Class Notes on Financial Mathematics. No-Arbitrage Pricing Model

Transcription

1 Class Notes on No-Arbitrage Pricing Model April 18, 2016 Dr. Riyadh Al-Mosawi Department of Mathematics, College of Education for Pure Sciences, Thiqar University References: 1. Stochastic Calculus for Finance I: The binomial Asset Pricing Model by Steven E. Shreve, Spinger, Stochastic Calculus for Finance II: Continuous-Time Model by Steven E. Shreve, Spinger, Brownian Motion Calculus by Ubbo F Wiersema, John Wiley & Sons, Ltd

2 1 No-Arbitrage Pricing Model 1.1 Introduction An option is a tickets which gives its holder the right, but not the obligation, to buy or sell a financial asset at or before a certain date (maturity) T at a predetermined price (strike price) K. The financial asset include, but are not limited to stocks, currencies, commodities (gold, copper, oil,...) and options. An option to buy the financial asset is referred to as a Call Option while an option to sell is called at Put Option. If you can exercise the option only at time of expiration T, this is called a European Option. If you can exercise at the time T or any time before the time T, it is called an American Option. The holder of a call option is not obliged to exercise it. The writer (seller) of European call option needs to pay the buyer of the option an amount of (value of the option) { (S T K) + ST K, if S T > K; = max(0, S T K) = 0, if S T K. where S T is stock price ate time T. Similarly, the value of European put option is given by { K (K S T ) + ST, if S T < K; = max(0, K S T ) = 0, if S T K. It is clear that the writer (seller) of the option bears an unlimited risk. The natural question, then, is how much the writer should charge as Pricing the Option at time t = 0. 1

3 The fundamental condition for establishing the price of an option is that it should not permit an Arbitrage. Arbitrage ( No free Lunch ) opportunity is a situation where an investor can make a granted profit without incurring any risk (riskless profit). Hedging the Option. The procedure to find the no-arbitrage price of the option is to determine how much money should be invested in the bond (riskless investment like bank account) and stock market to meet payment obligations (with respect to the seller of the option) at maturity time T. The main reason of using the options is that the option is more sensitive to price changes than the underlying asset itself. Consider for example a European call with strike price K = 0. At a time t let S t = 60 then the value of the option is (S t K) + = (60 0) + = 10. If S t increases by 10% to 66 then the value of the option will be (66 0) + = 16, an increase of 60%. Similarly if S t decreases by 10% to 4 then the value of the option decreases from 10 to (4 0) + = 4, a loss of 60%. A portfolio means a collection of financial assets such as stocks and bonds and cash equivalents. Self-financing portfolio means that the change in the portfolio value over time should only come from the change in the value of the stock and the change in borrowing i.e. no many is withdrawn or added freely. Compound Interest Rate. Consider a simple savings account with initial investment D$ invested at an annual interest rate of r compound monthly. After T years the account will be () T D$ After n months (or T = n/12 years ), the account will be ( ) nd$ ( = ) 12T D$ Suppose we compound m times a year at an annual rate of r Then after T years ( the account will be 1 + m) r mt D$. 2

4 Then the account is compound continuously if m which yields ( lim ) mt ( ( D = lim m ) T D = e m m m m) rt D$. Forward Contract. The buyer of a forward contract with maturity T and delivery price K is obliged to buy the underlying asset at price K on date T. The payments at time T will be S T K. 1.2 One Period Binomial Model Let t 0 =Begin time of the period t 1 =End time of the period S 0 =Price of a one stock at t 0 S 1 =Price of a one stock at t 1, where { S(H), if the result of a coin is Head; S 1 = S(T ), if the result of a coin is Tail. We assume that P(H) = p > 0, P(T ) = q = 1 p > 0 and p q. S 1 is known at time t 1 but not at time t 0 so it is random quantity. Suppose u = S 1(H) and d = S 1(T ), where S 0 S 0 3

5 d < u, u is called the up-factor and d is called the down-factor. S 1 (H) = us 0 S 0 S 1 (T ) = ds 0 Suppose r be the interest rate of the money market so 1$ invested (or borrowed) in the money market at time t 0 will yield (or indebt of) ()$ at time t 1. Arbitrage as a trading strategy that begins with no money at time t 0, has zero probability of losing money and has a positive probability of making money at time t 1. Let us consider the following two cases: Case(I): Suppose d. At time t 0 one could begin with zero money borrow m$ dollars from the money market in order to buy a share of stock. At time t 1 his profit from the stock market is at least dm$ his dept of the money market is ()m$ since d then dm ()m the net profit is (d 1 r)m 0. This is an arbitrage. Case(II):If u. At time t 0 one could sell a stock by m$ dollars in the stock market invest the money m dollars in the money market. At time t 1 4

6 his loss from the stock market is at most um$ his profit from the money market is ()m$ since u then um ()m the net profit is ( u)m 0. This is an arbitrage. Thus to roll out (remove) the arbitrage from the market, we should assume d < < u Let us consider a European Call Option. The owner of the call option has the right but not obligation to buy one share of the stock at time t 1 for strike price K. Assume S 1 (T ) < K < S 1 (H). Case-I: If at time t 1, we get a Head on the coin then the option can be exercised and yields the profit S 1 (H) K. Case-II: If at time t 1, we get a Tail on the coin then no need to exercise the option since S 1 (T ) < K. Hence, at time t 1, the owner of the option will get max{s 1 K, 0} = (S 1 K) +. The fundamental question of option pricing is what is the price of the option at time t 0 the coin toss results head or tail. before we know whether This question is called the Option Valuation. Example 1.1 Hedging (Replicating) the Option. Let S 0 = 4, u = 2, d = 1 and r = 1. Then, the price of stock will be 2 4 { S1 (H) = us 0 = 2 4 = 8, if Head; S 0 = 4 S 1 (T ) = ds 0 = = 2, if Tail. Consider a European Call Option with strike price be K = and maturity at t 1. Assume the initial money is X 0 = 1.20$ and we want to buy, at time t 0, 0 = 1 2 shares of stock.

7 At time t 0. The cost of 0 = 1 2 shares of stock is 0 S 0 = = 2$ then have to borrow X 0 0 S 0 = = 0.8$. So we have indebt of 0.8 to the money market. At time t 1. We will be in dept of 1$ to the money market because We will also have stock value at either or ()(X 0 0 S 0 ) = (1 + 1 )( 0.8) = 1$ S 1(H) = = S 1(T ) = = 1. The portfolio of stock and money market account at time t 1 will be or X 1 (H) = 1 2 S 1(H) + ()(X 0 0 S 0 ) = 4 1 = 3 X 1 (T ) = 1 2 S 1(T ) + ()(X 0 0 S 0 ) = 1 1 = 0. The value of the option at time t 1 is either or (S 1 (H) K) + = max(8, 0) = max(3, 0) = 3 = X 1 (H) (S 1 (T ) K) + = max(2, 0) = max( 3, 0) = 0 = X 1 (T ). Hence the no-arbitrage price of the option is 1.2$. Example 1.2 Consider Example 1.1. Suppose the price of the option is 1.21$. At time t 0 : 6

8 the seller of the option could invest 1.21 X 0 = = 0.01$ in the money market the seller of the option will invest X 0 = 1.2 to replicate the option as in Example 1.1. At time t 1 : the seller of the option will get a granted profit 0.01 () = 0.01 (1.2) = 0.012$ from the money market This is an arbitrage because the seller of the option needs no money initially and without risks of loss has profit 0.012$ at time t 1. Example 1.3 Consider Example 1.1. Suppose the price of the option is 1.19$. At time t 0 : the buyer of the option could sell 0 = 1 shares of stock from his stocks in the stock 2 market with price 0 S 0 = = 2$. the buyer of the option uses 1.19$ to buy the option, put 0.80$ in the many market and in a separate account puts the remaining 0.01$. At time t 1. If there a Head in the coin, the buyer of the option will need 4 dollars to get (restore) 1 2 share of stocks get (S 1 (H) K) + = (8, 0) + = (3, 0) + = 3$ from the option get 0.80 () = 0.80 (1.2) = 1$ from the money market 7

