Applied Probability and Mathematical Finance Theory

Transcription

1 Applied Probability and Mathematical Finance Theory Hiroshi Toyoizumi 1 January 16, toyoizumi@waseda.jp

2 Contents 1 Introduction 5 2 Basic Probability Theory Why Probability? Probability Space Conditional Probability and Independence Random Variables Expectation, Variance and Standard Deviation Covariance and Correlation How to Make a Random Variable References Exercises Normal Random Variables What is Normal Random Variable? Lognormal Random Variables References Exercises Useful Probability Theorems The Law of Large Numbers Poisson Random Variables and the Law of Small Numbers The Central Limit Theorem Useful Estimations References Exercises Brownian Motions and Poisson Processes Geometric Brownian Motions Discrete Time Tree Binomial Process Brownian Motions Poisson Processes References Exercises

3 CONTENTS 3 6 Simulating Brownian Motions Mathematica As an Advanced Calculator Generating Random Variables Two-Dimensional Brownian Motions Random Variable on Unit Circle Generating Brownian Motion Geometric Brownian Motions Generating Bernouilli Random Variables Generating geometric brownian motions Present Value Analysis Interest Rates Continuously Compounded Interest Rates Present Value Rate of Return References Exercises Risk Neutral Probability and Arbitrage Option to Buy Stocks Risk Neutralization Arbitrage and Price of Option Duplication and Law of One Price Arbitrage Theorem References Exercises Black-Scholes Formula Risk-neutral Tree Binomial Model Option Price on the Discrete Time Black-Scholes Model Examples of Option Price via Black-Scholes Formula References Exercises Delta Hedging Strategy Binomial Hedging Model Hedging in Black-Scholes Model Partial Derivative References Exercises

4 4 CONTENTS 11 More Vanilla Options in Detail American Call Options Put Options Pricing American Put Options Stock with Continuous Dividend Stock with Fixed-time-Dividend Forward and Futures Contract Exercises

5 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

6 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

7 2.2. PROBABILITY SPACE 7 An event ω can be first prize, second prize,..., lose, and so on. Sometimes, it is easy to use sets of events in sample space Ω. Example 2.3 (Sets in Lottery). The followings are examples in Ω of Example 2.2. W = {win} = {first prize, second prize,..., sixth prize} (2.1) L = {lose} (2.2) Thus, we can say that what is the probability of win?, instead of saying what 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) = For any sequence of mutually exclusive A 1,A 2... P( A i ) = i=1 i=1 In addition, P is said to be the probability measure on Ω. P(A i ). (2.3) 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 P(n) = P(win n-th prize) = n 100 (2.4) P(0) = P(lose) = (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) = 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 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(A) = P(B A). (2.6) P(A) 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 W) P(W) = P(F) P(W) = 10/ /1000 = = P(F). 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(A) where we used A and B are independent. = P(B)P(A) P(A) = P(B),

9 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) = 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) 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 history 1. 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 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) dx = dp{x x}. (2.10) dx 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} = A dp{x x} = f (x)dx. (2.11) A 2.5 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). E[X] = xdp{x x} = x f (x)dx. (2.12) Remark 2.4. For a discrete random variable, we can rewrite (2.12) as E[X] = x n P[X = x n ]. (2.13) n Lemma 2.2. Let (X n ) n=1,...,n be the sequence of random variables. change the order of summation and the expectation. Then we can E[X X N ] = E[X 1 ] + + E[X N ] (2.14)

11 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. (2.18) Var[X] = E[X 2 ] E[X] 2 = E[X] E[X] 2 = 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) (2.20) Lemma 2.4. For any pair of independent random variables X and Y, we have

12 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, E[XY ] = xydp(x x,y y). Since X and Y are independent, we have P(X x,y y) = P(X x)p(y y). Thus, E[XY ] = = xydp(x x)dp(y y) xdp(x x) ydp(x y) = E[X]E[Y ]. Using the first part, it is easy to check the second part (see Exercise 2.9.) Example 2.11 (Binomial random variable). Let X be a random variable with X = n i=1 X i, (2.21) where X i 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, (2.22) Var[X] = np(1 p). (2.23) 2.6 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.

