Martingales, Lévy s continuity theorem, and the martingale central limit theorem

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1 Martigales, Lévy s cotiuity theorem, ad the martigale cetral limit theorem Jorda Bell jorda.bell@gmail.com Departmet of Mathematics, Uiversity of Toroto May 29, Itroductio I this ote, ay statemet we make about filtratios ad martigales is about filtratios ad martigales idexed by the positive itegers, rather tha the oegative real umbers. We take if =, ad for m >, we take = 0. k=m (Defied rightly, these are ot merely coveiet ad hoc defiitios. 2 Coditioal expectatio Let (Ω, A, P be a probability space ad let B be a sub-σ-algebra of A. For each f L 1 (Ω, A, P, there is some g : Ω R such that (i g is B-measurable ad (ii for each B B, B gdp = fdp, ad if h : Ω R satisfies (i ad (ii B the h(ω = g(ω for almost all ω Ω. 1 We deote some g : Ω R satisfyig (i ad (ii by E(f B, called the coditioal expectatio of f with respect to B. I other words, E(f B is the uique elemet of L 1 (Ω, B, P such that for each B B, E(f BdP = fdp. The map f E(f B satisfies the followig: B 1. f E(f B is positive liear operator L 1 (Ω, A, P L 1 (Ω, B, P with orm 1. 1 Mafred Eisiedler ad Thomas Ward, Ergodic Theory: with a view towards Number Theory, p. 121, Theorem 5.1. B 1

2 2. If f L 1 (Ω, A, P ad g L (Ω, B, P, the for almost all ω Ω, E(gf B(ω = g(ωe(f B(ω. 3. If C is a sub-σ-algebra of B, the for almost all ω Ω, E(E(f B C (ω = E(f C (ω. 4. If f L 1 (Ω, B, P the for almost all ω Ω, E(f B(ω = f(ω. 5. If f L 1 (Ω, A, P, the for almost all ω Ω, E(f B(ω E( f B(ω. 6. If f L 1 (Ω, A, P is idepedet of B, the for almost all ω Ω, 3 Filtratios E(f B(ω = E(f. A filtratio of a σ-algebra A is a sequece F, 1, of sub-σ-algebras of A such that F m F if m. We set F 0 = {, Ω}. A sequece of radom variables ξ : (Ω, A, P R is said to be adapted to the filtratio F if for each, ξ is F -measurable. Let ξ : (Ω, A, P R, 1, be a sequece of radom variables. The atural filtratio of A correspodig to ξ is F = σ(ξ 1,..., ξ. It is apparet that F is a filtratio ad that the sequece ξ is adapted to F. 4 Martigales Let F be a filtratio of a σ-algebra A ad let ξ : (Ω, A, P R be a sequece of radom variables. We say that ξ is a martigale with respect to F if (i the sequece ξ is adapted to the filtratio F, (ii for each, ξ L 1 (P, ad (iii for each, for almost all ω Ω, E(ξ +1 F (ω = ξ (ω. I particular, i.e. E(ξ 1 = E(ξ 2 =, E(ξ m = E(ξ, m. 2

3 We say that ξ is a submartigale with respect to F if (i ad (ii above are true, ad if for each, for almost all ω Ω, I particular, i.e. E(ξ +1 F (ω ξ (ω. E(ξ 1 E(ξ 2, E(ξ m E(ξ, m. We say that ξ is a supermartigale with respect to F if (i ad (ii above are true, ad if for each, for almost all ω Ω, I particular, i.e. ξ (ω E(ξ +1 F (ω. E(ξ 1 E(ξ 2, E(ξ m E(ξ, m. If we speak about a martigale without specifyig a filtratio, we mea a martigale with respect to the atural filtratio correspodig to the sequece of radom variables. 5 Stoppig times Let F be a filtratio of a σ-algebra A. A stoppig time with respect to F is a fuctio τ : Ω {1, 2,...} { } such that for each 1, {ω Ω : τ(ω = } F. It is straightforward to check that a fuctio τ : Ω {1, 2,...} { } is a stoppig time with respect to F if ad oly if for each 1, {ω Ω : τ(ω } F. The followig lemma shows that the time of first etry ito a Borel subset of R of a sequece of radom variables adapted to a filtratio is a stoppig time. 2 Lemma 1. Let ξ be a sequece of radom variables adapted to a filtratio F ad let B B R. The is a stoppig time with respect to F τ(ω = if{ 1 : ξ (ω B} 2 Zdzis law Brzeźiak ad Tomasz Zastawiak, Basic Stochastic Processes, p. 55, Exercise 3

