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Chapter 4 The Markov Property 4.1 Binomial Model Pricing and Hedging Recall that V m is the given simple European derivative security, and the value and portfolio pro- cesses are given by: V k =1+r k f IE1 + r ,m V m jF k ; k =0;1;::: ;m, 1: k ! 1 ;::: ;! k = V k+1 ! 1 ;::: ;! k ;H, V k+1 ! 1 ;::: ;! k ;T S k+1 ! 1 ;::: ;! k ;H, S k+1 ! 1 ;::: ;! k ;T ; k =0;1;::: ;m, 1: Example 4.1 (Lookback Option) u =2;d =0:5;r =0:25;S 0 =4;~p= 1+r,d u,d =0:5;~q=1,~p=0:5: Consider a simple European derivative security with expiration 2, with payoff given by (See Fig. 4.1): V 2 = max 0k2 S k , 5 + : Notice that V 2 HH=11;V 2 HT =36=V 2 TH=0;V 2 TT=0: The payoff is thus “path dependent”. Working backward in time, we have: V 1 H = 1 1+r ~pV 2 HH+ ~qV 2 HT = 4 5 0:5 11 + 0:5 3 = 5:60; V 1 T = 4 5 0:5 0+0:50 = 0; V 0 = 4 5 0:5 5:60 + 0:5 0=2:24: Using these values, we can now compute: 0 = V 1 H , V 1 T S 1 H , S 1 T =0:93; 1 H = V 2 HH , V 2 HT S 2 HH , S 2 HT =0:67; 67 68 S = 4 0 S (H) = 8 S (T) = 2 S (HH) = 16 S (TT) = 1 S (HT) = 4 S (TH) = 4 1 1 2 2 2 2 Figure 4.1: Stock price underlying the lookback option. 1 T = V 2 TH , V 2 TT S 2 TH , S 2 TT =0: Working forward in time, we can check that X 1 H = 0 S 1 H + 1 + rX 0 , 0 S 0 =5:59; V 1 H =5:60; X 1 T = 0 S 1 T + 1 + rX 0 , 0 S 0 =0:01; V 1 T =0; X 1 HH= 1 HS 1 HH + 1 + rX 1 H , 1 H S 1 H =11:01; V 1 HH=11; etc. Example 4.2 (European Call) Let u =2;d = 1 2 ;r = 1 4 ;S 0 =4;~p=~q= 1 2 , and consider a European call with expiration time 2 and payoff function V 2 =S 2 ,5 + : Note that V 2 HH=11;V 2 HT =V 2 TH=0;V 2 TT=0; V 1 H= 4 5 1 2 :11 + 1 2 :0=4:40 V 1 T = 4 5 1 2 :0+ 1 2 :0 = 0 V 0 = 4 5 1 2 4:40 + 1 2 0 = 1:76: Define v k x to be the value of the call at time k when S k = x .Then v 2 x=x,5 + v 1 x= 4 5 1 2 v 2 2x+ 1 2 v 2 x=2; v 0 x= 4 5 1 2 v 1 2x+ 1 2 v 1 x=2: CHAPTER 4. The Markov Property 69 In particular, v 2 16 = 11;v 2 4 = 0;v 2 1 = 0; v 1 8 = 4 5 1 2 :11 + 1 2 :0 = 4:40; v 1 2 = 4 5 1 2 :0+ 1 2 :0 = 0; v 0 = 4 5 1 2 4:40 + 1 2 0 = 1:76: Let k x be the number of shares in the hedging portfolio at time k when S k = x .Then k x= v k+1 2x , v k+1 x=2 2x , x=2 ;k=0;1: 4.2 Computational Issues For a model with n periods (coin tosses), has 2 n elements. For period k , we must solve 2 k equations of the form V k ! 1 ;::: ;! k = 1 1+r ~pV k+1 ! 1 ;::: ;! k ;H+ ~qV k+1 ! 1 ;::: ;! k ;T: For example, a three-month option has 66 trading days. If each day is taken to be one period, then n =66 and 2 66 7 10 19 . There are three possible ways to deal with this problem: 1. Simulation. We have, for example, that V 0 =1+r ,n f IEV n ; and so we could compute V 0 by simulation. More specifically, we could simulate n coin tosses ! =! 1 ;::: ;! n under the risk-neutral probability measure. We could store the value of V n ! . We could repeat this several times and take the average value of V n as an approximation to f IEV n . 2. Approximate a many-period model by a continuous-time model. Then we can use calculus and partial differential equations. We’ll get to that. 3. Look for Markov structure. Example 4.2 has this. In period 2, the option in Example 4.2 has threepossiblevalues v 2 16;v 2 4;v 2 1 , ratherthan four possiblevalues V 2 HH;V 2 HT ;V 2 TH;V 2 TT . If there were 66 periods, then in period 66 there would be 67 possible stock price values(since the final price depends only on the number of up-ticks of the stock price – i.e., heads – so far) and hence only 67 possible option values, rather than 2 66 7 10 19 . 