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Chapter 8 Random Walks 8.1 First Passage Time Toss a coin infinitely many times. Then the sample space is the set of all infinite sequences ! =! 1 ;! 2 ;::: of H and T . Assume the tosses are independent, and on each toss, the probability of H is 1 2 , as is the probability of T .Define Y j != 1 if ! j = H; ,1 if ! j = T; M 0 = 0; M k = k X j=1 Y j ;k=1;2;::: The process fM k g 1 k=0 is a symmetric random walk (see Fig. 8.1) Its analogue in continuous time is Brownian motion. Define = minfk 0; M k =1g: If M k never gets to 1 (e.g., ! =TTTT : : : ), then = 1 . The random variable is called the first passage time to 1. It is the first time the number of heads exceeds by one the number of tails. 8.2 is almost surely finite ItisshowninaHomeworkProblemthat fM k g 1 k=0 and fN k g 1 k=0 where N k = exp M k , k log e + e , 2 ! = e M k 2 e + e , k 97 98 M k k Figure 8.1: The random walk process M k e + e 2 θ −θ θ 1 θ 1 2 e + e θ −θ Figure 8.2: Illustrating two functions of are martingales. (Take M k = ,S k in part (i) of the Homework Problem and take = , in part (v).) Since N 0 =1 and a stopped martingale is a martingale, we have 1=IEN k^ = IE " e M k^ 2 e + e , k^ (2.1) for every fixed 2 IR (See Fig. 8.2 for an illustration of the various functions involved). We want to let k!1 in (2.1), but we have to worry a bit that for some sequences ! 2 , ! =1 . We consider fixed 0 ,so 2 e + e , 1: As k!1 , 2 e + e , k^ ! 2 e +e , if 1; 0 if = 1 Furthermore, M k^ 1 , because we stop this martingale when it reaches 1, so 0 e M k^ e CHAPTER 8. Random Walks 99 and 0 e M k^ 2 e + e , k^ e : In addition, lim k! 1 e M k^ 2 e + e , k^ = e 2 e +e , if 1; 0 if = 1: Recall Equation (2.1): IE " e M k^ 2 e + e , k^ =1 Letting k!1 , and using the Bounded Convergence Theorem, we obtain IE e 2 e + e , I f1g =1: (2.2) For all 2 0; 1 ,wehave 0 e 2 e + e , I f1g e; so we can let 0 in (2.2), using the Bounded Convergence Theorem again, to conclude IE h I f1g i =1; i.e., IP f1g =1: We know there are paths of the symmetric random walk fM k g 1 k=0 which never reach level 1. We have just shown that these paths collectively have no probability. (In our infinite sample space , each path individually has zero probability). We therefore do not need the indicator I f1g in (2.2), and we rewrite that equation as IE 2 e + e , = e , : (2.3) 8.3 The moment generating function for Let 2 0; 1 be given. We want to find 0 so that = 2 e + e , : Solution: e + e , , 2=0 e , 2 ,2e , +=0 100 e , = 1 p 1 , 2 : We want 0 ,sowemusthave e , 1 .Now 0 1 ,so 0 1 , 2 1 , 1 , 2 ; 1 , p 1, 2 ; 1, p 1, 2 ; 1, p 1, 2 1 We take the negative square root: e , = 1 , p 1 , 2 : Recall Equation (2.3): IE 2 e + e , = e , ;0: With 2 0; 1 and 0 related by e , = 1 , p 1 , 2 ; = 2 e + e , ; this becomes IE = 1 , p 1 , 2 ; 0 1: (3.1) We have computed the moment generating function for the first passage time to 1. 