4 Proof. Let 1. The where {ω Ω : τ(ω = } = = ( 1 {ω Ω : ξ k (ω B} {ω Ω : ξ (ω B} k=1 ( 1 k=1 A c k A, A k = {ω Ω : ξ k (ω B}. Because the sequece ξ k is adapated to the filtratio F k, A c k F k ad A F, ad because F k is a filtratio, the right-had side of the above belogs to F. If ξ is a sequece of radom variables adapted to a filtratio F ad a stoppig time τ with respect to F, for 1 we defie ξ τ : Ω R by ξ τ (ω = ξ τ(ω (ω, ω Ω. ξ τ is called the sequece ξ stopped at τ. 3 Lemma 2. ξ τ : (Ω, A, P R is a sequece of radom variables adapted to the filtratio F. Proof. Let 1 ad let B B R. Because ad for ay k, we get {ω : ξ τ (ω B, τ(ω > } = {ω : ξ (ω B, τ(ω > } {ω : ξ τ (ω B, τ(ω = k} = {ω : ξ k B, τ(ω = k}, {ω : ξ τ (ω B} = {ω : ξ (ω B, τ(ω > } But ad ad therefore {ω : ξ k (ω B, τ(ω = k}. k=1 {ξ B, τ > } = {ξ B} {τ > } F {ξ k B, τ = k} = {ξ k B} {τ = k} F k, {ξ τ B} F. I particular, {ξ τ B} A, amely, ξ τ is a radom variable, ad the above shows that this sequece is adapted to the filtratio F Zdzis law Brzeźiak ad Tomasz Zastawiak, Basic Stochastic Processes, p. 55, Exercise 4

5 We ow prove that a stopped martigale is itself a martigale with respect to the same filtratio. 4 Theorem 3. Let F be a filtratio of a σ-algebra A ad let τ be a stoppig time with respect to F. 1. If ξ is a submartigale with respect to F the so is ξ τ. 2. If ξ is a supermartigale with respect to F the so is ξ τ. 3. If ξ is a martigale with respect to F the so is ξ τ. Proof. For 1, defie α (ω = { 1 τ(ω 0 τ(ω < ; we remark that τ(ω if ad oly if τ(ω > 1 ad τ(ω < if ad oly if τ(ω 1. For B B R, (i if 0, 1 B the (ii if 0, 1 B the (iii if 0 B ad 1 B the {ω Ω : α (ω B} = F 1, {ω Ω : α (ω B} = Ω F 1, {ω Ω : α (ω B} = {ω Ω : α (ω = 0} = {ω Ω : τ(ω 1} F 1, ad (iv if 1 B ad 0 B the {ω Ω : α (ω B} = {ω Ω : α (ω = 1} = {ω Ω : τ(ω > 1} F 1, Therefore {α B} F 1. Set ξ 0 = 0, ad we check that ξ τ = α k (ξ k ξ k 1. k=1 It is apparet from this expressio that if ξ is adapted to F the ξ τ is adapted to F, ad that if each ξ belogs to L 1 (P the each ξ τ belogs to L 1 (P. As each of α 1,..., α +1 is F -measurable ad is bouded, E(ξ τ (+1 F = E(α k (ξ k ξ k 1 F = α k E(ξ k ξ k 1 F. (1 k=1 4 Zdzis law Brzeźiak ad Tomasz Zastawiak, Basic Stochastic Processes, p. 56, Propositio 3.2. k=1 5

