investment strategy to get some positive gain whenthe stock price is b, i.e.

x1r1(b)+ x2r2(b) > 0. (5.31)

Now assume unfortunately our stock price turn out to be a. We have the gain as

R = x1r1(a)+ x2r2(a)

=p−1

p{x1r1(b)+ x2r2(b)}

However, because of (5.31) and the fact (p− 1)/p ≤ 0, the gain when X = a isalways negative (non-positive). The same argument can be applied to the casewhen there is a strategy to have positive gain at X = b. Thus there is no sure-winwhen there is risk-neutral probability.

Next we show that whenever there is no sure-win, there exists a risk-neutralprobability. Define two vectors by

r(a) = (r1(a),r2(a)), (5.32)r(b) = (r1(b),r2(b)). (5.33)

Then, the gain is nothing but the inner product of these vectors and the vectorx = (x1,x2), i.e.,

R = (x,r) =

{x1r1(a)+ x2r2(a),x1r1(b)+ x2r2(b).

. (5.34)

If there is no sure-win, the vectors r(a) and r(b) should be in line with the oppositedirection, i.e.,

r(a) =−αr(b) (5.35)

for some constant α ≥ 0. Indeed, if not, we can always find some vector x havingpositive inner product with both r(a) and r(b). It is easy to see that

|r1(b)||r1(a)|+ |r1(b)|

=|r2(b)|

|r2(a)|+ |r2(b)|=

1α +1

. (5.36)

Set

p =1

α +1. (5.37)


By (5.35),

pr1(a)+(1− p)r1(b) =1

|r1(a)|+ |r1(b)|{|r1(b)|r1(a)+ |r1(a)|r1(b)}

=1

|r1(a)|+ |r1(b)|{−|r1(b)|αr1(b)+ |αr1(b)|r1(b)}

= 0.

Similarly, we have

pr2(a)+(1− p)r2(b) = 0. (5.38)

Thus, p is the risk-neutral probability.

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 theexperiment result is J = j, we can get ri( j). Note that the amount of bet can benegative.

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

n

∑i=1

xiri(J). (5.39)

Theorem 5.3 (arbitrage theorem). Either there exists risk neutral probability, orthere is the sure-win strategy (arbitrage). More precisely, either one of the follow-ings 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, (5.40)

where we regard p j = P{J = j}.

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

n

∑i=1

xiri( j) > 0 for all outcome j. (5.41)

5.5. REFERENCES 33

5.5 Referencesreference here!


Chapter 6

Tools for Modeling Dynamics ofStock Prices

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

6.1 Geometric Brownian MotionsDefinition 6.1 (Geometric Brownian motion). We say S(t) is a geometric Brow-nian motion with the drift parameter µ and the volatility parameter σ if for anyy≥ 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 µy and itsvariance σ2y.

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

• the future price growth rate is independent of the past price, and

• the distribution of the growth rate is distributed as the lognormal with theparameter µt and σ2t..

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

34

6.1. GEOMETRIC BROWNIAN MOTIONS 35

Lemma 6.1. If S(t) is a geometric Brownian motion, we have

E[S(t)|S(0) = s0] = s0et(µ+σ2/2). (6.1)

Proof. Set

Y =S(t)S(0)

=S(t)s0

.

Then, Y is lognormal with (tµ, tσ2). Thus by using Theorem 2.3 we have

E[Y ] = et(µ+σ2/2).

On the other hand, we have

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

s0.

Hence, we have (6.1).

Remark 6.1. 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 log-normal distribution has some bias, σ affects the average exponential growth rate.

Theorem 6.1. Let S(t) be a geometric Brownian motion with its drift µ andvolatility σ , then for a fixed t ≥ 0, S(t) can be represented as

S(t) = S(0)eW , (6.2)

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

Proof. We need to check that (6.2) satisfies the second part of Definition 6.1,which is easy by taking log on both sides in (6.2) and by seeing

log(

S(t)S(0)

)= W. (6.3)

36 CHAPTER 6. TOOLS FOR MODELING DYNAMICS OF STOCK PRICES

Sometimes instead of Definition 6.1, we use the following stochastic differen-tial equation to define geometric Brownian motions,

dS(t) = µS(t)dt +σS(t)dB(t), (6.4)

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

Note that the standard Brownian motion is continuous function but nowheredifferentiable, so the terms like dB(t)/dt should be treated appropriately. Thus,the following parts is not mathematically rigorous. The solution to this equationis

S(t) = S(0)e(µ−σ2/2)t+σB(t). (6.5)

To see this is the solution, we need “unordinary” calculus. We will see the detailin Theorem 11.2.

Problem 6.1. Use “ordinary” calculus to obtain the derivative,

dS(t)dt

, (6.6)

formally. Check if it satisfies (6.4), or not.

6.2 Discrete Time Tree Binomial ProcessIn this section, we imitate a stock price dynamic on the discrete time, which issubject to geometric Brownian motion.

Definition 6.2 (Tree binomial process). The process Sn is said to be a Tree Bino-mial process, if the dynamics is express as

Sn =

{uSn−1 with probability p,

dSn−1 with probability 1− p,(6.7)

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

Problem 6.2. Figure 6.1 depicts a sample path of Tree Binomial Process. Do youthink it is reasonably similar to Figure 6.2, which is a exchange rate of yen?

6.2. DISCRETE TIME TREE BINOMIAL PROCESS 37

20 40 60 80 100

80

90

100

110

120

130

Figure 6.1: An example of Tree binomial process with d = 0.970446,u =1.03045, p = 0.516667 and starting from S0 = 100.

Figure 6.2: Exchange rate of yen at 2005/10/19. Adopted fromhttp://markets.nikkei.co.jp/kawase/index.cfm


Theorem 6.2 (Approximation of geometric brownian motion). A geometric Brow-nian motion S(t) with the drift µ and the volatillity σ can be approximated by thelimit of a tree binomial process with the parameters:

d = e−σ√

∆, (6.8)

u = eσ√

∆, (6.9)

p =12

(1+

µ

σ

√∆

). (6.10)

Proof. Let ∆ be a small interval and n = t/∆. Define a tree binomial process onthe discrete time {i∆} such as

S(i∆) =

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

(6.11)

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

Yi =

{1 up at time i∆,0 down at time i∆.