9 get a granted profit from the money market () = 0.80 (1.2) = 0.012$ If there a Tail in the coin, the buyer of the option will need 1$ to get (restore) 1 2 share of stocks get (S 1 (T ) K) + = (2, 0) + = ( 3, 0) + = 0$ from the option get 0.80 () = 0.80 (1.2) = 1$ from the money market get a granted profit 0.01 () = 0.80 (1.2) = 0.012$ from the money market. This is an arbitrage. Example 1.4 Consider a European call option with strike price K = 98 and maturity t 1. Suppose at time t 0, the stock price is S 0 = 100 and the price at time t 1 is either S 1 (H) = 112 or S 1 (T ) = 84. Suppose the interest rate is r = %. We want to obtain a no-arbitrage price for the call option. t 0 t 1 t 1 Money Stock 100 S 1 (T )= 84 S 1 (H)=112 Option? (S 1 (T ) 98) + = (84 98) + = 0 (S 1 (H) 98) + = (112 98) + = 14 Solution : To find the no-arbitrage pricing for this option we should construct a portfolio that satisfies the equation Value of portfolio at time t 1 =Value of option at time t 1. 8

10 Let a denote money invested in the money market and b is money invested in the stock market. Then the no-arbitrage price satisfies the equations 1.0a + 84b = 0 1.0a + 112b = 14. Hence a = 40 and b = 0.. This means that we borrow 40$ from bank and buy a half share of stocks stock. At time t 0, this portfolio has a value of = 10$. Hence the no-arbitrage price of the option is 10$. Any other price will be arbitrage price. Case-1 Suppose the price is $. At time t 0. sell 0. share of stock by = 0$. buy the option by $ put 4$ in the money market. At time t 1. If there a Head in the coin, the buyer of the option will need 6$ to get (restore) 0. share of stocks get 14$ from the option get = 47.2$ from the money market get a granted profit =.2$. If there a Tail in the coin, the buyer of the option will need 42$ to get (restore) 0. share of stocks get 0$ from the option get = 47.2$ from the money market get a granted profit =.2$ rom the money market. 9

11 This is an arbitrage. Case-2 Suppose the price is 1$. This is an arbitrage At time t 0. sell the option by 1$ borrow 3$ from the money market buy 0. of stock by 0$. At time t 1. If there a Head in the coin, the seller of the option will sell 0. of stock by 6$ give = 36.7$ to the bank give 14$ to the buyer of the option get a granted profit =.2$. If there a Tail in the coin, the seller of the option will sell 0. of stock by 42$ give = 36.7$ to the bank give 0$ to the buyer of the option get a granted profit =.2$ This is an arbitrage Value of the Derivative Security The derivative is a form of an agreement to buy or sell an financial asset at a fixed price on or before a certain date. The value derivative securities varies with the value of the financial assets. Forward contracts, call and put options are some common types of derivatives. Suppose a derivative security V of amounts at time t 1 are given as V 1 (H) and V 1 (T ). Now we want to determine the value for a derivative security at time t 0 10

12 by replicating it. Let X 0 =initial wealth (money at time zero) 0 =share of stocks to be purchased at time zero. The value of portfolio of stock and money market at time one is X 1 = 0 S 1 + ()(X 0 0 S 0 ) =()X (S 1 ()S 0 ). that is mean X 1 (H) =()X (S 1 (H) ()S 0 ) X 1 (T ) =()X (S 1 (T ) ()S 0 ). Note that V 1, S 1, S 0, r are known and X 0, 0, V 0 are unknown We want to choose X 0 and 0 so that Now X 1 (H) = V 1 (H) and X 1 (T ) = V 1 (T ). Also X 1 (H) = V 1 (H) ()X (S 1 (H) ()S 0 ) = V 1 (H) ( ) S1 (H) X S 0 = V 1(H). (1.1) X 1 (T ) = V 1 (T ) ()X (S 1 (T ) ()S 0 ) = V 1 (T ) ( ) S1 (T ) X S 0 = V 1(T ). (1.2) Multiplying (1.1) by p and (1.2) by q = 1 p i.e. p + q = 1 gives us ( ) S1 (H) px 0 + p 0 S 0 = pv 1(H) and (1.3) ( ) S1 (T ) qx 0 + q 0 S 0 = qv 1(T ). (1.4) 11

13 and by adding (1.3) to (1.4) we get ( p + q)x ( ps1 (H) + qs 1 (T ) Since p + q = 1 then Take Then (1.) becomes X ( ps1 (H) + qs 1 (T ) ( p + q)s 0 ) S 0 ) = pv 1(H) + qv 1 (T ). = pv 1(H) + qv 1 (T ). (1.) S 0 = ps 1(H) + qs 1 (T ). (1.6) X 0 = pv 1(H) + qv 1 (T ). (1.7) By subtracting (1.2) from (1.1) we get ( ) S1 (H) X S 0 ( ) S1 (T ) X 0 0 S 0 0 S 1 (H) 0 0 = V 1(H) V 1 (T ) S 1 (H) S 1 (T ) = V 1(H) = V 1(T ) S 1 (T ) = V 1(H) V 1(T ) (1.8) From (1.17), we have which implies ( ) 1 pu + qd S 0 = pus 0 + qds 0 = S 0 ( pu + qd ) = 1 = 1 ( ( ) u d = p = p = d u d ) pu + (1 p)d + d = 1 = 1 (1.9) (1.10) 12

14 and q = 1 p = 1 d u d = u 1 r u d. Then p = d u d and q = u 1 r u d. (1.11) Equation (1.8) is called Delta-Hedging formula. Let P = ( p, q) and P = (p, q). Hence the no-arbitrage price of the derivative security at time zero is X 0 = V 0 V 0 = p V 1(H) + q V 1(T ) and any other price would introduce an arbitrage. Remark 1.1 (1.12) We call ( p, q) risk-neutral probabilities while we call (p, q) the actual probabilities. Since d < < u, then ()S 0 < ps 1 (H) + qs 1 (T ) = pus 0 + qds 0 < p u + q d and from Equation (1.17) we get ()S 0 = ps 1 (H) + qs 1 (T ) = pus 0 + qds 0 = p u + q d. Equation (1.12) is called the risk-neutral pricing formula for the one period binomial model. Note that the value of the derivative security at time zero (Eq. 1.12) does not depend upon the probability of up (p) and down factor (q) but it depends of the size of the two possible moves (u and d). The quantities V 1 /() is called the discounted derivative security at time t 1, and S 1 /() is called the discounted stock price at time t 1, where r is the interest rate of money market. 13

15 1.3 Multiperiod Binomial Model Consider the initial stock price S 0 > 0 and two-period (T = t 2 ) binomial model. At time t 1 (after the first toss): The stock price S 1 will be { S1 (H) = us 0, if there is a Head in the coin; S 1 = S 1 (T ) = ds 0, if there is a Tail in the coin. At time t 2 (after the second toss): The stock price S 2 will be S 2 (HH) = u 2 S 0, S 2 = S 2 (HT ) = S 2 (T H) = dus 0, S 2 (T T ) = d 2 S 0, if there are two Heads in the two coins; if there is one Head in the two coins. if there is no Head in the two coins. S 2 (HH) = u 2 S 0 S 1 (H) = us 0 S 0 S 2 (HT ) = uds 0 = S 2 (T H) S 1 (T ) = ds 0 S 2 (T T ) = d 2 S 0 Note that V 2, S 0, S 1, S 2, r are known and X 0, 0, V 0 are unknown The portfolio X 2 at time t 2 is X 2 = 1 S 2 + ()(X 1 1 S 1 ) From the equation V 2 = X 2 we have V 2 (HH) = 1 (H)S 2 (HH) + ()(X 1 (H) 1 (H)S 1 (H)), (1.13) V 2 (HT ) = 1 (H)S 2 (HT ) + ()(X 1 (H) 1 (H)S 1 (H)), (1.14) V 2 (T H) = 1 (T )S 2 (T H) + ()(X 1 (T ) 1 (T )S 1 (T )), (1.1) V 2 (T T ) = 1 (T )S 2 (T T ) + ()(X 1 (T ) 1 (T )S 1 (T )). (1.16) 14

