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

1 Introduction

In this note, any statement we make about filtrations and martingales is about filtrations and martingales indexed by the positive integers, rather than the nonnegative real numbers.

We take

$$inf\mathrm{\varnothing }=\mathrm{\infty },$$

and for $m>n$, we take

$$\sum _{k=m}^n=0.$$

(Defined rightly, these are not merely convenient ad hoc definitions.)

2 Conditional expectation

Let $(\mathrm{\Omega },𝒜,P)$ be a probability space and let $ℬ$ be a sub-$\sigma$-algebra of $𝒜$. For each $f\in L^1(\mathrm{\Omega },𝒜,P)$, there is some $g:\mathrm{\Omega }\to ℝ$ such that (i) $g$ is $ℬ$-measurable and (ii) for each $B\in ℬ$, ${\int }_{B}g𝑑P={\int }_{B}f𝑑P$, and if $h:\mathrm{\Omega }\to ℝ$ satisfies (i) and (ii) then $h(\omega )=g(\omega )$ for almost all $\omega \in \mathrm{\Omega }$. We denote some $g:\mathrm{\Omega }\to ℝ$ satisfying (i) and (ii) by $E(f|ℬ)$, called the conditional expectation of $f$ with respect to $\mathrm{B}$. In other words, $E(f|ℬ)$ is the unique element of $L^1(\mathrm{\Omega },ℬ,P)$ such that for each $B\in ℬ$,

$${\int }_{B}E(f|ℬ)𝑑P={\int }_{B}f𝑑P.$$

The map $f\mapsto E(f|ℬ)$ satisfies the following:

1. $f\mapsto E(f|ℬ)$ is positive linear operator $L^1(\mathrm{\Omega },𝒜,P)\to L^1(\mathrm{\Omega },ℬ,P)$ with norm $1$.

2. If $f\in L^1(\mathrm{\Omega },𝒜,P)$ and $g\in L^{\mathrm{\infty }}(\mathrm{\Omega },ℬ,P)$, then for almost all $\omega \in \mathrm{\Omega }$,

   $$E(gf|ℬ)(\omega )=g(\omega )E(f|ℬ)(\omega ).$$

3. If $𝒞$ is a sub-$\sigma$-algebra of $ℬ$, then for almost all $\omega \in \mathrm{\Omega }$,

   $$E(E(f|ℬ)|𝒞)(\omega )=E(f|𝒞)(\omega ).$$

4. If $f\in L^1(\mathrm{\Omega },ℬ,P)$ then for almost all $\omega \in \mathrm{\Omega }$,

   $$E(f|ℬ)(\omega )=f(\omega ).$$

5. If $f\in L^1(\mathrm{\Omega },𝒜,P)$, then for almost all $\omega \in \mathrm{\Omega }$,

   $$|E(f|ℬ)(\omega )|\le E(|f||ℬ)(\omega ).$$

6. If $f\in L^1(\mathrm{\Omega },𝒜,P)$ is independent of $ℬ$, then for almost all $\omega \in \mathrm{\Omega }$,

   $$E(f|ℬ)(\omega )=E(f).$$

3 Filtrations

A filtration of a $\sigma$-algebra $𝒜$ is a sequence ${ℱ}_n$, $n\ge 1$, of sub-$\sigma$-algebras of $𝒜$ such that ${ℱ}_m\subset {ℱ}_n$ if $m\le n$. We set ${ℱ}_0=\{\mathrm{\varnothing },\mathrm{\Omega }\}$.

A sequence of random variables ${\xi }_n:(\mathrm{\Omega },𝒜,P)\to ℝ$ is said to be adapted to the filtration ${\mathrm{F}}_n$ if for each $n$, ${\xi }_n$ is ${ℱ}_n$-measurable.

Let ${\xi }_n:(\mathrm{\Omega },𝒜,P)\to ℝ$, $n\ge 1$, be a sequence of random variables. The natural filtration of $\mathrm{A}$ corresponding to ${\xi }_n$ is

$${ℱ}_n=\sigma ({\xi }_1,\mathrm{\dots },{\xi }_n).$$

It is apparent that ${ℱ}_n$ is a filtration and that the sequence ${\xi }_n$ is adapted to ${ℱ}_n$.