8.4 Expectation of Recall that IE = 1 , p 1 , 2 ; 0 1; so d d IE = IE ,1 = d d 1 , p 1 , 2 ! = 1 , p 1 , 2 2 p 1 , 2 : CHAPTER 8. Random Walks 101 Using the Monotone Convergence Theorem, we can let "1 in the equation IE ,1 = 1, p 1, 2 2 p 1, 2 ; to obtain IE = 1: Thus in summary: 4 = minfk ; M k =1g; IPf1g =1; IE = 1: 8.5 The Strong Markov Property The random walk process fM k g 1 k=0 is a Markov process, i.e., IE random variable depending only on M k+1 ;M k+2 ;:::jF k =IE same random variable jM k : In discrete time, this Markov property implies the Strong Markov property: IE random variable depending only on M +1 ;M +2 ;:::jF =IE same random variable j M : for any almost surely finite stopping time . 8.6 General First Passage Times Define m 4 = minfk 0; M k = mg;m=1;2;::: Then 2 , 1 is the number of periods between the first arrival at level 1 and the first arrival at level 2. The distribution of 2 , 1 is the same as the distribution of 1 (see Fig. 8.3), i.e., IE 2 , 1 = 1 , p 1 , 2 ;20; 1: 102 τ 1 τ 1 τ 2 τ 2 M k k − Figure 8.3: General first passage times. For 2 0; 1 , IE 2 jF 1 = IE 1 2 , 1 jF 1 = 1 IE 2 , 1 jF 1 (taking out what is known) = 1 IE 2 , 1 jM 1 (strong Markov property) = 1 IE 2 , 1 M 1 =1; not random = 1 1 , p 1 , 2 ! : Take expectations of both sides to get IE 2 = IE 1 : 1 , p 1 , 2 ! = 1 , p 1 , 2 ! 2 In general, IE m = 1 , p 1 , 2 ! m ;20; 1: 8.7 Example: Perpetual American Put Consider the binomial model, with u =2;d = 1 2 ;r = 1 4 , and payoff function 5 , S k + .Therisk neutral probabilities are ~p = 1 2 , ~q = 1 2 , and thus S k = S 0 u M k ; CHAPTER 8. Random Walks 103 where M k is a symmetric random walk under the risk-neutral measure, denoted by f IP . Suppose S 0 =4 . Here are some possible exercise rules: Rule 0: Stop immediately. 0 =0;V 0 =1 . Rule 1: Stop as soon as stock price falls to 2, i.e., at time ,1 4 = minfk ; M k = ,1g: Rule 2: Stop as soon as stock price falls to 1, i.e., at time ,2 4 = minfk ; M k = ,2g: Because the random walk is symmetric under f IP , ,m has the same distribution under f IP as the stopping time m in the previous section. This observation leads to the following computations of value. ValueofRule1: V ,1 = f IE 1 + r , ,1 5 , S ,1 + = 5 , 2 + IE h 4 5 ,1 i = 3: 1 , q 1 , 4 5 2 4 5 = 3 2 : ValueofRule2: V ,2 = 5 , 1 + f IE h 4 5 ,2 i = 4: 1 2 2 = 1: This suggests that the optimal rule is Rule 1, i.e., stop (exercise the put) as soon as the stock price falls to 2, and the value of the put is 3 2 if S 0 =4 . Suppose instead we start with S 0 =8 , and stop the first time the price falls to 2. This requires 2 down steps, so the value of this rule with this initial stock price is 5 , 2 + f IE h 4 5 ,2 i =3: 1 2 2 = 3 4 : In general, if S 0 =2 j for some j 1 , and we stop when the stock price falls to 2, then j , 1 down steps will be required and the value of the option is 5 , 2 + f IE h 4 5 ,j,1 i =3: 1 2 j,1 : We define v 2 j 4 =3: 1 2 j,1 ;j=1;2;3;::: 104 If S 0 =2 j for some j 1 , then the initial price is at or below 2. In this case, we exercise immediately, and the value of the put is v 2 j 4 =5,2 j ;j=1;0;,1;,2;::: Proposed exercise rule: Exercise the put whenever the stock price is at or below 2. The value of this rule is given by v 2 j as we just defined it. Since the put is perpetual, the initial time is no different from any other time. This leads us to make the following: Conjecture 1 The value of the perpetual put at time k is v S k . How do we recognize the value of an American