Applied Probability
and Mathematical Finance Theory
Hiroshi Toyoizumi 1
January 16, 2006
1 E-mail:
[email protected]
Contents
1
Introduction
2
Basic Probability Theory
2.1 Why Probability? . . . . . . . . . . . . . . .
2.2 Probability Space . . . . . . . . . . . . . . .
2.3 Conditional Probability and Independence . .
2.4 Random Variables . . . . . . . . . . . . . . .
2.5 Expectation, Variance and Standard Deviation
2.6 Covariance and Correlation . . . . . . . . . .
2.7 How to Make a Random Variable . . . . . . .
2.8 References . . . . . . . . . . . . . . . . . . .
2.9 Exercises . . . . . . . . . . . . . . . . . . .
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6
6
6
8
9
10
12
13
14
14
Normal Random Variables
3.1 What is Normal Random Variable?
3.2 Lognormal Random Variables . .
3.3 References . . . . . . . . . . . . .
3.4 Exercises . . . . . . . . . . . . .
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16
16
17
20
20
Useful Probability Theorems
4.1 The Law of Large Numbers . . . . . . . . . . . . . . . . .
4.2 Poisson Random Variables and the Law of Small Numbers
4.3 The Central Limit Theorem . . . . . . . . . . . . . . . . .
4.4 Useful Estimations . . . . . . . . . . . . . . . . . . . . .
4.5 References . . . . . . . . . . . . . . . . . . . . . . . . . .
4.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . .
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22
22
23
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24
27
27
Brownian Motions and Poisson Processes
5.1 Geometric Brownian Motions . . . .
5.2 Discrete Time Tree Binomial Process
5.3 Brownian Motions . . . . . . . . . .
5.4 Poisson Processes . . . . . . . . . . .
5.5 References . . . . . . . . . . . . . . .
5.6 Exercises . . . . . . . . . . . . . . .
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28
28
30
33
36
37
37
3
4
5
5
2
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CONTENTS
6
7
8
9
3
Simulating Brownian Motions
6.1 Mathematica . . . . . . . . . . . . . . . . . . .
6.1.1 As an Advanced Calculator . . . . . . .
6.2 Generating Random Variables . . . . . . . . . .
6.3 Two-Dimensional Brownian Motions . . . . . . .
6.3.1 Random Variable on Unit Circle . . . . .
6.3.2 Generating Brownian Motion . . . . . .
6.4 Geometric Brownian Motions . . . . . . . . . .
6.4.1 Generating Bernouilli Random Variables
6.4.2 Generating geometric brownian motions .
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38
38
38
39
39
39
39
41
41
41
Present Value Analysis
7.1 Interest Rates . . . . . . . . . . . . . . .
7.2 Continuously Compounded Interest Rates
7.3 Present Value . . . . . . . . . . . . . . .
7.4 Rate of Return . . . . . . . . . . . . . . .
7.5 References . . . . . . . . . . . . . . . . .
7.6 Exercises . . . . . . . . . . . . . . . . .
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42
42
43
44
45
47
47
Risk Neutral Probability and Arbitrage
8.1 Option to Buy Stocks . . . . . . . . .
8.1.1 Risk Neutralization . . . . . .
8.1.2 Arbitrage and Price of Option
8.2 Duplication and Law of One Price . .
8.3 Arbitrage Theorem . . . . . . . . . .
8.4 References . . . . . . . . . . . . . . .
8.5 Exercises . . . . . . . . . . . . . . .
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48
48
48
49
50
51
54
54
Black-Scholes Formula
9.1 Risk-neutral Tree Binomial Model . . . . . . . . . .
9.2 Option Price on the Discrete Time . . . . . . . . . .
9.3 Black-Scholes Model . . . . . . . . . . . . . . . . .
9.4 Examples of Option Price via Black-Scholes Formula
9.5 References . . . . . . . . . . . . . . . . . . . . . . .
9.6 Exercises . . . . . . . . . . . . . . . . . . . . . . .
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55
55
56
57
61
62
62
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63
63
67
68
70
70
10 Delta Hedging Strategy
10.1 Binomial Hedging Model . . . .
10.2 Hedging in Black-Scholes Model
10.3 Partial Derivative ∆ . . . . . . .
10.4 References . . . . . . . . . . . .
10.5 Exercises . . . . . . . . . . . .
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4
11 More Vanilla Options in Detail
11.1 American Call Options . . . . .
11.2 Put Options . . . . . . . . . . .
11.3 Pricing American Put Options .
11.4 Stock with Continuous Dividend
11.5 Stock with Fixed-time-Dividend
11.6 Forward and Futures Contract .
11.7 Exercises . . . . . . . . . . . .
CONTENTS
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71
71
72
73
77
79
80
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 you
find any differences between the following probabilities?
1. the probability that the both flips land on heads given that the first flip lands on
heads.
2. the probability that the both flips land on heads given that at least one of flips
lands on heads.
Example 1.2 (Wait or Not Wait). Assume you would like to enter a building, but you
realized that you forgot your ID card to enter the building. You have been waiting for
your colleagues with ID card 10 minutes so far. Is it reasonable to think that you would
enter the building more likely next 1 minute?
Example 1.3 (Easy Money). Imagine you are betting on a gamble. Here’s the best
scheme [2] to get easy money on gambling. When you lose, you bet twice as much
next 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 to
answer the above question, we need to study probability theory. The aim of this lecture
is 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: Options and Other Topics[6].
Note that the items covered in [6] are not complete enough to cover all of this
lecture. 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, we
need to handle them using extreme care in rigorous mathematical framework, called
probability theory. (See Exercise 2.1).
2.2
Probability Space
Be patient to learn the basic terminology in probability theory. To determine the probabilistic 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 we
write 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}
L = {lose}
(2.1)
(2.2)
Thus, we can say that “what is the probability of win?”, instead of saying “what is
the 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 each
set 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 ) = ∑ P(Ai ).
(2.3)
i=1
In addition, P is said to be the probability measure on Ω.
Mathematically, all function f which satisfies Definition 2.1 can regarded as probability. 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 10
first prizes, 20 second prizes · · · 60 sixth prizes out of total 1000 tickets, then we have
a probability measure P defined by
n
100
79
P(0) = P(lose) =
.
100
P(n) = P(win n-th prize) =
(2.4)
(2.5)
It is easy to see that P satisfies Definition 2.1. According to the definition P, we can
calculate 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 instead
of 10 first prizes. Then your customer might have a different P satisfying Definition
2.1. Thus it is pretty important to select an appropriate probability measure. Selecting
the 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 familiar
with measure theory and advanced integral theory, you may proceed to read [3].
2.3
Conditional Probability and Independence
Now we introduce the most uselful and probably most difficult concepts of probability
theory.
Definition 2.2 (Conditional Probability). Define the probability of B given A by
P(B | A) =
P(B & A) P(B ∩ A)
=
.
P(A)
P(A)
(2.6)
We can use the conditional probability to calculate complex probability. It is actually the only tool we can rely on. Be sure that the conditional probability P(B|A) is
different with the regular probability P(B).
Example 2.5 (Lottery). Let W = {win} and F = {first prize} in Example 2.4. Then
we have the conditional probability that
P(F | W ) = the probability of winning 1st prize given you win the lottery
P(F)
P(F ∩W )
=
P(W )
P(W )
10/1000
1
=
=
210/1000 21
10
6=
= P(F).
1000
=
Remark 2.2. Sometimes, we may regard Definition 2.2 as a theorem and call Bayse
rule. 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 independent
if
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 are
independent, then the conditional probability of B given A is equal to the probability of
B.
Proof. By Definition 2.2, we have
P(B | A) =
P(B ∩ A) P(B)P(A)
=
= P(B),
P(A)
P(A)
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)
1 1
= · .
2 2
Example 2.7 (Dependent events on two dice). Even though the two dices are independent, 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 Variables
The name random variable has a strange and stochastic history1 . Although its fragile
history, the invention of random variable certainly contribute a lot to the probability
theory.
Definition 2.4 (Random Variable). The random variable X = X(ω) is a real-valued
function 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 variables 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
(
1 with probability p.
X=
0 with probability 1 − p.
(2.8)
for some p ∈ [0, 1]. The random variable X is said to be a Bernouilli random variable.
1 J. Doob quoted in Statistical Science. (One of the great probabilists who established probability as
a 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.” We
obviously had to use the same name in our books, so we decided the issue by a stochastic procedure. That
is, we tossed for it and he won.
10
CHAPTER 2. BASIC PROBABILITY THEORY
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) is
defined by
F(x) = P{X ≤ x}.
(2.9)
The probability distribution function fully-determines the probability structure of
a random variable X. Sometimes, it is convenient to consider the probability density
function instead of the probability distribution.
Definition 2.6 (probability density function). The probability density function f (t) is
defined by
f (x) =
dF(x) dP{X ≤ x}
=
.
dx
dx
(2.10)
Sometimes we use dF(x) = dP{X ≤ x} = P(X ∈ (x, x + dx]) even when F(x) has
no derivative.
Lemma 2.1. For a (good) set A,
P{X ∈ A} =
Z
dP{X ≤ x} =
Z
A
2.5
f (x)dx.
(2.11)
A
Expectation, Variance and Standard Deviation
Let X be a random variable. Then, we have some basic tools to evaluate random variable X. First we have the most important measure, the expectation or mean of X.
Definition 2.7 (Expectation).
Z ∞
E[X] =
xdP{X ≤ x} =
−∞
Z ∞
x f (x)dx.