(6.12)

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

n−∑ni=1Yi. 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, (6.13)

and show that the limit S(t) is geometric Brownian motion with the drift µ andthe volatility 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. (6.14)

6.2. DISCRETE TIME TREE BINOMIAL PROCESS 39

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. (6.15)

Consider taking the limit ∆ → 0, then the number of terms in the sum increases.Using Central Limit Theorem (Theorem 3.5), we have the approximation:

t/∆

∑i=1

Yi ∼ Normal distribution. (6.16)

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,

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], (6.17)


and moreover S(t + y)/S(t) is independent with the past because of the construc-tion of S(t). Hence the limiting distribution of S(t) with ∆ → 0 is a geometricBrownian motion, which means all the geometric Brownian motion S(t) with thedrift µ and the volatility σ can be approximated by an appropriate tree Binomialprocess.

Remark 6.2. In the following, to show the properties of geometric Brownian mo-tions, we sometimes check the properties holds for the corresponding tree Bino-mial Process and then taking appropriate limit to show the validity of the proper-ties of geometric Brownian motions.

Example 6.1 (Trajectories of geometric brownian motions). This is an exampleof different trajectories of Geometric Brownian Motions. Here we set the driftµ = 0.1, the volatility σ = 0.3 and the interval ∆ = 0.01 in the geometric brow-nian motion. As pointed out in Remark 6.2, we use tree binomial process as anapproximation of geometric brownian motion. Thus, the corresponding parame-ters are d = 0.970446,u = 1.03045 and p = 0.516667.

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,the volatility σ = 0.3 and the interval ∆ = 0.01 and starting from S0 = 100.

It is important to see that the path from geometric brownian motion will differtime by time, even we have the same parameters. It can rise and down. Figure 6.4

6.3. BROWNIAN MOTIONS 41

gathers those 10 different trajectories. Since we set the drift µ = 0.1 > 0, we cansee up-ward trend while each trajectories are radically different.

On the other hand, Figure 6.5 shows the geometric brownian motion with thenegative drift µ =−0.1, and the probability of the upward change p = 0.483333.Can you see the difference between Figure 6.5 and Figure 6.4?

20 40 60 80 100

80

100

120

140

160

180

200

Figure 6.4: Many trajectories of geometric brownian motion with the upwarddrift µ = 0.1, the volatility σ = 0.3 and the interval ∆ = 0.01 and starting fromS0 = 100.

6.3 Brownian MotionsDefinition 6.3 (Brownian motion). A stochastic process S(t) is said to be a Brow-nian motion, if S(t + y)− S(y) is a normal random variable N[µt,σ2t], which isindependent with the past history up to the time y.

Note that for a Brownian motion, we have

E[S(t)] = µt, (6.18)

Var[S(t)] = σ2t. (6.19)

Thus, both quantities are increasing as t increases.


20 40 60 80 100

100

120

140

Figure 6.5: Many trajectories of geometric brownian motion with the downwarddrift µ =−0.1.

Brownian motions is more widely used than geometric Brownian motions.However, to apply them in finance theory, Brownian motion has two major drawbacks. First, Brownian motions allows to have negative value. Second, in Brow-nian motion, the price differences on the same time intervals with same length isstochastically same, no matter the initial value, which is not realistic in the realfinance world.

An example of trajectroy of two-dimensional brownian motion can be foundin Figure 6.6.

6.4 Poisson ProcessesDefinition 6.4 (Counting process). Let N(t) be the number of event during [0, t).The process N(t) is called a counting process.

Definition 6.5 (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), (6.20)

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

6.4. POISSON PROCESSES 43

-20 20 40 60

-80

-60

-40

-20

Figure 6.6: An example of trajectory of two-dimensional brownian motion.

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

Definition 6.6 (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. P{N(t + s)−N(s) = n}= e−λ t (λ t)n

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

Theorem 6.3. Let N(t) be a Poisson process with its rate λ , then we have

E[N(t)] = λ t, (6.21)Var[N(t)] = λ t. (6.22)

Or rewriting this, we have

λ =E[N(t)]

t, (6.23)

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


Proof. By Definition 6.6, we have

E[N(t)] =∞

∑n=0

nP{N(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 6.2).

6.5 References

6.6 ExercisesExercise 6.1. Find E[Sn] and Var[Sn] for tree binomial process.

Exercise 6.2. Let N(t) be a Poisson process with its rate λ , and prove

Var[N(t)] = λ t. (6.24)

Chapter 7

On a Way to Black-Scholes Formula

7.1 The Only Possible Process to Avoid Arbitrage

We show the particular tree binomial process will emerge form arbitrage theorem.This is the process we will assume in the analysis of the stock dynamics.

Theorem 7.1 (Risk-neutral tree binomial process). Consider the stock price onthe discrete time. Let r be the interest rate, and let Sn be the stock price at time nand satisfies the following special dynamics:

Sn =

{uSn−1,

dSn−1,(7.1)

where we assume d < (1 + r) < u. If there is no arbitrage opportunity, then theprocess Sn 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,(7.2)

Note that (X1,X2, . . . ,Xn) determines the stock price Sn. By Arbitrage Theorem(Theorem 5.2), in order to avoid arbitrage, the expected return of all bet is equalto zero.

45

46 CHAPTER 7. ON A WAY TO BLACK-SCHOLES FORMULA

First, let us consider an investment scheme. The stock price data up to timen−1 are used only for the observation. According to some rule, we decide to buystock or not. More precisely, if

(X1,X2, . . . ,Xn−1) = (x1,x2, . . . ,xn−1), (7.3)

we buy the stock at time n−1 by borrowing money from a bank, and sell it at timen.

Problem 7.1. Describe why this is fairly common rule.

Let

α = P{(X1,X2, . . . ,Xn−1) = (x1,x2, . . . ,xn−1)}, (7.4)p = P{Xn = 1|(X1,X2, . . . ,Xn−1) = (x1,x2, . . . ,xn−1)}. (7.5)

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, (7.6)

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

E[profit from buying the stock] = puSn−1 +(1− p)dSn−1− (1+ r)Sn−1, (7.7)

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

E[Gn] = α{puSn−1 +(1− p)dSn−1− (1+ r)Sn−1}. (7.8)

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

0 = E[Gn] = α{puSn−1 +(1− p)dSn−1− (1+ r)Sn−1}, (7.9)

or,

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

u−d, (7.10)

which is independent of our specific rule (x1,x2, . . . ,xn−1). This means that Xnis independent of (X1,X2, . . . ,Xn−1). Accordingly, using inductive arguments, weconclude that (X1,X2, . . . ,Xn) are independent. Thus, the process Sn is a tree bi-nomial process.

7.2. OPTION PRICE ON THE DISCRETE TIME 47

7.2 Option Price on the Discrete TimeDefinition 7.1 (Option and its Strike price and expiration time). Let Sn be theprice of a stock at time n. Consider an option that we can buy a stock at the priceK at time n. The value K is called the strike price and n is called the expirationtime of the option.