6 Suppose that ξ is a submartigale. By (1, E(ξ τ (+1 F = α k (ξ k ξ k 1 + α +1 E(ξ +1 F α +1 ξ k=1 ξ τ + α +1 ξ α +1 ξ = ξ τ, which shows that ξ τ is a submartigale; the statemet that E(ξ τ (+1 F ξ τ meas that E(ξ τ (+1 F (ω ξ τ (ω for almost all ω Ω. We ow prove the optioal stoppig theorem. 5 Theorem 4 (Optioal stoppig theorem. Let F be a filtratio of a σ-algebra A, let ξ be a martigale with respect to F, ad let τ be a stoppig time with respect to F. Suppose that: The 1. For almost all ω Ω, τ(ω <. 2. ξ τ L 1 (Ω, A, P. 3. E(ξ 1 {τ>} 0 as. E(ξ τ = E(ξ 1. Proof. For each, Ω = {τ } {τ > }, ad therefore ξ τ = ξ τ + ξ τ 1 {τ>} ξ 1 {τ>} = ξ τ + ξ k 1 {τ=k} ξ 1 {τ>}. k=+1 Theorem 3 tells us that ξ τ is a martigale with respect to to F, ad hece so E(ξ τ = E(ξ 1 + But as ξ τ L 1 (P, (ξ τ (ωdp (ω = Ω E(ξ τ = E(ξ τ 1 = E(ξ 1, k=1 k=+1 {τ=k} E(ξ k 1 {τ=k} E(ξ 1 {τ>}. (2 ξ k (ωdp (ω = E(ξ k 1 {τ=k}, ad the fact that this series coverges meas that k=+1 E(ξ k1 {τ=k} 0. With the hypothesis E(ξ 1 {τ>} 0, as we have E(ξ 1 + k=+1 k=1 E(ξ k 1 {τ=k} E(ξ 1 {τ>} E(ξ 1. But (2 is true for each, so we get E(ξ τ = E(ξ 1, provig the claim Zdzis law Brzeźiak ad Tomasz Zastawiak, Basic Stochastic Processes, p. 58, Theorem 6

7 Suppose that η is a sequece of idepedet radom variables each with the Rademacher distributio: P (η = 1 = 1 2, P (η = 1 = 1 2. Let ξ = k=1 η k ad let F = σ(η 1,..., η. Because ξ 2 +1 = (ξ + η +1 2 = η η +1 ξ + ξ 2, we have, as ξ is F -measurable ad belogs to L (P ad as η +1 is idepedet of the σ-algebra F, E(ξ 2 +1 ( + 1 F = E(η η +1 ξ + ξ 2 ( + 1 F = E(η ξ E(η +1 + ξ 2 ( + 1 = ξ 2 ( + 1 = ξ 2. Therefore, ξ 2 is a martigale with respect to F. Let K be a positive iteger ad let τ = if{ 1 : ξ = K}. Namely, τ is the time of first etry i the Borel subset { K, K} of R, hece by Lemma 1 is a stoppig time with respect to the filtratio F. With some work, 6 oe shows that (i P (τ > 2K 0 as, (ii E( ξ 2 τ τ <, ad (iii E((ξ 2 1 {τ>} 0 as. The we ca apply the optioal stoppig theorem to the martigale ξ 2 : we get that Hece E(ξ 2 τ τ = E(ξ = E(ξ = E(η = 0. But ξ τ = K, so ξ 2 τ = K 2, hece 6 Maximal iequalities E(τ = E(ξ 2 τ. E(τ = E(K 2 = K 2. We ow prove Doob s maximal iequality. 7 6 Zdzis law Brzeźiak ad Tomasz Zastawiak, Basic Stochastic Processes, p. 59, Example Zdzis law Brzeźiak ad Tomasz Zastawiak, Basic Stochastic Processes, p. 68, Propositio

8 Theorem 5 (Doob s maximal iequality. Suppose that F is a filtratio of a σ-algebra A, that ξ is a submartigale with respect to F, ad that for each, ξ 0. The for each 1 ad λ > 0, ( λp max ξ k λ E ( ξ 1 {max1 k ξ k λ}. 1 k Proof. Defie ζ (ω = max 1 k ξ k (ω, which is F -measurable, ad defie τ : Ω {1,..., } by τ(ω = mi{1 k : ξ k (ω λ} if there is some 1 k for which ξ k (ω λ, ad τ(ω = otherwise. For 1 k, k 1 {τ = k} = {ξ k < λ} {ξ k λ} F k, ad for k >, j=1 {τ = k} = F k, showig that τ is a stoppig time with respect to the filtratio F k. For k 1, ξ k+1 ξ τ (k+1 = k 1 {τ=j} (ξ k+1 ξ τ (k+1 = j=1 k 1 {τ=j} (ξ k+1 ξ j, hece, because τ is a stoppig time with respect to the filtratio F k ad because ξ k is a submartigale with respect to this filtratio, E(ξ k+1 ξ τ (k+1 F k = = j=1 k 1 {τ=j} E((ξ k+1 ξ j F k j=1 k 1 {τ=j} (E(ξ k+1 F k ξ j j=1 k 1 {τ=j} (ξ k ξ j j=1 k 1 = 1 {τ=j} (ξ k ξ j j=1 = ξ k ξ τ k, from which we have that the sequece ξ k ξ τ k is a submartigale with respect to the filtratio F k. Therefore E(ξ k ξ τ k E(ξ 1 ξ τ 1 = E(ξ 1 E(ξ τ 1 = E(ξ 1 E(ξ 1 = 0, 8