16 Take S 1 (H) = ps 2(HH) + qs 1 (HT ), and S 1 (T ) = ps 2(T H) + qs 1 (T T ) (1.17) Now, multiplying (1.13) by p and (1.14) by q and adding them gives us pv 2 (HH) + qv 2 (HT ) = 1 (H)( ps 2 (HH) + qs 2 (HT )) + ()(X 1 (H) 1 (H)S 1 (H)) = 1 (H)()S 1 (H) + ()X 1 (H) 1 (H)()S 1 (H) =()X 1 (H) X 1 (H) = p V 2(HH) + q V 2(HT ) (1.18) where p = d u d and q = d 1 u u d. Similarly, by multiplying (1.1) by p and (1.16) by q and adding them gives us X 1 (T ) = p V 2(T H) + q V 2(T T ) (1.19) So be considering X 1 = V 1 we get V 1 = p V 2 + q V 2 (1.20) or V 1 (H) = p V 2(HH) V 1 (T ) = p V 2(T H) + q V 2(HT ) + q V 2(T T ) (1.21) (1.22) Equations (1.21) and (1.22) are called risk-neutral pricing formula. Subtracting (1.14) from (1.13) and (1.16) from (1.1), gives us 1 (H) = V 2(HH) V 2 (HT ) S 2 (HH) S 2 (HT ) and 1(T ) = V 2(T H) V 2 (T T ) S 2 (T H) S 2 (T T ) (1.23) The Equation (1.23) is called the delta-hedging formula. For one-period binomial model, we have V 0 = p V 1(H) + q V 1(T ) and 0 = V 1(H) V 1 (T ) S 1 (H) S 1 (T ) 1

17 Using Eq. (1.20) in the above equation for V 1, we obtain V 0 = p V 1(H) + q V 1(T ) = p [ p V 2(HH) + q V 2(HT ) ] + q [ p V 2(T H) + q ] = 1 () 2 [ p 2 V 2 (HH) + p qv 2 (HT ) + p qv 2 (T H) + q 2 V 2 (T T ) Remark 1.2 Note that we have three stochastic process { 0, 1 } share of stocks stochastic process {X 0, X 1, X 2 } portfolio stochastic process {V 0, V 1, V 2 } price of derivative security stochastic process ] V 2(T T ) (1.24) Notation: For simplicity, let w n = w 1 w 2 w n and w c n = w n+1 w n+2 w N where w i = H or w i = T. So w N = w 1 w 2 w N, w n H = w 1 w 2 w n H and w n T = w 1 w 2 w n T. Remark 1.3 Generally, for N-period binomial model, we have the following equations for n = 0, 1,, N 1 The delta-hedging formulas is given by V n (w n ) = 1 [ p Vn+1 (w n H) + q V n+1 (w n T ) ] (1.2) n (w n ) = V n+1(w n H) V n+1 (w n T ) S n+1 (w n H) S n+1 (w n T ) (1.26) Example 1. (2-Period Binomial Model) Suppose S 0 = 4, u = 2, d = 1 2, r = 1 4. Then p = d u d = = 1 2 and q = u 1 r u d = =

18 S 2 (HH) = 16 S 0 = 4 S 1 (H) = 8 S 1 (T ) = 2 S 2 (HT ) = 4 = S 2 (T H) S 2 (T T ) = 1 Consider the derivative security is a European call option call option with maturity at time t 2 and strike price 3 i.e. V 2 = [S 2 3] +. At time t 2 we have V 2 (HH) =[S 2 (HH) 3] + = [16 3] + = 13 V 2 (HT ) =[S 2 (HT ) 3] + = [4 3] + = 1 V 2 (T H) =[S 2 (T H) 3] + = [4 3] + = 1 V 2 (T T ) =[S 2 (T T ) 3] + = [1 3] + = 0. At time t 1 we have At time t 0 we have V 1 (H) = 1 ( pv 2(HH) + qv 2 (HT )) = 4 ( ) =.6 V 1 (T ) = 1 ( pv 2(T H) + qv 2 (T T )) = 4 ( ) = 0.4. V 0 = 1 [ pv1 (H) + qv 1 (T ) ] = 4 ( ) = 2.4 Thus the no-arbitrage price of the option at time t 0 is 2.4$ and using the delta-hedging formula we get 0 = V 1(H) V 1 (T ) S 1 (H) S 1 (T ) So that the replication process is as follows: At time t 0 = = shares of stock 17

19 - sell option by V 0 = X 0 = 2.4$ - buy 0 = shares of stock by 0 S 0 = = 3.467$ - borrow X 0 0 S 0 = = 1.067$ from the money market. At time t 1 - from money market we indept ()(X 0 0 S 0 ) = = 1.334$ - if there is Head in the coin, then from stock market we have 0 S 1 (H) = = 6.936$ so the portfolio is X 1 (H) = V 1 (H) = =.6$ - if there is Tail in the coin, then from stock market 0 S 1 (T ) = = 1.734$ so the portfolio is X 1 (T ) = V 1 (T ) = = 0.4$ Example 1.6 (Three-Period Binomial Model) Suppose S 0 = 4, u = 2, d = 1 2, r = 1 4. Then p = d u d = 1 2 and q = u 1 r u d = 1 2. Consider the derivative security is a lookback call option which entitled to a payment at time t 3 as V 3 = max 0 n 3 (S n) S 3 18

20 S 3 (HHH) = 32 S 0 S 1 (H) = 8 S 1 (T ) = 2 S 2 (HH) = 16 S 2 (HT ) = 4 = S 2 (T H) S 2 (T T ) = 1 S 3 (T HH) = 8 = S 3 (HT H) = S 3 (T HH) S 3 (T T H) = 2 = S 3 (T HT ) = S 3 (HT T ) S 3 (T T T ) = 0. Now, at time t 3 we have V 3 (HHH) = max{s 0, S 1 (H), S 2 (HH), S 3 (HHH)} S 3 (HHH) =S 3 (HHH) S 3 (HHH) = = 0 V 3 (HHT ) = max{s 0, S 1 (H), S 2 (HH), S 3 (HHT )} S 3 (HHT ) =S 2 (HH) S 3 (HHT ) = 16 8 = 8 V 3 (HT H) = max{s 0, S 1 (H), S 2 (HT ), S 3 (HT H)} S 3 (HT H) =S 1 (H) S 3 (HT H) = 8 8 = 0 V 3 (T HH) = max{s 0, S 1 (T ), S 2 (T H), S 3 (T HH)} S 3 (T HH) =S 3 (T HH) S 3 (T HH) = 8 8 = 0 V 3 (HT T ) = max{s 0, S 1 (H), S 2 (HT ), S 3 (HT T )} S 3 (HT T ) =S 1 (H) S 3 (HT T ) = 8 2 = 6 V 3 (T HT ) = max{s 0, S 1 (T ), S 2 (T H), S 3 (T HT )} S 3 (T HT ) =S 2 (T H) S 3 (T HT ) = 4 2 = 2 V 3 (T T H) = max{s 0, S 1 (T ), S 2 (T T ), S 3 (T T H)} S 3 (T T H) =S 0 S 3 (T T H) = 4 2 = 2 V 3 (T T T ) = max{s 0, S 1 (T ), S 2 (T T ), S 3 (T T T )} S 3 (T T T ) =S 0 S 3 (T T T ) = 4 0. =

21 At time t 2 we have V 2 (HH) = 1 ( pv 3(HHH) + qv 3 (HHT )) = 4 ( ) = 3.2 V 2 (HT ) = 1 ( pv 3(HT H) + qv 3 (HT T )) = 4 ( ) = 2.4 V 2 (T H) = 1 ( pv 3(T HH) + qv 3 (T HT )) = 4 ( ) = 0.8 V 2 (T T ) = 1 ( pv 3(T T H) + qv 3 (T T T )) = 4 ( ) = 2.2. At time t 1 we have At t 0 is given by V 1 (H) = 1 ( pv 2(HH) + qv 2 (HT )) = 4 ( ) = 2.24 V 1 (T ) = 1 ( pv 2(T H) + qv 2 (T T )) = 4 ( ) = 1.2. V 0 = 1 ( pv1 (H) + qv 1 (T ) ) = 4 ( ) = Now using the delta-hedging formula we get 0 = V 1(H) V 1 (T ) S 1 (H) S 1 (T ) = So that the replication process is as follows: At time t 0 = shares of stock. - sell option by V 0 = X 0 = 1.376$ - buy 0 = shares of stock by 0 S 0 = = $ - invest X 0 0 S 0 = = $ in the money market. At time t 1 20

