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Chapter 25 American Options This and the following chapters form part of the course Stochastic Differential Equations for Fi- nance II. 25.1 Preview of perpetual American put dS = rS dt + S dB Intrinsic value at time t :K,St + : Let L 2 0;K be given. Suppose we exercise the first time the stock price is L or lower. We define L = minft 0; S t Lg; v L x=IEe ,r L K , S L + = K , x if x L , K , LIEe ,r L if xL: Theplanistocomute v L x andthenmaximizeover L to find the optimal exercise price. We need to know the distribution of L . 25.2 First passage times for Brownian motion: first method (Based on the reflection principle) Let B be a Brownian motion under IP ,let x0 be given, and define = minft 0; B t=xg: is called the first passage time to x . We compute the distribution of . 247 248 K K x Intrinsic value Stock price Figure 25.1: Intrinsic value of perpetual American put Define M t = max 0ut B u: From the first section of Chapter 20 we have IP fM t 2 dm; B t 2 dbg = 22m , b t p 2t exp , 2m , b 2 2t dm db; m0;b m: Therefore, IP fM t xg = Z 1 x Z m ,1 22m , b t p 2t exp , 2m , b 2 2t db dm = Z 1 x 2 p 2t exp , 2m , b 2 2t b=m b=,1 dm = Z 1 x 2 p 2t exp , m 2 2t dm: We make the change of variable z = m p t in the integral to get = Z 1 x= p t 2 p 2 exp , z 2 2 dz : Now tM t x; CHAPTER 25. American Options 249 so IP f 2 dtg = @ @t IP f tg dt = @ @t IP fM t xg dt = " @ @t Z 1 x= p t 2 p 2 exp , z 2 2 dz dt = , 2 p 2 exp , x 2 2t : @ @t x p t dt = x t p 2t exp , x 2 2t dt: We also have the Laplace transform formula IEe , = Z 1 0 e ,t IP f 2 dtg = e ,x p 2 ; 0: (See Homework) Reference: Karatzas and Shreve, Brownian Motion and Stochastic Calculus, pp 95-96. 25.3 Drift adjustment Reference: Karatzas/Shreve, Brownian motion and Stochastic Calculus, pp 196–197. For 0 t1 ,define e B t=t + Bt; Z t = expf,B t , 1 2 2 tg; = expf, e B t+ 1 2 2 tg; Define ~ = minft 0; e B t=xg: We fix a finite time T and change the probability measure “only up to T ”. More specifically, with T fixed, define f IP A= Z A ZTdP; A 2FT: Under f IP , the process e B t; 0 t T , is a (nondrifted) Brownian motion, so f IP f ~ 2 dtg = IP f 2 dtg = x t p 2t exp , x 2 2t dt; 0 tT: 250 For 0 tT we have IP f ~ tg = IE h 1 f ~ tg i = f IE 1 f ~ tg 1 Z T = f IE h 1 f ~ tg expf e B T , 1 2 2 T g i = f IE 1 f ~ tg f IE expf e B T , 1 2 2 T g F ~ ^ t = f IE h 1 f ~ tg expf e B ~ ^ t , 1 2 2 ~ ^ tg i = f IE h 1 f~tg expfx , 1 2 2 ~ g i = Z t 0 expfx , 1 2 2 sg f IP f~ 2 dsg = Z t 0 x s p 2s exp x , 1 2 2 s , x 2 2s ds = Z t 0 x s p 2s exp , x , s 2 2s ds: Therefore, IP f ~ 2 dtg = x t p 2t exp , x , t 2 2t dt; 0 tT: Since T is arbitrary, this must in fact be the correct formula for all t0 . 25.4 Drift-adjusted Laplace transform Recall the Laplace transform formula for = minft 0; B t=xg for nondrifted Brownian motion: IEe , = Z 1 0 x t p 2t exp ,t , x 2 2t dt = e ,x p 2 ; 0;x 0: For ~ = minft 0; t + Bt=xg; CHAPTER 25. American Options 251 the Laplace transform is IEe ,~ = Z 1 0 x t p 2t exp ,t , x , t 2 2t dt = Z 1 0 x t p 2t exp ,t , x 2 2t + x , 1 2 2 t dt = e x Z 1 0 x t p 2t exp , + 1 2 2 t , x 2 2t dt = e x,x p 2+ 2 ; 0;x 0; where in the last step we have used the formula for IEe , with replaced by + 1 2 2 . If ~ ! 1 ,then lim 0 e , ~ ! =1; if ~ ! =1 ,then e , ~ ! =0 for every 0 ,so lim 0 e , ~ ! =0: Therefore, lim 0 e , ~ ! = 1 ~1 : Letting 0 and using the Monotone Convergence Theorem in the Laplace transform formula IEe ,~ = e x,x p 2+ 2 ; we obtain IP f ~1g = e x,x p 2 = e x,xj j : If 0 ,then IP f ~1g =1: If 0 ,then IP f ~1g = e 2x 1: (Recall that x0 ). 