Martingale Problems. Abhay G. Bhatt Theoretical Statistics and Mathematics Unit Indian Statistical Institute, Delhi

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1 s Abhay G. Bhatt Theoretical Statistics and Mathematics Unit Indian Statistical Institute, Delhi Lectures on Probability and Stochastic Processes III Indian Statistical Institute, Kolkata November 2008

2 Outline Introduction Associated Semigroups and Generators 1 Introduction Associated Semigroups and Generators 2 3 Time-Inhomogeneous Jump Perturbations

3 Introduction Associated Semigroups and Generators Let X be a E-valued process. X satisfies the Markov property if P (X t+r Γ σ(x s : 0 s t)) = P (X t+r Γ X t ) A function P(t, x, Γ) is called a transition function if P(t, x, ) is a probability measure for all (t, x) [0, ) E P(0, x, ) = δ x for all x E P(,, Γ) is measurable for all Γ E (Interpretation) P (X t Γ X 0 = x) = P(t, x, Γ) X is a Markov process if it admits a transition function so that ( ) P X t+s Γ Ft X = P(s, X t, Γ) t, s 0, Γ E Equivalently [ ] E f (X t+s ) Ft X = f (y)p(s, X t, dy) f C b (E)

4 Associated Semigroups Introduction Associated Semigroups and Generators Define T t f (x) = f (y)p(t, x, dy) = E [f (X t ) X 0 = x] T t T s = T t+s T t f 0 whenever f 0 T t f f [ E f (X t+s ) Ft X semigroup property positivity contraction ] = T s f (X t )

5 Generator Introduction Associated Semigroups and Generators A contraction semigroup {T t } satisfying lim T t f = T 0 f = f t 0 is called a Strongly continuous contraction semigroup The Generator L of a strongly continuous contraction semigroup is defined as follows. D(L) = T t f f = { f : lim t 0 T t f f t T t f f Lf = lim. t 0 t t 0 exists }. T s Lfds f D(L) (1)

6 Introduction Associated Semigroups and Generators Proposition 1 Let X, P, (T t ) and L be as above. Then X is a solution of the martingale problem for L. Proof. Fix f D(L) and let M t = f (X t ) t 0 Lf (X s)ds. [ ] [ ] t+s [ ] E M t+s Ft X = E f (X t+s ) Ft X E Lf (X u ) Ft X du 0 t+s t = T s f (X t ) T u t Lf (X t )du Lf (X u )du = T s f (X t ) t s 0 t T u Lf (X t )du Lf (X u )du 0 t = f (X t ) Lf (X u )du= M t (using (1)) 0 0

7 Outline 1 Introduction Associated Semigroups and Generators 2 3 Time-Inhomogeneous Jump Perturbations

8 Finite Dimensional Distributions Lemma 1 A process X is a solution to the martingale problem for A if and only if [ ( tn+1 ) n ] E f (X ) f (X t tn+1 n) Af (X s )ds h k (X tk ) = 0 (2) t n k=1 for all f D(A), 0 t 1 < t 2 <... < t n+1, h 1, h 2,..., h n B(E), and n 1. Thus, (being a) solution of the martingale problem is a finite dimensional property. Thus if X is a solution and Y is a modification of X, then Y is also a solution.

9 The space D([0, ), E) D([0, ), E) - the space of all E valued functions on [0, ) which are right continuous and have left limits Skorokhod topology on D([0, ), E) D([0, ), E) - complete, separable, metric space S E, the Borel σ-field on D([0, ), E). θ t (ω) = ω t - co-ordinate process r.c.l.l. process - process taking values in D([0, ), E)

10 r.c.l.l. Solutions Definition 2.1 A probability measure P P(D([0, ), E)) is solution of the martingale problem for (A, µ) if there exists a D([0, ), E)- valued process X with L(X ) = P and such that X is a solution to the martingale problem for (A, µ) Equivalently, P P(D([0, ), E)) is a solution if θ defined on (D([0, ), E), S E, P) is a solution For a r.c.l.l. process X defined on some (Ω, F, P), we will use the dual terminology X is a solution P X 1 is a solution

11 Well-posedness - Definitions Definition 2.2 The martingale problem for (A, µ) is well - posed in a class C of processes if there exists a solution X C of the martingale problem for (A, µ) and if Y C is also a solution to the martingale problem for (A, µ), then X and Y have the same finite dimensional distributions. i.e. uniqueness holds When C is the class of all measurable processes then we just say that the martingale problem is well - posed. Definition 2.3 The martingale problem for A is well - posed in C if the martingale problem for (A, µ) is well-posed for all µ P(E).