22 - from money market we get ()(X 0 0 S 0 ) = = 0.833$ - if there is Head in the coin, then from stock market 0 S 1 (H) = = $ so the portfolio is X 1 (H) = V 1 (H) = = 2.24$ - if there is Tail in the coin, then from stock market 0 S 1 (T ) = = $ so the portfolio is X 1 (T ) = V 1 (T ) = = 1.2$ and so on. 21

23 2 Finite Probability Space Definition 2.1 (Finite Probability space) A finite probability space consists of a sample space Ω and a probability measure P. The sample space Ω is a non-empty finite set i.e. Ω = {w 1, w 2, w n } and the probability measure P is a function that assigns to each element w of Ω a number in [0, 1] so that P(w) = 1. An event is a subset of Ω, and we define the probability of an event A to be P(A) = P(w). w A w Ω Example 2.1 Suppose a coin is tossed three times. The sample space is Ω = {w : w = HHH, HHT, HT H, T HH, HT T, T HT, T T H, T T T }. Suppose that on each toss the probability of a head P(H) = p and the probability of tail P(T ) = q = 1 p. We assume the tosses are independent. Then the probability measure P is p 3, if w = HHH; p 2 q, if w = HHT, HT H, T HH; P(w) = pq 2, if w = HT T, T HT, T HH; q 3, if w = T T T. Then the space (Ω, P) is a finite probability space. Let A be an event for the first toss is a head, then P(A) =P{w : w = HHH, HT H, HHT, HT T } =p 3 + p 2 q + p 2 q + pq 2 =p 2 (p + q) + pq(p + q) =p 2 + pq =p(p + q) = p. 22

24 Definition 2.2 (Random Variable) Let (Ω, P) be a finite probability space. A random variable X is a real-valued function defined on Ω. The distribution of the r.v. X is the probability of the X at each value X = x. Remark 2.1 A random variable is not a distribution and a distribution is not a random variable. Example 2.2 Consider Example 2.1. Let p = q = 1. Define the random variables 2 X = Total number of heads and Y = Total number of tails. Note that X Y because Y = 2 X. The probability distribution of X is P X (X = 0) = P{w : w = T T T } = 1 8 P X (X = 1) = P{w : w = T T H, T HT, HT T } = 3 8 The probability distribution of Y is P X (X = 2) = P{w : w = T HH, HT H, HHT } = 3 8 P X (X = 3) = P{w : w = HHH} = 1 8. P Y (Y = 0) = P{w : w = HHH} = 1 8 P Y (Y = 1) = P{w : w = T HH, HT H, HHT } = 3 8 P Y (Y = 2) = P{w : w = T T H, T HT, T T H} = 3 8 P Y (Y = 3) = P{w : w = T T T } = 1 8. Then P X P Y. So that P X P Y X Y. Example 2.3 Consider Example 2.2. Let p = 2 and q = 1. Define the random variables 3 3 X = Total number of heads. 23

25 The probability distribution of X is P X (X = 0) = P{w : w = T T T } = 1 27 P X (X = 1) = P{w : w = T T H, T HT, HT T } = 6 27 P X (X = 2) = P{w : w = T HH, HT H, HHT } = P X (X = 3) = P{w : w = HHH} = Note that the distribution of X under the probability measure (p, q) = ( 1, 1 ) is different 2 2 from the probability distribution under the probability measure (p, q) = ( 2, 1 ). So that 3 3 Definition 2.3 (Expectation) X Y P X P Y. Let X be a random variable defined on a finite probability space (Ω, P). The expected value of X is given by The variance of X is E P (X) = w Ω X(w)P(w). V P (X) = E P (X E P (X)) 2 = E P (X 2 ) (E P (X)) 2. Remark 2.2 Suppose P be another probability measure defined on Ω. Then the expected value of X is E P(X) = w Ω X(w) P(w). Note that P X P X = E P(X) E P (X). Homework : Find the mean and variance of X of the examples 2.2 and 2.3. Definition 2.4 (Convex Function) 1. A function g(x) is said to be convex over an interval (a, b) if for every x 1 and x 2 we have g(λx 1 + (1 λ)x 2 ) λg(x 1 ) + (1 λ)g(x 2 ), where 0 λ 1. 24

26 2. A function is said to be strictly convex if g(λx 1 + (1 λ)x 2 ) < λg(x 1 ) + (1 λ)g(x 2 ), where 0 < λ < A function g(x) is concave if g(x) is convex. Remark 2.3 The following are graphes of some convex and concave functions. Example 2.4 Let g(x) = x. g(λx + (1 λ)y) = λx + (1 λ)y λ x + 1 λ y =λ x + (1 λ) y λg(x) + (1 λ)g(y). Then g(x) is a convex function. 2

27 Example Convex functions: x log x for x > 0, x 2, e x. 2. Concave functions: log x for x > 0, x for x > Both convex and concave: ax + b. Theorem 2.1 A function is convex (or strictly convex) if and only if its second derivative is non-negative (positive) everywhere. Example Let g(x) = x 2. Since d2 g dx 2 = 2 > 0 then g(x) is convex function. 2. Let g(x) = 1. Since d2 g = 2 < 0 for x > 0 and d2 g = 2 x dx 2 x 3 dx 2 x 3 concave for x > 0 and convex function for x < 0 function. > 0 for x < 0 then g(x) is 3. Let g(x) = x 3. Since d2 g = 6x > 0 for x > 0 and d2 g = 6x < 0 for x < 0 then g(x) is dx 2 dx 2 convex function for x > 0 and concave function x < 0. Theorem 2.2 (Jensen s Inequality) Let X be a random variable on a finite probability space (Ω, P). Let ϕ(x) be a convex function of x. Then E P (ϕ(x)) ϕ(e P (X)). Remark 2.4 Since g(x) = x 2 is convex function, then using Jensen s inequality, we get E P (X 2 ) (E P (X)) 2 = V P (X) = E P (X 2 ) (E P (X)) 2 0. Remark 2. If g(x) is a linear function then E P (g(x)) = g(e P (X)). For example, if g(x) = ax + b then Example 2.7 E P (g(x)) = E P (ax + b) = ae P (X) + b = g(e P (X)). 1. Since e x is convex then E P (e X ) e E P(X). 2. Since log(x) is concave then E P (log(x)) log(e P (X)). 26

28 3. Since x 3 is convex for x > 0 and concave for x < 0 then Notation: For simplicity, let E P (X 3 ) (E P (X)) 3 for x > 0 and E P (X 3 ) (E P (X)) 3 for x < 0. w n = w 1 w 2 w n, and w c n = w n+1 w n+2 w N where w i = H or w i = T. So w N = w 1 w 2 w N, w n H = w 1 w 2 w n H and w n T = w 1 w 2 w n T. 2.1 Conditional Expectation Definition 2. (Conditional Expectation) Define #H(w n ) = Number of heads in the sequance w n, and #T (w n ) = Number of tails in the sequance w n. Suppose probability distribution is P = (p, q) where p = P(H) and q = 1 p = P(T ). Let X(w N ) be a random variable. For 1 n N, and for a given w n the conditional expectation of X given w n is given by E P [X w n ] = w c n p # H(wc n) q # T (wc n) X(w N ), where =. w c n w N =H or T w n+2 =H or T w n+1 =H or T Special Cases: (i) The conditional expectation given no information is given by E P [X w 0 ] = wn p # H(w N ) q # T (w N ) X(w N ) = E P [X] 27

29 (ii) The conditional expectation given all information is given by E P [X w N ] = X. (iii) If X depends on the first w n+1 i.e. X(w n+1 ) we get the following special case E P [X w n ] = w n+1 =H or T =px(w n H) + qx(w n T ). p # H(w n+1) q # T (w n+1) X(w n+1 ) Remark 2.6 Note that the unconditional expectation E P [X] is a numeric quantity while the conditional expectation E P [X w n ] is a random variable. Remark 2.7 Note that w 1 p # H(w1) q # T (w1) = p + q = 1. w 1 w 2 p # H(w1w2) q # T (w1w2) = p 2 + 2pq + q 2 = (p + q) 2 = 1 for 1 n 1 n 2 N. w n1 w n2 p # H(wn1 wn2 ) q # T (wn1 wn2 ) = (p + q) n2 n1 = 1, Remark 2.8 If Z(w m ) for 0 n < m N then E P [Z(w m ) w n ] = w c n p # H(wc n ) q # T (wc n ) Z(w m ) = { }}{ p # H(wn+1 wm) q # T (wn+1 wm) Z(w m ) p # H(wc m) q # T (wc m) w n+1 w m = E P [Z(w m ) w n ] = Example 2.8 Consider the following diagram w c m w n+1 w m p # H(wn+1 wm) q # T (wn+1 wm) Z(w m ). =1 28