4 Martingales

Let ${ℱ}_n$ be a filtration of a $\sigma$-algebra $𝒜$ and let ${\xi }_n:(\mathrm{\Omega },𝒜,P)\to ℝ$ be a sequence of random variables. We say that ${\xi }_n$ is a martingale with respect to ${\mathrm{F}}_n$ if (i) the sequence ${\xi }_n$ is adapted to the filtration ${ℱ}_n$, (ii) for each $n$, ${\xi }_n\in L^1(P)$, and (iii) for each $n$, for almost all $\omega \in \mathrm{\Omega }$,

$$E({\xi }_{n+1}|{ℱ}_n)(\omega )={\xi }_n(\omega ).$$

In particular,

$$E({\xi }_1)=E({\xi }_2)=\mathrm{\cdots },$$

i.e.

$$E({\xi }_m)=E({\xi }_n),m\le n.$$

We say that ${\xi }_n$ is a submartingale with respect to ${\mathrm{F}}_n$ if (i) and (ii) above are true, and if for each $n$, for almost all $\omega \in \mathrm{\Omega }$,

$$E({\xi }_{n+1}|{ℱ}_n)(\omega )\ge {\xi }_n(\omega ).$$

In particular,

$$E({\xi }_1)\le E({\xi }_2)\le \mathrm{\cdots },$$

i.e.

$$E({\xi }_m)\le E({\xi }_n),m\le n.$$

We say that ${\xi }_n$ is a supermartingale with respect to ${\mathrm{F}}_n$ if (i) and (ii) above are true, and if for each $n$, for almost all $\omega \in \mathrm{\Omega }$,

$${\xi }_n(\omega )\ge E({\xi }_{n+1}|{ℱ}_n)(\omega ).$$

In particular,

$$E({\xi }_1)\ge E({\xi }_2)\ge \mathrm{\cdots },$$

i.e.

$$E({\xi }_m)\ge E({\xi }_n),m\le n.$$

If we speak about a martingale without specifying a filtration, we mean a martingale with respect to the natural filtration corresponding to the sequence of random variables.

5 Stopping times

Let ${ℱ}_n$ be a filtration of a $\sigma$-algebra $𝒜$. A stopping time with respect to ${\mathrm{F}}_n$ is a function $\tau :\mathrm{\Omega }\to \{1,2,\mathrm{\dots }\}\cup \{\mathrm{\infty }\}$ such that for each $n\ge 1$,

$$\{\omega \in \mathrm{\Omega }:\tau (\omega )=n\}\in {ℱ}_n.$$

It is straightforward to check that a function $\tau :\mathrm{\Omega }\to \{1,2,\mathrm{\dots }\}\cup \{\mathrm{\infty }\}$ is a stopping time with respect to ${ℱ}_n$ if and only if for each $n\ge 1$,

$$\{\omega \in \mathrm{\Omega }:\tau (\omega )\le n\}\in {ℱ}_n.$$

The following lemma shows that the time of first entry into a Borel subset of $ℝ$ of a sequence of random variables adapted to a filtration is a stopping time.

Lemma 1.

Let ${\xi }_n$ be a sequence of random variables adapted to a filtration ${\mathrm{F}}_n$ and let $B\mathrm{\in }{\mathrm{B}}_{\mathrm{R}}$. Then

$$\tau (\omega )=inf\{n\ge 1:{\xi }_n(\omega )\in B\}$$

is a stopping time with respect to ${\mathrm{F}}_n$.

Proof.

Let $n\ge 1$. Then

$\mathrm{\{}\omega \in \mathrm{\Omega }:\tau (\omega )=n\}$ $=\left({\displaystyle \bigcap _{k=1}^{n-1}}\{\omega \in \mathrm{\Omega }:{\xi }_k(\omega )\notin B\}\right)\cap \{\omega \in \mathrm{\Omega }:{\xi }_n(\omega )\in B\}$

$=\left({\displaystyle \bigcap _{k=1}^{n-1}}{A}_k^c\right)\cap A_n,$

where

$$A_k=\{\omega \in \mathrm{\Omega }:{\xi }_k(\omega )\in B\}.$$

Because the sequence ${\xi }_k$ is adapated to the filtration ${ℱ}_k$, $A_k^c\in {ℱ}_k$ and $A_n\in {ℱ}_n$, and because ${ℱ}_k$ is a filtration, the right-hand side of the above belongs to ${ℱ}_n$. ∎

If ${\xi }_n$ is a sequence of random variables adapted to a filtration ${ℱ}_n$ and a stopping time $\tau$ with respect to ${ℱ}_n$, for $n\ge 1$ we define ${\xi }_{\tau \wedge n}:\mathrm{\Omega }\to ℝ$ by

$${\xi }_{\tau \wedge n}(\omega )={\xi }_{\tau (\omega )\wedge n}(\omega ),\omega \in \mathrm{\Omega }.$$

${\xi }_{\tau \wedge n}$ is called the sequence ${\xi }_n$ stopped at $\tau$.