12 Well-posedness in D([0, ), E) finite dimensional distributions characterize the probability measures on D([0, ), E) Definition 2.4 The D([0, ), E) - martingale problem for (A, µ) is well - posed if there exists a solution P P(D([0, ), E)) of the D([0, ), E) - martingale problem for (A, µ) and if Q is any solution to the D([0, ), E) - martingale problem for (A, µ) then P = Q.

13 Bounded-pointwise convergence Definition 2.5 Let f k, f B(E). f k converge bounded-ly and pointwise to f bp f k f if f k M and f k (x) f (x) for all x E. A class of functions U B(E) bp-closed if f k U, f k bp f implies f U. bp-closure(u) - the smallest class of functions in B(E) which contains U and is bp-closed. i.e. E 0 - a field; σ(e 0 ) = E H - class of all E 0 -simple functions Then bp-closure(h) = class of all bounded, E - measurable functions.

14 Separability condition Definition 2.6 The operator A satisfies the separability condition if There exists a countable subset {f n } D(A) such that bp closure({(f n, Af n ) : n 1}) {(f, Af ) : f D(A)}. Let A 0 = A {fn}, the restriction of A to {f n } Solution of martingale problem for A = solution of martingale problem for A 0

15 Separability condition Definition 2.6 The operator A satisfies the separability condition if There exists a countable subset {f n } D(A) such that bp closure({(f n, Af n ) : n 1}) {(f, Af ) : f D(A)}. Let A 0 = A {fn}, the restriction of A to {f n } Solution of martingale problem for A solution of martingale problem for A 0 from Lemma 1 Use Dominated convergence Theorem to show that the set of all {(g, Ag)} satisfying (2) is bp-closed

16 Markov Family of Solutions Theorem 1 Let A be an operator on C b (E) satisfying the separability condition. Suppose the D([0, ), E) - martingale problem for (A, δ x ) is well-posed for each x E. Then 1 x P x (C) is measurable for all C S E. 2 For all µ P(E), the D([0, ), E) - martingale problem for (A, µ) is well - posed, with the solution P µ given by P µ (C) = P x (C)µ(dx). 3 Under P µ, θ t is a Markov process with transition function E P(s, x, F ) = P x (θ s F ). (3)

17 Proof of (1) Choose M C b (E) - countable such that B(E) bp-closure(m). { Let H = η : tm+1 η(θ) = (f n (θ tm+1 ) f n (θ tm ) t m Af n (θ s )ds) m h k (θ tk ) k=1 } where h 1, h 2,..., h m M, 0 t 1 < t 2... < t m+1 Q H is countable Lemma 1 = M 1 = η H {P : ηdp = 0} is the set of solutions of the martingale problem for A. P ηdp is continuous. Hence M 1 is Borel set

18 Proof of (1) (Contd.) G : P(D([0, ), E)) P(E) G is continuous G(P) = P θ(0) 1. M = M 1 G 1 ({δ x : x E}) = {P x : x E} is Borel Well-posedness = G restricted to M is one-to-one mapping onto {δ x : x E}. G 1 : {δ x : x E} M is Borel G(P x ) = δ x. Hence δ x P x is measurable x P x = x δ x P x is measurable

19 Proof of (2) For F E P µ θ 1 0 (F ) = For η H, E D([0, ),E) P x θ 1 0 (F )µ(dx) = ηdp µ = E D([0, ),E) E δ x (F )µ(dx) = µ(f ). ηdp x µ(dx) = 0. Hence P µ is a solution to the martingale problem for (A, µ). Let Q be another solution of the D([0, ), E) martingale problem for (A, µ) Let Q ω be the regular conditional probability of Q given θ 0.