30 S 3 (HHH) = 32 S 0 = 4 S 1 (H) = 8 S 1 (T ) = 2 S 2 (HH) = 16 S 2 (HT ) = 4 = S 2 (T H) S 2 (T T ) = 1 S 3 (T HH) = 8 = S 3 (HT H) = S 3 (T HH) S 3 (T T H) = 2 = S 3 (T HT ) = S 3 (HT T ) S 3 (T T T ) = 0. Suppose the probability distribution is given by P = (p, q) = (0., 0.). E P [S 2 H] =ps 2 (HH) + qs 2 (HT ) = = 10 E P [S 2 T ] =ps 2 (T H) + qs 2 (T T ) = = 2. E P [S 3 HH] =ps 3 (HHH) + qs 3 (HHT ) = = 20 E P [S 3 HT ] =ps 3 (HT H) + qs 3 (HT T ) = = E P [S 3 T H] =ps 3 (T HH) + qs 3 (T HT ) = = E P [S 3 T T ] =ps 3 (T T H) + qs 3 (T T T ) = = E P [S 3 H] =p 2 S 3 (HHH) + pqs 2 (HHT ) + pqs 3 (HT H) + q 2 S 3 (HT T ) = = E P [S 3 T ] =p 2 S 3 (T HH) + pqs 2 (T HT ) + pqs 3 (T T H) + q 2 S 3 (T T T ) = =

31 Note that the unconditional expectations are E P [S 1 ] =ps 1 (H) + qs 1 (T ) = =. E P [S 2 ] =p 2 S 2 (HH) + pqs 2 (HT ) + pqs 2 (T H) + q 2 S 3 (T T ) = = 6.2. E P [S 3 ] =p 3 S 3 (HHH) + p 2 qs 2 (HHT ) + p 2 qs 3 (HT H) + pq 2 S 3 (HT T ) + p 2 qs 3 (T HH) + pq 2 S 2 (T HT ) + pq 2 S 3 (T T H) + q 3 S 3 (T T T ) = = Properties of Conditional Expectation Let N be a given integer, and let X(w N ) and Y (w N ) be random variables. Let 0 n N be given. Property-1: Linearity where c 1 and c 2 are constants. Proof : E P [c 1 X + c 2 Y w n ] = c 1 E P [X w n ] + c 2 E P [Y w n ], Recall the definition of the conditional expectation as E P [X w n ] = w c n p # H(wc n) q # T (wc n) X(w N ) Now, since X and Y be random variables depending on the first N coin tosses then c 1 X+c 2 Y is also a random variable depending on the first N coin tosses. Hence E P [c 1 X + c 2 Y w n ] = E P [c 1 X + c 2 Y w n ] = p # H(wc n) q # T (wc n) [c 1 X(w N ) + c 2 Y (w N )] wn c =c 1 p # H(wc n ) q # T (wc n ) X(w N ) + c 2 p # H(wc n ) q # T (wc n ) Y (w N ) w c n w c n =c 1 E P [X w n ] + c 2 E P [Y w n ]. 30

32 Property-2: Taking out what is known If X depends only on the first n coin tosses, then Proof : Homework E P [XY w n ] = XE P [Y w n ]. Property-3: Iterated Conditioning If 0 n < m N, then In particular E P [E P [X w n ]] = E P [X]. Proof : E P [E P [X w m ] w n ] = E P [X w n ] = E P [E P [X w n ] w m ] Denote Z = E P [X w m ]. Then Z actually depends on w m only and E P [E P [X w m ] w n ] =E P [Z w n ] = = w n+1 w N p # H(wc = n ) q # T (wc n ) Z(w m ) w n+1 w m p # H(wn+1 wm) q # T (wn+1 wm) Z(w m ) =1 {}}{ w m+1 w N p # H(wm+1 wn ) q # T (wm+1 wn ) p # H(wn+1 wm) q # T (wn+1 wm) Z(w m ) w n+1 w m = p # H(wn+1 wm) q # T (wn+1 wm) E P [X w m ] w n+1 w m = p # H(wn+1 wm) q # T (wn+1 wm) w n+1 w m m) q # T (wc m) X(w N ) = w m+1 w N p # H(wc w n+1 w N p # H(wc n ) q # T (wc n ) X(w N ) = E P [X w n ]. Property-4: Independence If X depends only on tosses n + 1, n + 2,, N, then E P [X w n ] = E P [X]. 31

33 Proof : Since X does not depend on the first n coin tosses then E P [X w n ] = w c n = w c n p # H(wc n ) q # T (wc n ) X(w N ) p # H(wc n ) q # T (wc n ) X(w c n) =1 { }}{ = p # H(wn) q # T (wn) p p # H(wc n ) q # T (wc n ) X(wn) c w n = wn w c n p # H(w N ) q # T (w N ) X(w c n) =E P [X]. Property-: Conditional Jensen s Inequality If g(x) is a convex function of x, then E P [g(x) w n ] g (E P [X w n ]). Example 2.9 Consider Example 2.8 with p = 2 and q = Linearity. Since E P [S 2 H] = = 12, E P [S 3 H] = = 18, then E P [S 2 H] + E P [S 3 H] = = 30. But also E P [S 2 + S 3 H] = 4 9 ( ) + 2 (16 + 8) (4 + 8) + 1 (4 + 2) 9 =30. Then E P [S 2 + S 3 H] = E P [S 2 H] + E P [S 3 H]. 32

34 Similarly, we can prove (Homework!!!) E P [S 2 + S 3 T ] = E P [S 2 T ] + E P [S 3 T ]. Hence E P [S 2 + S 3 w 1 ] = E P [S 2 w 1 ] + E P [S 3 w 1 ]. 2. Taking out what is known. Note that S 1 depends only on the first coin toss w 1. Since E P [S 2 H] = 12 then Also Then S 1 (H)E P [S 2 H] = 8 12 = 96. E P [S 1 S 2 H] = = E P [S 1 S 2 H] = S 1 (H)E P [S 2 H]. Similarly, we can prove (Homework!!!) E P [S 1 S 2 T ] = 6 = S 1 (T )E P [S 2 T ]. Hence E P [S 1 S 2 w 1 ] = S 1 E P [S 2 w 1 ]. 3. Iterated Conditioning. E P [S 3 HH] = = 24, E P [S 3 HT ] = = 6, E P [S 3 T H] = = 6, E P [S 3 T T ] = =

35 Now E P [E 2 [S 3 w 1 w 2 ] H] = 2 3 E P[S 3 HH] E P[S 3 HT ] = = 18, E P [E P [S 3 w 1 w 2 ] T ] = 2 3 E P[S 3 T H] E P[S 3 T T ] = = 4.. Also E P [S 3 H] = = 18, E P [S 3 T ] = = 4.. Hence E P [E P [S 3 w 1 w 2 ] w 1 ] = E P [S 3 w 1 ]. 4. Independence. Note that and S 2 (HH) S 1 (H) = 16 8 = 2 = S 2(T H) S 1 (T ) S 2 (HT ) S 1 (H) = 4 8 = 1 2 = S 2(T T ) S 1 (T ) that means S 2 S 1 depends only on the result of the second toss, w 2. [ ] S2 E P S 1 H = 2 S 2 (HH) + 1 S 2 (HT ) 3 S 1 (H) 3 S 1 (H) = = 3 [ ] 2 S2 E P T = 2 S 2 (T H) + 1 S 2 (T T ) 3 S 1 (T ) 3 S 1 (T ) S 1 = = 3 2. So that [ ] [ ] S2 E P H S2 = E P T = 3 2. S 1 S 1 34