Lemma 2.

${\xi }_{\tau \wedge n}:(\mathrm{\Omega },𝒜,P)\to ℝ$ is a sequence of random variables adapted to the filtration ${\mathrm{F}}_n$.

Proof.

Let $n\ge 1$ and let $B\in {ℬ}_{ℝ}$. Because

$$\{\omega :{\xi }_{\tau \wedge n}(\omega )\in B,\tau (\omega )>n\}=\{\omega :{\xi }_n(\omega )\in B,\tau (\omega )>n\}$$

and for any $k$,

$$\{\omega :{\xi }_{\tau \wedge n}(\omega )\in B,\tau (\omega )=k\}=\{\omega :{\xi }_k\in B,\tau (\omega )=k\},$$

we get

$$\{\omega :{\xi }_{\tau \wedge n}(\omega )\in B\}=\{\omega :{\xi }_n(\omega )\in B,\tau (\omega )>n\}\cup \bigcup _{k=1}^n\{\omega :{\xi }_k(\omega )\in B,\tau (\omega )=k\}.$$

But

$$\{{\xi }_n\in B,\tau >n\}=\{{\xi }_n\in B\}\cap \{\tau >n\}\in {ℱ}_n$$

and

$$\{{\xi }_k\in B,\tau =k\}=\{{\xi }_k\in B\}\cap \{\tau =k\}\in {ℱ}_k,$$

and therefore

$$\{{\xi }_{\tau \wedge n}\in B\}\in {ℱ}_n.$$

In particular, $\{{\xi }_{\tau \wedge n}\in B\}\in 𝒜$, namely, ${\xi }_{\tau \wedge n}$ is a random variable, and the above shows that this sequence is adapted to the filtration ${ℱ}_n$. ∎

We now prove that a stopped martingale is itself a martingale with respect to the same filtration.

Theorem 3.

Let ${\mathrm{F}}_n$ be a filtration of a $\sigma$-algebra $\mathrm{A}$ and let $\tau$ be a stopping time with respect to ${\mathrm{F}}_n$.

1. If ${\xi }_n$ is a submartingale with respect to ${ℱ}_n$ then so is ${\xi }_{\tau \wedge n}$.

2. If ${\xi }_n$ is a supermartingale with respect to ${ℱ}_n$ then so is ${\xi }_{\tau \wedge n}$.

3. If ${\xi }_n$ is a martingale with respect to ${ℱ}_n$ then so is ${\xi }_{\tau \wedge n}$.

Proof.

For $n\ge 1$, define

we remark that $\tau (\omega )\ge n$ if and only if $\tau (\omega )>n-1$ and if and only if $\tau (\omega )\le n-1$. For $B\in {ℬ}_{ℝ}$, (i) if $0,1\notin B$ then

$$\{\omega \in \mathrm{\Omega }:{\alpha }_n(\omega )\in B\}=\mathrm{\varnothing }\in {ℱ}_{n-1},$$

(ii) if $0,1\in B$ then

$$\{\omega \in \mathrm{\Omega }:{\alpha }_n(\omega )\in B\}=\mathrm{\Omega }\in {ℱ}_{n-1},$$

(iii) if $0\in B$ and $1\notin B$ then

$$\{\omega \in \mathrm{\Omega }:{\alpha }_n(\omega )\in B\}=\{\omega \in \mathrm{\Omega }:{\alpha }_n(\omega )=0\}=\{\omega \in \mathrm{\Omega }:\tau (\omega )\le n-1\}\in {ℱ}_{n-1},$$

and (iv) if $1\in B$ and $0\notin B$ then

$$\{\omega \in \mathrm{\Omega }:{\alpha }_n(\omega )\in B\}=\{\omega \in \mathrm{\Omega }:{\alpha }_n(\omega )=1\}=\{\omega \in \mathrm{\Omega }:\tau (\omega )>n-1\}\in {ℱ}_{n-1},$$

Therefore $\{{\alpha }_n\in B\}\in {ℱ}_{n-1}$.