(2.12)
−∞
Remark 2.4. For a discrete random variable, we can rewrite (2.12) as
E[X] = ∑ xn P[X = xn ].
(2.13)
n
Lemma 2.2. Let (Xn )n=1,...,N be the sequence of random variables. Then we can
change 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 we
need to measure the strength of this stochastic fluctuation. The natural choice may be
X − 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[X 2 ] − E[X]2 .
(2.16)
Proof. See Exercise 2.6.
Unfortunately, the variance Var[X] has the dimension of X 2 . So, in some cases, it
is inappropriate to use the variance. Thus, we need the standard deviation σ [X] which
has 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 variable
with P[X = 1] = p and P[X = 0] = 1 − p. Then we have
E[X] = 1p + 0(1 − p) = p.
2
2
(2.18)
2
Var[X] = E[X ] − E[X] = E[X] − E[X] = p(1 − p),
(2.19)
where we used the fact X 2 = X for Bernouille random variables.
In many cases, we need to deal with two or more random variables. When these
random 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 the
sets {X ≤ x} and {Y ≤ y} are independent for all x and y. In other words, when X and
Y are independent,
P(X ≤ x,Y ≤ y) = P(X ≤ x)P(Y ≤ y)
Lemma 2.4. For any pair of independent random variables X and Y , we have
(2.20)
12
CHAPTER 2. BASIC PROBABILITY THEORY
• 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,
Z
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,
Z
E[XY ] =
Z
=
xydP(X ≤ x)dP(Y ≤ y)
xdP(X ≤ x)
Z
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
n
X = ∑ Xi ,
(2.21)
i=1
where Xi are independent Bernouilli random variables with the mean p. The random
variable X is said to be a Binomial random variable. The mean and variance of X can
be obtained easily by using Lemma 2.4 as
E[X] = np,
Var[X] = np(1 − p).
2.6
(2.22)
(2.23)
Covariance and Correlation
When we have two or more random variables, it is natural to consider the relation of
these random variables. But how? The answer is the following:
Definition 2.10 (Covariance). Let X and Y be two (possibly not independent) random
variables. 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 their
mean. 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 X2
be the independent Binomial random variables with the same parameter n and p. The
covariance of X1 and X2 is
Cov[X1 , X2 ] = E[X1 X2 ] − E[X1 ]E[X2 ] = 0,
since X1 and X2 are independent. Thus, more generally, if the two random variables
are independent, their covariance is zero. (The converse is not always true. Give some
example!)
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[X12 ] − 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 order
of XY , which is not convenience to compare the strength among different pair of random variables. Don’t worry, we have the correlation function, which is normalized by
standard deviations.
Definition 2.11 (Correlation). Let X and Y be two (possibly not independent) random
variables. 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 Variable
Suppose we would like to simulate a random variable X which has a distribution F(x).
The following theorem will help us.
14
CHAPTER 2. BASIC PROBABILITY THEORY
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).
2.8
(2.29)
References
There are many good books which useful to learn basic theory of probability. The
book [5] is one of the most cost-effective book who wants to learn the basic applied
probability featuring Markov chains. It has a quite good style of writing. Those who
want more rigorous mathematical frame work can select [3] for their starting point. If
you want directly dive into the topic like stochatic integral, your choice is maybe [4].
2.9
Exercises
Exercise 2.1. Find an example that our intuition leads to mistake in random phenomena.
Exercise 2.2. Define a probability space according to the following steps.
1. Take one random phenomena, and describe its sample space, events and probability 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 are
dependent 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 the
graph 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 random
variables X and Y .
Exercise 2.10 (Binomial random variable). Let X be a random variable with
n
X = ∑ Xi ,
(2.30)
i=1
where Xi are independent Bernouilli random variables with the mean p. The random
variable 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 random
variables.
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 probability density function
dP{X ≤ x} = f (x)dx =
2
2
1
e−(x−µ) /2σ dx,
(2π)1/2 σ
(3.1)
for some µ and σ . The random variable is called the normal random variable with the
parameters µ and σ .
Theorem 3.1 (Mean and variance of normal random variables). Let X be a normal
random variable with the parameters µ and σ . Then, we have the mean
E[X] = µ,
(3.2)
Var[X] = σ 2 .
(3.3)
and the variance
Proof.
Definition 3.2 (Standard normal random variable). Let X be a normal random variable
with µ = 0 and σ = 1. The random variable is called the standard normal random
variable.
16
3.2. LOGNORMAL RANDOM VARIABLES
17
Lemma 3.1. Let X be a normal random variable with its mean µ and standard deviation σ . Set Z = (X − µ)/σ . Then, Z is the standard normal random variable.
Proof. See Exercise 3.5.
Theorem 3.2. Let (Xi )i=1,2,...,n are independent normal random variables with its mean
µi and standard deviation σi . Then the sum of these random variables X = ∑ni=1 Xi is
again a normal random variable with
n
µ = ∑ µi ,
(3.4)
σ 2 = ∑ σi2 .
(3.5)
i=1
n
i=1
Proof. We just prove X satisfies (3.4) and (3.5). By Lemma 2.2,
"
#
n
n
µ = E[X] = E
n
= ∑ E [Xi ] = ∑ µi .
∑ Xi
i=1
i=1
i=1
Also, by Lemma 2.4, we have
"
2
σ = E[X] = Var
n
∑ Xi
i=1
#
n
n
= ∑ Var [Xi ] = ∑ σi2 .
i=1
i=1
We can prove that X is a normal random variable by using so-called characteristic
function method (or Fourier transform).
So, it is very comfortable to be in the world of normal random variables. Very
closed!
3.2
Lognormal Random Variables
Definition 3.3 (lognormal random variable). The random variable Y is said to be lognormal 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 measure
role in finance theory!
18
CHAPTER 3. NORMAL RANDOM VARIABLES
Theorem 3.3 (lognormal). If X is a normal random variable having the mean µ and
the standard deviation σ 2 , the lognormal random variable Y = eX has the mean and
the variance as
E[Y ] = eµ+σ
2 /2
,
2µ+2σ 2
Var[Y ] = e
(3.7)
2µ+σ 2
−e
.
(3.8)
It is important to see that although the mean of the lognormal random variable is
subjected to not only the mean of the original normal random variable but also the
standard 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
m0 (0) =
d
m(t)|t=0 = E[XetX ]|t=0 = E[X].
dt
(3.10)
Further, we have
m00 (0) = E[X 2 ].
On the other hand, since X is the standard normal random variable, we have
1
m(t) = E[etX ] = √
2π
Z ∞
etx e−x
2 /2
dx.
−∞
Since tx − x2 /2 = {t 2 − (x − t)2 }/2, we have
2
1
m(t) = √ et /2
2π
Z ∞
2 /2
e−(x−t)
dx,
−∞
where the integrand of the right hand side is nothing but the density of the normal
random variable N(t, 1). Thus,
m(t) = et
2 /2
.
More generally, when X is a normal random variable with N(µ, σ ), we can obtain
m(t) = E[etX ] = eµt+σ
2 t 2 /2
.
(3.11)
(See exercise 3.6.) Since Y = eX , we have
E[Y ] = m(1) = eµ+σ
2 /2
,
(3.12)
3.2. LOGNORMAL RANDOM VARIABLES
19
and
2
E[Y 2 ] = m(2) = e2µ+2σ .
(3.13)
Thus,
2
2
Var[Y ] = E[Y 2 ] − E[Y ]2 = e2µ+2σ − e2µ+σ .
(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 identically distributed as the lognormal random variable. Taking log on both side of (3.15),
then we have
log S(n) = log S(n − 1) + X(n),
(3.16)
where if X(n) = logY (n) is regarded as the error term and normally distributed, the
above assumption is validated.
Example 3.1 (Stock price rises in two weeks in a row). Suppose Y (n) is the growth
rate of a stock at the n-th week, which is independent and lognormally distributed with
the parameters µ and σ . We will find the probability that the stock price rises in two
weeks in a row.
First, we will estimate P{the stock rises}. Since it is equivalent that y > 1 and
log y > 0, we have
P{the stock rises} = P{S(1) > S(0)}
S(1)
>1
=P
S(0)
S(1)
>0
= P log
S(0)
= P {X > 0}
where X = logY (1) is N(µ, σ ), and we can define
Z=
X −µ
,
σ
as the standard normal distribution. Hence, we have
0−µ
X −µ
>
P{S(1) > S(0)} = P
σ
σ
= P{Z > −µ/σ }
= P{Z < µ/σ },
(3.17)
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
P{the stock rises two week in a row} = P{Y (1) > 0,Y (2) > 0}
= P{Y (1) > 0}P{Y (2) > 0}
= P{Z < µ/σ }2
3.3
References
3.4
Exercises
Exercise 3.1. Prove the symmetry of the normal distribution. That is, let X be the
normal distribution with the mean µ and the standard deviation σ , then for any x, we
have
P{X > −x} = P{X < x}.
(3.18)
Exercise 3.2 (moment generating function). Let X be a random variable. The function
m(t) = E[etX ] is said to be the moment generating function of X.
1. Prove E[X] = m0 (0).
2. Prove E[X n ] =
dn
dt n 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) =
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.
3.4. EXERCISES
21
What can you say about these graphs, especially large x? Click help in your Excel to
find 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 deviation 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 ). Find
E[Y ] and Var[Y ].