36 Now Hence [ ] S2 E P =p 2 S 2(HH) + pq S 2(HT ) S 1 S 1 (H) S 1 (H) + pq S 2(T H) + q 2 S 2(T T ) S 1 (T ) S 1 (T ) = = 3 2. [ S2 ] [ ] w1 S2 E P = E P = 3 S 1 S Jensen s Inequality. Let g(x) = x 2 which is convex. E P (S 2 2 H) =ps 2 2(HH) + qs 2 2(HT ) = 2 3 (16) (4)2 = 176 E P (S 2 2 T ) =ps 2 2(T H) + qs 2 2(T T ) = 2 3 (4) (1)2 = 11 (E P (S 2 H)) 2 =(ps 2 (HH) + qs 2 (HT )) 2 = (E P (S 2 T )) 2 =(ps 2 (T H) + qs 2 (T T )) 2 = ( 2 ) 2 (16) + 1(4) 3 3 = 144 ( ) 2 2 (4) + 1(1) 3 3 = 9. Hence E P (S 2 2 H) > (E P (S 2 H)) 2 and E P (S 2 2 T ) > (E P (S 2 T )) 2 = E P (S 2 2 w 1 ) > (E P (S 2 w 1 )) 2. 3

37 3 Martingales Definition 3.1 (Adapted Stochastic Process) A stochastic process {M n } N n=1 is called an adapted stochastic process if M n depends only the first n coin tosses w n i.e is independent of n + 1, n + 2, N coin tosses w c n. Hence {M n } N n=1 is adapted iff = M 1 depends only on w 1 M 2 depends only on w 1, w 2. M n depends only on w 1, w 2,..., w n. Remark 3.1 If {M n } N n=1 is an adapted stochastic process then E[M n w n ] = M n. Definition 3.2 (Martingale) Let {M n } N n=0 be an adapted stochastic process and M 0 is constant. 1. If M n =E P [M n+1 w n ], n = 0, 1,, N 1 (3.1) we say this process is martingale (no tendency to rise or fall). 2. If M n E P [M n+1 w n ], n = 0, 1,, N 1 we say this process is submartingale (tendency to rise). 3. If M n E P [M n+1 w n ], n = 0, 1,, N 1 we say this process is supermartingale (tendency to fall). 36

38 Remark 3.2 Note that if {M n } is a martingale process then M n+1 = E P [M n+2 w n+1 ] = E P [M n+1 w n ] = E P [E P [M n+2 w n+1 ] w n ] = E P [M n+1 w n ] = E P [M n+2 w n ] = M n = E P [M n+2 w n ]. Then, the stochastic process {M n } is martingale if and only if M n = E P [M m w n ], 0 n < m N. Remark 3.3 The expectation of a martingale is constant over time i.e. if {M n } is martingale, then M 0 =E P (M n ), n = 0, 1,, N. (3.2) Observe that M 0 =E P (M 1 w 0 ) = E P (M 1 ) M 1 =E P (M 2 w 1 ) = E P (M 1 ) = E P (E P (M 2 w 1 )) = E P (M 2 ). M n =E P (M n+1 w n ) = E P (M n ) = E P (E P (M n+1 w n )) = E P (M n+1 ). Since M 0 is constant then M 0 = E P M 0 = E P M 1 = = E P M N. Theorem 3.1 Let Z be a random variable. Define Z n = E P (Z w n ), n = 0, 1,, N. Then {Z n } N n=0 is a martingale under P. Proof : For n = 0, 1,, N 1, we have E P [Z n+1 w n ] = E P [E P [Z w n+1 ] w n ] = E P [Z w n ] = Z n. This shows that {Z n } N n=0 is a martingale. 37

39 Remark 3.4 It is shown that S n (w n ) = 1 ( ps n+1(w n H) + qs n+1 (w n T )) = 1 E P[S n+1 (w n+1 ) w n ] Divide both sides of the above equation by () n, we get [ S n (w n ) () = Sn+1 (w n+1 ) ] wn n E P. (3.3) () n+1 We call S n the stock price at time n and S n we call the discounted stock price at time n, where r is the interest rate of () n money market. Theorem 3.2 Consider the general binomial model with 0 < d < < u. Let the risk neutral probabilities P = ( p, q), where p = d u d, q = u 1 r u d. Then, under the risk-neutral probabilities, the discounted stock price is a martingale i.e. [ S ] [ n+1 wn S ] n+1 Sn S n E P = () n+1 E P = () n+1 (). n Proof : Let n and w n be given. Then [ S ] n+1 wn 1 = () n+1 () 1 n [ ps n+1(w n H) + qs n+1 (w n T )] E P = = 1 () 1 n [ pus n(w n ) + qds n (w n )] S n pu + qd () n = S n pu + qd, since () n = 1. Example 3.1 Consider the following three period binomial model 38

40 S 3 (HHH) = 32 S 0 = 4 S 1 (H) = 8 S 1 (T ) = 2 S 2 (HH) = 16 S 2 (HT ) = 4 = S 2 (T H) S 2 (T T ) = 1 S 3 (T HH) = 8 = S 3 (HT H) = S 3 (T HH) S 3 (T T H) = 2 = S 3 (T HT ) = S 3 (HT T ) S 3 (T T T ) = 0. Suppose up and down probabilities are P 1 = (1/3, 2/3), P 2 = (2/3, 1/3) and P = (1/2, 1/2). For the probability distribution P 1 E P1 [S 1 S 0 ] = ps 1 (H) + qs 1 (T ) = = 4 = S 0 E P1 [S 2 H] = ps 2 (HH) + qs 2 (HT ) = = 8 = S 1(H) E P1 [S 2 T ] = ps 2 (T H) + qs 2 (T T ) = = 2 = S 1(T ) = E P1 [S 2 w 1 ] = S 1 E P1 [S 3 HH] = ps 3 (HHH) + qs 3 (HHT ) = = 16 = S 2(HH) E P1 [S 3 HT ] = ps 3 (HT H) + qs 3 (HT T ) = = 4 = S 2(HT ) E P1 [S 3 T H] = ps 3 (T HH) + qs 3 (T HT ) = = 4 = S 2(T H) E P1 [S 3 T T ] = ps 3 (T T H) + qs 3 (T T T ) = = 1 = S 2(T T ) 39

41 = E P1 [S 3 w 1 w 2 ] = S 2 E P1 [S 3 H] = p 2 S 3 (HHH) + pqs 2 (HHT ) + pqs 3 (HT H) + q 2 S 3 (HT T ) = = 8 = S 1(H) E P1 [S 3 T ] = p 2 S 3 (T HH) + pqs 2 (T HT ) + pqs 3 (T T H) + q 2 S 3 (T T T ) = = 2 = S 1(T ) = E P1 [S 3 w 1 ] = S 1. Thus E P1 [S m w n ] = S n, 0 n < m 3, and hence the process {S 0, S 1, S 2, S 3 } is martingale under the probability measure P 1. For the probability distribution P 2 E P2 [S 1 S 0 ] = ps 1 (H) + qs 1 (T ) = = 6 = 3 2 S 0 E P2 [S 2 H] = ps 2 (HH) + qs 2 (HT ) = = 12 = 3 2 S 1(H) E P2 [S 2 T ] = ps 2 (T H) + qs 2 (T T ) = = 3 = 3 2 S 1(T ) = E P2 [S 2 w 1 ] = 3 2 S 1 E P2 [S 3 HH] = ps 3 (HHH) + qs 3 (HHT ) = = 24 = 3 2 S 2(HH) E P2 [S 3 HT ] = ps 3 (HT H) + qs 3 (HT T ) = = 6 = 3 2 S 2(HT ) E P2 [S 3 T H] = ps 3 (T HH) + qs 3 (T HT ) = = 6 = 3 2 S 2(T H) E P2 [S 3 T T ] = ps 3 (T T H) + qs 3 (T T T ) = = 3 2 = 3 2 S 2(T T ) 40

