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Chapter 19 A two-dimensional market model Let B t=B 1 t;B 2 t; 0 t T; be a two-dimensional Brownian motion on ; F ; P .Let F t; 0 t T; be the filtration generated by B . In what follows, all processes can depend on t and ! , but are adapted to F t; 0 t T .To simplify notation, we omit the arguments whenever there is no ambiguity. Stocks: dS 1 = S 1 1 dt + 1 dB 1 ; dS 2 = S 2 2 dt + 2 dB 1 + q 1 , 2 2 dB 2 : We assume 1 0; 2 0;,11: Note that dS 1 dS 2 = S 2 1 2 1 dB 1 dB 1 = 2 1 S 2 1 dt; dS 2 dS 2 = S 2 2 2 2 2 dB 1 dB 1 + S 2 2 1 , 2 2 2 dB 2 dB 2 = 2 2 S 2 2 dt; dS 1 dS 2 = S 1 1 S 2 2 dB 1 dB 1 = 1 2 S 1 S 2 dt: In other words, dS 1 S 1 has instantaneous variance 2 1 , dS 2 S 2 has instantaneous variance 2 2 , dS 1 S 1 and dS 2 S 2 have instantaneous covariance 1 2 . Accumulation factor: t = exp Z t 0 rdu : The market price of risk equations are 1 1 = 1 , r 2 1 + q 1 , 2 2 2 = 2 , r (MPR) 203 204 The solution to these equations is 1 = 1 , r 1 ; 2 = 1 2 , r , 2 1 , r 1 2 p 1 , 2 ; provided ,1 1 . Suppose ,1 1 . Then (MPR) has a unique solution 1 ; 2 ;wedefine Z t = exp , Z t 0 1 dB 1 , Z t 0 2 dB 2 , 1 2 Z t 0 2 1 + 2 2 du ; f IP A= Z A ZTdIP; 8A 2F: f IP is the unique risk-neutral measure. Define e B 1 t= Z t 0 1 du + B 1 t; e B 2 t= Z t 0 2 du + B 2 t: Then dS 1 = S 1 h rdt+ 1 d e B 1 i ; dS 2 = S 2 rdt+ 2 d e B 1 + q 1 , 2 2 d e B 2 : We have changed the mean rates of return of the stock prices, but not the variances and covariances. 19.1 Hedging when ,1 1 dX = 1 dS 1 + 2 dS 2 + rX , 1 S 1 , 2 S 2 dt d X = 1 dX , rX dt = 1 1 dS 1 , rS 1 dt+ 1 2 dS 2 , rS 2 dt = 1 1 S 1 1 d e B 1 + 1 2 S 2 2 d e B 1 + q 1 , 2 2 d e B 2 : Let V be F T -measurable. Define the f IP -martingale Y t= f IE V T Ft ; 0tT: CHAPTER 19. A two-dimensional market model 205 The Martingale Representation Corollary implies Y t=Y0 + Z t 0 1 d e B 1 + Z t 0 2 d e B 2 : We have d X = 1 1 S 1 1 + 1 2 S 2 2 d e B 1 + 1 2 S 2 q 1 , 2 2 d e B 2 ; dY = 1 d e B 1 + 2 d e B 2 : We solve the equations 1 1 S 1 1 + 1 2 S 2 2 = 1 1 2 S 2 q 1 , 2 2 = 2 for the hedging portfolio 1 ; 2 . With this choice of 1 ; 2 and setting X 0 = Y 0 = f IE V T ; we have X t=Yt; 0t T; and in particular, X T = V: Every F T -measurable random variable can be hedged; the market is complete. 19.2 Hedging when =1 The case = ,1 is analogous. Assume that =1 .Then dS 1 = S 1 1 dt + 1 dB 1 dS 2 = S 2 2 dt + 2 dB 1 The stocks are perfectly correlated. The market price of risk equations are 1 1 = 1 , r 2 1 = 2 , r (MPR) The process 2 is free. There are two cases: 206 Case I: 1 ,r 1 6= 2 ,r 2 : There is no solution to (MPR), and consequently, there is no risk-neutral measure. This market admits arbitrage. Indeed d X = 1 1 dS 1 , rS 1 dt+ 1 2 dS 2 , rS 2 dt = 1 1 S 1 1 , r dt + 1 dB 1 + 1 2 S 2 2 , r dt + 2 dB 1 Suppose 1 ,r 1 2 ,r 2 : Set 1 = 1 1 S 1 ; 2 = , 1 2 S 2 : Then d X = 1 1 , r 1 dt + dB 1 , 1 2 , r 2 dt + dB 1 = 1 1 , r 1 , 2 , r 2 | z Positive dt Case II: 1 ,r 1 = 2 ,r 2 : The market price of risk equations 1 1 = 1 , r 2 1 = 2 , r have the solution 1 = 1 , r 1 = 2 , r 2 ; 2 is free; there are infinitely many risk-neutral measures. Let f IP be one of them. Hedging: d X = 1 1 S 1 1 , r dt + 1 dB 1 + 1 2 S 2 2 , r dt + 2 dB 1 = 1 1 S 1 1 1 dt + dB 1 + 1 2 S 2 2 1 dt + dB 1 = 1 1 S 1 1 + 1 2 S 2 2 d e B 1 : Notice that e B 2 does not appear. Let V be an F T -measurable random variable. If V depends on B 2 , then it can probably not be hedged. For example, if V = hS 1 T ;S 2 T; and 1 or 2 depend on B 2 , then there is trouble. CHAPTER 19. A two-dimensional market model 207 More precisely, we define the f IP -martingale Y t= f IE V T Ft ; 0tT: We can write Y t=Y0 + Z t 0 1 d e B 1 + Z t 0 2 d e B 2 ; so dY = 1 d e B 1 + 2 d e B 2 : To get d X to match dY ,wemusthave 2 =0: 208 [...]