Theorem 7.2 (Option price on binomial process). If there is no arbitrage oppor-tunity, the option price C should be

C = (1+ r)−nE[(Sn−K)+], (7.11)

where Sn is defined and proved to be the risk-neutral binomial tree process inTheorem 7.1.

Proof. We need to find out appropriate value of C, which gives no sure-win. ByTheorem 7.1, we know that Sn should be the risk-neutral binomial process whenthere is no arbitrage opportunities.

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

(Sn−K)+, (7.12)

where x+ = max(x,0). Thus, the present value of the expected return of this in-vestment R is

E[R] = (1+ r)−nE[(Sn−K)+]−C. (7.13)

By Arbitrage Theorem (Theorem 5.3), this should be zero, or we have an arbi-trage. Thus,

C = (1+ r)−nE[(Sn−K)+]. (7.14)

7.3 Black-Scholes ModelNow we proceed to the continuous time model. Let S(t) be a stock price at timet. We assume the stock price S(t) is a geometric Brownian motion with the drift


µ and the volatility σ as defined in Section 6.1. As noted in Theorem 6.2, if wedivide t into n interval, all geometric Brownian motion can be approximated bydiscrete-time binomial tree process with

Sn =

{uSn−1 with probability p,

dSn−1 with probability 1− p,(7.15)

where

d = e−σ√

∆, (7.16)

u = eσ√

∆, (7.17)

p =12

(1+

µ

σ

√∆

). (7.18)

and ∆ = t/n is the time interval. However, by Theorem 7.1. if we assume there isno arbitrage, the binomial process should be risk-neutral with

p = P{up}=1+ rt/n−d

u−d, (7.19)

since the nominal interest rate is rt/n. Using the Taylor expansion of the expo-nential function, we have

d = e−σ√

t/n = 1−σ√

t/n+σ2t2n

+O((t/n)3/2), (7.20)

u = eσ√

t/n = 1+σ√

t/n+σ2t2n

+O((t/n)3/2). (7.21)

Using these in (7.19), we have

p≈ 12

(1+

r−σ2/2σ

√t/n

), (7.22)

for a large n.

Problem 7.2. Derive (7.20) by using Taylor expansion and find that p shouldsatisfy (7.22) for large n.

Comparing this to (7.18), we can conclude that the original geometric Brown-ian motion has to have the drift

µ = r−σ2/2, (7.23)

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

7.3. BLACK-SCHOLES MODEL 49

Theorem 7.3 (Risk neutral geometric Brownian motion). Let r be the nominalinterest rate. If there is no arbitrage, the geometric Brownian motion representingstock price dynamics should have the drift µ = r−σ2/2 and the volatility σ . Thisprocess is called the risk neutral geometric Brownian motion.

Problem 7.3. Give intuitive explanation why the volatility σ affect the drift µ =r−σ2/2.

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

Theorem 7.4 (Black-Scholes formula). Consider a call option of the stock withstrike price K and the expiration time t. Let C be the price of this option. If weassume there is no arbitrage opportunity, then we have

C = e−rtE[(S(t)−K)+], (7.24)

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, includingpurchase of the option, should be zero. Thus, supposing to buy a option with costC, we have

E[gain at time t] = E[(S(t)−K)+−Cert ] = 0, (7.25)

which results (7.24).

Remark 7.1. Since the term e−rt represents mapping back to present value, (7.24)is nothing but the present value of expectation of option payoff.

Although, (7.24) is simple for our eyes, it is hard to evaluate.

Problem 7.4. Use Black-Scholes formula (7.24) to analyze the effect of changingparameters. What will happen to the option price if

1. the strike price K increases.

2. the interest rate r drops.


Corollary 7.1 (Original Black-Scholes formula). Given S(0) = s, the option priceC is obtained by

C = e−rtE[(seW −K)+] (7.26)

= sΦ(ω)−Ke−rtΦ(ω−σ

√t), (7.27)

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

ω =rt +σ2t/2− log(K/s)

σ√

t. (7.28)

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

S(t) = seW (7.29)

= se(r−σ2/2)t+σ√

tZ, (7.30)

where Z is the standard normal random variable.

Lemma 7.1. By using the representation in (7.29), the following two set of eventsare considered to be equivalent:

{S(t) > K}⇐⇒{

Z > σ√

t−ω}

. (7.31)

Proof. By (7.29), we have

{S(t) > K}⇐⇒{

e(r−σ2/2)t+σ√

tZ >Ks

}. (7.32)

Since log is increasing function, we have{e(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. (7.33)

7.3. BLACK-SCHOLES MODEL 51

Lemma 7.2. Let I be the indicator such as

I =

{1 if S(t) > K,

0 if S(t)≤ K.(7.34)

Then,

E[I] = P{S(t) > K}= Φ(σ√

t−ω). (7.35)

Proof. Using Lemma 7.1, we have

E[I] = P{I = 1}= P{S(t) > K}= P{Z > σ

√t−ω}

= P{Z < ω−σ√

t}.

Lemma 7.3.

e−rtE[IS(t)] = sΦ(ω). (7.36)

Proof. Set α = σ√

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

E[IS(t)] = E[1{Z>α}se(r−σ2/2)t+σ

√tZ

]=

∫∞

α

se(r−σ2/2)t+σx√

tdP{Z ≤ 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 standardnormal distribution, we have

= sertP{Z >−ω}= sert

Φ(ω),

by the symmetry of Φ.

Proof of Corollary 7.1. Since (S(t)−K)+ = I(S(t)−K), by Theorem 7.4 and ,we have

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 7.2 and 7.3.

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

Example 7.1 (Option price via Black-Scholes). Suppose the current price of astock s = S(0) = 30. The yearly interest rate r = 0.08 and the volatility σ = 0.20.Let C be the price of call option with the strike price K = 36 and its expirationdate is 3 month from now. We will estimate C.

Then, t = 1/4 and ω in (7.28) 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 (7.27) in Corollary 7.1, we have

C = sΦ(ω)−Ke−rtΦ(ω−σ

√t),

= 30 ·Φ(−1.57322)−36e0.08·1/4Φ(−1.57322−0.2

√1/4)

= 0.0715115

7.5. REFERENCES 53

Thus, the price of this option is 0.07. Table 7.1 shows the return of stock andoption purchase. You can get more profit when the stock price rises.

Table 7.1: 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 7.2,

P{S(t) > K}= Φ(σ√

t−ω) = 0.0471424. (7.37)

With high probability you will lose your money with this option. Thus, you canunderstand that option is high risk and high return.

In Figure 7.1, we show the simulation of the stock price dynamics. Youropinion?

20 40 60 80 100

25

30

35

40

Figure 7.1: 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.