42 Thus = E P2 [S 3 w 1 w 2 ] = 3 2 S 2 E P2 [S 3 H] = p 2 S 3 (HHH) + pqs 2 (HHT ) + pqs 3 (HT H) + q 2 S 3 (HT T ) = = 18 = [ 3 2] 2 S 1 (H) E P2 [S 3 T ] = p 2 S 3 (T HH) + pqs 2 (T HT ) + pqs 3 (T T H) + q 2 S 3 (T T T ) = = 9 2 = [ 3 2] 2 S 1 (T ) = E P2 [S 3 w 1 ] = S 1. E P2 [S m w n ] = ( ) m n 3 S n S n, 0 n < m 3, 2 and hence the process {S 0, S 1, S 2, S 3 } is submartingale under the probability measure P 2. Assume the interest rate r = 1/4. The discounted stock price at time n is ( ) n S n 4 () = S n n. Then [ (4 ] n+1 ( ) n+1 4 E P2 S n+1 w ) n = E P2 [S n+1 w n ] ( ) n = 2 S n ( ) n 4 = S n ( ) n 4 S n. Thus the discounted stock price is submartingale under the probability measure P 2. This is typically the case in the real markets. 41

43 For the risk neutral probability distribution P E P[S 1 S 0 ] = ps 1 (H) + qs 1 (T ) = = = 4 S 0 E P[S 2 H] = ps 2 (HH) + qs 2 (HT ) = = 10 = 4 S 1(H) E P[S 2 T ] = ps 2 (T H) + qs 2 (T T ) = = 2 = 4 S 1(T ) = E P[S 2 w 1 ] = 4 S 1 E P[S 3 HH] = ps 3 (HHH) + qs 3 (HHT ) = = 20 = 4 S 2(HH) E P[S 3 HT ] = ps 3 (HT H) + qs 3 (HT T ) = = = 4 S 2(HT ) E P[S 3 T H] = ps 3 (T HH) + qs 3 (T HT ) = = = 4 S 2(T H) E P[S 3 T T ] = ps 3 (T T H) + qs 3 (T T T ) = = 3 4 = 4 S 2(T T ) = E P[S 3 w 1 w 2 ] = 4 S 2 E P[S 3 H] = p 2 S 3 (HHH) + p qs 2 (HHT ) + p qs 3 (HT H) + q 2 S 3 (HT T ) = = 2 [ ] 2 2 = S 1 (H) 4 Thus E P[S 3 T ] = p 2 S 3 (T HH) + p qs 2 (T HT ) + p qs 3 (T T H) + q 2 S 3 (T T T ) = = 2 [ ] 2 8 = S 1 (T ) 4 = E P[S 3 w 1 ] = S 1. E P[S m w n ] = ( ) m n S n S n, 0 n < m 3, 4 and hence the process {S 0, S 1, S 2, S 3 } is submartingale under the risk neutral probabilities P = { p = 1/2, q = 1/2}. Assume the interest rate r = 1/4. The discounted stock price at time n is Now since S n () n = ( ) n 4 S n. E P[S n+1 w n ] = 4 S n, 42

44 then ( ) n+1 ( ) n Ẽn [S n+1] = = Ẽn [ (4 ) n+1 S n+1] = ) n S n. ( 4 4 S n Thus the discounted stock price is a martingale under the risk neutral probabilities P. Remark 3. The portfolio equation of N-period binomial model is X n+1 = n S n+1 + ()(X n n S n ), n = 0, 1, 2,, N 1. (3.4) Since n, the share of stock at time n, depends only on first n coin tosses and also S n, the stock price at time n, depends only on first n coin tosses then, the stochastic processes { n } and {S n } are adapted stochastic processes. Thus the portfolio process {X n } is also adapted stochastic process. Theorem 3.3 Consider N-binomial period model. Let the share of stocks { n } N 1 n=0 be an adapted stochastic process. Let the portfolio process {X n } N n=1 where X n+1 = n S n+1 + ()(X n n S n ), n = 0, 1, 2,, N 1 { Xn and X 0 be a constant. The discounted portfolio stochastic process is a () n n=0 martingale under the risk-neutral probabilities P i.e. [ X n () = X ] n+1 wn n E P, n = 0, 1, 2,, N 1. (3.) () n+1 } N 43

45 Proof : E P We compute [ X ] n+1 wn () n+1 =E P =E P = n E P [ n S n+1 () + X n n S ] n wn n+1 () [ n n S ] [ n+1 wn Xn n S ] n wn + () n+1 E P () [ n S ] n+1 wn + X n n S n () n+1 () n (Taking out what is known) (linearity) = ns n () n + X n n S n () n (Theorem 3.2) = X n () n. Corollary 3.1 Under the conditions of Theorem 3.3, we have [ X ] n wn E P = X () n 0, n = 0, 1, 2,, N. (3.6) Theorem 3.4 (First Fundamental Theorem of Asset Pricing) If we can find a risk-neutral probabilities P in a model then there is no arbitrage in the model. 44

46 3.1 Random Walk Process Definition 3.3 (Random Walk) Toss a coin repeatedly. Assume P(H) = p and P(T ) = q = 1 p. Let { 1, if the j-th toss is H; X j = 1, if the j-th toss is T. Consider the stochastic process M n, n = 0, 1, defined by M 0 = 0 and n M n = X j, n 1. This stochastic process is called a Random Walk Process. Example 3.2 Suppose we have w 1 = H, w 2 = T, w 3 = T, w 4 = T, w = H. j=1 Then X 1 = 1, X 2 = 1, X 3 = 1, X 4 = 1, X = 1 which implies that M 0 =0 M 1 =X 1 = 1 M 2 =X 1 + X 2 = 1 1 = 0 M 3 =X 1 + X 2 + X 3 = = 1 M 4 =X 1 + X 2 + X 3 + X 4 = = 2 M =X 1 + X 2 + X 3 + X 4 + X = = 1 4

47 1 M 1 M 0 1 M M 1 3 1M 1 M 4 46

48 Remark With each toss, the process either steps up one unite with probability p or steps down one unite with probability q. 2. If p = 1, the process M 2 n is called a symmetric random walk. 3. If p 1 the process M 2 n is called asymmetric random walk. 4. The symmetric random walk is a martingale process. Proof : Now Note that n+1 M n+1 = X i = i=1 n X i + X n+1 = M n + X n+1. i=1 E(M n+1 w n ) =E(M n + X n+1 w n ) =E(M n w n ) + E(X n+1 w n ) (linearity) =M n + E(X n+1 ) (adaptivity and independence) =M n (1) ( 1) =M n. Remark 3.7 For a symmetric random walk {M n }, the mean is E(M n ) = 0 and the variance is V(M n ) = n. (Prove.) 4 Markov Process Definition 4.1 (Markov Process) Let X 0, X 1,, X N, be an adapted process. If for every n between 0 and N 1 and for every function f(x), there is another function g(x) (depending on n and f) such that E P [f(x n+1 ) w n )] = g(x n ) (4.1) we say that X 0, X 1,, X N, is a Markov Process. 47

49 Example 4.1 Random walk process {M n } N n=0 is an example of Markov process. Note that M n+1 = X n+1 + M n. E P [f(m n+1 (w n+1 )) w n )] = E P [f(x n+1 (w n+1 ) + M n (w n )) w n ] [due to independent of X 1,, X n+1 ] = p #H(wn+1) q #T (wn+1) f(x n+1 (w n+1 ) + M n (w n )) w n+1 = pf(x n+1 (H) + M n (w n )) + qf(x n+1 (T ) + M n (w n )) = pf(m n + 1) + qf(m n 1) = g(m n ), where g(y) = pf(y + 1) + qf(y 1). Remark 4.1 (i) By definition, E P [f(x n+1 ) w n )] is random; it depends on the first n coin tosses. The Markov property says that this dependence on the coin tosses occurs through X n and does not depend on X 0, X 1,..., X n 1 i.e., the information about X n is sufficient to evaluate E P [f(x n+1 ) w n ]. (ii) The Markov process reduces to a martingale process when f(x) = g(x) = x. (iii) Not every martingale is Markov. For example, E(X n+1 w n ) = X n is martingale does not imply that we can find a function g such that E(Xn+1 w 2 n ) = g(x n ). (iv) Not every Markov process is a martingale. When considering the function f(x) = x, the Markov property requires finding a function g such that E[X n+1 w n ] = g(x n ) it does not require that the function g be given by g(x) = x. Lemma 4.1 (Independence) Suppose the random variables X 1...,, X K depend only on coin tosses w 1,..., w n and the random variables Y 1,..., Y L depend only on coin tosses w n+1,..., w N. Let f(x 1,, x K, y 1,, y L ) be a function of variables x l,, x K and y l,, y L. Then E P f(x 1,, X K, Y 1, Y L w n ) = E P f(x 1,, x K, Y 1, Y L ). Example 4.2 Consider a random walk process M n. Let X = M n which depends on w n and 48