13 2.7. HOW TO MAKE A RANDOM VARIABLE 13 Example 2.12 (The covariance of a pair of Binomial random variables). Let X 1 and X 2 be the independent Binomial random variables with the same parameter n and p. The covariance of X 1 and X 2 is Cov[X 1,X 2 ] = E[X 1 X 2 ] E[X 1 ]E[X 2 ] = 0, since X 1 and X 2 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 = X 1 + X 2. How about the covariance of X 1 and Y? Cov[X 1,Y ] = E[X 1 Y ] E[X 1 ]E[Y ] = E[X 1 (X 1 + X 2 )] E[X 1 ]E[X 1 + X 2 ] = E[X 2 1 ] E[X 1 ] 2 = Var[X 1 ] = np(1 p) > 0. Thus, the covariance of X 1 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 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 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). (2.29) 2.8 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.

15 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 X = n i=1 X i, (2.30) where X i 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 Prove for any pair of random variables, we have 1 ρ[x,y ] 1. (2.31)

16 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 = 1 (2π) 1/2 σ e (x µ)2 /2σ 2 dx, (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 and the variance E[X] = µ, (3.2) Var[X] = σ 2. (3.3) 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

17 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 (X i ) i=1,2,...,n are independent normal random variables with its mean µ i and standard deviation σ i. Then the sum of these random variables X = n i=1 X i is again a normal random variable with µ = σ 2 = n i=1 n i=1 µ i, (3.4) σ 2 i. (3.5) Proof. We just prove X satisfies (3.4) and (3.5). By Lemma 2.2, µ = E[X] = E Also, by Lemma 2.4, we have σ 2 = E[X] = Var [ n i=1x i ] [ n i=1x i ] = = n i=1 n i=1 E [X i ] = n i=1 Var [X i ] = We can prove that X is a normal random variable by using so-called characteristic function method (or Fourier transform). µ i. n i=1 σ 2 i. 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 = e X, (3.6) where X is a normal random variable. Lognormal random variables plays a measure role in finance theory!

18 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 = e X has the mean and the variance as E[Y ] = e µ+σ 2 /2, (3.7) Var[Y ] = e 2µ+2σ 2 e 2µ+σ 2. (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[e tx ]. (3.9) Then, by differentiate the left hand side and setting t = 0, we have Further, we have m (0) = d dt m(t) t=0 = E[Xe tx ] t=0 = E[X]. (3.10) m (0) = E[X 2 ]. On the other hand, since X is the standard normal random variable, we have m(t) = E[e tx ] = 1 2π Since tx x 2 /2 = {t 2 (x t) 2 }/2, we have m(t) = 1 e t2 /2 2π e tx e x2 /2 dx. e (x t)2 /2 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) = e t2 /2. More generally, when X is a normal random variable with N(µ,σ), we can obtain (See exercise 3.6.) Since Y = e X, we have m(t) = E[e tx ] = e µt+σ 2 t 2 /2. (3.11) E[Y ] = m(1) = e µ+σ 2 /2, (3.12)

19 3.2. LOGNORMAL RANDOM VARIABLES 19 and E[Y 2 ] = m(2) = e 2µ+2σ 2. (3.13) Thus, Var[Y ] = E[Y 2 ] E[Y ] 2 = e 2µ+2σ 2 e 2µ+σ 2. (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 logs(n) = logs(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 logy > 0, we have P{the stock rises} = P{S(1) > S(0)} { } S(1) = P S(0) > 1 { ( ) } S(1) = P log > 0 S(0) = P{X > 0} where X = logy (1) is N(µ,σ), and we can define as the standard normal distribution. Hence, we have { X µ P{S(1) > S(0)} = P σ Z = X µ σ, (3.17) = P{Z > µ/σ} = P{Z < µ/σ}, > 0 µ } σ

20 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} 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[e tx ] is said to be the moment generating function of X. 1. Prove E[X] = m (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 = 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.