20 Proof of (2) (Contd.) Fix η H, h C b (E). Define η (θ) = η(θ)h(θ 0 ). η H. Thus E Q [η(θ)h(θ 0 )] = E Q [η ] = 0. Since this holds for all h C b (E), E Qω [η] = E Q [η θ 0 ] = 0 a.s. - Q. Since H is countable, ONE Q-null set N 0 satisfying E Qω [η] = 0 ω N 0 Q ω is a solution of the martingale problem for A initial distribution δ θ0 (ω). Well - posedness implies Hence Q = P µ. Q ω = P θ0 (ω) a.s.[q]

21 Proof of (3) Fix s. Let θ t = θ t+s. Let Q ω be the regular conditional probability distribution of θ (under P x ) given F s. Q ω is a solution to the martingale problem for (A, δ θs(ω)). Well-posedness = Q ω(θ t F ) = P(t, θ s (ω), F ) (See (3)) Hence for f B(E), ] E Px f (θ t+s ) = E [E Px Px [f (θ t+s ) F s ] [ ] = E Px f (y)p(t, θ s ( ), dy) E = f (y 2 )P(t, y 1, dy 2 )P(s, x, dy 1 ). E E P(s + t, x, F ) = P x (θ t+s F ) = E P(t, y, F )P(s, x, dy)

22 One-dimensional equality Theorem 2 Suppose that for each µ P(E), any two solutions X and Y (defined respectively on (Ω 1, F 1, P 1 ) and (Ω 2, F 2, P 2 )) of the martingale problem for (A, µ) have the same one-dimensional distributions. Then X and Y have the same finite dimensional distributions, i.e. the martingale problem is well - posed. Proof. To show [ m ] [ m ] E P 1 f k (X tk ) = E P 2 f k (Y tk ) (4) k=1 k=1 for all 0 t 1 < t 2 <... t m, f 1, f 2,..., f m B(E) and m 1.

23 Induction argument Case: m = 1 - true by hypothesis Assume that the Induction hypothesis (4) is true for m = n Fix 0 t 1 < t 2 <... t n, f 1, f 2,..., f n B(E), f k > 0. Define Q 1 (F 1 ) = EP 1 [I F1 n k=1 f k(x tk )] E P 1 [ n k=1 f k(x tk )] Q 2 (F 2 ) = EP 2 [I F2 n k=1 f k(y tk )] E P 2 [ n k=1 f k(y tk )] F 1 F 1 F 2 F 2 Let X t = X tn+t, Ỹ t = Y tn+t. Fix 0 s 1 < s 2 <..., s m+1 = t, h 1, h 2,..., h m B(E) and f D(A).

24 Induction argument (Contd.) E P 1 η(θ) = [ η(x tn+ ) ( sm+1 ) m f (θ sm+1 ) f (θ sm ) Af (θ s )ds h k (θ tk ) s m k=1 ] n f k (X tk ) = k=1 E P 1[ ( f (X sm+1 +t n ) f (X sm+t n ) m h j (X tn+sj ) j=1 tn+s m+1 t n+s m ) Af (X u )du ] n f k (X tk ) = 0. k=1

25 Induction argument (Contd.) Hence E Q 1 [η( X )] = EP 1 [η(x n tn+ ) k=1 f k(x tk )] E P 1 [ n k=1 f = 0. k(x tk )] Similarly E Q 2 [η(ỹ )] = 0. X and Ỹ are solutions of the martingale problems for (A, L( X 0 )) and (A, L(Ỹ0)) respectively. E Q 1 [f ( X 0 )] = EP 1 [f (X tn ) n k=1 f k(x tk )] E P 1 [ n k=1 f k(x tk )] = EP 2 [f (Y tn ) n k=1 f k(y tk )] E P 2 [ n k=1 f = E Q 2 [f (Ỹ 0 )] f B(E). k(y tk )] This equality follows from induction hypothesis for m = n

26 Induction argument (Contd.) Hence X and Ỹ have the same initial distribution. One-dimensional uniqueness implies E Q 1 [f ( X t )] = E Q 2 [f (Ỹ t )] t 0, f B(E). E P 1 [f (X tn+t) n f k (X tk )] = E P 2 [f (X tn+t) k=1 Induction Hypothesis (4) is true for m = n + 1 set t n+1 = t n + t n f k (X tk )] k=1

27 Semigroup associated with the Suppose A satisfies the conditions of Theorem 1. Associate the Markov semigroup (T t ) t 0 with A - T t f (x) = f (y)p(t, x, dy) The following theorem can be proved exactly as the previous one. E

