Stochastic Differential Equations

Transcription

1 CHAPTER 1 Stochastic Differential Equations Consider a stochastic process X t satisfying dx t = bt, X t,w t dt + σt, X t,w t dw t. 1.1 Question. 1 Can we obtain the existence and uniqueness theorem for 1.1? What are the properties of the solution? 2 How to solve it? 1.1. Examples and some solution methods Example 1.1. Solving dx t = αx t dw t + σx t dt, 1.2 where the initial condition X is given and α, σ are constant. Rewrite this equation as dx t X t = αdw t + σ dt, and we can get dx u X u = αdw u + σdu= αw t + σt. 1.3 Due to the Itô s formula, df t, X t = f t ft, X t dt + f x ft, X t dx t f 2 x ft, X tdx 2 t 2, 237

2 STOCHASTIC DIFFERENTIAL EQUATIONS we may take ft, x =fx =lnx, then f x = 1 x, f x = 1 x 2. Then d lnx t = 1 X t dx t X 2 t dx t 2 = 1 X t dx t α2 2 dt, i.e., lnx t lnx = dx u X u α2 t Combining 1.3 and 1.4, we get ln Xt X Thus, the solution to 1.2 is given by + α2 t t 2 = dx u X u X t = X exp αw t + = αw t + σt. σ α2 t. 2 Definition 1.2. A stochastic process X t of the form X t = X expαw t + μt is called the geometric Brownian motion. Remark If W t is independent of X, then [ E[X t ]=E X exp αw t + 2 i If σ>α 2 /2, then X t as t P-a.s. ii If σ<α 2 /2, then X t ast P-a.s. ] σ α2 t = E[X ]e σt. 2 iii If σ = α 2 /2, then X t will fluctuate between arbitrary large and arbitrary small values as t P-a.s.

3 1.1. EXAMPLES AND SOME SOLUTION METHODS 239 Example 1.4 Hull-White interest-rate model. Consider dr t =a t b t R t dt + σ t dw t, with R = r, 1.5 where a t, b t, and σ t are deterministic function. Then dr t + b t R t dt = a t dt + σ t dw t, which implies that d R t exp b u du = a t exp b u du dt + σ t exp b u du dw t. Thus, the solution to 1.5 is given by R t = r exp b u du + + σ s exp a s exp s s b u du ds b u du dw s. Example 1.5. Consider the stochastic differential equation dx t = rx t K X t dt + βx t dw t, with X = x>. 1.6 Rewrite the equation as dx t X t + rx t dt = rk dt + βdw t. Taking integration on the both sides, we have dx u X u + r X u du = rkt + βw t. Using a similar argument as in Example 1.1, we get dx u = lnx t lnx d X X u 2 Xu 2 u Xt = ln β 2 Xu 2 du =ln x 2 X 2 u Xt x β2 t.

4 24 1. STOCHASTIC DIFFERENTIAL EQUATIONS Thus, which implies ln Xt x X t exp r + r X u du = βw t + rk 12 β2 t, [ X u du = x exp βw t + rk 12 ] β2 t. Integration with respect to t on the both sides, we obtain x Hence, i.e., [ exp βw s + rk 12 ] β2 s ds = exp r X u du =1+rx X u du = 1 r ln 1+rx = X s exp r exp r = 1 r exp r = 1 r [ exp r s s s X u du ds X u du t X u du s d X u du s= ] X u du 1. [ exp βw s + rk 12 ] β2 s ds, [ exp βw s + rk 12 ] β2 s ds. Taking derivative with respect to t, we see that the solution to 1.6 is given by X t = 1 r rx exp [ βw t + rk 1 β2 t ] 2 1+rx exp [ βw s + rk 1 β2 s ] ds 2 exp [ βw t + rk 1 2 = β2 t ] x 1 + r exp [ βw s + rk 1 β2 s ] ds. 2 Example 1.6. Solving the stochastic differential equation dx t = α t dt + b t X t dw t. Rewrite it as dx t b t X t dw t = α t dt.