... choice of 1 ; 2 and setting f V X 0 = Y 0 = IE T ; we have X t = Y t ; 0 t T; and in particular, X T = V: Every F T -measurable random variable can be hedged; the market is complete 19.2 Hedging when The case =1 = ,1 is analogous Assume that = 1 Then dS1 = S1 1 dt + dS2 = S2 2 dt + 1 dB1 2 dB1 The stocks are perfectly correlated The market price of risk equations are 1 1 = 1 ,... be an F T -measurable random variable If V depends on B2 , then it can probably not be hedged For example, if V = h S1 T ; S2 T ; and 1 or 2 depend on B2 , then there is trouble CHAPTER 19 A two-dimensional market model 207 f More precisely, we define the I -martingale P f V Y t = IE T F t ; We can write Y t = Y 0 + Zt 0 e 1 dB1 + so dY = e e 1 dB1 + 2 dB2 : To get d X to match... 2 2 2 1 f Let V be F T -measurable Define the I -martingale P f V Y t = IE T F t ; 0 t T: 2 2 e2 : dB CHAPTER 19 A two-dimensional market model 205 The Martingale Representation Corollary implies Y t = Y 0 + We have Zt 0 e 1 dB1 + Zt 0 e 2 dB2: X 1 1 S dB = 1 S1 1 + 2 2 2 e1 d q e + 1 2S2 1 , 2 2 dB2 ; e e dY = 1 dB1 + 2 dB2 : We solve the equations 1 S + 1 S 1 1 1 2 2...2 dB2 , 1 2 Zt 8A 2 F : Z T dIP; A Zt 0 2 2 1 + 2 du ; f IP is the unique risk-neutral measure Define e B1 t = e B2 t = Then Zt Z0 0 t 1 du + B1 t ; 2 du + B2 t : h dS1 = S1 r dt + dS2 = S2 r dt + e 1 dB1 i ; q e 2 dB1 + 1 , 2 2 dB2 e : We have changed the mean rates of return of the stock prices, but not the variances and covariances 19.1 Hedging when ,1 1 dX X =... Positive Case II: 1 ,r 1 = ,r : The market price of risk equations 2 2 1 1 2 1 have the solution = 1 , r = 2 , r 1 = 1 , r = 2 , r ; 1 2 f 2 is free; there are infinitely many risk-neutral measures Let IP be one of them Hedging: X 1 d = 1S1 1 , r dt + 1 dB1 + 1 2S2 2 , r dt + 2 dB1 = 1 1S1 1 1 dt + dB1 + 1 2 S2 2 1 dt + dB1 1 1 S dB : = 1 S1 1 + 2 2 2 e1 e Notice that B2 does not appear... stocks are perfectly correlated The market price of risk equations are 1 1 = 1 , r 2 1 = 2 , r The process 2 is free There are two cases: (MPR) 206 Case I: 1 ,r 6= 2 ,r : There is no solution to (MPR), and consequently, there is no risk-neutral 1 2 measure This market admits arbitrage Indeed X 1 = 1 dS1 , rS1 dt + 1 2 dS2 , rS2 dt d = 1 1S1 1 , r dt + Suppose 1 ,r 1 1 dB1 + 1 2S2 2 , r ... market model 207 f More precisely, we define the I -martingale P f V Y t = IE T F t ; We can write Y t = Y 0 + Zt 0 e 1 dB1 + so dY = e e 1 dB1 + 2 dB2 : To get d X to match dY , we must have 2 = 0: 0 t T: Zt 0 e 2 dB2 ; 208 . 1 has instantaneous variance 2 1 , dS 2 S 2 has instantaneous variance 2 2 , dS 1 S 1 and dS 2 S 2 have instantaneous covariance 1 2 . Accumulation. -measurable. Define the f IP -martingale Y t= f IE V T Ft ; 0tT: CHAPTER 19. A two-dimensional market model 205 The Martingale Representation
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