Chapter 5 StoppingTimesandAmericanOptions 5.1 American Pricing Let us first review the European pricing formula in a Markov model. Consider the Binomial model with n periods. Let V n = g S n be the payoff of a derivative security. Define by backward recursion: v n x = g x v k x = 1 1+r ~pv k+1 ux+ ~qv k+1 dx: Then v k S k is the value of the option at time k , and the hedging portfolio is given by k = v k+1 uS k , v k+1 dS k u , dS k ; k =0;1;2;::: ;n , 1: Now consider an American option. Again a function g is specified. In any period k , the holder of the derivative security can “exercise” and receive payment g S k . Thus, the hedging portfolio should create a wealth process which satisfies X k g S k ; 8k; almost surely. This is because the value of the derivative security at time k is at least g S k , and the wealth process value at that time must equal the value of the derivative security. American algorithm. v n x = g x v k x = max 1 1+r ~pv k+1 ux+ ~qv k+1 dx;gx Then v k S k is the value of the option at time k . 77 78 v (16) = 0 2 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 v (4) = 1 v (1) = 4 2 2 Figure 5.1: Stock price and final value of an American put option with strike price 5. Example 5.1 See Fig. 5.1. S 0 =4;u =2;d = 1 2 ;r = 1 4 ; ~p =~q= 1 2 ;n =2 .Set v 2 x=gx=5,x + . Then v 1 8 = max 4 5 1 2 :0+ 1 2 :1 ;5 , 8 + = max 2 5 ; 0 = 0:40 v 1 2 = max 4 5 1 2 :1+ 1 2 :4 ;5 , 2 + = maxf2; 3g = 3:00 v 0 4 = max 4 5 1 2 :0:4 + 1 2 :3:0 ; 5 , 4 + = maxf1:36; 1g = 1:36 Let us now construct the hedging portfolio for this option. Begin with initial wealth X 0 =1:36 . Compute 0 as follows: 0:40 = v 1 S 1 H = S 1 H 0 +1+rX 0 , 0 S 0 = 8 0 + 5 4 1:36 , 4 0 = 3 0 +1:70 = 0 = ,0:43 3:00 = v 1 S 1 T = S 1 T 0 +1+rX 0 , 0 S 0 = 2 0 + 5 4 1:36 , 4 0 = ,3 0 +1:70 = 0 = ,0:43 CHAPTER 5. StoppingTimesandAmericanOptions 79 Using 0 = ,0:43 results in X 1 H =v 1 S 1 H = 0:40;X 1 T=v 1 S 1 T=3:00 Now let us compute 1 (Recall that S 1 T =2 ): 1 = v 2 4 = S 2 TH 1 T + 1 + rX 1 T , 1 T S 1 T = 4 1 T + 5 4 3 , 2 1 T = 1:5 1 T +3:75 = 1 T =,1:83 4 = v 2 1 = S 2 TT 1 T + 1 + rX 1 T , 1 T S 1 T = 1 T + 5 4 3 , 2 1 T = ,1:5 1 T +3:75 = 1 T =,0:16 We get different answers for 1 T !Ifwehad X 1 T =2 , the value of the European put, we would have 1=1:5 1 T +2:5= 1 T=,1; 4=,1:5 1 T +2:5= 1 T=,1; 5.2 Value of Portfolio Hedging an American Option X k+1 = k S k+1 +1+rX k , C k , k S k = 1 + rX k + k S k+1 , 1 + rS k , 1 + rC k Here, C k is the amount “consumed” at time k . The discounted value of the portfolio is a supermartingale. The value satisfies X k g S k ;k =0;1;::: ;n . The value process is the smallest process with these properties. When do you consume? If f IE 1 + r ,k+1 v k+1 S k+1 jF k 1 + r ,k v k S k ; or, equivalently, f IE 1 1+r v k+1 S k+1 jF k v k S k 80 and the holder of the American option does not exercise, then the seller of the option can consume to close the gap. By doing this, he can ensure that X k = v k S k for all k ,where v k is the value defined by the American algorithm in Section 5.1. In the previous example, v 1 S 1 T =3;v 2 S 2 TH = 1 and v 2 S 2 TT = 4 . Therefore, f IE 1 1+ r v 2 S 2 jF 1 T = 4 5 h 1 2 :1+ 1 2 :4 i = 4 5 5 2 = 2; v 1 S 1 T = 3; so there is a gap of size 1. If the owner of the option does not exercise it at time one in the state ! 1 = T , then the seller can consume 1 at time 1. Thereafter, he uses the usual hedging portfolio k = v k+1 uS k , v k+1 dS k u , dS k In the example, we have v 1 S 1 T = g S 1 T . It is optimal for the owner of the American option to exercise whenever its value v k S k agrees with its intrinsic value g S k . Definition 5.1 (Stopping Time) Let ; F ; P be a probability space and let fF k g n k=0 be a filtra- tion. A stopping time is a random variable :!f0; 1; 2;::: ;ng f1g with the property that: f! 2 ; != kg2F k ; 8k=0;1;::: ;n;1: Example 5.2 Consider the binomial model with n =2;S 0 =4;u =2;d = 1 2 ;r = 1 4 ,so ~p =~q= 1 2 .Let v 0 ;v 1 ;v 2 be the value functions defined for the American put with strike price 5. Define ! = minfk; v k S k =5,S k + g: The stopping time corresponds to “stopping the first time the value of the option agrees with its intrinsic value”. It is an optimal exercise time. We note that != 1 if ! 