50 let Y = M n+1 M n which depends on w n+1. Then E P [f(m n+1 ) w n ] =E P [f ( ) M wn ] n+1 M n M n =E P [f(xy ) w n ] =E P [f(xy )] =pf(x) + qf( x) =pf(m n ) + qf( M n ) =g(m n ). Definition 4.2 (2 dimensional Markov process) Let {(X n, Y n ); n = 0, 1,, N} be a 2 dimensional adapted process; i.e., X n and Y n are adapted processes. If, for every n between 0 and N 1 and for every function f(x, y), there is another function g(x, y) (depending on n and f) such that E P [f(x n+1, Y n+1 ) w n ] = g(x n, Y n ) (4.2) we say that {(X n, Y n ); n = 0, 1,, N} is a 2 dimensional Markov process. Remark 4.2 In the binomial model, the stock price at time n + 1 given the stock price at time n by { usn (w S n+1 (w n+1 ) = n ), if w n+1 = H; ds n (w n ), if w n+1 = T. Therefore, E P [f(s n+1 ) w n ] = pf(us n (w n )) + qf(ds n (w n )), and we can rewrite the above equation as E P [f(s n+1 ) w n ] = g(s n ), where the function g is defined by g(x) = pf(ux) + qf(dx). This shows that the stock price process is Markov. Remark 4.3 The stock price process is Markov under either the actual or the risk-neutral probability measure. To determine the price V n at time n of a derivative security whose 49

51 payoff at time N is a function v N of the stock price S N i.e., V N = v N (S N ), we use the risk-neutral pricing formula V n = 1 E P[V n+1 w n ], n = 0, 1,..., N 1. But V N = v N (S N ) and the stock price process is Markov, so for some function v N 1. Similarly, V N 1 = 1 E P[V N w N 1 ] = 1 E P[v N (S N ) w N 1 ] = p v N(uS N 1 ) =v N 1 (S N 1 ) + q v N(dS N 1 ) V N 2 = 1 E P[v N 1 (S N 1 )] = p v N 1(uS N 2 ) + q v N 1(dS N 2 ) for some function v N 2. = v N 2 (S N 2 ) In general, V n = v n (S n ) for some function v n. Moreover, we can compute the functions v N, v N 1,..., v 0 by v n (s) = 1 [ pv n+1(us) + qv n+1 (ds)], n = N 1, N 2,, 0 (4.3) Example 4.3 Consider the three-period model Suppose S 0 = 4, u = 2, d = 1 2, r = 1 4. Assume a European put option with strike price K = and maturity time T = t 3 V 3 = ( S 3 ) +. It can be seen that the price of option depends on the current stock prince so it is a function of the current stock price i.e. V n = v n (S n ). Then Note that p = d u d = 1 2 and q = u 1 r u d V 3 (HHH) =( S 3 (HHH)) + = ( 32) + = 0 = 1 2. V 3 (HHT ) =( S 3 (HHT )) + = ( 8) + = 0 = V 3 (HT H) = V 3 (T HH) V 3 (HT T ) =( S 3 (HT T )) + = ( 2) + = 3 = V 3 (T HT ) = V 3 (T T H) V 3 (T T T ) =( S 3 (T T T )) + = ( 0.) + = 4.. i.e 0

52 Let v 3 (S) be the value of the option at time t 3 when the stock price at time t 3 is S. Then 0, if S = 32, 8; V 3 =v 3 (S) = 3, if S = 2; 4., if S = 0.. So at time t 2 we have v 2 (S) = 1 ( p v 3(uS) + q v 3 (ds)) = 4 (0. v 3(2S) + 0. v 3 (0.S)) = 2 (v 3(2S) + v 3 (0.S)) Hence Similarly, at time t 1 we have Finally, at time t 0 we have v 2 (16) = 2 (v 3(32) + v 3 (8)) = 0 v 2 (4) = 2 (v 3(8) + v 3 (2)) = 2 (0 + 3) = 1.2 v 2 (1) = 2 (v 3(2) + v 3 (0.)) = 2 (3 + 4.) = 3. v 1 (S) = 2 (v 2(2S) + v 2 (0.S)) = v 1 (8) = 2 (v 2(16) + v 2 (4)) = 2 ( ) = 0.48 = v 1 (2) = 2 (v 2(4) + v 2 (1)) = 2 ( ) = v 0 (S) = 2 (v 1(2S) + v 1 (0.S)) = v 0 (4) = 2 (v 1(8) + v 1 (2)) = 2 ( ) = Therefor, the no-arbitrage price of the option at time t 0 is The delta-hedging formula for this method is n (S) = v n+1(us) v n+1 (ds) us ds = v n+1(2s) v n+1 (0.S). 1.S 1

53 Remark 4.4 To calculate V 0 of Example 4.3 using the alternative procedure, we can use the formula v 0 (s) = 1 () N N i=0 [ ( ) N ] p N i q i v N (u N i d i s) i For N = 3, u = 2, d = 1, s = S 2 0 = 4, r = 1, we get 4 [ [ v 0 (s) = ] ( ) 3 (1 3 i ( 1 ) i ( v3 2 i 2) 3 i 1 ] 2 2 s) i i=0 [ 2 3 [ ] = v 3 (8s) + 3v 3 (2s) + 3v 3 (s/2) + +v 3 (s/8) ] [ 2 3 [ ] v 0 (4) = v 3 (32) + 3v 3 (8) + 3v 3 (2) + v 3 (1/2) ] [ 2 3 [ ] = ] =0.864 Example 4.4 (Non-Markov process) Case I: Consider three binomial model with the random process M n = max 0 k n S k, n = 0, 1, 2, 3. S 3 (HHH) = 32 (4.4) S 0 = 4 S 1 (H) = 8 S 1 (T ) = 2 S 2 (HH) = 16 S 2 (HT ) = 4 = S 2 (T H) S 2 (T T ) = 1 S 3 (T HH) = 8 = S 3 (HT H) = S 3 (T HH) S 3 (T T H) = 2 = S 3 (T HT ) = S 3 (HT T ) S 3 (T T T ) = 0. 2

54 Note that M 0 = S 0 = 4 M 1 (H) = max{s 0, S 1 (H)} = {4, 8} = 8 M 1 (T ) = max{s 0, S 1 (T )} = {4, 2} = 4 M 2 (HH) = max{s 0, S 1 (H), S 2 (HH)} = {4, 8, 16} = 16 M 2 (HT ) = max{s 0, S 1 (H), S 2 (HT )} = {4, 8, 4} = 8 M 2 (T H) = max{s 0, S 1 (T ), S 2 (T H)} = {4, 2, 4} = 4 M 2 (T T ) = max{s 0, S 1 (T ), S 2 (T T )} = {4, 2, 1} = 4 M 3 (HHH) = max{s 0, S 1 (H), S 2 (HH), S 3 (HHH)} = {4, 8, 16, 32} = 32 M 3 (HHT ) = max{s 0, S 1 (H), S 2 (HH), S 3 (HHT )} = {4, 8, 16, 8} = 16 M 3 (HT H) = max{s 0, S 1 (H), S 2 (HT ), S 3 (HT H)} = {4, 8, 4, 8} = 8 M 3 (T HH) = max{s 0, S 1 (T ), S 2 (T H), S 3 (T HH)} = {4, 2, 4, 8} = 8 M 3 (HT T ) = max{s 0, S 1 (H), S 2 (HT ), S 3 (HT T )} = {4, 8, 4, 2} = 8 M 3 (T HT ) = max{s 0, S 1 (T ), S 2 (T H), S 3 (T HT )} = {4, 2, 4, 2} = 4 M 3 (T T H) = max{s 0, S 1 (T ), S 2 (T T ), S 3 (T T H)} = {4, 2, 1, 2} = 4 M 3 (T T T ) = max{s 0, S 1 (T ), S 2 (T T ), S 3 (T T T )} = {4, 2, 1, 0.} = 4 3