Chapter 4
Useful Probability Theorems
4.1
The Law of Large Numbers
In many cases, we need to evaluate the average of large number of independent and
identically-distributed random variables. Perhaps, you can intuitively assume the average will be approximated by its mean. This intuition can be validated by the following
theorems.
Theorem 4.1 (Weak law of large numbers). Let X1 , X2 , .... be an i.i.d. sequence of
random variables with
µ = E[Xn ].
(4.1)
P{|Sn /n − µ| > ε} → 0 as n → ∞.
(4.2)
Let Sn = ∑ni=1 Xi . Then, for all ε > 0,
Or, if we take sufficiently large number of samples, the average will be close to µ with
high probability.
You can say, “OK. I understand, for the large sample, we can expect the average
will be close to the mean most of the times. So, it might be possible to have some
exceptions...” In that case, we have another answer.
Theorem 4.2 (Strong law of large numbers). Let X1 , X2 , .... be an i.i.d. sequence of
random variables with
µ = E[Xn ].
(4.3)
Sn
→ µ as n → ∞.
n
(4.4)
With probability 1, we have
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 Small
Numbers
Compare to the the theorems of large numbers, the law of small numbers are less
famous. 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 Poisson
random variable, when
P{N = n} =
(λ )n −λ
e ,
n!
(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 independent
rare 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 customer
visits this shop independently and its frequency to visit there is relatively rare, then N
can be approximated by a Poisson random variable.
4.3
The Central Limit Theorem
Let X1 , X2 , .... be an i.i.d. sequence of random variables with
µ = E[Xn ],
2
σ = Var[Xn ].
(4.7)
(4.8)
Now we would like to estimate the sum of this random variables, i.e.
n
Sn = ∑ Xi .
(4.9)
i=1
Theorem 4.5 (Central limit theorem). For a large n, we have
Sn − nµ
√ ≤ x ≈ Φ(x),
P
σ n
(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 is
like, the sum Sn can be regarded
Instead of stating lengthy and technically advanced proof of laws of large numbers
and ccentral limit theorem, we give some examples of how Sn converges to a Normal
random variable.
Example 4.3 (Average of Bernouilli Random Variables). Let Xi be i.i.d Bernoulli random variables with p = 1/2. Suppose we are going to evaluate the sample average A
of Xi ’s;
A=
1 n
∑ Xi .
n i=1
(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 mean
1/2. On the other hand, when we have more samples, the sample average A tends to
be 1/2 as we see in Figure 4.2 and 4.3, which can be a demonstration of Weak Law of
Large Numbers (Theorem 4.1).
As we see in in Figure 4.1 to 4.3, we may still have occasional high and low averages. How about individual A for large sample n. Figure 4.4 shows the sample path
level convergence of the sample average A to E[X] = 1/2, as it can be expected from
Strong 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 Normal distribution. Figure 4.5 show the detail of the histogram of A. The histogram is
quite similar to the corresponding Normal distribution.
4.4
Useful Estimations
We have the following two estimation of the distribution, which is sometimes very
useful.
Theorem 4.6 (Markov’s Inequality). Let X be a nonnegative random variable, then we
have
P{X ≥ 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
aI{X≥a} ≤ X.
(4.14)
Taking expectation on the both side, we have
aP{X ≥ a} ≤ E[X].
(4.15)
4.4. USEFUL ESTIMATIONS
25
8
6
4
2
0.2
0.4
0.6
0.8
1
Figure 4.1: Histogram of the sample average A = n1 ∑ni=1 Xi when n = 10, where Xi is a
Bernouilli random variable with E[Xi ] = 1/2.
8
6
4
2
0.2
0.4
0.6
0.8
1
Figure 4.2: The sample average A when n = 100.
17.5
15
12.5
10
7.5
5
2.5
0.2
0.4
0.6
0.8
Figure 4.3: The sample average A when n = 1000.
1
26
CHAPTER 4. USEFUL PROBABILITY THEOREMS
1
0.8
0.6
0.4
0.2
20
40
60
80
100
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.
30
25
20
15
10
5
0.46
0.48
0.5
0.52
0.54
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 corresponding Normal distribution.
4.5. REFERENCES
27
Theorem 4.7 (Chernov Bounds). Let X be the random variable with moment generating function M(t) = E[etX ]. Then, we have
P{X ≥ a} ≤ e−ta M(t) for all t > 0,
−ta
P{X ≤ a} ≥ e
M(t) for all t < 0.
(4.16)
(4.17)
Proof. For t > 0 we have
P{X ≥ a} = P{etX ≥ eta }.
(4.18)
P{etX ≥ eta } ≤ e−ta E[etX ].
(4.19)
Using Theorem 4.6,
Thus, we have the result.
4.5
References
The proofs omitted in this chapter can be found in those books like ITO KIYOSHI and
Durret [3].
4.6
Exercises
Exercise 4.1. Something here!
Chapter 5
Brownian Motions and Poisson
Processes
Brownian Motions and Poisson Processes are the most useful and practical tools to
understand stochastic processes appeared in the real world.
5.1
Geometric Brownian Motions
Definition 5.1 (Geometric Brownian motion). We say S(t) is a geometric Brownian
motion 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 variance
σ 2t.
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 a
geometric 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 parameter µt and σ 2t..
The future price is probabilistically decided by the present price. Sometimes, this
kind 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 ] = s0 et(µ+σ
28
2 /2)
.
(5.1)
5.1. GEOMETRIC BROWNIAN MOTIONS
29
Proof. Set
Y=
S(t)
S(t)
=
.
S(0)
s0
Then, Y is lognormal with (tµ,tσ 2 ). Thus by using Theorem 3.3 we have
E[Y ] = et(µ+σ
2 /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 + σ 2 /2,
not the rate of µ. The parameter σ represents the fluctuation, but since the lognormal
distribution 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 volatility
σ , 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, σ 2t).
Proof. We need to check that (5.2) satisfies the second part of Definition 5.1, which is
easy by taking log on both sides in (5.2) and by seeing
S(t)
= W.
(5.3)
log
S(0)
Sometimes instead of Definition 5.1, we use the following stochastic differential
equation to define geometric Brownian motions,
dS(t) = µS(t)dt + σ S(t)dB(t),
(5.4)
where B(t) is the standard Brownian motion and dB(t) is define by so-called Ito calculus.
Note that the standard Brownian motion is continuous function but nowhere differentiable, so the terms like dB(t)/dt should be treated appropriately. But these treatment
includes the knowledge beyond this book. Thus, the following parts is not mathematically rigorous. The solution to this equation is
S(t) = S(0)eµt+σ B(t) .
(5.5)
Formally, if we differentiate (5.5), we can verify this satisfy (5.4). It easy to see that
S(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 differential
equation,
dS(t) = µS(t)dt + σ S(t)dB(t),
(5.6)
with the initial condition S(0).
5.2
Discrete Time Tree Binomial Process
In this section, we imitate a stock price dynamic on the discrete time, which is subject
to geometric Brownian motion.
Definition 5.2 (Tree binomial process). The process Sn is said to be a Tree Binomial
process, if the dynamics is express as
(
uSn−1 with probability p,
(5.7)
Sn =
dSn−1 with probability 1 − p,
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 is
reasonably similar to Figure 5.2, which is a exchange rate of yen?
Theorem 5.3 (Approximation of geometric brownian motion). A geometric Brownian
motion S(t) with the drift µ and the volatillity σ can be regardede as the limit of a tree
binomial process with the parameters:
d = e−σ
√
∆
√
σ ∆
,
u=e
,
µ√ 1
1+
p=
∆ .
2
σ
(5.8)
(5.9)
(5.10)
Proof. Let ∆ be a small interval and n = t/∆. Define a tree binomial process on the
discrete time {i∆} such as
(
uS((i − 1)∆) with probability p,
S(i∆) =
(5.11)
dS((i − 1)∆) with probability 1 − p,
for all n. The multiplication factors d and u as well as the probability p have some
specific value, as described in the following, to imitate the dynamics of the Brownian
motion. Let Yi be the indicator of “up”s (Bernouilli random variable), i.e.,
(
1 up at time i∆,
Yi =
(5.12)
0 down at time i∆.
5.2. DISCRETE TIME TREE BINOMIAL PROCESS
31
130
120
110
100
90
80
20
40
60
80
100
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
http://markets.nikkei.co.jp/kawase/index.cfm
2005/10/19.
Adopted
from
32
CHAPTER 5. BROWNIAN MOTIONS AND POISSON PROCESSES
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∑ Yi d n−∑ Yi
u ∑ Yi
.
= S(0)d n
d
Now set
t/∆
S(t) = S(0)d t/∆ (u/d)∑i=1 Yi ,
(5.13)
and show that the limit S(t) is geometric Brownian motion with the drift µ and the
volatility σ as ∆ → 0. Taking log on the both side, we have
S(t)
log
S(0)
=
u t/∆
t
log d + log
∑ Yi .
∆
d i=1
(5.14)
To imitate the dynamics of geometric Brownian motion, here we artificially set
d = e−σ
√
∆
,
√
σ ∆
u=e
,
1
µ√ ∆ .
p=
1+
2
σ
Note that log d and log u are symmetric, while d and u are asymmetric, and p → 1/2 as
∆ → 0. Now using these, we have
S(t)
log
S(0)
√ t/∆
−tσ
= √ + 2σ ∆ ∑ Yi .
∆
i=1
(5.15)
Consider taking the limit ∆ → 0, then the number of terms in the sum increases. Using
Central Limit Theorem (Theorem 4.5), we have the approximation:
t/∆
∑ Yi ∼ Normal distribution.