2 A T 2 if ! 2 A H We verify that is indeed a stopping time: f! ; !=0g = 2F 0 f!;!=1g = A T 2F 1 f!;!=2g = A H 2F 2 Example 5.3 (A random time which is not a stopping time) In the same binomialmodel as in the previous example, define ! = minfk; S k !=m 2 !g; CHAPTER 5. StoppingTimesandAmericanOptions 81 where m 2 4 = min 0j 2 S j .Inotherwords, stops when the stock price reaches its minimum value. This random variable is given by ! = 8 : 0 if ! 2 A H ; 1 if ! = TH; 2 if ! = TT We verify that is not a stopping time: f!; !=0g = A H 62 F 0 f!; !=1g = fTHg62 F 1 f!;!=2g = fTTg2F 2 5.3 Information up to a Stopping Time Definition 5.2 Let be a stopping time. We say that a set A is determined by time provided that A f!;!=kg2F k ;8k: The collection of sets determined by is a -algebra, which we denote by F . Example 5.4 In the binomial model considered earlier, let = minfk; v k S k =5,S k + g; i.e., != 1 if ! 2 A T 2 if ! 2 A H The set fHT g is determined by time ,buttheset fTHg is not. Indeed, fHT gf!;!=0g = 2F 0 fHT gf!;!=1g = 2F 1 fHT gf!;!=2g = fHT g2F 2 but fTHgf!;!=1g=fTHg62F 1 : The atoms of F are fHT g; fHHg;A T =fTH;TTg: Notation 5.1 (Value of Stochastic Process at a Stopping Time) If ; F ; P is a probabilityspace, fF k g n k=0 is a filtration under F , fX k g n k=0 is a stochastic process adapted to this filtration, and is a stopping time with respect to the same filtration, then X is an F -measurable random variable whose value at ! is given by X ! 4 = X ! ! : 82 Theorem 3.17 (Optional Sampling) Suppose that fY k ; F k g 1 k=0 (or fY k ; F k g n k=0 ) is a submartin- gale. Let and be bounded stopping times, i.e., there is a nonrandom number n such that n; n; almost surely. If almost surely, then Y IE Y jF : Taking expectations, weobtain IEY IEY , and in particular, Y 0 = IEY 0 IEY .If fY k ; F k g 1 k=0 is a supermartingale, then implies Y IE Y jF . If fY k ; F k g 1 k=0 is a martingale, then implies Y = IE Y jF . Example 5.5 In theexample 5.4 considered earlier, we define ! =2 for all ! 2 . Under the risk-neutral probability measure, the discounted stock price process 5 4 ,k S k is a martingale. We compute e IE " 4 5 2 S 2 F : The atoms of F are fHHg; fHT g; and A T . Therefore, e IE " 4 5 2 S 2 F HH = 4 5 2 S 2 HH; e IE " 4 5 2 S 2 F HT = 4 5 2 S 2 HT ; and for ! 2 A T , e IE " 4 5 2 S 2 F ! = 1 2 4 5 2 S 2 TH+ 1 2 4 5 2 S 2 TT = 1 2 2:56 + 1 2 0:64 = 1:60 In every case we have gotten (see Fig. 5.2) e IE " 4 5 2 S 2 F ! = 4 5 ! S ! !: CHAPTER 5. StoppingTimesandAmericanOptions 83 S = 4 0 1 2 2 2 2 S (HH) = 10.24 S (HT) = 2.56 S (TH) = 2.56 S (TT) = 0.64 1 S (T) = 1.60 (4/5) S (H) = 6.40(4/5) (16/25) (16/25) (16/25) (16/25) Figure 5.2: Illustrating the optional sampling theorem. 84 [...]... 4 2 5 2 5 S2 HH ; 4 2 5 S2 HT ; 4 1 5 S2 TH + 2 5 = 1 2:56 + 1 0:64 2 2 = 1:60 2 S2 F ! = e IE 4 4 1 2 ! S2 F ! = 4 S ! ! : 5 2 S2 TT CHAPTER 5 StoppingTimes and American Options 83 (16/25) S (HH) = 10.24 2 (4/5) S (H) = 6.40 1 (16/25) S2 (HT) = 2.56 S =4 0 (16/25) S2 (TH) = 2.56 (4/5) S (T) = 1.60 1 (16/25) S2 (TT) = 0.64 Figure 5.2: Illustrating the optional . Chapter 5 Stopping Times and American Options 5.1 American Pricing Let us first review the European pricing. 5. Stopping Times and American Options 81 where m 2 4 = min 0j 2 S j .Inotherwords, stops when the stock price reaches its minimum value. This random