Martingale representation theorem

Transcription

1 Martingale representation theorem Ω = C[, T ], F T = smallest σ-field with respect to which B s are all measurable, s T, P the Wiener measure, B t = Brownian motion M t square integrable martingale with respect to F t Then there exists σ(t, ω) which is 1 progressively measurable 2 square integrable 3 B([, )) F mble such that M t = M + σ(s)db s () Stochastic Calculus March 3, 27 1 / 1

2 Lemma A = set of all linear combinations of random variables of the form er T hdb 1 2 A is dense in L 2 (Ω, F T, P) Proof R T h2 dt, h L 2 ([, T ]) Suppose g L 2 (Ω, F T, P) is orthogonal to all such functions We want to show that g = By an easy choice of simple functions h we find that for any λ 1,..., λ n R and t 1,..., t n [, T ], E P [ge λ 1B t1 + +λ nb t n ] = lhs real analytic in λ and hence has an analytic extension to λ C n () Stochastic Calculus March 3, 27 2 / 1

3 Since E P [ge λ 1B t1 + +λ nb t n ] is analytic and vanishes on the real axis, it is zero everywhere. In particular E P [ge i(y 1B t1 + +y nb t n ) ] = Suppose φ C (Rn ) ˆφ(y) = (2π) R n/2 φ(x)e ix y dx n Fourier inversion: φ(x) = (2π) R n/2 ˆφ(y)e ix y dy n E P [gφ(b t1,..., B tn )] = (2π) n/2 R n ˆφ(y)E P [e iy 1B t1 + +y nb t n ) ]dy = Hence g is orthogonal to fns of form φ(b t1,..., B tn ) where φ C (Rn ) Dense in L 2 (Ω, F T, P) g = () Stochastic Calculus March 3, 27 3 / 1

4 Lemma F L 2 (Ω, F T, P) There exists a unique f (t, ω) which is 1 progressively measurable 2 square integrable 3 B([, )) F measurable such that T F(ω) = E[F] + fdb. Proof of Uniqueness suppose T T F = E[F] + f 1 db = E[F] + f 2 db T (f 2 f 1 )db = T E[(f 2 f 1 ) 2 ]dt = f 2 = f 1 () Stochastic Calculus March 3, 27 4 / 1

5 Proof of existence First we prove it if F is of the form F = e R T Defining F t = e R t hdb 1 R t 2 h2ds gives hdb 1 R T 2 h2 ds so Plugging in t = T gives the result. df = hfdb, F = 1, F t = 1 + F s hdb. If F is a linear combination of such functions the result follows by linearity () Stochastic Calculus March 3, 27 5 / 1

6 Proof of existence for F L 2 (Ω, F T, P) F n L 2 (Ω, F T, P) with F n F and T F n = E[F n ] + f n db. E[F n ] E[F], so wlog E[F n ] = E[F] = E[(F n F m ) 2 ] = T f n Cauchy in L 2 ([, T ] Ω, dx dp). Let f be the limit. Taking limits we have E[(f n f m ) 2 ]dt as n, m T F = E[F] + fdb. () Stochastic Calculus March 3, 27 6 / 1

7 Proof of the martingale representation theorem By previous lemma, for each t we have σ t (s, ω) such that M t = E[M t ] + σ t (s)db s Let t 2 > t 1 M t1 = E[M t2 F t1 ] 1 σ t2 (s)db s = 1 σ t1 (s)db s Uniqueness σ t1 = σ t2 () Stochastic Calculus March 3, 27 7 / 1

8 Quadratic variation of X t = σ(s)db s e λ R t R λ2 σ(s)dbs t 2 σ2 (s)ds =martingale E[e λ R t i+1 t σ(s)db i s λ2 R ti+1 2 t σ 2 (s)ds i Fti ] = i+1 i+1 E[Z (t i, t i+1 ) F ti ] =, Z (t i, t i+1 ) = ( σ(s)db s ) 2 σ 2 (s)ds t i t i i+1 E[{Z (t i, t i+1 )} 2 4( σ 2 (s)ds) 2 F ti ] = t i E[( i Z (t i, t i+1 )) 2 ] 4E[( i i+1 ( σ 2 (s)ds) 2 ] t i X t, X t = σ 2 (s, ω)ds () Stochastic Calculus March 3, 27 8 / 1