(5.16)
i=1
Thus, log(S(t)/S(0)) has also the normal distribution with its mean and variance:
√ t/∆
−tσ
E[log(S(t)/S(0)] = √ + 2σ ∆ ∑ E[Yi ]
∆
i=1
√ t
−tσ
= √ + 2σ ∆ p,
∆
∆
√ t −tσ
µ√ = √ +σ ∆
1+
∆
∆
σ
∆
= µt,
5.3. BROWNIAN MOTIONS
33
and
t/∆
Var[log(S(t)/S(0)] = 4σ 2 ∆ ∑ Var[Yi ]
i=1
2
= 4σ t p(1 − p)
→ σ 2t, as ∆ → 0,
since p → 1/2. Thus,
log(S(t)/S(0)) ∼ N[µt, σ 2t],
(5.17)
and moreover S(t + y)/S(t) is independent with the past because of the construction
of S(t). Hence the limiting distribution of S(t) with ∆ → 0 is a geometric Brownian
motion, which means all the geometric Brownian motion S(t) with the drift µ and the
volatility 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 Process
and then taking appropriate limit to show the validity of the properties of geometric
Brownian motions.
Example 5.1 (Trajectories of geometric brownian motions). This is an example of different trajectories of Geometric Brownian Motions. Here we set the drift µ = 0.1, the
volatility σ = 0.3 and the interval ∆ = 0.01 in the geometric brownian motion. As
pointed out in Remark 5.2, we use tree binomial process as an approximation of geometric 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 time
by time, even we have the same parameters. It can rise and down. Figure 5.4 gathers
those 10 different trajectories. Since we set the drift µ = 0.1 > 0, we can see up-ward
trend while each trajectories are radically different.
On the other hand, Figure 5.5 shows the geometric brownian motion with the negative drift µ = −0.1, and the probability of the upward change p = 0.483333. Can you
see the difference between Figure 5.5 and Figure 5.4?
5.3
Brownian Motions
Definition 5.3 (Brownian motion). A stochastic process S(t) is said to be a Brownian
motion, if S(t + y) − S(y) is a normal random variable N[µt, σ 2t], which is independent
with 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)] = σ 2t.
(5.19)
34
CHAPTER 5. BROWNIAN MOTIONS AND POISSON PROCESSES
130
120
110
100
90
80
20
40
60
80
100
Figure 5.3: 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.
200
180
160
140
120
100
80
20
40
60
80
100
Figure 5.4: Many trajectories of geometric brownian motion with the upward drift
µ = 0.1, the volatility σ = 0.3 and the interval ∆ = 0.01 and starting from S0 = 100.
5.3. BROWNIAN MOTIONS
35
140
120
100
20
40
60
80
100
Figure 5.5: Many trajectories of geometric brownian motion with the downward drift
µ = −0.1.
Thus, both quantities are increasing as t increases.
Brownian motions is more widely used than geometric Brownian motions. However, 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 in
Figure 5.6.
-20
20
40
60
-20
-40
-60
-80
Figure 5.6: An example of trajectory of two-dimensional brownian motion.
36
CHAPTER 5. BROWNIAN MOTIONS AND POISSON PROCESSES
5.4
Poisson Processes
Definition 5.4 (Counting process). Let N(t) be the number of event during [0,t). The
process N(t) is called a counting process.
Definition 5.5 (Independent increments). A stochastic process N(t) is said to have
independent 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 increment are independent with its past history.
Definition 5.6 (Poisson Processes). Poisson process N(t) is a counting process of
events, which has the following features:
1. N(0) = 0,
2. N(t) has independent increments,
n
3. P{N(t + s) − N(s) = n} = e−λt (λt)
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)] = λt,
Var[N(t)] = λt.
(5.21)
(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)] =
∑ nP{N(t) = n}
n=0
∞
=
∑ ne−λt
n=0
= λte−λt
(λt)n
n!
∞
(λt)n−1
∑ (n − 1)!
n=0
= λte−λt eλt
= λt.
The other is left for readers to prove (see Exercise 5.2).
5.5. REFERENCES
5.5
References
5.6
Exercises
37
Exercise 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)] = λt.
(5.24)
Chapter 6
Simulating Brownian Motions
We simulate Brownian motions using Mathematica, following the steps below.
6.1
6.1.1
Mathematica
As an Advanced Calculator
Mathematica can handle “mathematical formulas” and derive the result as much precision as possible. You can write an appropriate formula after the prompt “In[ ]:=”, and
hit “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]=
7
12
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
Log[x] = loge x
(6.2)
(6.3)
Sin[θ ] = sin(θ ).
(6.4)
(6.5)
38
6.2. GENERATING RANDOM VARIABLES
6.2
39
Generating Random Variables
Generating random variables is a core of simulation. The function Random[ ] will
create 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 generate
a 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 SeedRandom
is set, Mathematica set the seed from the current time, so as to generate the different
random 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
6.3.1
Two-Dimensional Brownian Motions
Random Variable on Unit Circle
Generate a pair of random variable (X,Y ) on unit circle as shown in Figure 6.1. The
pair should have the relation X 2 +Y 2 = 1.
Check your pairs of random variables drawing a graph like Figure 6.1. You can use
the function ListPlot to display the results.
6.3.2
Generating Brownian Motion
Using the pairs of random variables as the movement, simulate a brownian motion as
shown in Figure 6.2.
40
CHAPTER 6. SIMULATING BROWNIAN MOTIONS
1
0.5
-1
-0.5
1
0.5
-0.5
-1
Figure 6.1: An example of Random Variables on Unit Circle.
-20
20
40
60
-20
-40
-60
-80
Figure 6.2: An example of trajectroy of two-dimensional brownian motion.
6.4. GEOMETRIC BROWNIAN MOTIONS
6.4
41
Geometric Brownian Motions
Simulate geometric brownian motions with the drift µ and the volatility σ . Use an
approximation of tree-binomial process.
6.4.1
Generating Bernouilli Random Variables
Generate a Bernouilli random variable T with its mean E[T ] = p. You may use the
package 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 process 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 motions
Take ∆ = 0.01 and set appropriate u, d and p for the tree-binomial process. Generate
an approximation of geometric brownian motion with the drift µ and the volatility σ
as in Figure 6.3.
130
120
110
100
90
80
20
40
60
80
100
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.
Chapter 7
Present Value Analysis
7.1
Interest Rates
Let r be the interest rate on some term, and P be the amount of money you saved in the
bank. 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 compounded
semi-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 interest
rate.
Definition 7.1 (Effective interest rate). Let P0 be the amount of money you invested
initially, P1 be the amount of money you obtained at the end of the year. Then, the
value re is said to be the effective interest rate, where
P1 − P0
re =
.
(7.3)
P0
In other words, given you invested P0 initially, the amount you get in the end of a
year 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
P1 − P0
re =
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 rate
not including the effect of inflation, while the word “effective interest rate” is used as
indicated to be adjusted to the inflation factor.
7.2
Continuously Compounded Interest Rates
First, let us recall the definition of exponential, or Euler constant.
Definition 7.2 (exponential). We define e by the following:
e = lim (1 + 1/n)n .
n→∞
(7.5)
The following is a useful expression of the so-called exponential function Exp(x) =
ex .
Lemma 7.1 (exponential).
ex = lim (1 + x/n)n .
n→∞
(7.6)
Proof. By the Definition 7.2, we have
n
ox
ex = lim (1 + 1/n)n
n→∞
= lim (1 + 1/n)xn
n→∞
= lim (1 + x/m)m ,
m→∞
where we set m = xn.
Suppose we divide one year into n equal intervals, and consider n-compounded
yearly 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 the
continuous compounding.
lim P(1 + r/n)n = Per ,
n→∞
(7.8)
where we used Lemma 7.1.
Theorem 7.1 (Continuous compounding). Let r be the nominal interest rate. The
effective 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 er − P0
=
= er − 1.
P0
P0
Corollary 7.1. The effective interest rate is always larger than the nominal interest
rate.
Proof. By Theorem 7.1, we have
re = er − 1 > r,
(7.10)
for r ≥ 0.
7.3
Present Value
Suppose we can get be the amount of money v at the end of the period i, when r is the
nominal interest rate of the period. What can you say about the “value” of this money
now? Or more precisely, when someone is offered to us the above investment, how
much 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 time
i. 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 to
v(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 cash
stream;
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 value
PV (a) such as
n
ai
.
(1
+
r)i
i=1
PV (a) = ∑
(7.13)
Or, if we deposit the amount PV (a) at time 0 at a bank, we can obtain the equivalent
cash 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, taking
into the interest rate r, we have as the remainder,
a2
an
(1 + r)PV (a) − a1 =
+···+
.
(7.14)
1+r
(1 + r)n−1
Continue this procedure to the end of the year i, and we have
ai+1
an
+···+
,
1+r
(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, we
pay 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 cash
stream 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
∞
c
(1
+
r)i
i=1
PV (a) = ∑
=
c ∞
1
∑
1 + r i=0 (1 + r)i
1
c
1 + r 1 − 1/(1 + r)
c
= .
r
=
Remark 7.2. The result of Corollary 7.2 is not surprising, if we consider to deposit c/r
in a bank, and we can get c as annual interest without affecting the original principal.
7.4
Rate of Return
Definition 7.3 (Rate of return). Suppose we invest a as the initial payment of an investment, and let b be its return after 1 period. Then we can define the rate of return
as
r=
b
− 1,
a
(7.17)
which is the interest rate that makes the present value of the return equal to the initial
payment.