70 4.3 Markov Processes Technical condition always present: We consider only functions on IR and subsets of IR which are Borel-measurable, i.e., we only consider subsets A of IRthatarein B and functions g : IR!IR such that g ,1 is a function B!B . Definition 4.1 () Let ; F ; P be a probability space. Let fF k g n k=0 be a filtration under F .Let fX k g n k=0 be a stochastic process on ; F ; P . This process is said to be Markov if: The stochastic process fX k g is adapted to the filtration fF k g ,and (The Markov Property). For each k =0;1;::: ;n , 1 , the distribution of X k+1 conditioned on F k is the same as the distribution of X k+1 conditioned on X k . 4.3.1 Different ways to write the Markov property (a) (Agreement of distributions). For every A 2B 4 =BIR ,wehave IP X k+1 2 AjF k = IE I A X k+1 jF k = IE I A X k+1 jX k = IP X k+1 2 AjX k : (b) (Agreement of expectations of all functions). For every (Borel-measurable) function h : IR!IR for which IE jhX k+1 j 1 ,wehave IE hX k+1 jF k =IEhX k+1 jX k : (c) (Agreement of Laplace transforms.) For every u 2 IR for which IEe uX k+1 1 ,wehave IE e uX k+1 F k = IE e uX k+1 X k : (If we fix u and define hx=e ux , then the equations in (b) and (c) are the same. However in (b) we have a condition which holds for every function h , and in (c) we assume this condition onlyfor functions h of the form hx=e ux . A main resultin the theory of Laplace transforms is that if the equation holds for every h of this special form, then it holds for every h , i.e., (c) implies (b).) (d) (Agreement of characteristic functions) For every u 2 IR ,wehave IE h e iuX k+1 jF k i = IE h e iuX k+1 jX k i ; where i = p ,1 .(Since je iux j = j cos x + sin xj1 we don’t need to assume that IE je iux j 1 .) CHAPTER 4. The Markov Property 71 Remark 4.1 In every case of the Markov properties where IE :::jX k appears, we could just as well write g X k for some function g . For example, form (a) of the Markov property can be restated as: For every A 2B ,wehave IP X k+1 2 AjF k =gX k ; where g is a function that depends on the set A . Conditions (a)-(d) are equivalent. The Markov property as stated in (a)-(d) involves the process at a “current” time k and one future time k +1 . Conditions (a)-(d) are also equivalent to conditions involving the process at time k and multiple future times. We write these apparently stronger but actually equivalent conditions below. Consequences of the Markov property. Let j be a positive integer. (A) For every A k+1 IR;::: ;A k+j IR , IPX k+1 2 A k+1 ;::: ;X k+j 2 A k+j jF k =IPX k+1 2 A k+1 ;::: ;X k+j 2 A k+j jX k : (A’) For every A 2 IR j , IP X k+1 ;::: ;X k+j 2 AjF k =IPX k+1 ;::: ;X k+j 2 AjX k : (B) For every function h : IR j !IR for which IE jhX k+1 ;::: ;X k+j j 1 ,wehave IE hX k+1 ;::: ;X k+j jF k =IEhX k+1 ;::: ;X k+j jX k : (C) For every u =u k+1 ;::: ;u k+j 2 IR j for which IE je u k+1 X k+1 +:::+u k+j X k+j j 1 ,wehave IE e u k+1 X k+1 +:::+u k+j X k+j jF k =IEe u k+1 X k+1 +:::+u k+j X k+j jX k : (D) For every u =u k+1 ;::: ;u k+j 2 IR j we have IE e iu k+1 X k+1 +:::+u k+j X k+j jF k =IEe iu k+1 X k+1 +:::+u k+j X k+j jX k : Once again, every expression of the form IE :::jX k can also be written as g X k ,wherethe function g depends on the random variable represented by ::: in this expression. Remark. All these Markov properties have analogues for vector-valued processes. 