28 Strong Markov Property Theorem 3 Suppose that the D([0, ), E)- martingale problem for A is well - posed with associated semigroup T t Let X, defined on (Ω, F, P), be a solution of the martingale problem for A (with respect to (G t ) t 0 ). Let τ be a finite stop time. Then for f B(E), t 0, In particular E[f (X τ+t ) G τ ] = T t f (X τ ) P((X τ+t Γ) G τ ) = P(t, X τ, Γ) Γ E

29 r.c.l.l. modification Definition 2.7 Let D be a class of functions on E 1 D is measure determining if fdp = fdq forall f D implies P = Q. 2 D separates points in E if x y g D such that g(x) g(y). Theorem 4 Let E be a compact metric space. Let A be an operator on C(E) such that D(A) is measure determining and contains a countable subset that separates points in E. Let X, defined on (Ω, F, P), be a solution to the martingale problem for A. Then X has a modification with sample paths in D([0, ), E).

30 Proof Proof. Let {g k : k 1} D(A) separate points in E. Define t M k (t) = g k (X t ) Ag k (X s )ds 0 M k is a martingale for all k. Then for all t lim s t s Q M k (s), lim M k (s) exist a.s. s t s Q Hence Ω Ω with P(Ω ) = 1 and lim s t s Q g k (X s (ω)), lim g k (X s (ω)) exist ω Ω, t 0, k 1 s t s Q

31 Proof (contd.) Fix t 0, {s n } Q with s n > t, lim n s n = t and ω Ω. Since E is compact, a subsequence {s ni } such that lim i X sni (ω) exists. Clearly ( ) g k lim X s ni (ω) i = lim g k (X s (ω)) k. s t s Q Since {g k : k 1} separate points in E, lim s t X s (ω) exists. s Q Similarly lim s t X s (ω) exists s Q Define Y t (ω) = lim X s (ω). s t s Q

32 Proof (contd.) For ω Ω, Y t (ω) is r.c.l.l. & Define Y suitably for ω Ω Yt (ω) = lim X s (ω) s t s Q Then Y has sample paths in D([0, ), E). Since X is a solution to the martingale problem for A, for f D(A), a measure determining set = X = Y a.s. E[f (Y t ) Ft X ] = lim E[f (X s ) Ft X ] = f (X t ). s t s Q

33 Outline Time-Inhomogeneous Jump Perturbations 1 Introduction Associated Semigroups and Generators 2 3 Time-Inhomogeneous Jump Perturbations

34 Definitions Time-Inhomogeneous Jump Perturbations For t 0, let (A t ) t 0 be linear operators on M(E) with a common domain D M(E). Definition 3.1 A measurable process X defined on some probability space (Ω, F, P) is a solution to the martingale problem for (A t ) t 0 with respect to a filtration (G t ) t 0 if for any f D t f (X t ) A s f (X s )ds 0 is a (G t ) - martingale. Let µ P(E). The martingale problem for ((A t ) t 0, µ) is well-posed if there exists an unique solution for the martingale problem

35 Space-Time Process Time-Inhomogeneous Jump Perturbations Let E 0 = [0, ) E. Let Xt 0 = (t, X t ) Define { } k D(A 0 ) = g(t, x) = h i (t)f i (x) h i Cc 1 ([0, )), f i D i=1 A 0 g(t, x) = k [f i (x) t h i (t) + h i (t)a t f i (x)] i=1 Theorem 5 X is a solution to the martingale problem for (A t ) t 0 if and only if X 0 is a solution to the martingale problem for A 0.

36 Time-Inhomogeneous Jump Perturbations Proof. Let X be a solution (with respect to a filtration (G t ) t 0 ) to the martingale problem for (A t ) t 0. Let fh D(A 0 ) For 0 < s < t, let g(t) = E[f (X t ) G s ]. Then g(t) g(s) = g(t)h(t) g(s)h(s) = = = t s t s t s t s E[A u f (X u ) G s ]du u [g(u)h(u)] du {h(u)e[a u f (X u ) G s ] + g(u) u h(u)} du E [ A 0 (fh)(x 0 u ) G s ] du.

37 Proof (Contd.) Time-Inhomogeneous Jump Perturbations fh(x 0 (t)) t 0 A0 fh(x 0 (s))ds is a martingale X 0 is a solution to the martingale problem for A 0. The converse follows by taking h = 1 on [0, T ], T > 0.