9 ad so E(ξ τ k E(ξ k. Because τ = τ, this yields We have E(ξ τ E(ξ. E(ξ τ = E(ξ τ 1 {ζ λ} + E(ξ τ 1 {ζ<λ}. If ω {ζ λ} the (ξ τ (ω λ, ad if ω {ζ < λ} the τ(ω = ad so (ξ τ (ω = ξ (ω. Therefore E(ξ τ E(λ 1 {ζ λ} + E(ξ 1 {ζ<λ} = λp (ζ λ + E(ξ 1 {ζ<λ}. Therefore But ξ = ξ 1 ζ<λ + ξ 1 ζ λ, hece which proves the claim. λp (ζ λ + E(ξ 1 {ζ<λ} E(ξ. λp (ζ λ E(ξ 1 {ζ λ}, The followig is Doob s L 2 maximal iequality, which we prove usig Doob s maximal iequality. 8 Theorem 6 (Doob s L 2 maximal iequality. Suppose that F is a filtratio of a σ-algebra A ad that ξ is a submartigale with respect to F such that for each 1, ξ 0 ad ξ L 2 (P. The for each 1, E ( max 1 k ξ k 2 4E(ξ. 2 Proof. Defie ζ (ω = max 1 k ξ k (ω. It is a fact that if η L 2 (P ad η 0 the E(η 2 = 2 0 tp (η tdt. Usig this, Doob s maximal iequality, Fubii s theorem, ad the Cauchy Zdzis law Brzeźiak ad Tomasz Zastawiak, Basic Stochastic Processes, p. 68, Theorem 9

10 Schwarz iequality, E(ζ 2 = 2 2 = = 2 = 2 Ω Ω tp (ζ > tdt E(ξ 1 {ζ t}dt ( ξ (ωdp (ω dt {ζ t} ( ζ(ω dt ξ (ωdp (ω 0 ζ (ωξ (ωdp (ω 2(E(ζ 2 1/2 (E(ξ 2 1/2. If E(ζ 2 = 0 the claim is immediate. Otherwise, we divide this iequality by (E(ζ 2 1/2 ad obtai ad so provig the claim. 7 Upcrossigs (E(ζ 2 1/2 2(E(ξ 2 1/2, E(ζ 2 4E(ξ 2, Suppose that ξ is a sequece of radom variables that is adapted to a filtratio F ad let a < b be real umbers. Defie ad by iductio for m 1, ad τ 0 = 0, σ m (ω = if{k τ m 1 (ω : ξ k (ω a} τ m (ω = if{k σ m (ω : ξ k (ω b}, where if =. For each m, τ m ad σ m are each stoppig times with respect to the filtratio F k. For 0 we defie For x R, we write U [a, b](ω = sup{m 0 : τ m (ω }. x = max{0, x} = mi{0, x}. amely, the egative part of x. We ow prove the upcrossigs iequality. 10

11 Theorem 7 (Upcrossigs iequality. If ξ, 1, is a supermartigale with respect to a filtratio F ad a < b, the for each 1, (b ae(u [a, b] E((ξ a. Proof. For 1 ad ω Ω, ad writig N = U [a, b](ω, for which N, we have (ξ τm (ω ξ σm (ω = = = m=1 N (ξ τm (ω ξ σm (ω + ξ τn+1 (ω ξ σn+1 (ω m=1 + m=n+2 (ξ τm (ω ξ σm (ω N (ξ τm (ω ξ σm (ω + ξ (ω ξ σn+1 (ω + m=1 m=n+1 N (ξ τm (ω ξ σm (ω + 1 {σn+1 }(ω(ξ (ω ξ σn+1 (ω m=1 N (b a + 1 {σn+1 }(ω(ξ (ω ξ σn+1 (ω. m=1 Because ξ σn+1 (ω a, we have (b an 1 {σn+1 }(ω(a ξ (ω + Oe proves that 9 Thus (ξ (ω ξ (ω (ξ τm (ω ξ σm (ω. m=1 1 {σn+1 }(ω(a ξ (ω mi{0, a ξ (ω} = (ξ (ω a. (b ae(u [a, b] E((ξ a + E(ξ τm ξ σm. m=1 Usig that ξ is a supermartigale, for each 1 m, Therefore E(ξ τm ξ σm 0. (b ae(u [a, b] E((ξ a. 9 I am ot this oe. I have ot sorted out why this iequality is true. I every proof of the upcrossigs iequality I have see there are pictures ad thigs like this are asserted to be obvious. I am ot satisfied with that reasoig; oe should ot have to iterpret a iequality visually to prove it. 11