7.6 ExercisesExercise 7.1. Estimate the value of following call options, given the current priceof a stock s = S(0) = 30, and compare the result with Example 7.1. What can yousay about 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 7.2. Give two examples where buying options is more appropriate thanbuying its stocks.

Chapter 8

Delta Hedging Strategy

We have already known that we can duplicate option for one term in Section 5.3.This idea can be extended to multiple terms.

8.1 Bernouilli Hedging ModelAs usual, we use the discrete-time model of stock price. Let S0 = s be the initialprice and

S1 =

{us,ds.

(8.1)

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

P1 =

{a up,

b down.(8.2)

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

R1 =

{ys ·u+(x− ys)(1+ r) up,

ys ·d +(x− ys)(1+ r) down,(8.3)

55

56 CHAPTER 8. DELTA HEDGING STRATEGY

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

ys ·u+(x− ys)(1+ r) = a, (8.4)ys ·d +(x− ys)(1+ r) = b, (8.5)

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

y(us−ds) = a−b.

y =a−b

s(u−d).

Substituting this y into (8.4), 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.

](8.6)

Here we use the risk-neutral probability p defined in Theorem 7.1 as

p =1+ r−d

u−d, (8.7)

1− p =u−1− r

u−d. (8.8)

Then,

x = pa

1+ r+(1− p)

b1+ r

=1

1+ r{pa+(1− p)b}, (8.9)

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.Note that in this case, we should buy the stock as much as

y =a−b

s(u−d), (8.10)

which is the ratio of the difference of payoff to the stock price change.

8.2. BINOMIAL HEDGING MODEL 57

8.2 Binomial Hedging ModelNow we proceed to the time 2. In this case, the possibilities of payoff are not onlytwo like a and b in previous discussion, but three cases.

Problem 8.1. Explain why we have three cases, not four.

Let xi,2 be the money required to cover the payoff at time 2, given that

S2 = uid2−is, (8.11)

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 time2, we know that the money required at the time 1, say x1,1, should be

x1,1 =1

1+ r{px2,2 +(1− p)x1,2}. (8.12)

Note that p = (1 + r− d)/(u− d) depends on the stock price dynamics d,u butdoesn’t depend on the payoff. In order to achieve the payoff, we need to buy stockas much as

y1,1 =x2,2− x1,2

us(u−d), (8.13)

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

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

x0,1 =1

1+ r{px1,2 +(1− p)x0,2}, (8.14)

with

y0,1 =x1,2− x0,2

ds(u−d). (8.15)

Summarize the results (8.12) and (8.14), we have

xi,1 = E[ the time-1 present value of payoff at time 2|S(1) = uis], (8.16)


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 thispayoff, we need the money as

x0,0 =1

1+ r{px1,1 +(1− p)x0,1}. (8.17)

Using (8.12) and (8.14), we have

x0,0 =1

(1+ r)2 [p{px2,2 +(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

]. (8.18)

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

y0,0 =x1,1− x0,1

s(u−d). (8.19)

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

x0,0 = E[the present value of payoff at time n|S(0) = s]. (8.20)

Here’s an example how to meet the requirement by adjusting our stock posi-tion.

1. Prepare the money equal to x0,0, and buy the stock as much as y0,0 at time0. The rest is kept at the bank.

2. At time 1, we happen to have S1 = us (up). The value of our investmentat time 1 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 sellingor buying, and put the rest at the bank.

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

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

8.2. BINOMIAL HEDGING MODEL 59

Sk = uidk−is. Then

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

=1

1+ r{pxi+1,k+1 +(1− p)xi,k+1} (8.21)

= (1+ r)−(n−k)n−k

∑j=0

(n− k

j

)p j(1− p)n−k− jxi+ j,n, (8.22)

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

yi,k =xi+1,k+1− xi,k+1

Sk(u−d), (8.23)

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)+. (8.24)

for i = 1,2, . . . ,n. Clearly, this is nothing but the payoff of the call option withstrike price K and expiration time n. Thus, if we follow our investment strategy,we successfully replicate the option payoff.

By the law of one price (Theorem 5.1), the initial cost of our investment strat-egy x0,0 should be equal to the option price C which was derived from Theorem7.2).

Thus, we obtained the following theorem.

Theorem 8.1 (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)+], (8.25)

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+ r{pxi+1,k+1 +(1− p)xi,k+1} (8.26)

= (1+ r)−(n−k)n−k

∑j=0

(n− k

j

)p j(1− p)n−k− jxi+ j,n. (8.27)

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

yi,k = (xi+1,k+1− xi,k+1)/(Sk(u−d)), (8.28)

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


8.3 Hedging in Black-Scholes ModelProblem 8.2. What will happen when we let the time interval ∆→ 0 in the equa-tion below?

yi,k = (xi+1,k+1− xi,k+1)/(Sk(u−d)), (8.29)

Here’s a powerful tool to investigate limits.

Lemma 8.1 (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 fixedpoint c ∈ (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)

. (8.30)

Proof. See Exercise 8.1.

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 be the expiration time and σ be the volatility of the geometric brownian motion.

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

S(h) =

{seσ

√h,

se−σ√

h.(8.31)

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

C(seσ√

h, t−h), (8.32)

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

h and theexpiration time t−h. Similarly, if stock price goes down, we must prepare

C(se−σ√

h, t−h). (8.33)

Thus, by the argument in Section 8.2 and (8.10), we know that buying the stockas much as D,

D(h) =C(seσ

√h, t−h)−C(se−σ

√h, t−h)

seσ√

h− se−σ√

h, (8.34)

8.3. HEDGING IN BLACK-SCHOLES MODEL 61

we can cover the money required to buy the option at time h, given that we haveC(s, t) at the time 0. Now take h→ 0 in (8.34), 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 . (8.35)

where a =√

h. Since both the denominator and numerator are converges to 0 ash→ 0, we need to apply l’Hospital’s Rule (Lemma 8.1). Then,

D(0) = lima→0

dda

[C(seσa, t−a2)−C(se−σa, t−a2)

]dda [seσa− se−σa]

(8.36)

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, (8.36) can be rewritten as

D(0) = limh→0

D(h) =2sσCy(s.t)

2sσ

=∂C(s, t)

∂ s. (8.37)

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

Thus, we have the following theorem:


Theorem 8.2 (Delta hedge in geometric brownian process). Let S(t) be the risk-neutral geometric brownian process. Set the initial investment money as

C(s, t) = e−rtE[(S(t)−K)+], (8.38)

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

∆ =∂C(s, t)

∂ s, (8.39)

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

8.4 Partial Derivative ∆

As in Section 8.3, the partial derivative called ∆ is important for hedging.