21 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 = e X be a lognormal random variable where X is N(µ,σ). Find E[Y ] and Var[Y ].

22 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 X 1,X 2,... be an i.i.d. sequence of random variables with Let S n = n i=1 X i. Then, for all ε > 0, µ = E[X n ]. (4.1) P{ S n /n µ > ε} 0 as n. (4.2) 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 X 1,X 2,... be an i.i.d. sequence of random variables with With probability 1, we have S n n µ = E[X n ]. (4.3) µ as n. (4.4) So, now we cay say that almost all trials, the average will be converges to µ. Example 4.1. Gambler example 22

23 4.2. POISSON RANDOM VARIABLES AND THE LAW OF SMALL NUMBERS 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 where λ is a constant equal to its mean E[N]. Theorem 4.3. Let N be a Poisson random variable. P{N = n} = (λ)n n! e λ, (4.5) 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 X 1,X 2,... be an i.i.d. sequence of random variables with µ = E[X n ], (4.7) σ 2 = Var[X n ]. (4.8) Now we would like to estimate the sum of this random variables, i.e. S n = n i=1 X i. (4.9) Theorem 4.5 (Central limit theorem). For a large n, we have { } Sn nµ P σ n x Φ(x), (4.10) where Φ(x) is the distribution function of the standard normal distribution N(0,1). Or S n N(nµ,σ n). (4.11)

24 24 CHAPTER 4. USEFUL PROBABILITY THEOREMS The central limit theorem indicate that no matter what the random variable X i is like, the sum S n 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 S n converges to a Normal random variable. Example 4.3 (Average of Bernouilli Random Variables). Let X i be i.i.d Bernoulli random variables with p = 1/2. Suppose we are going to evaluate the sample average A of X i s; A = 1 n n i=1 X i. (4.12) Of course, we expect that A is close to the mean E[X i ] = 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 for all a > 0. P{X a} E[X]/a, (4.13) Proof. Let I A be the indicator function of the set A. It is easy to see Taking expectation on the both side, we have ai {X a} X. (4.14) ap{x a} E[X]. (4.15)

25 4.4. USEFUL ESTIMATIONS Figure 4.1: Histogram of the sample average A = 1 n n i=1 X i when n = 10, where X i is a Bernouilli random variable with E[X i ] = 1/ Figure 4.2: The sample average A when n = Figure 4.3: The sample average A when n = 1000.

26 26 CHAPTER 4. USEFUL PROBABILITY THEOREMS Figure 4.4: The sample path of the sample average A = 1 n n i=1 X i up to n = 100, where X i is a Bernouilli random variable with E[X i ] = 1/ Figure 4.5: The detailed histgram of the sample average A = 1 n n i=1 X i when n = 10, where X i is a Bernouilli random variable with E[X i ] = 1/2. The solid line is the corresponding Normal distribution.

27 4.5. REFERENCES 27 Theorem 4.7 (Chernov Bounds). Let X be the random variable with moment generating function M(t) = E[e tx ]. Then, we have P{X a} e ta M(t) for all t > 0, (4.16) P{X a} e ta M(t) for all t < 0. (4.17) Proof. For t > 0 we have P{X a} = P{e tx e ta }. (4.18) Using Theorem 4.6, P{e tx e ta } e ta E[e tx ]. (4.19) 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!

