13 The martingale problem

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2 Notations Ω complete metric space of all continuous functions from [0, + ) to R d endowed with the distance d(ω 1, ω 2 ) = k=1 ω 1 ω 2 C([0,k];H) 2 k (1 + ω 1 ω 2 C([0,k];H) ), ω 1, ω 2 Ω. F σ-algebra of all Borel subsets of Ω. X(t), t 0, random variable in (Ω, F ) defined by X(t)(ω) = ω(t), ω Ω, t 0.

3 C algebra of all cylindrical sets in Ω, C t1,t 2,...,t n;i = {ω Ω : (ω(t 1 ),..., ω(t n )) I}, where n N, 0 t 1 < < t n and I B(R dn ). As well known, the σ-algebra generated by C coincides with F.

4 F t, t 0, σ-algebra generated by all cylindrical sets of the form C t1,t 2,...,t n;i, 0 t 1 < < t n t, B(R dn ). Then X(t), t 0, is a stochastic process on (Ω, F ) adapted to the filtration (F t ) t 0. It is called the canonical stochastic process on Ω.

5 Marginals of P Let P be a probability measure on (Ω, F ). For any t 0 we consider the marginal p t := X(t) # P, ϕ(x)p t (dx) = ϕ(ω(t))p(dω), ϕ B b (R d ). (1) R d Ω

6 Moreover, for 0 < s < t we set p s,t := (X(s), X(t)) # P, so that ϕ 1 (x 1 )ϕ 2 (x 2 )p s,t (dx 1, dx 2 ) = ϕ 1 (ω(s))ϕ 2 (ω(t))p(dω), R 2d Ω (2) for all ϕ 1, ϕ 2 B b (R d ). For any sequence 0 t 1 < < t n we define similarly p t1,...,t n. The set of all these marginals uniquely determines P.

7 Kolmogorov operator We consider the differential operator defined by L (ϕ) := 1 2 Tr [a(x)d2 ϕ] + b(x), D x ϕ, ϕ C 2 b (Rd ), (3) where b : R d R d and σ : R d L(R d ) are continuous and bounded and a = σσ. Lemma L is dissipative in C b (R d ).

8 Proof Let λ > 0, ϕ C 1 b (Rd ) and set λϕ L ϕ = f. We have to show that We can assume that ϕ 0 1 λ f 0. (4) sup ϕ(x) = ϕ 0. x R d (Otherwise we replace ϕ with ϕ.)

9 Let ɛ > 0 and set ψ ɛ (x) = ϕ(x) ɛ x 2, x R d. Then lim ψ ɛ(x) = so that ψ ɛ attains the maximum at a x ± point x ɛ R d. Now L ( x 2 ) = Tr [a(x)] + 2 b(x), x L ψ ɛ (x) = L ϕ(x) ɛtr [a(x)] 2ɛ b(x), x.

10 Therefore λψ ɛ (x) L ψ ɛ (x) = λϕ(x) L ϕ(x) λɛ x 2 + 2ɛ b(x), x + ɛ Tr [a(x)], and so λψ ɛ (x) L ψ ɛ (x) = f (x) λɛ x 2 + 2ɛ b(x), x + ɛ Tr [a(x)].

11 Therefore λψ ɛ (x) L ψ ɛ (x) f 0 + 2ε b 0 x + ɛd a 0. But for x = x ɛ we have L ψ ɛ (x ɛ ) 0 so that λψ ɛ (x ɛ ) f 0 + 2ɛ b 0 x + ɛd a 0. Consequently ϕ(x) = ψ ɛ (x) + ɛ x 2 ψ ɛ (x ɛ ) + ɛ x 2 1 λ f λ ɛ(λ x b 0 x + d a 0 ).

12 Therefore λψ ɛ (x) L ψ ɛ (x) f 0 + 2ε b 0 x + ɛd a 0. But for x = x ɛ we have L ψ ɛ (x ɛ ) 0 so that λψ ɛ (x ɛ ) f 0 + 2ɛ b 0 x + ɛd a 0. Consequently ϕ(x) = ψ ɛ (x) + ɛ x 2 ψ ɛ (x ɛ ) + ɛ x 2 1 λ f λ ɛ(λ x b 0 x + d a 0 ).

13 Letting ɛ 0 yields ϕ(x) 1 λ f 0, x R d. So, ϕ 1 λ f 0, which proves dissipativity of L. By the lemma it follows that L is closable in C b (R d ); we shall denote by L its closure.

14 Definition A probability measure P on (Ω, F ) solves the martingale problem related to L with the initial point x R d if (i) P(Ω x ) = 1, where Ω x = {ω Ω : ω(0) = x}. (ii) For any ϕ C 2 b (Rd ) the process t M ϕ (t) := ϕ(x(t)) L ϕ(x(s))ds, t 0, (5) 0 is an (F t ) t 0 martingale.

15 Example Let us consider the stochastic differential equation dx = b(x)dt + σ(x)dw (t), t 0, X(0) = x, (6) where b : R d R d and σ : R d L(R r ; R d ) are Lipschitz continuous and W is an R r -valued standard Brownian motion on some filtered probability space.