Theorem 8.3 (Partial Derivative ∆). The partial derivative ∆ is obtained by

∆ =∂C(s, t)

∂ s= Φ(ω), (8.40)

where Φ(ω) is defined as in Lemma 7.3.

Proof. First, recall that

C(s, t) = E[e−rt(S(t)−K)+]. (8.41)

Set the indicator function I = 1{S(t)>K} as in Lemma 7.2, then

e−rt(S(t)−K)+ = e−rt(S(t)−K)I. (8.42)

Thus,

∆ =∂C(s, t)

∂ s

=∂

∂ sE[e−rt(S(t)−K)I]

= E[

∂

∂ s{e−rt(S(t)−K)I}

].

8.5. REFERENCES 63

Now, we have

∂

∂ s{e−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, (8.43)

on {S(t) 6= K}. Since P{S(t) = K}= 0, we have

∆ = E[

∂


∣∣∣∣S(t) 6= K]

P{S(t) 6= K}

+E[

∂


∣∣∣∣S(t) = K]

P{S(t) = K}

= E[

I∂{e−rt(S(t)−K)}

∂ s

](8.44)

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

S(t) = se(r−σ2)t+σ√

tZ, (8.45)

for some standard normal random variable Z (see (7.29)). Thus,

∂

∂ se−rt(S(t)−K) = e−rt ∂S(t)

∂ s

=S(t)e−rt

s. (8.46)

Using (8.46) in (8.44), we have

∆ =e−rt

sE[IS(t)] (8.47)

In Lemma 7.3, we have already shown that

e−rtE[IS(t)] = sΦ(ω). (8.48)



8.6 ExercisesExercise 8.1. Use

f ′(x) = limx−c

f (x)− f (c)x− c

(8.49)

to prove l’Hospital’s Rule (Lemma 8.1).

Chapter 9

Stochastic Integral

Consider the asset price on continuous time. Naturally, we need to extend thenotion of integral to stochastic integral, which is quite different with normal (Rie-mann) integral.

9.1 Diffusion ProcessDiffusion process is a building block of stochastic integral.

Definition 9.1 (Diffusion process). A stochastic process X(t) is said to be a dif-fusion process if

X(t +∆t)−X(t) = µ(t,X(t))∆t +σ(t,X(t))Z(t). (9.1)

Here, Z(t) is independent normally distributed disturbance term, µ is drift term,and σ is so-called diffusion term.

Definition 9.2 (Wiener process). A stochastic process W (t) is said to be a Wienerprocess if

1. W (0) = 0.

2. Independent increment, i.e., for r < s≤ t < u

W (u)−W (t),W (s)−W (r) (9.2)

are independent.

65

66 CHAPTER 9. STOCHASTIC INTEGRAL

3. For s < t, W (t)−W (s) is normally distributed as N[0, t− s].

4. W(t) has a continuous trajectory.

Remark 9.1. Sometimes, Wiener process is called by Brownian motion.

Using the Wiener process W (t) in (9.1), we can rewrite it as

X(t +∆t)−X(t) = µ(t,X(t))∆t +σ(t,X(t))∆W (t), (9.3)

where ∆W (t) =W (t +∆t)−W (t). Dividing both sides with ∆t and letting ∆t → 0,we have

dX(t)dt

= µ +σv(t). (9.4)

Here,

v(t) =dW (t)

dt. (9.5)

Remark 9.2. The process v(t) cannot be exist, while the Wiener process W (t)exists mathematically. Indeed W (t) cannot be differentiable almost everywhere.

Instead of dividing both sides with ∆t, just take ∆t → 0 in (9.1), then we have{dX(t) = µdt +σdW (t),X(0) = a.

(9.6)

Integrating (9.28) over [0, t], we have

X(t) = a+∫ t

0µds+

∫ t

0σdW (s). (9.7)

The term∫ t

0 µds is normal integral, so it is OK. However, we have a problem forthe term

∫ t0 σdW (s), which cannot be defined by ordinary integral.

Thus, in the following section, we need to study the integral like∫ t

0g(s)dW (s). (9.8)

9.2. INFORMATION 67

9.2 InformationDefinition 9.3 (Information of X(t)). Define F X

t by the information generated byX(s) on s ∈ [0, t]. Roughly speaking, F X

t is “what has happened to X(s) up totime t.”

When we can decide an event A happened or not by the the trajectory of X(t),we say that A is F X

t -measurable, and

A ∈F Xt . (9.9)

When a random variable Z can be completely determined by the trajectory of X(t),we write

Z ∈F Xt . (9.10)

Definition 9.4. A stochastic Y is said to be adopted to the filtration {F Xt } when

Y (t) ∈F Xt , (9.11)

for all t.

Example 9.1. These are examples of information.

1.

A = {X(s)≤ 3.14 for all s≤ 9}. (9.12)

Then, A ∈F X9 .

2.

A = {X(10) > 8}. (9.13)

Then, A ∈F X10, but A 6∈F X

9 .

3.

Z =∫ 5

0X(s)ds. (9.14)

Then, Z ∈F X5 .


4. Let W be the Wiener process, and

X(t) = sups≤t

W (s). (9.15)

Then, X is adopted to F Xt . However, when

Y (t) = sups≤t+1

W (s), (9.16)

Y is not adopted to F Xt .

9.3 Definition of Stochastic IntegralDefinition 9.5.

g ∈L 2[a,b], (9.17)

when

1.∫ b

a E[g(s)2]ds < ∞,

2. g is adopted FWt .

Our objective is to define the stochastic integral∫ b

a g(s)dW (s) on g∈L 2[a,b].Assume g is simple, i.e.,

g(s) = g(tk) for s ∈ [tk, tk+1). (9.18)

Then, the stochastic integral can be defined by

∫ b

ag(s)dW (s) =

n−1

∑k=0

g(tk)[W (tk+1−W (tk)]. (9.19)

Remark 9.3. Note that this stochastic integral is forward incremental.

Problem 9.1. Prove g ∈L 2[a,b].

When g is not simple, we will proceed as follows:

1. Approximate g with simple functions gn.

9.3. DEFINITION OF STOCHASTIC INTEGRAL 69

2. Take n→ ∞ in ∫ b

agn(s)dW (s)→

∫ b

ag(s)dW (s). (9.20)

Theorem 9.1. Here are some important properties for our stochastic integral.When g ∈L 2, we have

1.

E[∫ b

ag(s)dW (s)

]= 0, (9.21)

2.