5 1.1. EXAMPLES AND SOME SOLUTION METHODS 241 ddddd Example 1.4 dddd. ddddd ρ t such that ρ t dx t b t ρ t X t dw t = α t ρ t dt. 1.7 By integration by parts, dρ t X t =ρ t dx t + X t dρ t + d ρ, X t. Idea: Suppose X t dρ t = b t ρ t X t dw t. dddddd, dddd idea, dddddddddddd. Hence, we want to find ρ such that By Itô formula, Hence, We may take ρ = 1. Thus, b u dw u = ρ t = ρ exp dρ t ρ t = b t dw t. dρ u ρ u = lnρ t lnρ = lnρ t lnρ b u dw u ρ 2 u dρ u 2 b 2 u du. b 2 u du. dρ t X t = ρ t dx t + X t dρ t +dx t dρ t = ρ t dx t b t ρ t X t dw t +α t dt + b t X t dw t b t ρ t X t dw t = ρ t dx t b t ρ t X t dw t b 2 t ρ t X t dt. Plugging this result into 1.7, we have dρ t X t +b 2 t ρ t X t dt = α t ρ t dt.

6 STOCHASTIC DIFFERENTIAL EQUATIONS Using integrating factor again, let G t = exp b 2 u du, we get dg t ρ t X t =G t dρ t X t +ρ t X t dg t = α t ρ t G t dt. Set Thus, F t =ρ t G t = exp b u dw u b 2 u du. i.e., Hence, its solution is given by X t = F 1 t X + F 1 t df t X t =α t F t dt, F t X t F X = α u F u du = X exp b u dw u 1 2 b 2 u du + α u F u du. α u exp b v dw v 1 u 2 Example 1.7. Solving the stochastic differential equation u b 2 v dv du. LQ t + RQ t + 1 C Q t = G t + α W t, 1.8 where W t is the white noise. Introduce the vector then Thus, we may rewrite 1.8 as X t = X1 t X 2 t X 1 t = X 2 t, = Q t Q t, LX 2 t + RX 2 t + 1 C X1 t = G t + α W t. dx t = AX t dt + H t dt + KdW t,

7 1.2. AN EXISTENCE AND UNIQUENESS RESULT 243 where W t is a 1-dimensional Brownian motion, Thus, X t = A = H t = K = X1 t X 2 t 1 1 CL R L G t L α L. d exp [exp AtX t ] = exp AtH t dt + KdW t. The solution is of the form X t = expatx + expat dddddd coordinate dd. exp AsH s ds + KdW s An existence and uniqueness result Theorem 1.8 Existence and uniqueness theorem for stochastic differential equation. Let T> and let b, σ b :[,T] R n R n,σ :[,T] R n R n m be measurable functions satisfying bt, x + σt, x C1 + x for all x R n,t [,T] 1.9

8 STOCHASTIC DIFFERENTIAL EQUATIONS for some constant C, where σ 2 = n m σ ij 2, and i=1 j=1 bt, x bt, y + σt, x σt, y D x y for all x, y R n,t [,T] 1.1 for some constant D. Let Z be a random variable which is independent of F W T, the σ- algebra generated by W s : s T, and E Z 2 <. Then the stochastic differential equation dx t = bt, X t dt + σt, X t dw t, t T,X = Z has a unique t-continuous solution X t with the properties that i X t is adapted to Ft W σz; [ T ] ii E X t 2 dt <. Remark 1.9. The conditions 1.9 and 1.1 are natural in view of the following two simple examples from deterministic differential equations. a Consider dx t dt = X 2 t, X =1, t [, corresponding to bt, x =x 2 not satisfying 1.9. This differential equation has the unique solution X t = 1 1 t, t<1. But no global solution in this case. More generally, 1.9 ensures that the solution X t does not explode, i.e., X t does not tend to in a finite time. b Consider the differential equation dx t dt =3X 2 3 t, X = 1.11

9 1.3. WEAK AND STRONG SOLUTIONS 245 which has more than one solution. For any a>, the function for t a X t = t a 3 for t>a is the solution of In this case, bx =3x 2 3 does not satisfy the condition 1.1 at x = Weak and strong solutions Definition A strong solution of the stochastic differential equation dx t = bt, X t dt + σt, X t dw t 1.12 on a given probability space Ω, F, P and with respect to the fixed Brownian motion motion W and initial value Z, is a stochastic process X with continuous sample paths and with the following properties: i X is adapted to the filtration F t ; ii P[X = Z] =1; [ ] iii P b i s, x s + σijs, 2 X s ds < = 1 for all 1 i d, 1 j r, and t< ; iv the integral version of 1.12: X t = X + bs, X s ds + σs, X s dw s, t<. 2 A weak solution of 1.12 is a triple X, W, Ω, F, P, F t, where i Ω, F, P is a probability space, and F t is a filtration of sub-σ-algebra of F satisfying the usual conditions.