Set ${\xi }_0=0$, and we check that

$${\xi }_{\tau \wedge n}=\sum _{k=1}^n{\alpha }_k({\xi }_k-{\xi }_{k-1}).$$

It is apparent from this expression that if ${\xi }_n$ is adapted to ${ℱ}_n$ then ${\xi }_{\tau \wedge n}$ is adapted to ${ℱ}_n$, and that if each ${\xi }_n$ belongs to $L^1(P)$ then each ${\xi }_{\tau \wedge n}$ belongs to $L^1(P)$. As each of ${\alpha }_1,\mathrm{\dots },{\alpha }_{n+1}$ is ${ℱ}_n$-measurable and is bounded,

$$E({\xi }_{\tau \wedge (n+1)}|{ℱ}_n)=\sum _{k=1}^{n+1}E({\alpha }_k({\xi }_k-{\xi }_{k-1})|{ℱ}_n)=\sum _{k=1}^{n+1}{\alpha }_{k}E({\xi }_k-{\xi }_{k-1}|{ℱ}_n).$$ (1)

Suppose that ${\xi }_n$ is a submartingale. By (1),

$E({\xi }_{\tau \wedge (n+1)}|{ℱ}_n)$ $={\displaystyle \sum _{k=1}^n}{\alpha }_k({\xi }_k-{\xi }_{k-1})+{\alpha }_{n+1}E({\xi }_{n+1}|{ℱ}_n)-{\alpha }_{n+1}{\xi }_n$

$\ge {\xi }_{\tau \wedge n}+{\alpha }_{n+1}{\xi }_n-{\alpha }_{n+1}{\xi }_n$

$={\xi }_{\tau \wedge n},$

which shows that ${\xi }_{\tau \wedge n}$ is a submartingale; the statement that $E({\xi }_{\tau \wedge (n+1)}|{ℱ}_n)\ge {\xi }_{\tau \wedge n}$ means that $E({\xi }_{\tau \wedge (n+1)}|{ℱ}_n)(\omega )\ge {\xi }_{\tau \wedge n}(\omega )$ for almost all $\omega \in \mathrm{\Omega }$. ∎

We now prove the optional stopping theorem.

Theorem 4 (Optional stopping theorem).

Let ${\mathrm{F}}_n$ be a filtration of a $\sigma$-algebra $\mathrm{A}$, let ${\xi }_n$ be a martingale with respect to ${\mathrm{F}}_n$, and let $\tau$ be a stopping time with respect to ${\mathrm{F}}_n$. Suppose that:

1. For almost all $\omega \in \mathrm{\Omega }$, .

2. ${\xi }_{\tau }\in L^1(\mathrm{\Omega },𝒜,P)$.

3. $E({\xi }_n{1}_{\{\tau >n\}})\to 0$ as $n\to \mathrm{\infty }$.

Then

$$E({\xi }_{\tau })=E({\xi }_1).$$

Proof.

For each $n$, $\mathrm{\Omega }=\{\tau \le n\}\cup \{\tau >n\}$, and therefore

$${\xi }_{\tau }={\xi }_{\tau \wedge n}+{\xi }_{\tau }{1}_{\{\tau >n\}}-{\xi }_n{1}_{\{\tau >n\}}={\xi }_{\tau \wedge n}+\sum _{k=n+1}^{\mathrm{\infty }}{\xi }_k{1}_{\{\tau =k\}}-{\xi }_n{1}_{\{\tau >n\}}.$$

Theorem 3 tells us that ${\xi }_{\tau \wedge n}$ is a martingale with respect to to ${ℱ}_n$, and hence

$$E({\xi }_{\tau \wedge n})=E({\xi }_{\tau \wedge 1})=E({\xi }_1),$$

so

$$E({\xi }_{\tau })=E({\xi }_1)+\sum _{k=n+1}^{\mathrm{\infty }}E({\xi }_k{1}_{\{\tau =k\}})-E({\xi }_n{1}_{\{\tau >n\}}).$$ (2)

