Chapter 24 AnoutsidebarrieroptionBarrier process: dY t Y t = dt+ 1 dB 1 t: Stock process: dS t S t = dt+ 2 dB 1 t+ q 1, 2 2 dB 2 t; where 1 0; 2 0;,11 ,and B 1 and B 2 are independent Brownian motions on some ; F ; P . The option pays off: S T , K + 1 fY T Lg at time T ,where 0 S0 K; 0Y0 L; Y T = max 0tT Y t: Remark 24.1 The option payoff depends on both the Y and S processes. In order to hedge it, we will need the money market and two other assets, which we take to be Y and S . The risk-neutral measure must make the discounted value of every traded asset be a martingale, which in this case means the discounted Y and S processes. We want to find 1 and 2 and define d e B 1 = 1 dt + dB 1 ; d e B 2 = 2 dt + dB 2 ; 239 240 so that dY Y = rdt+ 1 d e B 1 =rdt+ 1 1 dt + 1 dB 1 ; dS S = rdt+ 2 d e B 1 + q 1 , 2 2 d e B 2 = rdt+ 2 1 dt + q 1 , 2 2 2 dt + 2 dB 1 + q 1 , 2 2 dB 2 : We must have = r + 1 1 ; (0.1) = r + 2 1 + q 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 is that (0.1) and (0.2) uniquelydetermine 1 and 2 . This implies the existence and uniquenessof the risk-neutralmeasure. We define Z T = exp n , 1 B 1 T , 2 B 2 T , 1 2 2 1 + 2 2 T o ; f IP A= Z A ZTdIP; 8A 2F: Under f IP , e B 1 and e B 2 are independent Brownian motions (Girsanov’s Theorem). f IP is the unique risk-neutral measure. Remark 24.2 Under both IP and f IP , Y has volatility 1 , S has volatility 2 and dY dS YS = 1 2 dt; i.e., the correlation between dY Y and dS S is . The value of the option at time zero is v 0;S0;Y0 = f IE h e ,rT ST , K + 1 fY T Lg i : We need to work out a density which permits us to compute the right-hand side. CHAPTER 24. Anoutsidebarrieroption 241 Recall that the barrier process is dY Y = rdt+ 1 d e B 1 ; so Y t=Y0 exp n rt + 1 e B 1 t , 1 2 2 1 t o : Set b = r= 1 , 1 =2; b B t= b t + e B 1 t; c M T = max 0tT b B t: Then Y t=Y0 expf 1 b B tg; Y T =Y0 expf 1 c M T g: The joint density of b B T and c M T , appearing in Chapter 20, is f IP f b B T 2 d ^ b; c M T 2 d ^mg = 22 ^m , ^ b T p 2T exp , 2 ^m , ^ b 2 2T + b ^ b , 1 2 b 2 T d ^ bd^m; ^m0; ^ b ^m: The stock process. dS S = rdt+ 2 d e B 1 + q 1 , 2 2 d e B 2 ; so S T = S0 expfrT + 2 e B 1 T , 1 2 2 2 2 T + q 1 , 2 2 e B 2 T , 1 2 1 , 2 2 2 T g = S 0 expfrT , 1 2 2 2 T + 2 e B 1 T + q 1, 2 2 e B 2 Tg From the above paragraph we have e B 1 T = , b T + b BT ; so S T = S0 expfrT + 2 b B T , 1 2 2 2 T , 2 b T + q 1 , 2 2 e B 2 T g 242 24.1 Computing the option value v 0;S0;Y0 = f IE h e ,rT ST , K + 1 fY T Lg i = e ,rT f IE S0 exp r , 1 2 2 2 , 2 b T + 2 b B T + q 1, 2 2 e B 2 T ,K + :1 fY0 exp 1 b M T Lg We know the joint density of b B T ; c M T . The density of e B 2 T is f IP f e B 2 T 2 d ~ bg = 1 p 2T exp , ~ b 2 2T d ~ b; ~ b 2 IR: Furthermore, the pair of random variables b B T ; c M T is independent of e B 2 T because e B 1 and e B 2 are independent under f IP . Therefore, the joint densityof the randomvector e B 2 T ; b B T ; c M T is f IP 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 option value