derivative security when we see it? There are three parts to the proof of the conjecture. We must show: (a) v S k 5 , S k + 8k; (b) n 4 5 k v S k o 1 k=0 is a supermartingale, (c) fv S k g 1 k=0 is the smallest process with properties (a) and (b). Note: To simplify matters, we shall only consider initial stock prices of the form S 0 =2 j ,so S k is always of the form 2 j , with a possibly different j . Proof: (a). Just check that v 2 j 4 =3: 1 2 j,1 5 , 2 j + for j 1; v 2 j 4 =5,2 j 5 , 2 j + for j 1: This is straightforward. Proof: (b). We must show that v S k f IE h 4 5 v S k+1 jF k i = 4 5 : 1 2 v 2S k + 4 5 : 1 2 v 1 2 S k : By assumption, S k =2 j for some j . We must show that v 2 j 2 5 v 2 j +1 + 2 5 v2 j ,1 : If j 2 ,then v 2 j =3: 1 2 j,1 and 2 5 v 2 j +1 + 2 5 v2 j ,1 = 2 5 :3: 1 2 j + 2 5 :3: 1 2 j ,2 = 3: 2 5 : 1 4 + 2 5 1 2 j ,2 = 3: 1 2 : 1 2 j ,2 = v 2 j : CHAPTER 8. Random Walks 105 If j =1 ,then v 2 j =v2 = 3 and 2 5 v 2 j +1 + 2 5 v2 j ,1 = 2 5 v 4 + 2 5 v 1 = 2 5 :3: 1 2 + 2 5 :4 = 3=5+ 8=5 = 2 1 5 v2 = 3 There is a gap of size 4 5 . If j 0 ,then v 2 j =5,2 j and 2 5 v 2 j +1 + 2 5 v2 j ,1 = 2 5 5 , 2 j +1 + 2 5 5 , 2 j ,1 = 4 , 2 5 4 + 12 j ,1 = 4 , 2 j v2 j =5,2 j : There is a gap of size 1. This concludes the proof of (b). Proof: (c). Suppose fY k g n k=0 is some other process satisfying: (a’) Y k 5 , S k + 8k; (b’) f 4 5 k Y k g 1 k=0 is a supermartingale. We must show that Y k v S k 8k: (7.1) Actually, since the put is perpetual, every time k is like every other time, so it will suffice to show Y 0 v S 0 ; (7.2) providedwe let S 0 in (7.2) be any number of the form 2 j . With appropriate(but messy) conditioning on F k , the proof we give of (7.2) can be modified to prove (7.1). For j 1 , v 2 j =5,2 j =5,2 j + ; so if S 0 =2 j for some j 1 , then (a’) implies Y 0 5 , 2 j + = v S 0 : Suppose now that S 0 =2 j for some j 2 , i.e., S 0 4 .Let = minfk ; S k =2g = minfk ; M k = j , 1g: 106 Then v S 0 = v 2 j =3: 1 2 j,1 = IE h 4 5 5 , S + i : Because f 4 5 k Y k g 1 k=0 is a supermartingale Y 0 IE h 4 5 Y i IE h 4 5 5 , S + i = v S 0 : Comment on the proof of (c): If the candidate value process is the actual value of a particular exercise rule, then (c) will be automatically satisfied. In this case, we constructed v so that v S k is the value of the put at time k if the stock price at time k is S k and if we exercise the put the first time ( k , or later) that the stock price is 2 or less. In such a situation, we need only verify properties (a) and (b). 8.8 Difference Equation If we imagine stock prices which can fall at any point in 0; 1 , not just at points of the form 2 j for integers j , then we can imagine the function v x , defined for all x0 , which gives the value of the perpetual American put when the stock price is x . This function should satisfy the conditions: (a) v x K , x + ; 8x , (b) v x 1 1+r ~pv ux+ ~qvdx ; 8x; (c) At each x , either (a) or (b) holds with equality. In the example we worked out, we have For j 1:v2 j =3: 1 2 j,1 = 6 2 j ; For j 1:v2 j =5,2 j : This suggests the formula v x= 6 x ; x3; 5,x; 0 x3: We then have (see Fig. 8.4): (a) v x 5 , x + ; 8x; (b) v x 4 5 h 1 2 v 2x+ 1 2 v x 2 i for every x except for 2 x4 .