But as ${\xi }_{\tau }\in L^1(P)$,

$${\int }_{\mathrm{\Omega }}({\xi }_{\tau })(\omega )𝑑P(\omega )=\sum _{k=1}^{\mathrm{\infty }}{\int }_{\{\tau =k\}}{\xi }_k(\omega )𝑑P(\omega )=\sum _{k=1}^{\mathrm{\infty }}E({\xi }_k{1}_{\{\tau =k\}}),$$

and the fact that this series converges means that ${\sum }_{k=n+1}^{\mathrm{\infty }}E({\xi }_k{1}_{\{\tau =k\}})\to 0$. With the hypothesis $E({\xi }_n{1}_{\{\tau >n\}})\to 0$, as $n\to \mathrm{\infty }$ we have

$$E({\xi }_1)+\sum _{k=n+1}^{\mathrm{\infty }}E({\xi }_k{1}_{\{\tau =k\}})-E({\xi }_n{1}_{\{\tau >n\}})\to E({\xi }_1).$$

But (2) is true for each $n$, so we get $E({\xi }_{\tau })=E({\xi }_1)$, proving the claim. ∎

Suppose that ${\eta }_n$ is a sequence of independent random variables each with the Rademacher distribution:

$$P({\eta }_n=1)=\frac{1}{2},P({\eta }_n=-1)=\frac{1}{2}.$$

Let ${\xi }_n={\sum }_{k=1}^n{\eta }_k$ and let ${ℱ}_n=\sigma ({\eta }_1,\mathrm{\dots },{\eta }_n)$. Because

$${\xi }_{n+1}^2={({\xi }_n+{\eta }_{n+1})}^2={\eta }_{n+1}^2+2{\eta }_{n+1}{\xi }_n+{\xi }_n^2,$$

we have, as ${\xi }_n$ is ${ℱ}_n$-measurable and belongs to $L^{\mathrm{\infty }}(P)$ and as ${\eta }_{n+1}$ is independent of the $\sigma$-algebra ${ℱ}_n$,

$E({\xi }_{n+1}^2-(n+1)|{ℱ}_n)$ $=E({\eta }_{n+1}^2+2{\eta }_{n+1}{\xi }_n+{\xi }_n^2-(n+1)|{ℱ}_n)$

$=E({\eta }_{n+1}^2)+2{\xi }_{n}E({\eta }_{n+1})+{\xi }_n^2-(n+1)$

$=1+0+{\xi }_n^2-(n+1)$

$={\xi }_n^2-n.$

Therefore, ${\xi }_n^2-n$ is a martingale with respect to ${ℱ}_n$.

Let $K$ be a positive integer and let

$$\tau =inf\{n\ge 1:|{\xi }_n|=K\}.$$

Namely, $\tau$ is the time of first entry in the Borel subset $\{-K,K\}$ of $ℝ$, hence by Lemma 1 is a stopping time with respect to the filtration ${ℱ}_n$. With some work, one shows that (i) $P(\tau >2Kn)\to 0$ as $n\to \mathrm{\infty }$, (ii) , and (iii) $E(({\xi }_n^2-n)1_{\{\tau >n\}})\to 0$ as $n\to \mathrm{\infty }$. Then we can apply the optional stopping theorem to the martingale ${\xi }_n^2-n$: we get that

$$E({\xi }_{\tau }^2-\tau )=E({\xi }_1^2-1)=E({\xi }_1^2)-1=E({\eta }_1^2)-1=0.$$

Hence

$$E(\tau )=E({\xi }_{\tau }^2).$$

But $|{\xi }_{\tau }|=K$, so ${\xi }_{\tau }^2=K^2$, hence

$$E(\tau )=E(K^2)=K^2.$$

6 Maximal inequalities

We now prove Doob’s maximal inequality.

Theorem 5 (Doob’s maximal inequality).

Suppose that ${\mathrm{F}}_n$ is a filtration of a $\sigma$-algebra $\mathrm{A}$, that ${\xi }_n$ is a submartingale with respect to ${\mathrm{F}}_n$, and that for each $n$, ${\xi }_n\mathrm{\ge }\mathrm{0}$. Then for each $n\mathrm{\ge }\mathrm{1}$ and $\lambda \mathrm{>}\mathrm{0}$,

$$\lambda P(\underset{1\le k\le n}{\max }{\xi }_k\ge \lambda )\le E\left({\xi }_n{1}_{\{{\max }_{1\le k\le n}{\xi }_k\ge \lambda \}}\right).$$

Proof.