12 8 Doob s martigale covergece theorem We ow use the uprossigs iequality to prove Doob s martigale covergece theorem. 10 Theorem 8 (Doob s martigale covergece theorem. Suppose that ξ, 1, is a supermartigale with respect to a filtratio F ad that M = sup E( ξ <. The there is some ξ L 1 (Ω, A, P such that for almost all ω Ω, ad with E( ξ M. lim ξ (ω = ξ(ω Proof. For ay a < b ad 1, the upcrossigs iequality tells us that E(U [a, b] E(ξ a b a E( ξ a b a E( ξ + a b a M + a b a. For each ω Ω, the sequece U [a, b](ω [0, is odecreasig, so by the mootoe covergece theorem, This implies that Let ( E lim U [a, b] = lim E(U [a, b] M + a b a. A = P ( ω Ω : lim U [a, b](ω < = 1. a,b Q,a<b { } ω Ω : lim U [a, b](ω <. This is a itersectio of coutably may sets each with measure 1, so P (A = 1. Let B = {ω Ω : lim if ξ (ω < lim sup ξ (ω}. If ω B, the there are a, b Q, a < b, such that lim if ξ (ω < a < b < lim sup ξ (ω. It follows from this lim U [a, b](ω =. Thus ω A, so B A =, ad because P (A = 1 we get P (B = Zdzis law Brzeźiak ad Tomasz Zastawiak, Basic Stochastic Processes, p. 71, Theorem 12

13 We defie ξ : Ω R by ξ(ω = { lim ξ (ω ω B 0 ω B, which is Borel measurable. Furthermore, sice ξ = lim if ξ almost everywhere, by Fatou s lemma we obtai E( ξ = E(lim if ξ lim if E( ξ sup E( ξ = M. 9 Uiform itegrability Let ξ : (Ω, A, P R be a radom variable. It is a fact 11 that ξ L 1 if ad oly if for each ɛ > 0 there is some M such that ξ dp < ɛ. { ξ >M} (Oe s istict might be to try to use the Cauchy-Schwarz iequality to prove this. This does t work. Thus, if ξ is a sequece i L 1 (Ω, A, P the for each ɛ > 0 there are M such that, for each, ξ dp < ɛ. { ξ >M } A sequece of radom variables ξ is said to be uiformly itegrable if for each ɛ > 0 there is some M such that, for each, ξ dp < ɛ. { ξ >M} If a sequece ξ is uiformly itegrable, the there is some M such that for each, ad so E( ξ = 4.3. { ξ M} ξ dp < 1, { ξ >M} ξ dp + ξ dp < MdP + 1 M + 1. { ξ >M} { ξ M} 11 Zdzis law Brzeźiak ad Tomasz Zastawiak, Basic Stochastic Processes, p. 73, Exercise 13

14 The followig lemma states that the coditioal expectatios of a itegrable radom variable with respect to a filtratio is a uiformly itegrable martigale with respect to that filtratio. 12 Lemma 9. Suppose that ξ L 1 (Ω, A, P ad that F is a filtratio of A. The E(ξ F is a martigale with respect to F ad is uiformly itegrable. We ow prove that a uiformly itegrable supermartigale coverges i L Theorem 10. Suppose that ξ is a supermartigale with respect to a filtratio F, ad that the sequece ξ is uiformly itegrable. The there is some ξ L 1 (Ω, A, P such that ξ ξ i L 1. Proof. Because the sequece ξ is uiformly itegrable, there is some M such that for each 1, E( ξ M + 1. Thus, because ξ is a supermartigale, Doob s martigale covergece theorem tells us that there is some ξ L 1 (Ω, A, P such that for almost all ω Ω, lim ξ (ω = ξ(ω. Because ξ is uiformly itegrable ad coverges almost surely to ξ, the Vitali covergece theorem 14 tells us that ξ ξ i L 1. The above theorem shows i particular that a uiformly itegrable martigale coverges to some limit i L 1. The followig theorem shows that the terms of the sequece are equal to the coditioal expectatios of this limit with respect to the atural filtratio. 15 Theorem 11. Suppose that a sequece of radom variables ξ is uiformly itegrable ad is a martigale with respect to its atural filtratio F = σ(ξ 1,..., ξ. The there is some ξ L 1 (Ω, A, P such that ξ ξ i L 1 ad such that for each 1, for almost all ω Ω, ξ (ω = E(ξ F (ω. 12 Zdzis law Brzeźiak ad Tomasz Zastawiak, Basic Stochastic Processes, p. 75, Exercise Zdzis law Brzeźiak ad Tomasz Zastawiak, Basic Stochastic Processes, p. 76, Theorem V. I. Bogachev, Measure Theory, volume I, p. 268, Theorem 4.5.4; utoroto.ca/jordabell/otes/l0.pdf, p. 8, Theorem Zdzis law Brzeźiak ad Tomasz Zastawiak, Basic Stochastic Processes, p. 77, Theorem