E

[(∫ b

ag(s)dW (s)

)2]

=∫ b

aE[g2(s)]ds, (9.22)

3. ∫ b

ag(s)dW (s):F X

b -measurable. (9.23)

Proof. We prove the case only when g(t) is simple. Set g(t) = A1[a,b] with A ∈F X

a . Since A is independent with the future W (b)−W (a), we have

E[∫ b

ag(s)dW (s)

]= E[A(W (b)−W (a))]

= E[A]E[W (b)−W (a)] = 0.

Similarly, we have

E

[(∫ b

ag(s)dW (s)

)2]

= E[A2(W (b)−W (a))2]

= E[A2]E[(W (b)−W (a))2]

= E[A2]t

=∫ b

aE[g2(s)]ds.

We can prove the general case using the approximation with simple functions.


9.4 MartingaleDefinition 9.6. Given a filtration Ft (the history observed up time t), a stochasticprocess Xt is said to be martingale, when

1. Xt is adopted to Ft ,

2. E[|X(t)|] for all t,

3. for all s≤ t,

E[X(t)|Ft ] = X(s). (9.24)

Theorem 9.2. Every stochastic integral is martingale. More precisely,

X(t) =∫ t

0g(s)dW (s), (9.25)

is an FWt -martingale.

Theorem 9.3. A stochastic process X(t) is martingale if and only if

dX(t) = g(t)dW (t), (9.26)

which means X(t) has no dt-term.

9.5 Calculus of Stochastic IntegralWe consider a stochastic process X(t) satisfying following stochastic integral:

X(t) = a+∫ t

0µ(s)ds+

∫ t

0σ(s)dW (s). (9.27)

Equivalently, we write {dX(t) = µ(t)dt +σ(t)dW (t),X(0) = a.

(9.28)

(9.28) is called stochastic differential equation.

Remark 9.4. We can see the infinitesimal dynamics more easily in the stochasticdifferential equations.

9.5. CALCULUS OF STOCHASTIC INTEGRAL 71

Remark 9.5. Wiener process W (t) is continuous but perfectly rugged. Thus, weneed unusual Ito-calculus.

Since W (t)−W (s) is N[0, t− s],

E[∆W ] = 0, (9.29)

E[(∆W )2] = ∆t, (9.30)Var[∆W ] = ∆t, (9.31)

Var[(∆W )2] = 2(∆t)2. (9.32)

For small ∆t, Var[(∆W )2] rapidly vanish, so deterministically we have

(∆W )2 ∼ ∆t. (9.33)

Thus, we can set

(dW )2 = dt, (9.34)

or ∫ t

0(dW )2 = t. (9.35)

Remark 9.6. Intuitive proof can be found in text Bjork [2004] p26.

Theorem 9.4 (Ito’s formula). Given a stochastic process X(t) with

dX(t) = µdt +σdW (t), (9.36)

set

Z(t) = f (t,X(t)). (9.37)

Then, we have the following change of variables rule:

dZ(t) = d f (t,X(t)) (9.38)

={

∂ f∂ t

+ µ∂ f∂x

+12

σ2 ∂ 2 f

∂x2

}dt +σ

∂ f∂x

dW (t), (9.39)

or more conveniently,

d f =∂ f∂ t

dt +∂ f∂x

dX +12

∂ 2 f∂x2 (dX)2. (9.40)


Roughly speaking, Ito’s formula reveals that the dynamics of a function ofstochastic process depends not only on the first order but also the second order ofthe underlying stochastic process.

Proof. Take Taylor expansion of f ,

d f =∂ f∂ t

dt +∂ f∂x

dX +12

∂ 2 f∂x2 (dX)2 +

12

∂ 2 f∂ t2 (dt)2 +

∂ 2 f∂x∂ t

dtdX . (9.41)

Note that the the term (dX)2 should be ignored in normal differential, but instochastic differential this terms survive as we see in the following.

By (9.36), formally,

(dX)2 = µ2(dt)2 +2µσdtdW +σ

2(dW )2. (9.42)

Now we can igonre those terms with (dt)2 and dtdW which converges much fasterthan dt. Also, by (9.34), (dW )2 = dt. Substituting these, we have the result.

Lemma 9.1. You can use the following conventions:

(dt)2 = 0, (9.43)dtdW = 0, (9.44)

(dW )2 = 0. (9.45)

Theorem 9.5 (n-dimensional Ito’s formula). Given an n-dimensional stochasticprocess

X(t) = (X1(t),X2(t), . . . ,Xn(t)), (9.46)

with n-dimensional diffusion equation,

dX(t) = µdt +σdW (t), (9.47)

set

Z(t) = f (t,X(t)). (9.48)

Then, we have the following change of variables rule:

d f =∂ f∂ t

dt +∑i

∂ f∂xi

dXi(t)+12 ∑

i, j

∂ 2 f∂xix j

dXi(t)dX j(t). (9.49)

9.5. CALCULUS OF STOCHASTIC INTEGRAL 73

Corollary 9.1. Let X(t) and Y (t) be diffusion processes, then

d(X(t)Y (t)) = X(t)dY (t)+Y (t)dX(t)+dX(t)dY (t). (9.50)

Proof. Use Ito’s formula (Theorem 9.5) and

dWi(t)dWj(t) = δi jdt, (9.51)dWidt = 0. (9.52)

Chapter 10

Examples of Stochastic Integral

For pricing via arbitrage theory, we need to evaluate the expectation of Wienerprocess. We can use Ito’s formula for the evaluation.

10.1 Evaluation of E[W (t)4]

We already know that the mean and variance of W (t) by definition, i.e.,

E[W (t)] = 0, (10.1)Var[W (t)] = t. (10.2)

But how about the higher moments?

Theorem 10.1. Let W (t) be a Wiener process, then

E[W (t)4] = 3t2. (10.3)

Proof. Let f (t,x) = x4 and

Z(t) = f (t,W (t)) = W (t)4. (10.4)

In Ito’s formula (Theorem 9.4), we can set µ = 0 and σ = 1 and,

dX(t) = 0dt +1dW (t) = dW (t). (10.5)

Then,

dZ(t) =∂ f∂ t

dt +∂ f∂x

dX +12

∂ 2 f∂x2 (dX)2. (10.6)

74

10.1. EVALUATION OF E[W (T )4] 75

Check∂ f∂ t

= 0, (10.7)

∂ f∂x

= 4x3, (10.8)

∂ 2 f∂x2 = 12x2, (10.9)

∂ 2 f∂ t2 = 0, (10.10)

and we have

dZ(t) = 6W (t)2dW (t)2 +4W (t)3dW (t). (10.11)

Since dW (t)2 = dt by Lemma 9.1, we can rewrite the dynamics for Z(t) = W (t)4

as {d(W (t)4) = 6W (t)2dt +4W (t)3dW (t),Z(0) = 0.