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 extending
Definition 7.3.
Definition 7.4 (Rate of return for cash stream). Suppose we invest a as the initial
payment of an investment, and let bi be the cash stream gained by this investment.
Define a function P(r) as
n
P(r) = −a + ∑ bi (1 + r)−i ,
(7.18)
i=1
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 cash
stream (a, b1 , b2 , . . . , bn ) with a > 0, bi ≥ 0 and bn > 0. Then, there is a unique rate of
return r∗.
Proof. First, we will show that P(r) is strictly decreasing. Take r < r0 , and we have
n
P(r) − P(r0 ) = ∑ bi {(1 + r)−i − (1 + r0 )−i } > 0,
(7.20)
i=1
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) is
continuous, 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 two
year. We are suppose to obtain $60 at the end of each year. We will put $100 as our
investment. Then,
P(r) = −100 +
60
60
+
.
1 + r (1 + r)2
(7.21)
Thus, the rate of return r∗ is the solution of the equation:
100 = +
60
60
+
.
1 + r (1 + r)2
(7.22)
Now take x = 1/(1 + r), then
3x2 + 3x − 5 = 0.
(7.23)
√
−3 ± 69
≈ 0.8844.
6
(7.24)
Solving this, we have
x=
Hence r∗ = 1/x − 1 = 0.131.
7.5. REFERENCES
7.5
References
reference here!
7.6
Exercises
Exercise 7.1. Something here!
47
Chapter 8
Risk Neutral Probability and
Arbitrage
8.1
Option to Buy Stocks
Let r be the interest rate. We consider to give the price of an option to purchase a stock
at a future time at 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 for simplicity.
We can consider an option to buy the stock at $150 at time 1. Note that this option
is 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 we
pay for this option?
Here’s the solution. Surpassingly, regardless of the probability of the stock’s up
and 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 follow
the simplest way.
8.1.1
Risk Neutralization
We 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 allow
both x and y to be negative. For example, when x < 0, the position of this stock is
short, 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 the
stock 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 the
stock at $50 which is far less expensive than exercising the option obtaining the stock
at $ 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 of
X1 and is
(
200x + 50y, if X1 = 200,
(8.2)
V1 =
50x
if X1 = 50.
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. By
solving (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 money
needed 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, but
at 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)x{3c − 100 + 50(1 + r)−1 }.
Consider the following situations:
(8.7)
50
CHAPTER 8. RISK NEUTRAL PROBABILITY AND ARBITRAGE
1. Suppose 3c > 100 − 50(1 + r)−1 . Set some positive x, then G1 > 0. That means
our investment guaranteed to be profitable. Actually, in this case, the option price
c satisfies
c>
100 − 50(1 + r)−1
.
3
(8.8)
Compared to (8.1), the option price is expensive. So, we can exploit the situation
by selling the options in short.
2. Suppose 3c < 100 − 50(1 + r)−1 . In this case by setting negative x, we can also
get 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, you
can 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 sorry
but you cannot obtain easy money.
Of course, in some temporary cases, we may experience case 1 and 2. We call these
sure-win situation arbitrage. However, for example, if the option price is expensive
compared to (8.1), as soon as many investors and the dealer of options realized that
they 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 the
price of option will converge to (8.1).
8.2
Duplication and Law of One Price
The 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 c2
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 8.1. If there is no arbitrage, the price of the investments with the same return
should be identical.
Now using Theorem 8.1, we will derive the option cost as in the same situation as
in 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
(
50 if the price of stock is $200,
P1 =
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. At
time 1 you need to pay back (1 + r)x, while selling the y share. Thus, the payoff
Q1 is
(
200y − x(1 + r) if the price of stock is $200,
(8.11)
Q1 =
50y − x(1 + r)
if the price of stock is $50.
By choosing appropriate x and y, we have the same pay off P1 = Q1 for these two
investment, i.e.,
200y − x(1 + r) = 50,
50y − x(1 + r) = 0,
(8.12)
(8.13)
or
1
y= ,
3
x=
50
.
3(1 + r)
(8.14)
(8.15)
In this case, two investments have the identical payoff. In other word, we could duplicate the option by buying stocks at spot and borrowing money from bank. Thus, using
Theorem 8.1, the initial costs of two investments are identical. Thus the option price c
is
c = 100y − x =
100 − 50(1 + r)−1
,
3
(8.16)
which is identical to (8.1).
8.3
Arbitrage Theorem
Here’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 experiment
result 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. If
we bet on the strategy x, we can get the amount of money
n
∑ xi ri (J).
i=1
(8.17)
52
CHAPTER 8. RISK NEUTRAL PROBABILITY AND ARBITRAGE
Definition 8.1 (risk neutral probability). The probability measure is said to be risk
neutral when the expectation of outcome on all bets is fair.
Theorem 8.2 (arbitrage theorem). Either there exists risk neutral probability, or there
is the sure-win strategy (arbitrage). More precisely, either one of the followings true.
1. There exists a probability vector p = (p1 , ..., pm ) for which
m
E[ri (J)] =
∑ p j ri ( j) = 0 for all wager i,
(8.18)
j=1
where we regard p j = P{J = j}.
2. There is a betting strategy x = (x1 , ..., xn ) for which
n
∑ xi ri ( j) > 0 for all outcome j.
(8.19)
i=1
Assume we need to estimate the price of a call option of a stock. Suppose the initial
price of the stock is s. After a week, the stock can be either a > 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 have the
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
(8.20)
taking into account of the initial payment s. Thus, the expected return of this investment
is
E[R1 ] = p{a(1 + r)−1 − s} + (1 − p){b(1 + r)−1 − s}
a−b
b
=p
+
− s.
1+r 1+r
(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 should
should 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 investment
Q1 is
(
(a − K)(1 + r)−1 −C if X1 = a,
(8.23)
Q1 =
−C
if X1 = b.
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 }
−C
a−b
(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 = P{X1 = 200}
100(1 + r) − 50
150
2r + 1
,
=
3
=
from (8.22). Thus,
E[X1 ] = 200p + 50(1 − p)
= 50 + 150p
2r + 1
= 50 + 150
3
= 100(1 + r).
This means that the only “rational” probability structure for the future stock price
should has the same expected present value. Of course, given that this probability
structure, we have the option price,
c=
100 − 50(1 + r)−1
.
3
54
8.4
CHAPTER 8. RISK NEUTRAL PROBABILITY AND ARBITRAGE
References
reference here!
8.5
Exercises
Exercise 8.1. Something here!
Chapter 9
Black-Scholes Formula
9.1
Risk-neutral Tree Binomial Model
We show the tree binomial process introduced in Section 5.2 can be derived form arbitrage theorem. This is the process we assume for the stock dynamics.
Theorem 9.1 (Risk-neutral tree binomial process). Consider the stock price on the
discrete time. Let r be the interest rate, and let Sn be the stock price at time n and
satisfies the following special dynamics:
(
uSn−1 ,
(9.1)
Sn =
dSn−1 ,
where we assume d < (1 + r) < u. If there is no arbitrage opportunity, then the process
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.,
(
1 up at n,
Xn =
0 down at n,
(9.2)
Note that (X1 , X2 , . . . , Xn ) determines the stock price Sn . By Arbitrage Theorem (Theorem 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 − 1
are used only for the observation. According to some rule, we decide to buy stock or
not. 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 )},
p = P{Xn = 1|(X1 , X2 , . . . , Xn−1 ) = (x1 , x2 , . . . , xn−1 )}.
(9.4)
(9.5)
According to the stock price change, the gain of this investment differs. By conditioning the event (X1 , X2 , . . . , Xn−1 ), we can get the expected gain at time n of this
investment 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 ] = α{puSn−1 + (1 − p)dSn−1 − (1 + r)Sn−1 }.
(9.8)
Thus, setting the expected gain to be zero, we have
0 = E[Gn ] = α{puSn−1 + (1 − p)dSn−1 − (1 + r)Sn−1 },
(9.9)
or,
p = P{Xn = 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 is
independent of (X1 , X2 , . . . , Xn−1 ). Accordingly, using inductive arguments, we conclude that (X1 , X2 , . . . , Xn ) are independent. Thus, the process Sn is a tree binomial
process.
9.2
Option Price on the Discrete Time
Definition 9.1 (Strike price and expiration time). Let Sn be the price of a stock at time
n. Consider an option that we can buy a stock at the price K at time n. The value K is
called 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)−n E[(Sn − K)+ ],
(9.11)
where Sn is defined and proved to be the risk-neutral binomial tree process in Theorem
9.1.
9.3. BLACK-SCHOLES MODEL
57
Proof. We need to find out appropriate value of C, which gives no sure-win. By Theorem 9.1, we know that Sn should be the risk-neutral binomial process when there is no
arbitrage opportunities.
Suppose we buy one option at cost C. The option is worthless if the stock price Sn
is less than K. On the other hand, if Sn is larger than the strike price K, then the value
of this option is Sn − K, since we can sell the stock obtained by exercising the option at
the 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 investment
R is
E[R] = (1 + r)−n E[(Sn − K)+ ] −C.
(9.13)
By Arbitrage Theorem (Theorem 8.2), this should be zero, or we have an arbitrage.
Thus,
C = (1 + r)−n E[(Sn − K)+ ].