9 Levy s Theorem Let X t be a process adapted to a filtration F t which 1 has continuous sample paths 2 is a martingale 3 has quadratic variation t Then X t is a Brownian motion () Stochastic Calculus March 3, 27 9 / 1

10 Proof of Levy s theorem Enough to show that for each λ, E[e iλ(x t X s) F s ] = e 1 2 λ2 (t s) Call M t = e iλx t λ2t, t j = s + j 2 n (t s) M t M s = M tj M tj 1 2 n j=1 = iλm tj 1 (X tj X tj 1 ) 1 2 λ2 M ξj [(X tj X tj 1 ) 2 (t j t j 1 )] 2 n j=1 E[M tj 1 (X tj X tj 1 ) F s ] = E[E[M tj 1 (X tj X tj 1 ) F tj 1 ] F s ] = E[M tj 1 E[(X tj X tj 1 ) F tj 1 ] F s ] = () Stochastic Calculus March 3, 27 1 / 1

11 Proof of Levy s theorem Fix m. Let ξ m = max{ i 2 m : i 2 m ξ} lim n M ξ m j [(X tj X tj 1 ) 2 (t j t j 1 )] = 2 n j=1 So we only have to show Would follow from lim n lim n [M ξ m j M ξj ](X tj X tj 1 ) 2 = 2 n j=1 (X ξ m j X ξj )(X tj X tj 1 ) 2 = 2 n j=1 Left hand side = t lim n max 1 j 2 n X ξ m j X ξj = a.s. () Stochastic Calculus March 3, / 1

12 Note the same proof gives Itô formula for semimartingales Let Mt 1,..., Md t be martingales with respect to a filtration F t, t, A 1 t,..., Ad t adapted processes of bounded variation, X t = x + A t + M t where x F, and f (t, x) C 1,2. Then f d f (t, X t ) = f (, X ) + t (s, X f s)ds + (s, X s )da i s + dms i x i + d i,j=1 i=1 2 f x i x j (s, X s )d M i, M j s Multidimensional Levy s theorem Let Mt 1,..., Md t F t, t, with Then M 1 t,..., Md t be continuous martingales with respect to a filtration M i, M j t = δ ij t is a Brownian motion in R d () Stochastic Calculus March 3, / 1

13 Time change Let Y t be a stochastic integral Y t = gds + fdb where f and g are adapted square integrable processes Let c t > be another adapted process and define β t = c s ds. Then β t is adapted and strictly increasing. We call α t its inverse. We can check that f t Y αt = c ds + g d B c for some Brownian motion B. In particular, if we are given a stochastic integral fdb we can choose f 2 = c as the rate of our time change and the resulting Y αt is a Brownian motion () Stochastic Calculus March 3, / 1

14 Time change Theorem Let B t be Brownian motion and F t its canonical σ-field Suppose that M t is a square integrable martingale with respect to F t Let M t = M + f (s)db s be its representation in terms of Brownian motion. Suppose that f 2 > (i.e. its quadratic variation is strictly increasing) Let c = f 2 and define α t as above Then M αt is a Brownian motion () Stochastic Calculus March 3, / 1

15 Example. Stochastic growth model dx = rxdt + σ XdB Solution is X t = rτ t + B(τ t ) where τ t = X t Because if dy = rdt + σdb then by time change X t = Y τt satisfies dx = rτ dt + σ τ db () Stochastic Calculus March 3, / 1

16 Here s a funny trick to solve dx = σ XdB Let φ = φ(t) be deterministic and look at X(t)φ(T t) Y (t) = e So if φ = σ2 2 φ2, φ() = λ then Called Duality Great if it works. dy = ( φ + σ2 2 φ2 )Ydt φy σ XdB E[e λx(t ) ] = e X()φ(T ) () Stochastic Calculus March 3, / 1

17 Example. Cox-Ingersol-Ross model The interest rate r(t) is assumed to satisfy the equation dr(t) = (α βr(t))dt + σ r(t)db(t). Note that the Lipschitz condition is not satisfied, but existence/uniqueness holds by the stronger theorem we did not prove If d = 4α/σ 2 is a positive integer, then we can find a solution as follows: Let B 1 (t), B 2 (t),..., B d (t) be d independent Brownian motions and let X 1 (t), X 2 (t),..., X d (t) be the solutions of the Langevin equations dx i = αx i dt + σdb i, i = 1,..., d. In other words, X 1 (t), X 2 (t),..., X d (t) are d independent Ornstein-Uhlenbeck processes () Stochastic Calculus March 3, / 1