10 STOCHASTIC DIFFERENTIAL EQUATIONS ii X = X t, F t t< is a continuous, adapted R n -valued process. W = W t, F t t< is a standard Brownian motion, and iii,iv of 1 are satisfied. dddd, ddddddddd strong solution ddddd weak solution. Remark There are stochastic differential equations which has no strong solution, but still has a weak solution. Example 1.12 Tanaka equation. Consider the stochastic differential equation dx t = signx t dw t 1.13 where 1, if x signx = 1, if x<. Note that σt, x = signx does not satisfy the Lipschitz condition 1.1. This implies that we cannot apply Theorem Claim that 1.13 has no strong solution. Suppose X is a strong solution of Then X is a Brownian motion by Lévy Theorem. Moreover, dw t = signx t dx t. This implies that the stochastic process W t given by W t = signx u dx u is a Brownian motion with respect to F X t. By the Tanaka equation, W t = X t 2L x t,

11 1.4. FEYNMAN-KAC FORMULA 247 where L x t is the local time of X t. This means that Ft W Ft X. This contradicts to X is a strong solution. 2 Find a weak solution of 1.13 Choose X t to be a Brownian motion B t. Then we can define B t by B t := i.e., d B t = signx t dx t. Thus, signb u db u = signx u dx u, dx t = signx t d B t. Hence, X t is a weak solution Feynman-Kac formula dddddddddddddddddd. Theorem 1.13 Feynman-Kac formula. Consider the stochastic differential equation dx t = βt, X t dt + γt, X t dw t. Let f be a Borel-measurable function. Fix T>and let t [,T] be given. Define the function gt, x =E t,x [fx T ] = E[fX T X t = x]. Assume that gt, x < for all t, x. Then gt, x satisfies the partial differential equation g t t, x+βt, xg x t, x+ 1 2 γ2 t, xg xx t, x = with the terminal condition gt,x=fx for all x.

12 STOCHASTIC DIFFERENTIAL EQUATIONS Remark gt, X t t T is a martingale. Theorem 1.15 Discounted Feynman-Kac formula. Consider the stochastic differential equation ds t = rβt, S t dt + σγt, S t dw t Let f be a Borel-measurable function and let r be constant. Fix T> and let t [,T] be given. Define the function ht, x =E t,x [e rt t fx T ] = E[e rt t fx T X t = x] Assume that ht, x < for all t, x. Then ht, x satisfies the partial differential equation h t t, x+βt, xh x t, x+ 1 2 γ2 t, xh xx t, x =rht, x with the terminal condition ht,x=fx for all x. d stochastic analysis ddd Feynmann-Kac Theorem ddddddd. ddddd dddd, ddddddddd finance ddddd. Example Suppose that S t satisfies the stochastic differential equation ds t = rs t dt + σs t dw t. Let ft, x =E t,x [hs t ]. Then ft, x satisfies the partial differential equation f t t, x+rxf x t, x+ 1 2 σ2 x 2 f xx t, x =. with terminal condition ht,x = fx. This equation can be solved numerically.

13 1.4. FEYNMAN-KAC FORMULA 249 Theorem dimensional Feynman-Kac formula. Let W t = W 1 t,w 2 t be a 2-dimensional Brownian motion. Consider two stochastic differntial equations dx t = β 1 t, X t,y t dt + γ 11 t, X t,y t dw 1 t + γ 12 t, X t,y t dw 2 t, dy t = β 2 t, X t,y t dt + γ 21 t, X t,y t dw t + γ 22 t, X t,y t dw 2 t. Let fx, y be a Borel measurable function, define gt, x, y = E t,x,y [fx T,Y T ], ht, x, y = E t,x,y [e rt t fx T,Y T ]. Then g and h satisfy the partial differential equations g t + β 1 g x + β 2 g y γ γ 2 12g xx +γ 11 γ 21 + γ 12 γ 22 g xy γ γ 2 22g yy =, h t + β 1 h x + β 2 h y γ γ 2 12h xx +γ 11 γ 21 + γ 12 γ 22 h xy γ γ 2 22h yy = rh with terminal conditions gt, x, y =ht, x, y =fx, y, for all x, y.