at time zero is v 0;S0;Y0 = e ,rT 1 1 log L Y 0 Z 0 ^m Z ,1 1 Z ,1 S 0 exp r , 1 2 2 2 , 2 b T + 2 ^ b + q 1 , 2 2 ~ b , K + : 1 p 2T exp , ~ b 2 2T : 22 ^m , ^ b T p 2T exp , 2 ^m , ^ b 2 2T + b ^ b , 1 2 b 2 T :d ~ bd ^ bd^m: The answer depends on T; S0 and Y 0 . It also depends on 1 ; 2 ;;r;K and L . It does not depend on ; ; 1 ; nor 2 . The parameter b appearing in the answer is b = r 1 , 1 2 : Remark 24.3 If we had not regarded Y as a traded asset, then we would not have tried to set its mean return equal to r . We would have had only one equation (see Eqs (0.1),(0.2)) = r + 2 1 + q 1 , 2 2 2 (1.1) to determine 1 and 2 . The nonuniqueness of the solution alerts us that some options cannot be hedged. Indeed, any option whose payoff depends on Y cannot be hedged when we are allowed to trade only in the stock. CHAPTER 24. Anoutsidebarrieroption 243 If we have anoption whose payoff depends only on S ,then Y is superfluous. Returning to the original equation for S , dS S = dt+ 2 dB 1 + q 1 , 2 2 dB 2 ; we should set dW = dB 1 + q 1, 2 dB 2 ; so W is a Brownian motion under IP (Levy’s theorem), and dS S = dt+ 2 dW: Now we have only Brownian motion, there will be only one , namely, = , r 2 ; so with d f W = dt+dW; we have dS S = rdt+ 2 d f W; and we are on our way. 24.2 The PDE for the outsidebarrieroption Returning to the case of the option with payoff S T , K + 1 fY T Lg ; we obtain a formula for v t; x; y = e ,rT,t f IE t;x;y h S T , K + 1 fmax tuT Y u Lg ; i by replacing T , S 0 and Y 0 by T , t , x and y respectively in the formula for v 0;S0;Y0 . Now start at time 0 at S 0 and Y 0 . Using the Markov property, we can show that the stochastic process e ,rt vt; S t;Yt is a martingale under f IP . We compute d h e ,rt vt; S t;Yt i = e ,rt ,rv + v t + rS v x + rY v y + 1 2 2 2 S 2 v xx + 1 2 SY v xy + 1 2 2 1 Y 2 v yy dt + 2 Sv x d e B 1 + q 1 , 2 2 Sv x d e B 2 + 1 Yv y d e B 1 244 L v(t, 0, 0) = 0 x y v(t, x, L) = 0, x >= 0 Figure 24.1: Boundary conditions for barrier option. Note that t 2 0;T is fixed. Setting the dt term equal to 0, we obtain the PDE , rv + v t + rxv x + ryv y + 1 2 2 2 x 2 v xx + 1 2 xy v xy + 1 2 2 1 y 2 v yy =0; 0tT; x0; 0yL: The terminal condition is v T ; x; y = x, K + ; x 0; 0 yL; and the boundary conditions are v t; 0; 0 = 0; 0 t T; vt; x; L= 0; 0 t T; x 0: CHAPTER 24. Anoutsidebarrieroption 245 x =0 y =0 ,rv + v t + ryv y + 1 2 2 1 y 2 v yy =0 ,rv + v t + rxv x + 1 2 2 2 x 2 v xx =0 This is the usual Black-Scholes formula in y . This is the usual Black-Scholes formula in x . The boundary conditions are The boundary condition is v t; 0;L= 0;vt; 0; 0 = 0; v t; 0; 0 = e ,rT ,t 0 , K + =0; the terminal condition is the terminal condition is v T; 0;y= 0, K + =0; y0: vT; x; 0 = x , K + ; x 0: On the x =0 boundary, the option value is v t; 0;y= 0; 0 y L: On the y =0 boundary, the barrier is ir- relevant, and the option value is given by the usual Black-Scholes formula for a Eu- ropean call. 24.3 The hedge After setting the dt term to 