18 Example. Cox-Ingersol-Ross model r(t) = X 2 1 (t) + + X 2 d (t), dr(t) = (α βr(t))dt + σ r(t) d B(t) = X 1(t), X 2 (t),..., X d (t) indep O-U. d i=1 { d i=1 X i (t) r(t) db i (t) B(t) is a martingale with quadratic variation d Bd B = d i,j=1 B(t) Brownian motion X i (t)x j (t) db i (t)db j (t) = r(t) } X i (t) db i (t). r(t) d i=1 X 2 i (t) r(t) dr(t) = (α βr(t))dt + σ r(t)d B(t) dt = dt () Stochastic Calculus March 3, / 1

19 The type of solutions dealt with in the existence/uniqueness theorem are strong solutions This means that you are given the Brownian motion B(t) and asked to come up with a solution X(t) of the stochastic differential equation dx = bdt + σdb. A weak solution is when you just find some Brownian motion B(t) for which you can solve the equation dx = bdt + σd B. Example. Cox-Ingersol-Ross. If d = 1 then B(t) = B 1 (t) so we have a strong solution But if d = 2, 3,... then all we have is a weak solution, because given a Brownian motion B(t) it is not at all clear how to find B 1 (t),..., B d (t) for which db(t) = d X i (t) i=1 db i (t). r(t) () Stochastic Calculus March 3, / 1

20 Diffusion process as probability measure on C([, )) Brownian motion=collection of rv s B t, t on (Ω, F, P) or Brownian motion=probability measure P on C([, )) If A is a (measurable) subset of continuous functions, then P(A) is just the probability that a Brownian path falls in that subset Same for any diffusion.if X(t) is the solution of the stochastic differential equation dx(t) = b(t, X(t))dt + σ(t, X(t))dB(t), X() = x then we can let Px a,b denote the probability measure on the space of continuous functions with P a,b x (A) = Prob(X( ) A) a = σσ T Question:What is the relation of P a,b x for different x, a, b? () Stochastic Calculus March 3, 27 2 / 1

21 1 If x 1 x 2 then Px a,b 1 Px a,b 2. 2 The quadratic variation 2 n T lim X( i + 1 n 2 n ) X( i T 2 n ) 2 = a(t, X(t))dt i= a.s. P a,b x Hence if a 1 a 2, P a 1,b x P a 2,b x 3 To see what happens if we change b, let dx i (t) = b i dt + σdb(t), i = 1, 2 P(X 1 (t 1 ) dx 1,..., X 1 (t n ) dx n ) P(X 2 (t 1 ) dx 1,..., X 2 (t n ) dx n ) = e P n 1 i= = e Z dx 1 dx n (x i+1 x i b 1 (t i+1 t i )) 2 (x i+1 x i b 2 (t i+1 t i )) 2 2σ 2 (t i+1 t i ) dx 1 dx n () Stochastic Calculus March 3, / 1

22 P(X 1 (t 1 ) dx 1,..., X 1 (t n ) dx n ) P(X 2 (t 1 ) dx 1,..., X 2 (t n ) dx n ) = ez dx 1 dx n n 1 Z = i= (x i+1 x i b 1 (t i+1 t i )) 2 (x i+1 x i b 2 (t i+1 t i )) 2 2σ 2 (t i+1 t i ) n 1 = σ 1 (b 2 b 1 )σ 1 (x i+1 x i b 2 (t i+1 t i )) n i= 1 n 1 (σ 1 (b 2 b 1 )) 2 (t i+1 t i ) 2 i= σ 1 (b 2 b 1 )db(s) 1 2 (σ 1 (b 2 b 1 )) 2 ds () Stochastic Calculus March 3, / 1

23 Cameron-Martin-Girsanov formula For each x the measure Px a,b is absolutely continuous on F t with respect to the measure Px a, and dpx a,b dpx a, Ft { = exp a 1 (X s )b(x s )dx s 1 } b(x s )a 1 (X s )b(x s )ds 2 Proof We want to show is if we define a measure { Q(A) = exp a 1 bdx s 1 2 A } a 1 b 2 ds dpx a, then Q is a diffusion with parameters a and b in other words, for each λ, exp{λ(x t X bds) λ2 ads} 2 is a martingale with respect to Q () Stochastic Calculus March 3, / 1

