Chapter 24 An outsde barrer opton Barrer process: dy (t) Y (t) = dt+ 1 db 1 (t): Stock process: (t) S(t) = dt+ 2 db 1 (t) + 1 ; 2 2 db 2 (t) where 1 > 0 2 > 0 ;1 <<1, and B 1 and B 2 are ndependent Brownan motons on some ( F P). The opton pays off: (S(T ) ; K) + 1 fy (T )<Lg at tme T, where 0 <S(0) <K 0 <Y(0) <L Y (T ) = max Y (t): 0tT Remark 24.1 The opton payoff depends on both the Y and S processes. In order to hedge t, we wll need the money market and two other assets, whch we take to be Y and S. The rsk-neutral measure must make the dscounted value of every traded asset be a martngale, whch n ths case means the dscounted Y and S processes. We want to fnd 1 and 2 and defne d e B 1 = 1 dt + db 1 d e B 2 = 2 dt + db 2 239
240 so that dy Y = rdt+ 1d e B 1 = rdt+ 1 1 dt + 1 db 1 S = rdt+ 2 db e 1 + 1 ; 2 2 db e 2 = rdt+ 2 1 dt + + 2 db 1 + 1 ; 2 2 2 dt 1 ; 2 2 db 2 : We must have = r + 1 1 (0.1) = r + 2 1 + 1 ; 2 2 2 : (0.2) We solve to get 1 = ; r 1 2 = ; r ; 2 1 p 1 ; 2 2 : We shall see that the formulas for 1 and 2 do not matter. What matters s that (0.1) and (0.2) unuely determne 1 and 2. Ths mples the exstence and unueness of the rsk-neutral measure. We defne o Z(T )=exp n; 1 B 1 (T ) ; 2 B 2 (T ) ; 12 (2 1 + 2 2 )T Z fip (A) = Z(T ) dip 8A 2F: A Under f IP, e B 1 and e B 2 are ndependent Brownan motons (Grsanov s Theorem). f IP s the unue rsk-neutral measure. Remark 24.2 Under both IP and f IP, Y has volatlty 1, S has volatlty 2 and dy YS = 1 2 dt.e., the correlaton between dy Y and S s. The value of the opton at tme zero s v(0 S(0) Y(0)) = IE f h e ;rt (S(T ) ; K) + 1 fy (T )<Lg : We need to work out a densty whch permts us to compute the rght-hand sde.
CHAPTER 24. An outsde barrer opton 241 Recall that the barrer process s so dy Y = rdt+ 1 d e B 1 n Y (t) =Y (0) exp rt + e o 1 B 1 (t) ; 1 2 2 1t : Set b = r= 1 ; 1 =2 bb(t) = b t + e B 1 (t) cm (T )= max 0tT bb(t): Then Y (t) =Y (0) expf 1 b B(t)g Y (T )=Y (0) expf 1 c M (T )g: The jont densty of b B(T ) and c M (T ), appearng n Chapter 20, s fip fb(t b ) 2 d^b M c (T ) 2 d ^mg ( 2 2(2 ^m ; ^b) = T p 2T exp (2 ^m ; ^b) ; 2T ^m >0 ^b< ^m: + b ^b ; 1 2 b 2 T ) d^b d^m The stock process. so S = rdt+ 2dB e 1 + 1 ; 2 2 db e 2 S(T )=S(0) expfrt + e 2 B 1 (T ) ; 1 2 2 2 2T + 1 ; 2 e 2 B 2 (T ) ; 1 2 (1 ; 2 ) 2 2T g = S(0) expfrt ; 1 2 2 2T + e 2 B 1 (T )+ 1 ; 2 e 2 B 2 (T )g From the above paragraph we have eb 1 (T )=; b T + b B(T ) so S(T )=S(0) expfrt + 2 b B(T ) ; 1 2 2 2T ; 2 b T + 1 ; 2 2 e B 2 (T )g
242 24.1 Computng the opton value h v(0 S(0) Y(0)) = IE f e ;rt (S(T ) ; K) + 1 fy (T )<Lg = e ;rt IE f S(0) exp (r ; 1 2 2 2 ; b 2)T + b 2 B(T )+ :1 fy (0) exp[ 1M(T b )]<Lg + 1 ; 2 e 2 B 2 (T ) ; K We know the jont densty of ( B(T b ) M c (T )). The densty of B e 2 (T ) s ( fip fb e 2 (T ) 2 d ~ bg = p2t 1 ~ ) exp b 2 ; d 2T ~ b ~ b 2 IR: Furthermore, the par of random varables ( b B(T ) c M (T )) s ndependent of e B 2 (T ) because e B 1 and eb 2 are ndependent under f IP. Therefore, the jont densty of the random vector ( e B 2 (T ) b B(T ) c M (T )) s fip f e B 2 (T ) 2 d ~ b b B(T ) 2 d^b c M (T ) 2 d ^m g = f IP f e B 2 (T ) 2 d ~ bg: f IP f b B(T ) 2 d^b c M (T ) 2 d ^mg The opton value at tme zero s v(0 S(0) Y(0)) = e ;rt 1 log L 1 Y (0) Z Z^m Z1 S(0) exp 0 ;1 ;1 ( 1 : p2t ~ ) exp b 2 ; 2T ( ) 2 2(2 ^m ; ^b) : T p 2T exp (2 ^m ; ^b) ; + ^b 2T b ; 1 b2 T 2 :d ~ bd^b d^m: (r ; 1 2 2 2 ; b + 2)T + 2^b + 1 ; 2 ~ 2 b ; K The answer depends on T S(0) and Y (0). It also depends on 1 2 r K and L. It does not depend on 1 nor 2. The parameter b appearng n the answer s b = r ; 1 1 2 : Remark 24.3 If we had not regarded Y as a traded asset, then we would not have tred to set ts mean return eual to r. We would have had only one euaton (see Es (0.1),(0.2)) = r + 2 1 + 1 ; 2 2 2 (1.1) to determne 1 and 2. The nonunueness of the soluton alerts us that some optons cannot be hedged. Indeed, any opton whose payoff depends on Y cannot be hedged when we are allowed to trade only n the stock.