16 It is well known that equation (6) has a unique solution X(, x), which is almost surely continuous. Let ϕ C 2 b (Rd ), then by Itô s formula we have ϕ(x(t)) = ϕ(x) + + t 0 t 0 (L ϕ)(x(r))dr D x ϕ(x(r)), σ(x(r))dw (r)

17 Therefore is a martingale. M ϕ (t) := ϕ(x(t)) t 0 (L ϕ)(x(r))dr, t 0, Consequently the law of X(, x) in C([0, ); R d ) solve the martingale problem related to L with initial point x.

18 Uniqueness We are going to prove a sufficient condition for the uniqueness of the martingale problem. We start with basic lemma. Lemma Assume that P solves the martingale problem related to L with initial point x. Then for any λ > 0, s 0 and any ϕ Cb 2(Rd ) we have [ ] E P e λr (λϕ L ϕ)(x(r))dr F s = e λs ϕ(x(s)). (7) s

19 Proof Let t > s. Since M ϕ is a martingale, we have t E P [ϕ(x(t)) 0 which is equivalent to t E P [ϕ(x(t)) s ] L ϕ(x(r))dr F s = ϕ(x(s)) s 0 L ϕ(x(r))dr, ] L ϕ(x(r))dr F s = ϕ(x(s)). (8)

20 Multiplying both sides of (8) by λe λt and integrating over [s, + ) with respect to t, yields [ E P λe λt ϕ(x(t))dt s s λe λt dt t s ] L ϕ(x(s))ds = e λs ϕ(x(s)). (9)

21 Since s = 0 λe λt dt we obtain t s L ϕ(x(r))dr = e λr L ϕ(x(r))dr, s L ϕ(x(r))dr r λe λt dt [ t ] E P e λr (λϕ(x(r)) L ϕ(x(r)))dr F s = e λs ϕ(x(s)), s as claimed.

22 Theorem Assume that the closure L of L is m-dissipative in C b (H). Then for any x H there is at most one solution of the martingale problem related to L with initial point x H.

23 Proof Let us denote by P t the semigroup generated by L. Let x H and let P be a solution of the martingale problem related to L with initial point x H. We are going to show that P t, t 0, determines all finite dimensional distributions of P.

24 Step 1. For any f C b (H) and any s 0 we have [ ] E P e λr f (X(r))dr F s = e λs R(λ, L )f (X(s)). (10) s

25 Let f C b (H), λ > 0. Set ϕ = R(λ, L )f. Since Cb 2 (H) is a core for L, there exists a sequence (ϕ n ) Cb 1 (H) such that ϕ n ϕ, L ϕ n L ϕ λϕ n L ϕ n f in C b (H). Now by the basic lemma we have [ ] E P e λr (λϕ n L ϕ n )(X(r))dr F s = e λs ϕ n (X(s)). s Letting n, yields (10) and Step 1 is proved.

26 Step 2. We have E P [f (X(r))] = P r f (x), f C b (H), r 0. (11) which determines the one-dimensional distributions of P. In fact, setting s = 0 in (10), yields [ ] E P e λr f (X(r))dr = R(λ, L )f (x), 0 from which, taking the inverse of Laplace transform, yields E P [f (X(r))dr] = P r f (x), r 0.

27 Step 3 For all t > s, f 1, f 2 C b (H) we have E P [f 1 (X(t))f 2 (X(s))] = P s [f 2 P t s f 1 ](x). (12) which determines the two-dimensional distributions of P.

28 Write E P [f 1 (X(t))f 2 (X(s))] = E P [E P f 1 (X(t))f 2 (X(s)) F s ] = E P [f 2 (X(s))E P [f 1 (X(t))] F s ]. (13) Taking the inverse of Laplace transform in (10), yields E P [f (X(t)) F s ] = P t s f (X(s)), r 0. Substituting this in (13) (with f 1 replacing f ), yields E P [f 1 (X(t))f 2 (X(s))] = E P [f 2 (X(s))P t s f 1 (X(s))], which coincides with (12). In a similar way one can determine all finite dimensional distributions of P. The proof is complete.

29 Write E P [f 1 (X(t))f 2 (X(s))] = E P [E P f 1 (X(t))f 2 (X(s)) F s ] = E P [f 2 (X(s))E P [f 1 (X(t))] F s ]. (13) Taking the inverse of Laplace transform in (10), yields E P [f (X(t)) F s ] = P t s f (X(s)), r 0. Substituting this in (13) (with f 1 replacing f ), yields E P [f 1 (X(t))f 2 (X(s))] = E P [f 2 (X(s))P t s f 1 (X(s))], which coincides with (12). In a similar way one can determine all finite dimensional distributions of P. The proof is complete.

