Chapter 23 Recognizinga Brownian Motion Theorem 0.62 (Levy) Let B t; 0 t T; be a process on ; F ; P , adapted to a filtration F t; 0 t T , such that: 1. the paths of B t are continuous, 2. B is a martingale, 3. hB it=t; 0 t T , (i.e., informally dB t dB t=dt ). Then B is a Brownian motion. Proof: (Idea) Let 0 stT be given. We need to show that B t , B s is normal, with mean zero and variance t , s ,and B t , B s is independent of F s . We shall show that the conditional moment generating function of B t , B s is IE e uB t,B s F s = e 1 2 u 2 t,s : Since the moment generating function characterizes the distribution, this shows that B t , B s is normal with mean 0 and variance t , s , and conditioning on F s does not affect this, i.e., B t , B s is independent of F s . We compute (this uses the continuity condition (1) of the theorem) de uB t = ue uB t dB t+ 1 2 u 2 e uB t dB t dB t; so e uB t = e uB s + Z t s ue uB v dB v + 1 2 u 2 Z t s e uB v dv : |z uses cond. 3 233 234 Now R t 0 ue uB v dB v is a martingale (by condition 2), and so IE Z t s ue uB v dB v F s = , Z s 0 ue uB v dB v + IE Z t 0 ue uB v dB v F s =0: It follows that IE e uB t F s = e uB s + 1 2 u 2 Z t s IE e uB v F s dv : We define 'v =IE e uB v F s ; so that 's=e uB s and 't=e uB s + 1 2 u 2 Z t s 'v dv ; ' 0 t= 1 2 u 2 't; 't=ke 1 2 u 2 t : Plugging in s ,weget e uB s = ke 1 2 u 2 s =k = e uB s, 1 2 u 2 s : Therefore, IE e uB t F s = 't=e uB s+ 1 2 u 2 t,s ; IE e uB t,B s F s = e 1 2 u 2 t,s : CHAPTER 23. Recognizinga Brownian Motion 235 23.1 Identifying volatility and correlation Let B 1 and B 2 be independent Brownian motions and dS 1 S 1 = rdt+ 11 dB 1 + 12 dB 2 ; dS 2 S 2 = rdt+ 21 dB 1 + 22 dB 2 ; Define 1 = q 2 11 + 2 12 ; 2 = q 2 21 + 2 22 ; = 11 21 + 12 22 1 2 : Define processes W 1 and W 2 by dW 1 = 11 dB 1 + 12 dB 2 1 dW 2 = 21 dB 1 + 22 dB 2 2 : Then W 1 and W 2 have continuous paths, are martingales, and dW 1 dW 1 = 1 2 1 11 dB 1 + 12 dB 2 2 = 1 2 1 2 11 dB 1 dB 1 + 2 12 dB 2 dB 2 = dt; and similarly dW 2 dW 2 = dt: Therefore, W 1 and W 2 are Brownian motions. The stock prices have the representation dS 1 S 1 = rdt+ 1 dW 1 ; dS 2 S 2 = rdt+ 2 dW 2 : The Brownian motions W 1 and W 2 are correlated. Indeed, dW 1 dW 2 = 1 1 2 11 dB 1 + 12 dB 2 21 dB 1 + 22 dB 2 = 1 1 2 11 21 + 12 22 dt = dt: 236 23.2 Reversing the process Suppose we are given that dS 1 S 1 = rdt+ 1 dW 1 ; dS 2 S 2 = rdt+ 2 dW 2 ; where W 1 and W 2 are Brownian motions with correlation coefficient .Wewanttofind = " 11 12 21 22 so that 0 = " 11 12 21 22 " 11 21 12 22 = " 2 11 + 2 12 11 21 + 12 22 11 21 + 12 22 2 21 + 2 22 = " 2 1 1 2 1 2 2 2 A simple (but not unique) solution is (see Chapter 19) 11 = 1 ; 12 =0; 21 = 2 ; 22 = q 1 , 2 2 : This corresponds to 1 dW 1 = 1 dB 1 =dB 1 = dW 1 ; 2 dW 2 = 2 dB 1 + q 1 , 2 2 dB 2 = dB 2 = dW 2 , dW 1 p 1, 2 ; 6=1 If = 1 , then there is no B 2 and dW 2 = dB 1 =dW 1 : Continuing in the case 6= 1 ,wehave dB 1 dB 1 = dW 1 dW 1 = dt; dB 2 dB 2 = 1 1 , 2 dW 2 dW 2 , 2dW 1 dW 2 + 2 dW 2 dW 2 = 1 1 , 2 dt , 2 2 dt + 2 dt = dt; CHAPTER 23. Recognizinga Brownian Motion 237 so both B 1 and B 2 are Brownian motions. Furthermore, dB 1 dB 2 = 1 p 1 , 2 dW 1 dW 2 , dW 1 dW 1 = 1 p 1 , 2 dt,dt=0: We can now apply an Extension of Levy’s Theorem that says that Brownian motions with zero cross-variation are independent, to conclude that B 1 ;B 2 are independent Brownians. 238 . We can now apply an Extension of Levy’s Theorem that says that Brownian motions with zero cross-variation are independent, to conclude that B 1 ;B 2 are. the moment generating function characterizes the distribution, this shows that B t , B s is normal with mean 0 and variance t , s , and conditioning