9.3
(9.14)
Black-Scholes Model
Now 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 and
the volatility σ as defined in Section 5.1. As noted in Theorem 5.3, if we divide t
into n interval, all geometric Brownian motion can be approximated by discrete-time
binomial tree process with
(
uSn−1 with probability p,
Sn =
(9.15)
dSn−1 with probability 1 − p,
where
d = e−σ
√
∆
,
(9.16)
√
u = eσ ∆ ,
µ√ 1
p=
1+
∆ .
2
σ
(9.17)
(9.18)
and ∆ = t/n is the time interval. However, by Theorem 9.1. if we assume there is no
arbitrage, the binomial process should be risk-neutral with
p = P{up} =
1 + rt/n − d
,
u−d
(9.19)
58
CHAPTER 9. BLACK-SCHOLES FORMULA
since the nominal interest rate is rt/n. Using the Taylor expansion of the exponential
function, we have
√
p
σ 2t
+ O((t/n)3/2 ),
t/n +
2n
√
p
σ 2t
u = eσ t/n = 1 + σ t/n +
+ O((t/n)3/2 ),
2n
d = e−σ
t/n
= 1−σ
(9.20)
(9.21)
Using these in (9.19), we have
1
r − σ 2 /2 p
p≈
1+
t/n ,
2
σ
(9.22)
for a large n. Comparing this to (9.18), we can conclude that the original geometric
Brownian motion has to have the drift
µ = r − σ 2 /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 interest
rate. If there is no arbitrage, the geometric Brownian motion representing stock price
dynamics should have the drift µ = r −σ 2 /2 and the volatility σ . This process is called
the 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 strike
price K and the expiration time t. Let C be the price of this option. If we assume there
is no arbitrage opportunity, then we have
C = e−rt E[(S(t) − K)+ ],
(9.24)
where S(t) is the geometric Brownian motion with the drift µ = r − σ 2 /2 and the
volatility σ .
Proof. Since we assume no arbitrage, the expected gain of all bets, including purchase
of 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) is
nothing 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 is
obtained by
C = e−rt E[(seW − K)+ ]
−rt
= sΦ(ω) − Ke
(9.26)
√
Φ(ω − σ t),
(9.27)
where W is a normal random variable with N((r − σ 2 /2)t, σ 2t), Φ(x) is the distribution function of the standard normal random variable and
ω=
rt + σ 2t/2 − log(K/s)
√
.
σ t
(9.28)
To prove this Corollary, we need the following Lemmas. Note that S(t) can be
expressed in
S(t) = seW
(9.29)
√
(r−σ 2 /2)t+σ tZ
= se
,
(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 are
considered to be equivalent:
√
{S(t) > K} ⇐⇒ Z > σ t − ω .
(9.31)
Proof. By (9.29), we have
√
2
K
.
{S(t) > K} ⇐⇒ e(r−σ /2)t+σ tZ >
s
(9.32)
Since log is increasing function, we have
√
(r−σ 2 /2)t+σ tZ
e
K
>
s
log(K/s) − (r − σ 2 /2)t
√
⇐⇒ Z >
σ t
√
⇐⇒ Z > σ t − ω ,
where we used the fact:
√
log(K/s − (r − σ 2 /2)t
√
σ t −ω =
.
σ t
(9.33)
60
CHAPTER 9. BLACK-SCHOLES FORMULA
Lemma 9.2. Let I be the indicator such as
(
1 if S(t) > K,
I=
0 if S(t) ≤ K.
(9.34)
Then,
√
E[I] = P{S(t) > K} = Φ(σ t − ω).
(9.35)
Proof. Using Lemma 9.1, we have
E[I] = P{I = 1} = P{S(t) > K}
√
= P{Z > σ t − ω}
√
= P{Z < ω − σ t}.
Lemma 9.3.
e−rt E[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 random
variable Z, we have
h
√ i
2
E[IS(t)] = E 1{Z>α} se(r−σ /2)t+σ tZ
Z ∞
=
se(r−σ
2 /2)t+σ x
√
t
se(r−σ
2 /2)t+σ x
√
t
dP{Z ≤ x}
α
Z ∞
=
α
2
1
√ e−x /2 dx
2π
√
∞
2
2
1
= √ se(r−σ /2)t
e−(x −2σ x t)/2 dx
α
2π
√
√
Since x2 − 2σ x t = (x − σ t)2 − σ 2t, we have
Z
√ 2
∞
1
e−(x−σ t) /2 dx
= √ sert
α
2π
Z ∞
2
1
rt
= √ se
e−y /2 dy,
−ω
2π
Z
√
where we put y = x − σ t. Since the integrant is indeed the density of standard normal
distribution, we have
= sert P{Z > −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 , we
have
C = e−rt E[(S(t) − K)+ ]
= e−rt E[I(S(t) − K)]
= e−rt E[I(S(t)] − Ke−rt E[I]
√
= sΦ(ω) − Ke−rt Φ(σ t − ω),
where we used Lemma 9.2 and 9.3.
9.4
Examples of Option Price via Black-Scholes Formula
Example 9.1 (Option price via Black-Scholes). Suppose the current price of a stock
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 expiration date is 3 month
from now. We will estimate C.
Then, t = 1/4 and ω in (9.28) is
rt + σ 2t/2 − log(K/s)
√
σ t
0.08 · 1/4 + 0.22 · 1/4 · 1/2 − log(36/30)
p
=
0.2 1/4
ω=
≈ −1.57322.
Thus, by (9.27) in Corollary 9.1, we have
√
C = sΦ(ω) − Ke−rt Φ(ω − σ t),
p
= 30 · Φ(−1.57322) − 34e0.08·1/4 Φ(−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 option
purchase. You can get more profit when the stock price rises.
Table 9.1: Return when the stock price rises
value at 0 value at 1 return
Stock
30
40
4/3
Option
0.07
4
4/0.07
However, by Lemma 9.2,
√
P{S(t) > K} = Φ(σ t − ω) = 0.0471424.
(9.37)
62
CHAPTER 9. BLACK-SCHOLES FORMULA
With high probability you will lose your money with this option. Thus, you can understand that option is high risk and high return.
In Figure 9.1, we show the simulation of the stock price dynamics. Your opinion?
40
35
30
25
20
40
60
80
100
Figure 9.1: Example of risk-neutral geometric brownian motion with the volatility
σ = 0.2 starting from S0 = 30, when the interest rate r = 0.08. Here we used the
tree-binomial approximation with the interval ∆ = 0.01.
9.5
References
reference here!
9.6
Exercises
Exercise 9.1. Estimate the value of following call options, given the current price of
a stock s = S(0) = 30, and compare the result with Example 9.1. What can you say
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 9.2. Give two examples where buying options is more appropriate than buying its stocks.
Chapter 10
Delta Hedging Strategy
10.1
Binomial Hedging Model
As usual, we use the discrete-time model of stock price. Let S0 = s be the initial price
and
(
us,
(10.1)
S1 =
ds.
Here’s our new assignment. Find the amount of money x at time 0 required to meet a
payment P1 below varying according to stock price.
(
a up,
(10.2)
P1 =
b down.
Actually, P1 is representing derivatives more general than option. Now assume we buy
y 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
(
ys · u + (x − ys)(1 + r) up,
R1 =
(10.3)
ys · d + (x − ys)(1 + r) down,
where r is the interest rate. Thus, if we can set (x, y) satisfying
ys · u + (x − ys)(1 + r) = a,
ys · d + (x − ys)(1 + r) = b,
then our assignment is solved. Subtracting both sides, we have
y(us − ds) = a − b.
a−b
y=
.
s(u − d)
63
(10.4)
(10.5)
64
CHAPTER 10. DELTA HEDGING STRATEGY
Substituting this y into (10.4), we have
a−b
a−b
· u + (x −
)(1 + r) = a,
u−d
u−d
a−b
· {u − (1 + r)} + x(1 + r) = a,
u−d
Solving this with respect to x, we have
1+r−d
u−1−r
1
a
+b
.
x=
1+r
u−d
u−d
(10.6)
Here we use the risk-neutral probability p defined in Theorem 9.1 as
1+r−d
,
u−d
u−1−r
1− p =
.
u−d
p=
(10.7)
(10.8)
Then,
a
b
+ (1 − p)
1+r
1+r
1
=
{pa + (1 − p)b},
1+r
x= p
(10.9)
which means that if we prepare x at time 0, which is just the present value of the
expected 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)
(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 money
required to cover the payoff at time 2, given that
S2 = ui d 2−i s,
(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 = u2 s, 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, we
know that the money required at the time 1, say x1,1 , should be
x1,1 =
1
{px2,2 + (1 − p)x1,2 }.
1+r
(10.12)
Note that p = (1 + r − d)/(u − d) depends on the stock price dynamics d, u but doesn’t
depend 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 1
as
x0,1 =
1
{px1,2 + (1 − p)x0,2 },
1+r
(10.14)
x1,2 − x0,2
.
ds(u − d)
(10.15)
with
y0,1 =
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) = ui s],
(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
{px1,1 + (1 − p)x0,1 }.
1+r
(10.17)
Using (10.12) and (10.14), we have
1
[p{px2,2 + (1 − p)x1,2 } + (1 − p){px1,2 + (1 − p)x0,2 }]
(1 + r)2
2
1
p x2,2 + 2p(1 − p)x1,2 + (1 − p)2 x0,2 .
(10.18)
=
2
(1 + r)
x0,0 =
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 the
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].
(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. The
rest is kept at the bank.
2. At time 1, we happen to have S1 = us (up). The value of our investment at 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 selling or buying, and put
the rest at the bank.