(10.12)

Integrating the both side of (10.12), we have

W (t)4 = 6∫ t

0W (s)2ds+4

∫ t

0W (s)3dW (s). (10.13)

Taking the expectation, we have

E[W (t)4] = 6E[∫ t

0W (s)2ds

]+4E

[∫ t

0W (s)3dW (s)

](10.14)

= 6∫ t

0E

[W (s)2]ds+4E

[∫ t

0W (s)3dW (s)

]. (10.15)

Since W (s) is N[0,√

s], we have

E[W (s)2] = s. (10.16)

Also, by Theorem 9.1,

E[∫ t

0W (s)3dW (s)

]= 0. (10.17)

Note that we used the fact W (s)3 ∈L 2. Now we can evaluate (10.14).

E[W (t)4] = 6∫ t

0sds = 3t2. (10.18)

76 CHAPTER 10. EXAMPLES OF STOCHASTIC INTEGRAL

10.2 Evaluation of E[eαW (t)]

Theorem 10.2. Let W (t) be a Wiener process, then

E[eαW (t)] = eα22 t . (10.19)

Proof. Set Z(t) = eαW (t). Then, by Ito’s forumula,

dZ(t) =12

α2eαW (t)dt +αeαW (t)dW (t). (10.20)

Problem 10.1. Show the above equation by using Ito’s formula.

Rewriting this, we have the stochastic differential equation,

dZ(t) =12

α2Z(t)dt +αZ(t)dW (t), (10.21)

Z(0) = 1. (10.22)

In integral form, we have

Z(t) = 1+12

α2∫ t

0Z(s)ds+α

∫ t

0Z(s)dW (s). (10.23)

Since E[∫ t

0 Z(s)dW (s)] = 0, taking the expectation, we have

m(t) = E[Z(t)] = 1+12

α2∫ t

0m(s)ds. (10.24)

Or equivalently,

m′(t) =12

α2m(t), (10.25)

m(0) = 1. (10.26)

By solving this, we have

E[eαW (t)] = m(t) = eα2t/2. (10.27)

Chapter 11

Differential Equations

11.1 Ordinary Differential EquationTheorem 11.1. Consider the following ordinary differential equation,

dB(t)dt

= rB(t), (11.1)

B(0) = B0. (11.2)

The solution of this equation is given by

B(t) = B0ert . (11.3)

Proof. By reordering (11.1), we have

dB(t)B(t)

= rdt. (11.4)

Integrating the both sides, we have∫ t

0

dB(s)B(s)

=∫ t

0rds. (11.5)

Thus,

logB(t) = rt +C, (11.6)

or

B(t) = Cert . (11.7)

77

78 CHAPTER 11. DIFFERENTIAL EQUATIONS

Using the initial condition (11.2), we finally have

B(t) = B0ert . (11.8)

11.2 Geometric Brownian MotionDefinition 11.1 (Geometric Brownian Motion). A stochastic process X(t) satis-fying the following dynamics is called geometric brownian motion.

dX(t) = αX(t)dt +σX(t)dW (t), (11.9)X(0) = x0, (11.10)

where α is called the drift and σ is called the volatility.

Theorem 11.2 (SDE of Geometric Brownian Motion). The solution of the equa-tion

dX(t) = αX(t)dt +σX(t)dW (t), (11.11)X(0) = x0, (11.12)

is given by

X(t) = x0e(α−σ2/2)t+σW (t). (11.13)

Proof. Here the tricks we used in Theorem 11.1 may not work here because dW (t)terms should be considered as the stochastic difference.

To simplify the proof, we assume X(t) > 0. Set

Z(t) = f (t,X(t)) = logX(t). (11.14)

Then by Ito’s formula (Theorem 9.4), we have

dZ(t) =∂ f∂ t

dt +∂ f∂x

dX(t)+12

∂ 2 f∂x2 (dX(t))2

=dX(t)X(t)

− 12

(dX(t))2

X(t)2 . (11.15)

11.3. STOCHASTIC PROCESS AND PARTIAL DIFFERENTIAL EQUATION79

By (11.11),

(dX(t))2 = α2X(t)2(dt)2 +2σX(t)dW (t)αX(t)dt +σ

2X(t)2(dW (t))2

= σ2X(t)2dt. (11.16)

Using this in (11.15) as well as (11.11), we have

dZ(t) = αdt +σdW (t)− 12

σ2dt. (11.17)

Luckily, the right hand side can be integrated directly here!

Z(t)−Z(0) =∫ t

0(α− 1

2σ

2)ds+σ

∫ t

0dW (s) (11.18)

= (α− 12

σ2)t +σdW (t). (11.19)

Using the initial condition (11.12), we have Z(0) = logx0. Thus,

Z(t) = (α− 12

σ2)t +σdW (t)+ logx0, (11.20)

or

X(t) = eZ(t) = x0e(α−σ2/2)t+σW (t). (11.21)

Remark 11.1. This proof has not prove the existence of the solution yet. Morerigorous treatment of the proof, see the textbook like Oksendal [2003].

11.3 Stochastic Process and Partial Differential Equa-tion

Theorem 11.3 (Feynman-Kac Represenation). Consider the following equationof a function F = F(t,x):

∂F∂ t

+ µ∂F∂x

+12

σ2 ∂ 2F

∂x2 − rF = 0, (11.22)

80 CHAPTER 11. DIFFERENTIAL EQUATIONS

with its boundary condition,

F(T,x) = Φ(x). (11.23)

In addition, consider the following stochastic differential equation,

dX(t) = σdt +σdW (t). (11.24)

Then, the solution of (11.22) has the following Feynman-Kac representation:

F(t,x) = e−r(T−t)E[Φ(XT )|X(t) = x]. (11.25)

Chapter 12

Portfolio Dynamics

12.1 Portfolio ModelTheorem 12.1. Assume there is a market including N different stocks. Considerwe have a portfolio consisted with these N stocks. Let Si(t) be the price of stock i.Let hi(t) be the number of stocks in our portfolio at time t. Then, the value of ourportfolio V (t) = S(t)h(t) has the following dynamics:

dV (t) = h(t)dS(t)− c(t)dt, (12.1)

which can be superficially understood as the change of value of our portfolio aredue to the change of the stock price and the consumption we made. Particularly,when c(t) = 0,

dV (t) = h(t)dS(t). (12.2)

Proof. Analyze the dynamics of the value of our portfolio V (t). The value of ourportfolio at just before t, V (t−∆t), can be estimated by

V (t−∆t) = h(t−∆t)S(t) =N

∑i=1

hi(t−∆t)Si(t), (12.3)

for a small ∆t.At time t, within our budget V (t−∆t), we will make change of our portfolio

from h(t−∆t) to h(t) based on the observation of the stock price S(t). When weallow to consume a part of our portfolio, we have

V (t−∆t) = h(t)S(t)+ c(t)∆t, (12.4)

81

82 CHAPTER 12. PORTFOLIO DYNAMICS

where c(t) is the consumption rate at time t. Thus, we have the budget equationas,

S(t)(h(t)−h(t−∆t))+ c(t)∆t = 0. (12.5)

Remark 12.1. It is wrong to conclude

S(t)dh(t)+ c(t)dt = 0, (12.6)

by letting ∆t → 0 in (12.5). Our stochastic calculus needs to have the forwarddifference as,

S(t)(h(t +∆t)−h(t))→ S(t)dh(t). (12.7)

We will rewrite (12.5) by subtracting the term S(t−∆t)(h(t)−h(t−∆t)), as

(S(t)−S(t−∆t))(h(t)−h(t−∆t))+S(t−∆t)(h(t)−h(t−∆t))+ c(t)∆t = 0.