24 Proof. Y t = (λ + a 1 b)dx s = (λ + a 1 b)σdb s. e Y t Y 1 2 R t a(λ+a 1 b) 2 ds = martingale w.r.t. P a, x e λ(x t X R t R λ2 bds) t 2 ads+r t a 1 bdx s 1 R t 2 a 1 b 2 ds = martingale w.r.t. P a, x e λ(x t X R t R λ2 bds) t 2 ads = martingale w.r.t. Q t dq = er a 1 bdx s 1 R t 2 a 1 b 2ds dpx a, () Stochastic Calculus March 3, / 1

25 Solution of one dimensional stochastic differential equations Suppose you want to solve the one dimensional stochastic differential equation dx(t) = σ(t, X(t))dB(t) + b(t, X(t))dt By Cameron-Martin-Girsanov, you could instead solve dz (t) = σ(t, Z (t))db(t) and then change to an equivalent measure which will correspond to the solution of X(t) Define t(τ) by τ(t) = σ 2 (u, Y (u))du B(τ) = Z (t(τ)) is a Brownian motion and Z (t) = B(τ(t)). Only works in d = 1 () Stochastic Calculus March 3, / 1

26 Brownian motion as the limit of random walks X 1, X 2,... iid Bernoulli P(X i = 1) = P(X i = 1) = 1/2 S n = X X n B n (t) = 1 n S tn Takes steps ± 1 n at times 1 n, 2 n,... Or B n (t) = polygonalized version. Almost the same but continuous B n (t) n Brownian motion B(t) What does it mean for stochastic processes to converge? dist(b n (t 1 ),..., B n (t k )) dist(b(t 1 ),..., B(t k )) k = 1, 2, 3,... Convergenence of finite dimesional distributions Immediate from (multidimensional) central limit theorem Same for B n (t) () Stochastic Calculus March 3, / 1

27 P n = measure on C[, T ] corresponding to B n (t), t T Invariance principle (Donsker s Theorem) P n P Much stronger than convergence of finite dimensional distributions Examples 1 1 dist( max S m ) dist( sup B(t)) m n n t 1 2 n dist(n 1 k 2 Sm) k dist( m=1 1 B k (t)dt) () Stochastic Calculus March 3, / 1

28 Brownian motion with variance σ 2 and drift b as the limit of random walks X n (t) jumps 1 n σ + 1 n b or 1 n σ + 1 nb with probabilities 1/2 at times 1 n, 2 n,... X n (t) b nt n = σb n(t) X n (t) σb(t) + bt General local diffusivity σ 2 (t, x) and drift b(t, x) X n (t) jumps 1 n σ( i n, X n( i n )) + 1 n b( i n, X n( i 1 n )) or n σ( i n, X n( i n )) + 1 n b( i n, X n( i n )) with probabilities 1/2 at times i n, i = 1, 2,... X n(t) X(t) But how to prove it? dx(t) = σ(t, X(t))dB(t) + b(t, X(t))dt () Stochastic Calculus March 3, / 1

29 Here s another proof that random walks converge to Brownian motions, which does generalize Recall B n (t) = 1 n S tn where S n = X X n Let f C 2 f (B n (t)) = 1 n tn 1 i= L n f (B n ( i n ) = Martingale L n f (x) = 1 2 n(f (x + n 1/2 ) 2f (x) + f (x n 1/2 )) f (B(t)) 1 2 L n f (x) 1 2 f (x) tn 1 1 L n f (B n ( i n n ) 1 2 i= f (B(s))ds f (B(s))ds = martingale B(t) Brownian motion () Stochastic Calculus March 3, / 1

30 Really one needs to show that P n are precompact as a set of probability measures. It is similar to the proof that Brownian motion is continuous, but you just use the martingale formulation directly. The details are long, but the final result is Theorem Suppose that 1 n y x 1 (y i x i )(y j x j )p 1/n (x, dy) a ij (x) uniformly on compact sets 2 n y x 1 (y i x i )p 1/n (x, dy) b i (x) uniformly on compact sets 3 np 1/n (x, B(x, ɛ) C ) uniformly on compact sets, for each ɛ > where a(x) and b(x) are continuous. Suppose that we have weak uniqueness for the stochastic differential equation dx = σ(x)db + b(x)dt and let P denote the measure on C[, T ] corresponding to X(t), t T. Then P n P () Stochastic Calculus March 3, 27 3 / 1