25.5 First passage times: Second method (Based on martingales) Let 0 be given. Then Y t = expfB t , 1 2 2 tg 252 is a martingale, so Y t ^ is also a martingale. We have 1=Y0 ^ = IEY t ^ = IE expfBt ^ , 1 2 2 t ^ g: = lim t!1 IE expfBt ^ , 1 2 2 t ^ g: We want to take the limit inside the expectation. Since 0 expfB t ^ , 1 2 2 t ^ ge x ; this is justified by the Bounded Convergence Theorem. Therefore, 1=IE lim t!1 expfBt ^ , 1 2 2 t ^ g: There are two possibilities. For those ! for which ! 1 , lim t!1 expfBt ^ , 1 2 2 t ^ g = e x, 1 2 2 : For those ! for which ! =1 , lim t!1 expfB t ^ , 1 2 2 t ^ g lim t!1 expfx , 1 2 2 tg =0: Therefore, 1=IE lim t!1 expfB t ^ , 1 2 2 t ^ g = IE e x, 1 2 2 1 1 = IEe x, 1 2 2 ; where we understand e x, 1 2 2 to be zero if = 1 . Let = 1 2 2 ,so = p 2 . We have again derived the Laplace transform formula e ,x p 2 = IEe , ; 0;x 0; for the first passage time for nondrifted Brownian motion. 25.6 Perpetual American put dS = rS dt + S dB S 0 = x S t=xexpfr , 1 2 2 t + Btg = x exp 8 : 2 6 6 6 4 r , 2 | z t + B t 3 7 7 7 5 9 = ; : CHAPTER 25. American Options 253 Intrinsic value of the put at time t : K , S t + . Let L 2 0;K be given. Define for x L , L = minft 0; S t=Lg = minft 0; t + Bt= 1 log L x g = minft 0; ,t , Bt= 1 log x L g Define v L =K,LIEe ,r L =K,L exp , log x L , 1 log x L p 2r + 2 =K,L x L , , 1 p 2r+ 2 : We compute the exponent , , 1 p 2r + 2 = , r 2 + 1 2 , 1 s 2r + r , =2 2 = , r 2 + 1 2 , 1 s 2r + r 2 2 , r + 2 =4 = , r 2 + 1 2 , 1 s r 2 2 + r + 2 =4 = , r 2 + 1 2 , 1 s r + =2 2 = , r 2 + 1 2 , 1 r + =2 = , 2r 2 : Therefore, v L x= 8 : K,x; 0x L; K , L , x L ,2r= 2 ; x L: The curves K , L , x L ,2r= 2 ; are all of the form Cx ,2r= 2 . We want to choose the largest possible constant. The constant is C =K,LL 2r= 2 ; 254 σ 2 -2r/ K K x Stock price K - x (K - L) (x/L) value Figure 25.2: Value of perpetual American put σ 2 -2r/ C 1 x σ 2 -2r/ x C 2 σ 2 -2r/ x C 3 x Stock price value Figure 25.3: Curves. CHAPTER 25. American Options 255 and @C @L = ,L 2r 2 + 2r 2 K , LL 2r 2 ,1 = L 2r 2 ,1+ 2r 2 K ,L 1 L = L 2r 2 , 1+ 2r 2 + 2r 2 K L : We solve , 1+ 2r 2 + 2r 2 K L =0 to get L = 2rK 2 +2r : Since 0 2r 2 +2r; we have 0 LK: Solution to the perpetual American put pricing problem (see Fig. 25.4): v x= 8 : K,x; 0x L ; K,L , x L ,2r= 2 ; x L ; where L = 2rK 2 +2r : Note that v 0 x= ,1; 0 xL ; , 2r 2 K,L L 2r= 2 x ,2r= 2 ,1 ; xL : We have lim xL v 0 x=,2 r 2 K,L 1 L =,2 r 2 K, 2rK 2 +2r 2 +2r 2rK = ,2 r 2 2 +2r,2r 2 +2r ! 2 +2r 2r = ,1 = lim x"L v 0 x: 256 σ 2 -2r/ K K x Stock price K - x value (K - L ) (x/L ) * * * L Figure 25.4: Solution to perpetual American put. 25.7 Value of the perpetual American put Set = 2r 2 ; L = 2rK 2 +2r = +1 K: If 0 xL ,then v x= K ,x .If L x1 ,then v x= K ,L L | z C x , (7.1) = IE x h e ,r K , L + 1 f1g i ; (7.2) where S 0 = x (7.3) = minft 0; S t=L g: (7.4) If 0 xL ,then ,rvx+ rxv 0 x+ 1 2 2 x 2 v 00 x=,rK,x+ rx,1 = ,rK: If L x1 ,then ,rvx+rxv 0 x+ 1 2 2 x 2 v 00 x = C,rx , , rxx ,,1 , 1 2 2 x 2 , , 1x , ,2 = Cx , ,r , r , 1 2 2 , , 1 = C , , 1x , r , 1 2 2 2r 2 =0: In other words, v solves the linear complementarity problem: (See Fig. 25.5). [...]... the Laplace transform formula for = minft 0; B t = xg for nondrifted Brownian motion: IEe, For Z1 x x2 dt = e,xp2 ; p exp , t , 2t = 0 t 2t ~ = minft 0; t + B t = xg; 0; x 0: CHAPTER 25 American Options the Laplace transform is 251 Z1 x x , t2 dt p exp , t , 2t 0 t 2t Z1 x x2 + x , 1 2 t dt p exp , t , 2t = 2 0 t 2t Z1 x 2 1 2 t , x x p exp , + 2 =e dt 2t t 2t IEe, ~ = 0 . Value of perpetual American put σ 2 -2r/ C 1 x σ 2 -2r/ x C 2 σ 2 -2r/ x C 3 x Stock price value Figure 25.3: Curves. CHAPTER 25. American Options 255 and. 1 x= p t 2 p 2 exp , z 2 2 dz : Now tM t x; CHAPTER 25. American Options 249 so IP f 2 dtg = @ @t IP f tg dt = @ @t IP fM t xg dt
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