72 Proof that (b) = (A). (with j =2 in (A)) Assume (b). Then (a) also holds (take h = I A ). Consider IP X k+1 2 A k+1 ;X k+2 2 A k+2 jF k = IE I A k+1 X k+1 I A k+2 X k+2 jF k (Definition of conditional probability) = IE IE I A k+1 X k+1 I A k+2 X k+2 jF k+1 jF k (Tower property) = IE I A k+1 X k+1 :IE I A k+2 X k+2 jF k+1 jF k (Taking out what is known) = IE I A k+1 X k+1 :IE I A k+2 X k+2 jX k+1 jF k (Markov property, form (a).) = IE I A k+1 X k+1 :g X k+1 jF k (Remark 4.1) = IE I A k+1 X k+1 :g X k+1 jX k (Markov property, form (b).) Now take conditional expectation on both sides of the above equation, conditioned on X k ,and use the tower property on the left, to obtain IP X k+1 2 A k+1 ;X k+2 2 A k+2 jX k =IEI A k+1 X k+1 :g X k+1 jX k : (3.1) Since both IP X k+1 2 A k+1 ;X k+2 2 A k+2 jF k and IP X k+1 2 A k+1 ;X k+2 2 A k+2 jX k are equal to the RHS of (3.1)), they are equal to each other, and this is property (A) with j =2 . Example 4.3 It is intuitively clear that the stock price process in the binomial model is a Markov process. We will formally prove this later. If we want to estimate the distributionof S k+1 based on the information in F k , the only relevant piece of information is the value of S k . For example, e IE S k+1 jF k =~pu +~qdS k =1+rS k (3.2) is a function of S k . Note however that form (b) of the Markov property is stronger then (3.2); the Markov property requires that for any function h , e IE hS k+1 jF k is a function of S k . Equation (3.2) is the case of hx=x . Consider a model with 66 periods and a simple European derivative security whose payoff at time 66 is V 66 = 1 3 S 64 + S 65 + S 66 : CHAPTER 4. The Markov Property 73 The value of this security at time 50 is V 50 = 1 + r 50 e IE 1 + r ,66 V 66 jF 50 = 1 + r ,16 e IE V 66 jS 50 ; because the stock price process is Markov. (We are using form (B) of the Markov property here). In other words, the F 50 -measurable random variable V 50 can be written as V 50 ! 1 ;::: ;! 50 =gS 50 ! 1 ;::: ;! 50 for some function g , which we can determine with a bit of work. 4.4 Showing that a process is Markov Definition 4.2 (Independence) Let ; F ; P be a probability space, and let G and H be sub- - algebras of F . We say that G and H are independent if for every A 2G and B 2H ,wehave IP A B =IPAIPB: We say that a random variable X is independent of a -algebra G if X ,the -algebra generated by X , is independent of G . Example 4.4 Consider the two-period binomial model. Recall that F 1 is the -algebra of sets determined by the first toss, i.e., F 1 contains the four sets A H 4 = fHH; HT g;A T 4 =fTH;TTg;;: Let H be the -algebra of sets determined by the second toss, i.e., H contains the four sets fHH; T Hg; fHT; T T g;;: Then F 1 and H are independent. For example, if we take A = fHH; HT g from F 1 and B = fHH; T Hg from H ,then IP A B =IPHH=p 2 and IP AIP B =p 2 +pqp 2 + pq= p 2 p+q 2 =p 2 : Note that F 1 and S 2 are not independent (unless p =1 or p =0 ). For example, one of the sets in S 2 is f!; S 2 != u 2 S 0 g=fHHg .Ifwetake A = fHH; HT g from F 1 and B = fHHg from S 2 ,then IP A B =IPHH=p 2 ,but IP AIP B =p 2 +pqp 2 = p 3 p + q=p 3 : The following lemma will be very useful in showing that a process is Markov: Lemma 4.15 (Independence Lemma) Let X and Y be random variables on a probability space ; F ; P .Let G be a sub- -algebra of F . Assume 74 X is independent of G ; Y is G -measurable. Let f x; y be a function of two variables, and define g y 4 = IEf X; y: Then IE f X; Y jG= gY: Remark. In this lemma and the following discussion, capital letters denote random variables and lower case letters denote nonrandom variables. Example 4.5 (Showing the stock price process is Markov) Consider an n -period binomial model. Fix a time k and define X 4 = S k+1 S k and G 4 = F k .Then X = u if ! k+1 = H and X = d if ! k+1 = T .Since X depends only on the k +1 st toss, X is independent of G .Define Y 4 = S k ,sothat Y is G -measurable. Let h be any function and set f x; y 4 = hxy .Then gy 4 = IEf X; y=IEhXy=phuy+qhdy: The Independence Lemma asserts that IE hS k+1 jF k = IE h S k+1 S k :S k jF k = IE f X; Y jG = gY = phuS k +qhdS k : This shows the stock price is Markov. Indeed, if we condition both sides of the above equation on S k and use the tower property on the left and the fact that the right hand side is S k -measurable, we obtain IE hS k+1 jS k =phuS k +qhdS k : Thus IE hS k+1 jF k and IE hS k+1 jX k are equal and form (b) of the Markov property is proved. Not only have we shown that the stock price process is Markov, but we have also obtained a formula for IE hS k+1 jF k as a function of S k . This is a special case of Remark 4.1. 4.5 Application to Exotic Options Consider an n -period binomial model. Define the running maximum of the stock price to be M k 4 = max 1j k S j : Consider a simple European derivative security with payoff at time n of v n S n ;M n . Examples: CHAPTER 4. The Markov Property 75 v n S n ;M n =M n ,K + (Lookback option); v n S n ;M n =I M n B S n ,K + (Knock-in Barrier option). Lemma 5.16 The two-dimensional process fS k ;M k g n k=0 is Markov. (Here we are working under the risk-neutral measure IP, although that does not matter). Proof: Fix k .Wehave M k+1 = M k _ S k+1 ; where _ indicates the maximum of two quantities. Let Z 4 = S k+1 S k ,so f IP Z = u= ~p; f IP Z = d= ~q; and Z is independent of F k .Let hx; y be a function of two variables. We have hS k+1 ;M k+1 = hS k+1 ;M k _ S k+1 = hZS k ;M k _ ZS k : Define g x; y 4 = f IEhZx; y _ Zx = ~phux; y _ ux + ~qhdx; y _ dx: The Independence Lemma implies f IE hS k+1 ;M k+1 jF k =gS k ;M k = ~phuS k ;M k _ uS k + ~qhdS k ;M k ; the second equality being a consequence of the fact that M k ^ dS k = M k . Since the RHS is a function of S k ;M k , we have proved the Markov property (form (b)) for this two-dimensional process. Continuingwith the exotic optionof the previousLemma . Let V k denote the value of the derivative security at time k .Since 1 + r ,k V k is a martingale under f IP ,wehave V k = 1 1+r f IEV k+1 jF k ;k =0;1;::: ;n , 1: At the final time, we have V n = v n S n ;M n : Stepping back one step, we can compute V n,1 = 1 1+ r f IEv n S n ;M n jF n,1 = 1 1+ r ~pv n uS n,1 ;uS n,1 _ M n,1 + ~qv n dS n,1 ;M n,1 : 76 This leads us to define v n,1 x; y 4 = 1 1+ r ~pv n ux; ux _ y + ~qv n dx; y so that V n,1 = v n,1 S n,1 ;M n,1 : The general algorithm is v k x; y = 1 1+r ~pv k+1 ux; ux _ y + ~qv k+1 dx; y ; and the value of the option at time k is v k S k ;M k . Since this is a simple European option, the hedging portfolio is given by the usual formula, which in this case is k = v k+1 uS k ; uS k _ M k , v k+1 dS k ;M k u , dS k [...]... vn,1 x; y = 1 + r pvn ux; ux _ y + qvn dx; y ~ ~ so that Vn,1 = vn,1 Sn,1; Mn,1 : The general algorithm is 1 ~ ~ vk x; y = 1 + r pvk+1 ux; ux _ y + qvk+1 dx; y ; and the value of the option at time k is vk Sk ; Mk Since this is a simple European option, the hedging portfolio is given by the usual formula, which in this case is _ Mk k = vk+1 uSk ; uSk u , d S, vk+1 dSk . said to be Markov if: The stochastic process fX k g is adapted to the filtration fF k g ,and (The Markov Property) . For each k =0;1;::: ;n , 1 , the distribution. function of S k . Note however that form (b) of the Markov property is stronger then (3.2); the Markov property requires that for any function h , e IE
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