CHAPTER 24. An outsde barrer opton 243 If we have an opton whose payoff depends only on S, then Y s superfluous. Returnng to the orgnal euaton for S, S = dt+ 2 db 1 + 1 ; 2 2 db 2 we should set dw = db 1 + 1 ; 2 db 2 so W s a Brownan moton under IP (Levy s theorem), and S = dt+ 2dW: Now we have only Brownan moton, there wll be only one, namely, so wth d f W = dt+ dw we have = ; r 2 and we are on our way. S = rdt+ 2 d f W 24.2 The PDE for the outsde barrer opton Returnng to the case of the opton wth payoff (S(T ) ; K) + 1 fy (T )<Lg we obtan a formula for v(t x y) =e ;r(t ;t)f h IE t x y (S(T ) ; K) + 1 fmaxtut Y (u) <Lg by replacng T, S(0) and Y (0) by T ; t, x and y respectvely n the formula for v(0 S(0) Y(0)). Now start at tme 0 at S(0) and Y (0). Usng the Markov property, we can show that the stochastc process e ;rt v(t S(t) Y(t)) s a martngale under IP f. We compute h d e ;rt v(t S(t) Y(t)) = e ;rt ;rv + v t + rsv x + ry v y + 1 2 2 2S 2 v xx + 1 2 SY v xy + 1 2 2 1Y 2 v yy + 2 Sv x db e 1 + 1 ; 2 2 Sv x db e 2 + 1 Yv y db e 1 dt
244 y L v(t, x, L) = 0, x >= 0 v(t, 0, 0) = 0 x Fgure 24.1: Boundary condtons for barrer opton. Note that t 2 [0 T] s fxed. Settng the dt term eual to 0, we obtan the PDE ; rv + v t + rxv x + ryv y + 1 2 2 2 x2 v xx + 1 2 xyv xy + 1 2 2 1y 2 v yy =0 0 t<t x 0 0 y L: The termnal condton s v(t x y)=(x ; K) + x 0 0 y<l and the boundary condtons are v(t 0 0) = 0 0 t T v(t x L) =0 0 t T x 0:
CHAPTER 24. An outsde barrer opton 245 x =0 y =0 ;rv + v t + ryv y + 1 2 2 1y 2 v yy =0 ;rv + v t + rxv x + 1 2 2 2x 2 v xx =0 Ths s the usual Black-Scholes formula n y. Ths s the usual Black-Scholes formula n x. The boundary condtons are The boundary condton s v(t 0 L)=0 v(t 0 0) = 0 v(t 0 0) = e ;r(t ;t) (0 ; K) + =0 the termnal condton s the termnal condton s v(t 0 y)=(0; K) + =0 y 0: v(t x 0) = (x ; K) + x 0: On the x =0boundary, the opton value s v(t 0 y)=0 0 y L: On the y =0boundary, the barrer s rrelevant, and the opton value s gven by the usual Black-Scholes formula for a European call. 24.3 The hedge After settng the dt term to 0, we have the euaton h d e ;rt v(t S(t) Y(t)) = e ;rt 2 Sv x db e 1 + 1 ; 2 2 Sv x db e 2 + 1 Yv y db e 1 where v x = v x (t S(t) Y(t)), v y = v y (t S(t) Y(t)), and B e 1 B e 2 S Y are functons of t. Note that h d e ;rt S(t) = e ;rt [;rs(t) dt + (t)] = e ;rt 2 S(t) db e 1 (t) + 1 ; 2 2 S(t) db e 2 (t) : h d e ;rt Y (t) = e ;rt [;ry (t) dt + dy (t)] = e ;rt 1 Y (t) d e B 1 (t): Therefore, h d e ;rt v(t S(t) Y(t)) = v x d[e ;rt S]+v y d[e ;rt Y ]: Let 2 (t) denote the number of shares of stock held at tme t, and let 1 (t) denote the number of shares of the barrer process Y. The value X(t) of the portfolo has the dfferental dx = 2 + 1 dy + r[x ; 2 S ; 1 Y ] dt:
246 Ths s euvalent to d[e ;rt X(t)] = 2 (t)d[e ;rt S(t)] + 1 (t)d[e ;rt Y (t)]: To get X(t) =v(t S(t) Y(t)) for all t, we must have X(0) = v(0 S(0) Y(0)) and 2 (t) =v x (t S(t) Y(t)) 1 (t) =v y (t S(t) Y(t)):