38 A more General Result Time-Inhomogeneous Jump Perturbations State spaces E 1 and E 2 Operators A 1 on M(E 1 ) and A 2 on M(E 2 ) Solutions X 1 and X 2 Define D(A) = {f 1 f 2 : f 1 D(A 1 ), f 2 D(A 2 )} A(f 1 f 2 ) = (A 1 f 1 )f 2 + f 1 (A 2 f 2 ) (X 1, X 2 ) is a solution of the martingale problem for A Theorem 6 Suppose uniqueness holds for the martingale problem for A 1, A 2. Then uniqueness holds for the martingale problem for A.

39 A perturbed operator Time-Inhomogeneous Jump Perturbations Let A be an operator with D(A) C b (E). Let λ > 0 and let η(x, Γ) be a transition function on E E. Let Bf (x) = λ (f (y) f (x))η(x, dy) f B(E). E Theorem 7 Suppose that for every µ P(E), there exists a solution to the D([0, ), E) martingale problem for (A, µ). Then for every µ P(E) there exists a solution to the martingale problem for (A + B, µ).

40 Proof Time-Inhomogeneous Jump Perturbations Proof. For k 1, let Ω k = D([0, ), E), Ω 0 k = [0, ) Let Ω = k=1 Ω k Ω 0 k Let θ k and ξ k denote the co-ordinate random variables Borel σ-fields - F k, F 0 k Let F be the product σ-field on Ω. Let G k the σ-algebra generated by cylinder sets C 1 i=k+1 (Ω i Ω 0 i ), where C 1 F 1 F F k Fk 0. Let G k be the σ-algebra generated by k i=1 (Ω i Ω 0 i ) C 2, where C 2 F k F 0 k....

41 Perturbed Solution X Time-Inhomogeneous Jump Perturbations X evolves in E as a solution to the martingale problem for A till an exponentially distributed time with parameter λ which is independent of the past. At this time if the process is at x, it jumps to y with probability η(x, dy) and then continues evolving as a solution to the martingale problem for (A, δ y ). To put this in a mathematical framework, we consider that between the k th and the (k + 1) th jump (dictated by B), the process lies in Ω k. The k th copy of the exponential time is a random variable in Ω 0 k.

42 Proof (Contd.) Time-Inhomogeneous Jump Perturbations Let P x, P µ be solutions of the martingale problems for (A, δ x ), (A, µ) respectively. Let γ be the exponential distribution with parameter λ. Fix µ P(E). Define, for Γ 1 F 1,..., Γ k F k, F 1 F 0 1,..., F k F 0 k, P 1 (Γ 1 ) = P µ (Γ 1 ) ; P 0 1 (θ 1, F 1 ) = γ(f 1 ).. P k (θ 1, ξ 1,..., θ k 1, ξ k 1, Γ k ) = P 0 k (θ 1,..., ξ k 1, θ k, F k ) = γ(f k ) E P x (Γ k )η(θ k 1 (ξ k 1 ), dx)

43 Proof(Contd.) Time-Inhomogeneous Jump Perturbations P 1 P(Ω 1 ) and P1 0, P 2, P2 0,... are transition probability functions. an unique P on (Ω, F) satisfying For C G k and C G k+1 [ P(C C ) = E C ] P(C θ k+1 (0) = x)η(θ k (ξ k ), dx). Define τ 0 = 0, τ k = k i=1 ξ i N t = k for τ k t < τ k+1. Note that N is a Poisson process with parameter λ.

44 Proof(Contd.) Time-Inhomogeneous Jump Perturbations Define F t = F X t F N t. For f D(A) X t = θ k+1 (t τ k ), τ k t < τ k+1 f (θ k+1 ((t τ k ) τ k+1 τ k )) f (θ k+1 (0)) is an (F t ) t 0 martingale. Summing over k we get f (X t ) f (X 0 ) t is an (F t ) t 0 martingale. 0 (t τk ) τ k+1 τ k Af (θ k+1 (s τ k ))ds N(t) Af (X (s))ds (f (θ k+1 (0)) f (θ k (ξ k ))) k=1

45 Proof(Contd.) Time-Inhomogeneous Jump Perturbations Also, the following are (F t ) t 0 martingales. Hence N t k=1 t 0 ( ) f (θ k+1 (0)) f (y)(η(θ k (ξ k ), dy)) E E (f (y) f (X s )) η(x s, dy)d(n s λs) t f (X t ) f (X 0 ) (Af (X s ) + Bf (X s )) ds 0 is an (F t ) t 0 martingale.