28 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 2 t. 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 2 t.. 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) = s 0 ] = s 0 e t(µ+σ 2 /2). (5.1) 28

29 5.1. GEOMETRIC BROWNIAN MOTIONS 29 Proof. Set Y = S(t) S(0) = S(t) s 0. Then, Y is lognormal with (tµ,tσ 2 ). Thus by using Theorem 3.3 we have E[Y ] = e t(µ+σ 2 /2). On the other hand, we have Hence, we have (5.1). E[Y ] = E[S(t)] s 0. 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 where W is the normal random variable N(µt,σ 2 t). S(t) = S(0)e W, (5.2) 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) log = W. (5.3) 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 + σs(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)e µt+σb(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 30 CHAPTER 5. BROWNIAN MOTIONS AND POISSON PROCESSES Theorem 5.2. The geometric Brownian motion S(t) satisfies the stochastic differential equation, with the initial condition S(0). ds(t) = µs(t)dt + σs(t)db(t), (5.6) 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 S n is said to be a Tree Binomial process, if the dynamics is express as { us n 1 with probability p, S n = (5.7) ds n 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, (5.8) u = e σ, (5.9) p = 1 ( 1 + µ ). (5.10) 2 σ 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 Y i be the indicator of up s (Bernouilli random variable), i.e., { 1 up at time i, Y i = (5.12) 0 down at time i.

31 5.2. DISCRETE TIME TREE BINOMIAL PROCESS Figure 5.1: An example of Tree binomial process with d = ,u = , p = and starting from S 0 = 100. Figure 5.2: Exchange rate of yen at 2005/10/19. Adopted from

32 32 CHAPTER 5. BROWNIAN MOTIONS AND POISSON PROCESSES Thus, the number of up s up to time n is n i=1 Y i, and the number of down s is n n i=1 Y i. Hence, the stock price at time n given that the initial price S(0) is Now set S(n ) = S(0)u Y i d n Y i ( = S(0)d n u ) Yi. d S(t) = S(0)d t/ (u/d) t/ i=1 Y i, (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 log ( ) S(t) = t ( u S(0) logd + log d ) t/ i=1 Y i. (5.14) To imitate the dynamics of geometric Brownian motion, here we artificially set d = e σ, u = e σ, p = 1 2 ( 1 + µ σ ). Note that logd and logu are symmetric, while d and u are asymmetric, and p 1/2 as 0. Now using these, we have log ( ) S(t) = tσ + 2σ t/ S(0) Y i. (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/ i=1 i=1 Y i Normal distribution. (5.16) Thus, log(s(t)/s(0)) has also the normal distribution with its mean and variance: E[log(S(t)/S(0)] = tσ + 2σ t/ E[Y i ] i=1 = tσ + 2σ t p, = tσ + σ t ( 1 + µ ) σ = µt,

33 5.3. BROWNIAN MOTIONS 33 and t/ Var[log(S(t)/S(0)] = 4σ 2 Var[Y i ] i=1 = 4σ 2 t p(1 p) σ 2 t, as 0, since p 1/2. Thus, log(s(t)/s(0)) N[µt,σ 2 t], (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 = , u = and p = 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 = 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 2 t], 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 2 t. (5.19)

34 34 CHAPTER 5. BROWNIAN MOTIONS AND POISSON PROCESSES 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 S 0 = 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 S 0 = 100.

35 5.3. BROWNIAN MOTIONS 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 Figure 5.6: An example of trajectory of two-dimensional brownian motion.

36 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(t 1 ) N(t 0 ),N(t 2 ) N(t 1 ),...,N(t n ) N(t n 1 ), (5.20) are independent for all choice of the time instants t 0 < t 1,<... < t n. 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, 3. P{N(t + s) N(s) = n} = e λt (λt) n n!, where λ > 0 is the rate of the events. Theorem 5.4. Let N(t) be a Poisson process with its rate λ, then we have Or rewriting this, we have which validates that we call λ the rate of the process. Proof. By Definition 5.6, we have E[N(t)] = E[N(t)] = λt, (5.21) Var[N(t)] = λt. (5.22) λ = E[N(t)], (5.23) t = n=0 n=0 np{n(t) = n} λt (λt)n ne n! = λte λt (λt) n 1 n=0 (n 1)! = λte λt e λt = λt. The other is left for readers to prove (see Exercise 5.2).