66
CHAPTER 10. DELTA HEDGING STRATEGY
3. At the time 2, we have S2 = uds (up and down), then the value of our investment
have x1,2 . which is our requirement.
Now it’s time to generalize the argument. Let xi,n be the payoff at time n given
that Sn = ui d n−i s. Also let xi,k be the money required at time k given that Sk = ui d k−i s.
Then
xi,k = E[time-k present value of payoff at time n|Sk = ui d k−i s]
1
=
{pxi+1,k+1 + (1 − p)xi,k+1 }
1+r
n−k n−k j
p (1 − p)n−k− j xi+ j,n ,
= (1 + r)−(n−k) ∑
j
j=0
(10.21)
(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 = (ui d n−i s − K)+ .
(10.24)
for i = 1, 2, . . . , n. Clearly, this is nothing but the payoff of the call option with strike
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 8.1), the initial cost of our investment strategy
x0,0 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 riskneutral binomial tree process. Set the initial investment money as
C = (1 + r)−n E[(Sn − K)+ ],
(10.25)
which is equal to the call option price of strike price K and expiration time n. Let
xi,n = (ui d n−i s − K)+ , and recursively set xi,k
1
{pxi+1,k+1 + (1 − p)xi,k+1 }
1+r
n−k n−k j
= (1 + r)−(n−k) ∑
p (1 − p)n−k− j xi+ j,n .
j
j=0
xi,k =
(10.26)
(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)),
when Sk = ui d k−i s for i = 0, 1, 2 . . . , k and k = 0, 1, 2, . . . , n.
(10.28)
10.2. HEDGING IN BLACK-SCHOLES MODEL
10.2
67
Hedging in Black-Scholes Model
Here’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 0 and g0 are continuous on (a, b), and g0 (c) for some fixed point
c ∈ (a, b).
If f (x) → 0 and g(x) → 0 as x → c, and there exists limx→c f 0 (x)/g0 (x), then
f (x)
f 0 (x)
= 0 .
x→c g(x)
g (x)
lim
(10.29)
Proof. See Exercise 10.1.
Let us consider a call option for a stock. Assume the stock price dynamics is
governed 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 brownian
motion. Let h be the time interval and S(t) be the stock price at time t. Then,
( √
h,
seσ √
S(h) =
(10.30)
−σ
h.
se
Let C(s,t) be the cost of call option with the current price s and the expiration time
t. After time h, if the stock price rises, we must prepare the money
√
C(seσ
h
,t − h),
(10.31)
√
seσ h
which is the Black-Schole’s option price when the stock price is
tion time t − h. Similarly, if stock price goes down, we must prepare
C(se−σ
√
h
,t − h).
and the expira-
(10.32)
Thus, by the argument in Section 10.1 and (10.10), we know that buying the stock as
much 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) = lim
h→0
C(seσ
√
h ,t − h) −C(se−σ h ,t − h)
√
√
seσ h − se−σ h
C(seσ a ,t − a2 ) −C(se−σ a ,t − a2 )
.
a→0
seσ a − se−σ a
= lim
(10.34)
68
CHAPTER 10. DELTA HEDGING STRATEGY
√
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
σa
2
−σ a ,t − a2 )
da C(se ,t − a ) −C(se
D(0) = lim
(10.35)
d
σa
−σ a ]
a→0
da [se − se
Now
d
∂C(y,t − a2 ) ∂C(seσ a , u) C(seσ a ,t − a2 ) =
+
da
∂a
∂a
y=seσ a
u=t−a2
σa
2
σa
= lim Ct (se ,t − a )(−2a) +Cs (se ,t − a2 )sσ eσ a
a→0
= sσCs (s.t),
where Ct (s,t) = ∂C(s,t)/∂t and Cs (s,t) = ∂C(s,t)/∂ s. Similarly,
−
d
C(se−σ a ,t − a2 ) = lim −Ct (se−σ a ,t − a2 )(−2a) −Cs (se−σ a ,t − a2 )(−sσ e−σ a )
a→0
da
= sσCs (s.t).
Thus, (10.35) can be rewritten as
2sσCy (s.t)
2sσ
∂C(s,t)
=
.
∂s
D(0) = lim D(h) =
h→0
(10.36)
The quantity D(0) = ∂C(s,t)/∂ s is the sensitivity of the option price with the initial
stock price and is called ∆.
Thus, we have the following theorem:
Theorem 10.2 (Delta hedge in geometric brownian process). Let S(t) be the riskneutral geometric brownian process. Set the initial investment money as
C(0,t) = e−rt E[(S(t) − K)+ ],
(10.37)
which is equal to the call option price of strike price K and expiration time n. We can
hedge the option payoff by continuously adjusting the position of stock by ∆
∆=
∂C(s,t)
,
∂s
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.
(10.38)
10.3. PARTIAL DERIVATIVE ∆
69
Theorem 10.3 (Partial Derivative ∆). The partial derivative ∆ is obtained by
∆=
∂C(s,t)
= Ψ(ω).
∂s
(10.39)
Proof. First, recall that
C(s,t) = E[e−rt (S(t) − K)+ ].
(10.40)
Set the indicator function I = 1{S(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
∂
= E[e−rt (S(t) − K)I]
∂ s
∂ −rt
{e (S(t) − K)I} .
=E
∂s
∆=
Now, we have
∂ −rt
∂ {e−rt (S(t) − K)}
∂I
{e (S(t) − K)I} = I
+ {e−rt (S(t) − K)}
∂s
∂s
∂s
∂ {e−rt (S(t) − K)}
=I
,
∂s
on {S(t) 6= K}. Since P{S(t) = K} = 0, we have
∂ −rt
∆=E
{e (S(t) − K)I} S(t) 6= K P{S(t) 6= K}
∂s
∂ −rt
+E
{e (S(t) − K)I} S(t) = K P{S(t) = K}
∂s
∂ {e−rt (S(t) − K)}
=E I
∂s
(10.42)
(10.43)
Since S(t) is the geometric brownian motion, S can be represented as
S(t) = se(r−σ
2 )t+σ
√
tZ
,
(10.44)
for some standard normal random variable Z (see (9.29)). Thus,
∂ −rt
∂ S(t)
e (S(t) − K) = e−rt
∂s
∂s
S(t)e−rt
=
.
s
(10.45)
70
CHAPTER 10. DELTA HEDGING STRATEGY
Using (10.45) in (10.43), we have
∆=
e−rt
E[IS(t)]
s
(10.46)
In Lemma 9.3, we have already shown that
10.4
e−rt E[IS(t)] = sΦ(ω).
(10.47)
f (x) − f (c)
x−c
(10.48)
References
reference here!
10.5
Exercises
Exercise 10.1. Use
f 0 (x) = lim
x−c
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 types
of options. The options or derivatives more complicated are called exotic options.
11.1
American Call Options
Let 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 option
of calling for the stock (or simply buy) at a strike price K with some specific expiration
time t.
Depending when the buyer of option can exercise the option, we have the following
two major types of options.
Definition 11.2 (European and American call option). When you buy an american type
call option, you can exercise your option at any time prior to the expiration time t. On
contrary, in european type call options, you can exercise your option only when the
expiration time.
Remark 11.1. We have been implicitly assume only european call options up to this
section. However, it will be shown that this assumption is valid!
The freedom when we can exercise the option seems to make option more attractive. The truth can be found in the following theorem.
Theorem 11.1 (No american call option). Never exercise an american option before
its expiration time.
Proof. Let S be the present price of the stock, K be the strike price, and t be the expiration time. Let us assume we exercise the option now. Of course, it should be S > K
to obtain the fixed profit Pa :
Pa = S − K.
71
(11.1)
72
CHAPTER 11. MORE VANILLA OPTIONS IN DETAIL
Consider that we sell the stock in short and wait exercising the option until the expiration time. Now you obtain S by selling the stock. Suppose at time t the stock price is
above 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 the
stock 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 your
option before the time t.
Remark 11.2. Suppose you have a call option. When the stock price rises beyond your
expectation, you might have the temptation to exercise the option now and obtain the
fixed profit. However, according to Theorem 11.1, what you should do is to sell short.
Thus, american call option means nothing.
11.2
Put Options
Here’s one other type of vanilla options.
Definition 11.3 (Put option). A put option of a stock is defined to give you the option
of putting the stock up for a sale (or simply sell) at a strike price K with some specific
expiration 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, in
european 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 expensive
than the european one.
Theorem 11.2 (Put-call option parity). Let S be a stock price at time 0, and let C be
the 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 time
t. 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 sell
the share we bought, and obtain K. On the other hand, when S(t) ≥ K, the put
option 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 time
t 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 time
0, which we can deposit in a bank. At time t, we need K independent of the
stock price (see Exercise 11.2). Meanwhile the money we deposited in our bank
is worth (S + P −C)ert > K, by the assumption.
Thus, in either cases, there is an arbitrage.
11.3
Pricing American Put Options
There is no closed form solution of pricing american put options. You need solve
dynamic programming.
Let S(t) be a stock price at time t which is the risk neutral geometric brownian
motion with the volatility sigma and the interest rate r. Let us consider the american
put option of the stock with expiration time t, which means the option of putting the
stock up for a sale (or simply sell) at a strike price K with some specific expiration time
t, and you can exercise your option any time up to time t.