15 Proof. By Theorem 10, there is some ξ L 1 (Ω, A, P such that ξ ξ i L 1. The hypothesis that the sequece ξ is a martigale with respect to F tells us that for that for 1 ad for ay m, ad so for A F, Thus A ξ m dp = E(ξ m F = ξ, A E(ξ m F dp = A ξ dp. (ξ ξdp = (ξ m ξdp A A ξ m ξ dp A E( ξ m ξ. But E( ξ m ξ 0 as m. Sice m does ot appear i the left-had side, we have (ξ ξdp = 0, ad thus A A ξ dp = A ξdp. But E(f F is the uique elemet of L 1 (Ω, F, P such that for each A F, E(f F dp = fdp, A ad because ξ satisfies this, we get that ξ = E(f F i L 1, i.e., for almost all ω Ω, ξ (ω = E(f F (ω, provig the claim. 10 Lévy s cotiuity theorem For a metrizable topological space X, we deote by P(X the set of Borel probability measures o X. The arrow topology o P(X is the coarsest topology such that for each f C b (X, the map µ fdµ is cotiuous P(X C. X A 15

16 A subset H of P(X is called tight if for each ɛ > 0 there is a compact subset K ɛ of X such that if µ H the µ(x \ K ɛ < ɛ, i.e. µ(k ɛ > 1 ɛ. (A elemet µ of P(X is called tight whe {µ} is a tight subset of P(X. For a Borel probability measure µ o R d, we defie its characteristic fuctio µ : R d C by µ(u = e ix u dµ(x, u R d. R d µ is bouded by 1 ad is uiformly cotiuous. Because µ(r d = 1, µ(0 = 1. Lemma 12. Let µ P(R. For δ > 0, ({ µ x R : x 2 } 1 δ δ δ δ (1 µ(udu; i particular, the right-had side of this iequality is real. Proof. Usig Fubii s theorem ad the fact that all real t, 1 si t t 0, δ δ ( (1 µ(udu = (1 e ixu dµ(x du δ δ R ( δ = 1 e iux du dµ(x R δ δ = (u eiux dµ(x R ix δ = (2δ eiδx R ix + e iδx dµ(x ix ( = 2δ 1 si(δx dµ(x R δx ( 2δ 1 si(δx dµ(x δx 2 δx ( 2δ 1 1 dµ(x δx 2 δx 1 2δ 2 dµ(x δx 2 = δµ({x R : δx 2}. The followig lemma gives a coditio o the characteristic fuctios of a sequece of Borel probability measures o R uder which the sequece is tight Krisha B. Athreya ad Soumedra N. Lahiri, Measure Theory ad Probability Theory, p. 329, Lemma