37 5.5. REFERENCES References 5.6 Exercises Exercise 5.1. Find E[S n ] and Var[S n ] 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)

38 Chapter 6 Simulating Brownian Motions We simulate Brownian motions using Mathematica, following the steps below. 6.1 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]= a 2 + 2ab + b 2 In[4]:=D[ax 2 + bx + c,x] Out[4]= b + 2ax Those mathematical symbols are built-in. Example 6.2 (Symbols). Here s examples of symbols in Mathematica. Pi = π (6.1) I = i imaginary number (6.2) Log[x] = log e x (6.3) Sin[θ] = sin(θ). (6.4) (6.5) 38

39 6.2. GENERATING RANDOM VARIABLES 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]= { , , } In[5]:= SeedRandom[128] In[6]:= {Random[],Random[],Random[]} Out[6]= { , , } 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 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 Generating Brownian Motion Using the pairs of random variables as the movement, simulate a brownian motion as shown in Figure 6.2.

40 40 CHAPTER 6. SIMULATING BROWNIAN MOTIONS Figure 6.1: An example of Random Variables on Unit Circle Figure 6.2: An example of trajectroy of two-dimensional brownian motion.

41 6.4. GEOMETRIC BROWNIAN MOTIONS Geometric Brownian Motions Simulate geometric brownian motions with the drift µ and the volatility σ. Use an approximation of tree-binomial process 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] = { , , , , , , , } 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 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 S 0 = 100.

42 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 P 0 be the amount of money you invested initially, P 1 be the amount of money you obtained at the end of the year. Then, the value r e is said to be the effective interest rate, where r e = P 1 P 0 P 0. (7.3) In other words, given you invested P 0 initially, the amount you get in the end of a year can be expressed as P 1 = P 0 (1 + r e ). (7.4) Example 7.2 (Compound interest rate). Assume the same situation as in Example 7.1. The effective interest rate r e in this case is r e = P 1 P 0 P 0 = P(1 + r/2)2 P P = (1 + r/2)

43 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 n (1 + 1/n) n. (7.5) The following is a useful expression of the so-called exponential function Exp(x) = e x. Lemma 7.1 (exponential). e x = lim n (1 + x/n) n. (7.6) Proof. By the Definition 7.2, we have { e x = lim (1 + 1/n)n} x n = lim (1 + 1/n) xn n = lim (1 + m x/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. where we used Lemma 7.1. lim P(1 + n r/n)n = Pe r, (7.8) Theorem 7.1 (Continuous compounding). Let r be the nominal interest rate. effective interest rate of continuous compounding on 1 year is The r e = e r 1. (7.9)

44 44 CHAPTER 7. PRESENT VALUE ANALYSIS Proof. By Definition 7.1, r e = P 1 P 0 P 0 = P 0e r P 0 P 0 = e r 1. Corollary 7.1. The effective interest rate is always larger than the nominal interest rate. Proof. By Theorem 7.1, we have for r 0. r e = e r 1 > r, (7.10) 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 = (a 1,a 2,...,a n ), (7.12) where a i is the return at the end of year i. Then, the cash stream a has the present value PV (a) such as PV (a) = n i=1 a i (1 + r) i. (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.

45 7.4. RATE OF RETURN 45 Proof. At the end of the first year, we withdraw a 1 from our bank account. Thus, taking into the interest rate r, we have as the remainder, (1 + r)pv (a) a 1 = a r + + a n. (7.14) (1 + r) n 1 Continue this procedure to the end of the year i, and we have a i r + + a n, (7.15) (1 + r) n i at our bank. At the end of year n 1, we have only a n /(1 + r) at our bank. Thus, we pay a n 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 PV (a) = i=1 = c 1 + r = c 1 + r = c r. c (1 + r) i i=0 1 (1 + r) i 1 1 1/(1 + 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, (7.17) a which is the interest rate that makes the present value of the return equal to the initial payment.