30 The non autonomous case We are given two continuous and bounded mappings F : [0, T ] H H, σ : [0, T ] H L(H) and consider the differential operator K F 0 (u) := D tu Tr [a(t, x)d2 u] + F(t, x), D x u, (14) where a = σσ and u belongs to the test functions space E := {u C([0, T ]; C 2 b (H)) C1 ([0, T ]; C b (H)) : u(t, ) = 0}.

31 The proof of the following result is similar to that of a previous lemma. Lemma K F 0 is dissipative in C T ([0, T ]; C b (H)). By the lemma it follows that K0 F by K0 F its closure. is closable. We shall denote

32 Definition A probability measure P on (Ω, F ) solves the martingale problem related to K0 F with initial point x H if (i) P(Ω x ) = 1. (ii) For any u E the process t M u (t) := u(t, X(t)) 0 is an (F t ) t 0 -martingale. K0 F u(r, X(r))dr, (15)

33 Example Let us consider the stochastic differential equation dx = b(t, X)dt + σ(t, X)dW (t), t 0, X(0) = x, (16) where b : [0, + ) R d R d and σ : [0, + ) R d L(R r ; R d ) are continuous. b and σ are Lipschitz continuous in x uniformly in t. W is an R r -valued standard Brownian motion on some filtered probability space.

34 Let K 0 be the corresponding Kolmogorov operator K 0 u = D t u Tr [a(t, x)d2 u] + b(t, x), D x u, u E, where a = σσ.

35 Equation (16) has a unique continuous solution X(, x). By Itô s formula we have for any ϕ C 2 b (Rd ), u(t, (X(t, x)) = u(0, x) + + t 0 t 0 (K 0 u)(r, X(r))dr D x u(r, X(r)), σ(r, X(r))dW (r), t 0.

36 Therefore M u (t, x) : = u(t, (X(t)) is a martingale. t 0 (K 0 u)(r, X(r))dr, t 0, In conclusion the law of X( ) in C([0, ); R d ) solves the martingale problem related to K 0 with initial point x.

37 Basic Lemma Lemma Assume that P solves the martingale problem related to K 0 with initial point x. Let u E and set f = K0 F u. Then we have T s E P [f (r, X(r)) F s ]dr = u(s, X(s)). (17)

38 Proof Since P is a solution of the martingale problem, we have for s [0, T ), ] T E P [u(t, X(T )) K0 F u(r, X(r))dr F s 0 which implies s = u(s, X(s)) 0 K0 F u(r, X(r))dr

39 ] T E P [u(t, X(T )) K0 F u(r, X(r))dr Fs = u(s, X(s)) s Taking into account that u(t, ) = 0, it follows that E P T s K0 F u(r, X(r))dr = u(s, X(s)). Since K0 F u = f, the conclusion follows.

40 Theorem Assume that the closure K0 F of K 0 F C T ([0, T ]; C b (H)). is m dissipative in Then for any x H there is at most one solution of the martingale problem related to K 0 with initial point x H.

41 Proof Step 1. If P solves the martingale problem with initial point x we have T t E P [f (r, X(r))]dr = u(t, x) = (K F 0 ) 1 f, (18) which determines the one dimensional distributions of P. In fact, since E is a core of K0 F, there exists a sequence (u n ) E such that, setting f n = K0 F u n, it results u n u, K F 0 u n f in C T ([0, T ]; C b (H)).

42 By the basic lemma it follows that T s E P [f n (r, X(r)) F s ]dr = u n (s, X(s)), (19) from which, letting n, T s E P [f (r, X(r)) F s ]dr = u(s, X(s)). (20) Now, setting s = t, Step 1 follows.

43 Step 2. Let r 1 < r 2. Then we have T 0 dr 1 T r 1 E P [f 1 (r 1, X(r 1 ))f 2 (r 2, X(r 2 ))]dr 2 (21) = E P {f 1 (r 1, X(r 1 ))(K F 0 ) 1 f 2 (x 1, X(r 1 ))}, which determines two dimensional distributions of P.

44 In fact write T 0 = = dr 1 T T 0 T 0 E P [f 1 (r 1, X(r 1 ))f 2 (r 2, X(r 2 ))]dr 2 r 1 T dr 1 E P {E[f 1 (r 1, X(r 1 ))f 2 (r 2, X(r 2 ))] F r1 }dr 2 r 1 T dr 1 E P {f 1 (r 1, X(r 1 ))E P [f 2 (r 2, X(r 2 ))] F r1 ]}dr 2 r 1

45 Taking into account (18), it follows that T 0 = dr 1 T T 0 r 1 E P [f 1 (r 1, X(r 1 ))f 2 (r 2, X(r 2 ))]dr 2 E P {f 1 (r 1, X(r 1 ))dr 1 T r 1 E P [f 2 (r 2, X(r 2 ))] F r1 ]}dr 2 T = E P {f 1 (r 1, X(r 1 ))dr 1 (K0 F ) 1 f 2 (x 1, X(r 1 ))}dr 2 0 = E P {f 1 (r 1, X(r 1 ))(K F 0 ) 1 f 2 (x 1, X(r 1 ))} and (21) follows. In a similar way we determine all finite distributions of P.