17 Lemma 13. Suppose that µ P(R ad that µ coverges poitwise to a fuctio φ : R C that is cotiuous at 0. The the sequece µ is tight. Proof. Write φ = µ. Because φ 1, for each δ > 0, by the domiated covergece theorem we have 1 δ δ δ(1 φ (tdt 1 δ δ δ (1 φ(tdt. O the other had, that φ is cotiuous at 0 implies that for ay ɛ > 0 there is some η > 0 such that whe t < η, φ(t 1 < ɛ, ad hece for δ < η, 1 δ δ δ (1 φ(tdt 2 sup 1 φ(t 2ɛ, t δ thus 1 δ (1 φ(tdt 0, δ 0. δ δ Let ɛ > 0. There is some δ > 0 for which 1 δ (1 φ(tdt δ < ɛ. The there is some δ such that whe δ, 1 δ φ (tdt δ δ(1 1 δ (1 φ(tdt δ δ < ɛ, whece 1 δ δ δ δ (1 φ (tdt < 2ɛ. Lemma 12 the says ({ µ x R : x 2 } 1 δ (1 φ (tdt < 2ɛ. δ δ δ Furthermore, ay Borel probability measure o a Polish space is tight (Ulam s theorem. 17 Thus, for each 1 < δ, there is a compact set K for which µ (R \ K < ɛ. Let { K ɛ = K 1 K δ 1 x R : x 2 }, δ which is a compact set, ad for ay 1, showig that the sequece µ is tight. µ (R \ K ɛ < 2ɛ, 17 Alexader S. Kechris, Classical Descriptive Set Theory, p. 107, Theorem

18 For metrizable spaces X 1,..., X d, let X = d i=1 X i ad let π i : X X i be the projectio map. We establish that if H is a subset of P(X such that for each 1 i d the family of ith margials of H is tight, the H itself is tight. 18 Lemma 14. Let X 1,..., X d be metrizable topological spaces, let X = d i=1 X i, ad let H P(X. Suppose that for each 1 i d, H i = {π i µ : µ H } is a tight set i P(X i. The H is a tight set i P(X. Proof. For µ H, write µ i = π i µ. Let ɛ > 0 ad take 1 i d. Because H i is tight, there is a compact subset K i of X i such that for all µ i H i, Let The for ay µ H, K = µ i (X i \ K i < ɛ d. d K i = i=1 µ(x \ K = µ which shows that H is tight. = µ = µ = < = ɛ, ( d i=1 X \ ( d i=1 ( d d i=1 i=1 π 1 i (K i. d i=1 π 1 i (K i π 1 i (K i c π 1 i (X i \ K i µ(π 1 i (X i \ K i d µ i (X i \ K i i=1 d i=1 ɛ d 18 Luigi Ambrosio, Nicola Gigli, ad Giuseppe Savare, Gradiet Flows: I Metric Spaces ad i the Space of Probability Measures, p. 119, Lemma 5.2.2; V. I. Bogachev, Measure Theory, volume II, p. 94, Lemma

19 We ow prove Lévy s cotiuity theorem, which we shall use to prove the martigale cetral limit theorem. 19 Theorem 15 (Lévy s cotiuity theorem. Suppose that µ P(R d, If µ P(R d ad µ µ arrowly, the for ay u R d, µ (u µ(u,. 2. If there is some φ : R d C to which µ coverges poitwise ad φ is cotiuous at 0, the there is some µ P(R d such that φ = µ ad such that µ µ arrowly. Proof. Suppose that µ µ arrowly. For each u R d, the fuctio x e ix u is cotiuous R d C ad is bouded, so µ (u = e ix u dµ (x R d e ix u dµ(x = µ(u. R d Suppose that µ coverges poitwise to φ ad that φ is cotiuous at 0. For 1 i d, let π i : R d R be the projectio map ad defie ι i : R R d by takig the ith etry of ι i (t to be t ad the other etries to be 0. Fix 1 i d ad write ν = π i µ P(R, ad for t R we calculate ν (t = e ist dν (s R = e iπi(xt dµ (x R d = e ix ιi(t dµ (x R d = µ (ι i (t, so ν = µ ι i. By hypothesis, ν coverges poitwise to φ ι i. Because φ is cotiuous at 0 R d, the fuctio φ ι i is cotiuous at 0 R. The Lemma 13 tells us that the sequece ν is tight. That is, for each 1 i d, the set {π i µ : 1} is tight i P(R. Thus Lemma 14 tells us that the set {µ : 1} is tight i P(R d. Prokhorov s theorem 20 states that if X is a Polish space, the a subset H of P(X is tight if ad oly if each sequece of elemets of H has a subsequece 19 cf. Jea Jacod ad Philip Protter, Probability Essetials, secod ed., p. 167, Theorem V. I. Bogachev, Measure Theory, volume II, p. 202, Theorem