This has the appropriate forms of (present value) × (future difference). Thus, wecan take the limit to have the proper budget equation as,

dS(t)dh(t)+S(t)dh(t)+ c(t)dt = 0. (12.8)

On the other hand, we can evaluate the current value of our portfolio as

V (t) = S(t)h(t). (12.9)

Using Ito’s formula (see Corollary 9.1) on this, we have

dV (t) = h(t)dS(t)+Sdh(t)+dS(t)dh(t). (12.10)

By (12.8),

dV (t) = h(t)dS(t)− c(t)dt, (12.11)

which can be superficially understood as the change of value of our portfolio aredue to the change of the stock price and the consumption we made. Particularly,when c(t) = 0,

dV (t) = h(t)dS(t). (12.12)

12.2. RATE OF RETURN OF STOCK AND RISK-FREE ASSET 83

12.2 Rate of Return of Stock and Risk-free AssetIf the dynamics of the value process of some asset does not have any dW (t)-terms,the value process inceases exponentially without any risk.

Definition 12.1 (Risk-free asset). The stochastic process B(t) is said to be risk-free with its rate of return r(t) if

dB(t)dt

= r(t)B(t). (12.13)

Assumption 12.1. Let S(t) be the price of a stock at time t, and the dynamics ofS(t) is governed by the stochastic differential equation as,

dS(t) = S(t)αdt +S(t)σdW (t). (12.14)

Remark 12.2. Let us compare the rate of return of the risk-free asset B(t) and thestock S(t). By Definition 12.1, formally, we have the rate of return of B(t) as

dB(t)B(t)dt

= r(t), (12.15)

which is observable at time t. On the other hand, the rate of return of S(t) isobtained by

dS(t)S(t)dt

= α +σdW (t)

dt. (12.16)

Here we have the term of dW (t) that is essentially future information.

Definition 12.1. Let B(t) be the value of a risk-free asset and S(t) be the stockprice at time t. The Black-Scholes model is composed by the (B(t),S(t)), whichhas the dynamics as

dB(t) = rB(t)dt, (12.17)dS(t) = αS(t)dt +σS(t)dW (t). (12.18)

12.3 Arbitrage and PortfolioAccording to the result in Section 12.1, we have the budget dynamics of the port-folio by

dV (t) = h(t)dS(t)− c(t)dt, (12.19)

84 CHAPTER 12. PORTFOLIO DYNAMICS

Generally, we have dW (t) terms in the dynamics of underlying stock price dy-namics dS(t). So, overall, portfolio dynamics dV (t) contain the dW (t) terms. Butin some special case, we can make our portfolio so cleverly that we can avoid thedW (t) terms, which means we have risk-free portfolio. In that case, the followinghappens.

Theorem 12.2. Suppose there is a risk free asset as

dB(t) = r(t)B(t)dt. (12.20)

Consider a portfolio h(t) having the dynamics as,

dV (t) = k(t)V (t)dt, (12.21)

for some function k(t). Apparently, the value of our portfolio V (t) does not haveany risk. Assume we have no arbitrage, then the rate of return of our risk-freeportfolio is r(t), i.e.,

k(t) = r(t). (12.22)

Proof. Assume r(t) and k(t) to be constant r and k for simplicity. First, let ussuppose k > r. Then, we can sell the risk-free asset B(t) and use the money in ourportfolio V (t). The rate of return r for selling the risk-free asset can be coveredby our portfolio, which has the rate of return k. This contradicts the no-arbitrageassumption.

In the case of k < r, we can reverse the argument.

Remark 12.3. Thus, there is a unique rate of return for risk-free asset, which turnsout to be the short-term interest rate of bank account.

Chapter 13

Pricing via Arbitrage

13.1 Way to Black-Scholes ModelWe would like to evaluate the price of an option at time t,

Π(t) = Π(t,S(t)), (13.1)

given that the option is expired at time T and have the value

Π(T ) = Φ(S(T )). (13.2)

Typically,

Φ(S(T )) = (S(T )−K)+ = max(S(T )−K,0) (13.3)

Here’s the procedure to derive Black-Scholes equation:

1. Consider the Black-Scholes model,

dB(t) = rB(t)dt, (13.4)dS(t) = αS(t)dt +σS(t)dW (t). (13.5)

2. Show we can build a risk-free portfolio h(t) using the Black-Scholes model,which has no dW (t)-terms.

3. The rate of return of our portfolio should be the short-term interest rate rbecause of Theorem 12.2.

85

86 CHAPTER 13. PRICING VIA ARBITRAGE

4. Based on the assumption above, evaluate the dynamics of the option price

Π(t), (13.6)

using Ito’s formula.

5. Solving the dynamics and use so-called Feyman-Kac representation (Theo-rem 11.3) to have

Π(t,S(t)) = e−r(T−t)E[(S(T )−K)+|S(t)]. (13.7)

Bibliography

T. Bjork. Arbitrage Theory in Continuous Time. Oxford Finance. Oxford Univ Pr,2nd edition, 2004.

R. Durrett. Probability: Theory and Examples. Thomson Learning, 1991.

K. Ito. Probability Theory. Iwanami, 1991.

B. Oksendal. Stochastic Differential Equations: An Introduction With Applica-tions. Springer-Verlag, 2003.

S. M. Ross. An Elementary Introduction to Mathematical Finance: Options andOther Topics. Cambridge Univ Pr, 2002.

H. Toyoizumi. Statistics and probability for business.http://www.f.waseda.jp/toyoizumi/classes/accounting/stat/2007/business.pdf,2007a.

H. Toyoizumi. Mathematics for management.http://www.f.waseda.jp/toyoizumi/classes/accounting/math/2007/matht.pdf,2007b.

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