74
CHAPTER 11. MORE VANILLA OPTIONS IN DETAIL
First, as usual, we will use tree-binomial process to approximate the risk-neutral
geometric brownian motion. Let n be arbitrary large number, and we split the time t
into n intervals, and let tk = kt/n be the discrete time instances. Thus,
(
uS(tk ) with probability p,
S(tk+1 ) =
(11.7)
dS(tk ) with probability 1 − p.
where
√
u = eσ
d=e
p=
−σ
n/t
√
,
n/t
,
1 + rt/n − d
.
u−d
With n → ∞, we have the above tree binomial process to be converged to the original
risk-neutral geometric brownian motion.
Further, for approximation of the american put option, we assume we can exercise
the 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 − i
downs, the stock price is
S(tk ) = ui d k−i s,
(11.8)
where s is the initial stock price, and i = 0, 1, . . . k. Thus, potentially, according to the
number of up’s, we have k + 1 different values of S(tk ). Consider when we are at the
time tk without exercising the put option. Depending the value of S(tk ), we should
determine 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 ) = ui d k−i s,
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 expected
return when we won’t do it now. Only when K − S(tk ) > Vk (i), we exercise the option
now. Starting from time tn . we will derive all those conditional expected return back to
time 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 of
an 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 − sui d n−i )+ ,
(11.9)
for i = 0, 1, . . . , n.
2. Descend from n to estimate
Vk (i) = max K − sui d k−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 the
american put option is approximated by V0 (0).
Proof. We start from time tn . Suppose we haven’t exercise the put option yet. The
optimal policy is to exercise the put option only when S(tn ) < K and obtain the return K − S(tn ), otherwise the put option is worthless. Thus, when S(tn ) = sui d n−i , the
conditional expected return Vn (i) is completely determined by
Vn (i) = (K − sui d n−i )+ .
(11.11)
Now suppose that the stock price at time tk is S(tk ) = sui d k−i , and we haven’t exercise
the put option yet. Here’s two choices you can make:
1. Do not exercise the put option now. Then, the present value of the expected
return 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 − sui d k−i ,
(11.13)
where of course K > sui d k−i .
Thus, the optimal policy to maximize the present value of the expected return is to
compare the above two and make a decision. The resulted return from this optimal
policy is
Vk (i) = max(K − sui d k−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 is
the price of the american put option.
Corollary 11.1 (When exercise your put option). You should exercise your put option
when you find your stock as
S(tk ) < K − e−rt/n {pVk+1 (i + 1) + (1 − p)Vk+1 (i)} .
(11.15)
76
CHAPTER 11. MORE VANILLA OPTIONS IN DETAIL
Example 11.1 (American put option). Consider an american put option with the initial
stock price s = 9, expiration time t = 0.25 (one quarter year), strike price K = 10. Let
the 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
= 0.505.
u−d
Now at time k = 5, we have 5 different cases;

9u5 > 10,





9u4 d = 9u3 > 10,



9u3 d 2 = 9u = 9.635,
S(t5 ) =

9u2 d 3 = 9d = 8.416,





9u1 d 4 = 9d 2 = 7.359,


 5
9d = 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 − 9u3 d 1 , e−rt/n {pV5 (4) + (1 − p)V5 (3)}) = e−rt/n {pV5 (4) + (1 − p)V5 (3)} = 0.181,
V4 (2) = max(10 − 9u2 d 2 , e−rt/n {pV5 (3) + (1 − p)V5 (2)}) = 10 − 9 = 1,
V4 (1) = max(10 − 9ud 3 , e−rt/n {pV5 (2) + (1 − p)V5 (1)}) = 10 − 9ud 3 = 2.130,
V4 (0) = max(10 − 9d 4 , e−rt/n {pV5 (1) + (1 − p)V5 (0)}) = 10 − 9d 4 = 3.119.
In other words, when you find that the stock price is less than 9 at time k = 4, we should
exercise 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 Dividend
First, recall the solution of the following elementary ordinary differential equation;
Lemma 11.1. The differential equation of
(
dN(t)
dt = αN(t),
N(0) = N0 ,
(11.17)
has an unique solution as
N(t) = N0 eαt .
(11.18)
Proof. Arranging and Integrating both side, we have
Z
dN(t)
=
N(t)
Z
αdt.
(11.19)
dx
= log x +C,
x
(11.20)
log N(t) = αt +C,
(11.21)
N(t) = Ceαt .
(11.22)
Since
Z
we have
or
Now taking t = 0, we can find the integral constant C = N0 .
78
CHAPTER 11. MORE VANILLA OPTIONS IN DETAIL
Now we consider the call option of a stock with dividend. According to the stock
price, 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 price
of a stock at time t whose dividend is continuously paid proportional to S(t). Then, the
stock price is geometric brownian motion with
S(t) = S(0)e−αt eW ,
(11.23)
where α is the ratio of dividend to the stock price and W is the normal random variable
of N[(r − σ 2 /2)t,tσ 2 ] and r is the interest rate.
Proof. Here we suppose we adopt the investment strategy that all the profit of dividend
will be use to purchase for the same stock. Let N(t) be the total number of stock
we have at time t, and let D(t) be the total dividend obtained up to time t. Thus
dD(t) = D(t + dt) − D(t) be the dividend paid for a small interval dt, which is assumed
to 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, the
number of stock we can add at time t is
dN(t) =
dD(t)
= αN(t)dt.
S(t)
(11.25)
Thus, we obtain a differential equation:
dN(t)
= αN(t),
dt
(11.26)
N(0) = 1.
(11.27)
with its initial condition
Thus, setting initially we owned one stock, i.e., N0 = 1 in Lemma 11.1, we can find the
number 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)
M(0) = S(0).
(11.30)
Note that we have initially,
11.5. STOCK WITH FIXED-TIME-DIVIDEND
79
By using the same argument as before, it is reasonable to assume M(t) should be
a risk-neutral geometric brownian motion. Otherwise, we have an arbitrage using this
investment 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 − σ 2 /2)t,tσ 2 ]. 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 with
dividend.
Theorem 11.5 (Call option of the stock with dividend). Let S(t) be the price of a
stock at time t whose dividend is continuously paid proportional to S(t) as in Theorem
11.4. Consider a call option of the stock with its expiration time t and the strike price
K. Then the price of this option C is evaluated by adjusting the initial stock price as
s = S(0)e−αt in the Black-Scholes formula (Corollary 9.1) as
C = e−rt E[(S(0)e−αt eW − K)+ ]
−αt
= S(0)e
−rt
Φ(ω) − Ke
√
Φ(ω − σ t).
(11.33)
(11.34)
Proof. Since the price of option is estimated by the present value of the expected payoff, we have
C = e−rt E[(S(t) − K)+ ].
(11.35)
By Theorem 11.4, we can rewrite the above as
C = e−rt E[(S(0)e−αt eW − 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-Dividend
The assumption of continuous paying is rather artificial. Now let us assume the dividend is paying at some specified time.
Suppose at time t0 the dividend αS(t0 ) will be paid. Then, the stock price will be
dropped as much as S(t0 ) at time t0 . Otherwise, we can buy the stock just before time
t0 and obtain the dividend, and then sell the stock. Thus, if we assume no arbitrage, the
stock will be dropped at the time t0 .
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CHAPTER 11. MORE VANILLA OPTIONS IN DETAIL
In this case, we cannot expect the stock price S(t) to be geometric brownian motion, since it has a jump. However, we consider an investment strategy whose value is
geometric brownian motion. Let S(t0 +) be the stock price just after t0 , i.e.,
S(t0 +) = lim S(t).
t↓t0
(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
α
αS(t0 )
=
.
S(t0 +) 1 − α
(11.39)
Also, the number of share we have after t0 is
1+
α
1
=
.
1−α
1−α
Let M(t) be the market value of our investment at time t.
(
S(t)
t < t0 ,
M(t) =
1
S(t)
t ≥ t0 .
1−α
(11.40)
(11.41)
Now we assume the market value of our investment M(t) follows geometric brownian motion with drift r − σ 2 /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 other
hand, 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 − σ 2 /2)t,tσ 2 ]. Since M(0) = S(0), we
have
S(t) = (1 − α)S(0)eW .
(11.44)
Thus, in this case, we can use Black-Scholes formula adjusting the initial stock price
to be s = S(0)(1 − α).
11.6
Forward and Futures Contract
Now consider pricing problem of forward contracts.
11.7. EXERCISES
81
Definition 11.4 (Forward contract). In a forward contract, you agree now to pay the
fixed money at time t for one share of a stock. Note that the deal is fixed now and you
cannot waive the purchase unlike the options, even when the stock price is descended
considerably.
Definition 11.5 (Futures contract). Futures contract is nothing but the standardized
format of forward contracts so as to dealt within the market. Thus, you can sell or buy
the futures contract before the expiration time.
Remark 11.3. The forward contract is mainly used among the bank and security company. While, the futures contracts are used for anyone. See the detailed difference and
explanation 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 buy
the forward contract, along with sell the stock in short. Thus at time 0, we obtain S and
put it in a bank. Then, at time t, we have Sert in cash. To cover the short-selling, we buy
a 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 the
bank, but we can sell the stock at the price F by the future contract. Thus, we can get
the amount F > Sert .
In either case, we have an arbitrage opportunity. Thus, by Arbitrage theorem (Theorem 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 we
need the extra storage cost for the commodities.
11.7
Exercises
Exercise 11.1. Consider the case when american put options have some advantage
over 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 Probability: 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 Applications. 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: Options
and Other Topics. Cambridge Univ Pr, 2002.
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