20 that coverges arrowly to some elemet of P(X. Thus, there is a subsequece µ a( of µ ad some µ P(R d such that µ a( coverges arrowly to µ. By the first part of the theorem, we get that µ a( coverges poitwise to µ. But by hypothesis µ coverges poitwise to φ, so φ = µ. Fially we prove that µ µ arrowly. Let µ b( be a subsequece of µ. Because {µ : 1} is tight, Prokhorov s theorem tells us that there is a subsequece µ c( of µ b( that coverges arrowly to some λ P(R d. By the first part of the theorem, µ c( coverges poitwise to λ. By hypothesis µ c( coverges poitwise to φ, so λ = φ = µ. The λ = µ. That is, ay subsequece of µ itself has a subsequece that coverges arrowly to µ, which implies that the sequece µ coverges arrowly to µ. (For a sequece x i a topological space X ad x X, x x if ad oly if each subsequece of x has a subsequece that coverges to x. 11 Martigale cetral limit theorem Let γ d be the stadard Gaussia measure o R d : γ d has desity 1 (2π d e 1 2 x 2 with respect to Lebesgue measure o R d. We ow prove the martigale cetral limit theorem. 21 Theorem 16 (Martigale cetral limit theorem. Suppose X j is a sequece i L 3 (Ω, A, P satisfyig the followig, with F k = σ(x 1,..., X k : 1. E(X j F j 1 = E(X 2 j F j 1 = There is some K for which E( X j 3 F j 1 K. The S coverges i distributio to some radom variable Z : Ω R with Z P = γ 1, where S = X j. j=1 Proof. For positive itegers ad j, defie φ,j (u = E(e iu 1 X j F j Jea Jacod ad Philip Protter, Probability Essetials, secod ed., p. 235, Theorem

21 For each ω Ω, by Taylor s theorem there is some ξ,j (ω betwee 0 ad X j (ω such that e iu 1 X j(ω = 1 + iu 1 X j (ω u2 2 X j(ω 2 iu3 6 ξ,j(ω 3. 3/2 Because f E(f F j 1 is a positive operator ad ξ,j 3 X j 3, we have, by the last hypothesis of the theorem, E( ξ,j 3 F j 1 E( X j 3 F j 1 K (3 we use this iequality later i the proof. Now, usig that E(X j F j 1 = 0 ad E(X 2 j F j 1 = 1, φ,j (u = 1 + iu 1 E(X j F j 1 u2 2 E(X2 j F j 1 iu3 6 3/2 E(ξ3,j F j 1 For p 1, = 1 u2 2 iu3 6 3/2 E(ξ3,j F j 1. E(e iu 1 S p = E(e iu 1 S p 1 e iu 1 X p which we write as ( E e i u S p = E(E(e iu 1 S p 1 e iu 1 X p F p 1 = E(e iu 1 S p 1 E(e iu 1 X p F p 1 = E(e iu 1 S p 1 φ,p (u = E (1 u2 2 ( ( e iu 1 S p 1 1 u2 2 Now usig (3 we get ( E e i u S p E iu3 6 E ( ξ,p F 3 3/2 p 1, ( e i u S p 1 = E e i u S p 1 iu 3 6 E ( ξ,p F 3 3/2 p 1. ( e i u S p 1 iu 3 (1 u2 e i u S p E ( ξ,p F 3 3/2 p 1 ( u 3 =E E ( ξ 3 6,p F 3/2 p 1 u 3 K. 63/2 Let u R ad let = (u be large eough so that 0 1 u For ( p 1 p, multiplyig the above iequality by 1 2 u2 yields p p+1 (1 u2 E(e iu 1 S p (1 u2 E(e iu 1 S p 1 K u 3. ( /2 21

22 Now, because p=1 (a p a p 1 = a a 0, ( ( p 1 u2 E(e iu 1 S p 2 p=1 =E(e iu 1 S (1 u2 2 =E(e iu 1 S (1 u2. 2 Usig this with (4 gives 1 E(eiu S E(e iu 1 S 0 (p 1 (1 u2 E(e iu 1 S p 1 2 (1 u2 K u 3 u 3 = K 2 63/2 6. 1/2 But if a b c, c 0, ad b b, the a b. As lim (1 u2 = e u2 2 2 ad K u 3 0, we therefore get that 6 1/2 as. Let µ = ( S E(e iu 1 S e u2 2 P ad let φ(u = e u2 2. We have just established that µ φ poitwise. The fuctio φ is cotiuous at 0, so Lévy s cotiuity theorem tells us that there is a Borel probability measure µ o R such that φ = µ ad such that µ coverges arrowly to µ. But φ(u = e u2 2 is the characteristic fuctio of γ 1, so we have that µ coverges arrowly to γ 1. 22