0, we have the equation d h e ,rt vt; S t;Yt i = e ,rt 2 Sv x d e B 1 + q 1 , 2 2 Sv x d e B 2 + 1 Yv y d e B 1 ; where v x = v x t; S t;Yt , v y = v y t; S t;Yt ,and e B 1 ; e B 2 ;S;Y are functions of t .Note that d h e ,rt S t i = e ,rt ,rSt dt + dS t = e ,rt 2 S t d e B 1 t+ q 1, 2 2 St d e B 2 t : d h e ,rt Y t i = e ,rt ,rY t dt + dY t = e ,rt 1 Y t d e B 1 t: Therefore, d h e ,rt vt; S t;Yt i = v x de ,rt S +v y de ,rt Y : Let 2 t denote the number of shares of stock held at time t ,andlet 1 t denote the number of “shares” of the barrier process Y .Thevalue X t of the portfolio has the differential dX = 2 dS + 1 dY + rX , 2 S , 1 Y dt: 246 This is equivalent to de ,rt X t = 2 tde ,rt S t + 1 tde ,rt Y t: To get X t= vt; S t;Yt for all t ,wemusthave X 0 = v 0;S0;Y0 and 2 t=v x t; S t;Yt; 1 t=v y t; S t;Yt: [...]... has volatility 1, S has volatility 2 and P dY dS = YS 1 2 dt; i.e., the correlation between dY and dS is Y S The value of the option at time zero is i fh v 0; S 0 ; Y 0 = IE e,rT S T , K + 1fY T Lg : We need to work out a density which permits us to compute the right-hand side f IP is the unique CHAPTER 24 Anoutsidebarrieroption 241 Recall that the barrier process is dY = r dt + Y so... for 1 and 2 do not matter What matters is that (0.1) and (0.2) uniquely determine 1 and 2 This implies the existence and uniqueness of the risk-neutral measure We define n o 2 1 2 Z T = exp , 1 B1 T , 2B2 T , 2 1 + 2 T ; Z f IP A = Z T dIP; 8A 2 F : A f e e Under I , B1 and B2 are independent Brownian motions (Girsanov’s Theorem) P risk-neutral measure f Remark 24.2 Under both IP and... pair of random variables B T ; M T is independent of B2 T because B1 and e2 are independent under IP Therefore, the joint density of the random vector B2 T ; B T ; M T f e b c B is f e f e IP fB2 T 2 d~; B T 2 d^; M T 2 dm; g = IP fB2 T 2 d~g:IP fB T 2 d^; M T 2 dmg b b b c ^ b f b b c ^ The option value at time zero is v 0; S 0 ; Y 0 = e,rT 1 1 log Y L Zm Z Z 0 ... 24.1 Computing the option value i fh v 0; S 0 ; Y 0 = IE e,rT S T , K + 1fY T Lg 1 f = e,rT IE S 0 exp r , 2 2 , 2 :1fY 0 exp M T Lg b b 2 T + q b 2 B T + 1 , 2 2 B2 T e ,K 1 b c e We know the joint density of B T ; M T The density of B2 T is b b b ffB2 T 2 d~g = p 1 exp , ~2 d~; ~ 2 IR: e IP b 2T 2T e e b c Furthermore, the pair of random variables B... rt + 1 B1 t , 2 2 t : 1 Set b = r= 1 , 1=2; b e b B t = t + B1 t ; c b M T = 0maxT B t : t Then b Y t = Y 0 expf 1B t g; c Y T = Y 0 expf 1M T g: b c The joint density of B T and M T , appearing in Chapter 20, is f b IP fB T 2 d^; M T 2 dmg b c ^ ^ 2m , ^ 2 b^ 1 b2 ^ m ^ = 2 2p , b exp , ^2T b + b , 2 T db dm; ^ T 2T m 0; ^ m: ^ b ^ The stock process dS = r dt . density which permits us to compute the right-hand side. CHAPTER 24. An outside barrier option 241 Recall that the barrier process is dY Y = rdt+ 1 d e B 1. determine 1 and 2 . The nonuniqueness of the solution alerts us that some options cannot be hedged. Indeed, any option whose payoff depends on Y cannot be