Define ${\zeta }_n(\omega )={\max }_{1\le k\le n}{\xi }_k(\omega )$, which is ${ℱ}_n$-measurable, and define $\tau :\mathrm{\Omega }\to \{1,\mathrm{\dots },n\}$ by

$$\tau (\omega )=\min \{1\le k\le n:{\xi }_k(\omega )\ge \lambda \}$$

if there is some $1\le k\le n$ for which ${\xi }_k(\omega )\ge \lambda$, and $\tau (\omega )=n$ otherwise. For $1\le k\le n$,

and for $k>n$,

$$\{\tau =k\}=\mathrm{\varnothing }\in {ℱ}_k,$$

showing that $\tau$ is a stopping time with respect to the filtration ${ℱ}_k$.

For $k\ge 1$,

${\xi }_{k+1}-{\xi }_{\tau \wedge (k+1)}$ $={\displaystyle \sum _{j=1}^k}{1}_{\{\tau =j\}}({\xi }_{k+1}-{\xi }_{\tau \wedge (k+1)})={\displaystyle \sum _{j=1}^k}{1}_{\{\tau =j\}}({\xi }_{k+1}-{\xi }_j),$

hence, because $\tau$ is a stopping time with respect to the filtration ${ℱ}_k$ and because ${\xi }_k$ is a submartingale with respect to this filtration,

$E({\xi }_{k+1}-{\xi }_{\tau \wedge (k+1)}|{ℱ}_k)$ $={\displaystyle \sum _{j=1}^k}{1}_{\{\tau =j\}}E(({\xi }_{k+1}-{\xi }_j)|{ℱ}_k)$

$={\displaystyle \sum _{j=1}^k}{1}_{\{\tau =j\}}(E({\xi }_{k+1}|{ℱ}_k)-{\xi }_j)$

$\ge {\displaystyle \sum _{j=1}^k}{1}_{\{\tau =j\}}({\xi }_k-{\xi }_j)$

$={\displaystyle \sum _{j=1}^{k-1}}{1}_{\{\tau =j\}}({\xi }_k-{\xi }_j)$

$={\xi }_k-{\xi }_{\tau \wedge k},$

from which we have that the sequence ${\xi }_k-{\xi }_{\tau \wedge k}$ is a submartingale with respect to the filtration ${ℱ}_k$. Therefore

$$E({\xi }_k-{\xi }_{\tau \wedge k})\ge E({\xi }_1-{\xi }_{\tau \wedge 1})=E({\xi }_1)-E({\xi }_{\tau \wedge 1})=E({\xi }_1)-E({\xi }_1)=0,$$

and so $E({\xi }_{\tau \wedge k})\le E({\xi }_k)$. Because $\tau \wedge n=\tau$, this yields

$$E({\xi }_{\tau })\le E({\xi }_n).$$

We have

If $\omega \in \{{\zeta }_n\ge \lambda \}$ then $({\xi }_{\tau })(\omega )\ge \lambda$, and if then $\tau (\omega )=n$ and so $({\xi }_{\tau })(\omega )={\xi }_n(\omega )$. Therefore

Therefore

But , hence

$$\lambda P({\zeta }_n\ge \lambda )\le E({\xi }_n{1}_{\{{\zeta }_n\ge \lambda \}}),$$

which proves the claim. ∎

The following is Doob’s $L^{\mathrm{2}}$ maximal inequality, which we prove using Doob’s maximal inequality.

Theorem 6 (Doob’s $L^2$ maximal inequality).

Suppose that ${\mathrm{F}}_n$ is a filtration of a $\sigma$-algebra $\mathrm{A}$ and that ${\xi }_n$ is a submartingale with respect to ${\mathrm{F}}_n$ such that for each $n\mathrm{\ge }\mathrm{1}$, ${\xi }_n\mathrm{\ge }\mathrm{0}$ and ${\xi }_n\mathrm{\in }{L}^{\mathrm{2}}\mathit{}\mathrm{(}P\mathrm{)}$. Then for each $n\mathrm{\ge }\mathrm{1}$,

$$E\left({\left|\underset{1\le k\le n}{\max }{\xi }_k\right|}^2\right)\le 4E({\xi }_n^2).$$

Proof.

Define ${\zeta }_n(\omega )={\max }_{1\le k\le n}{\xi }_k(\omega )$. It is a fact that if $\eta \in L^2(P)$ and $\eta \ge 0$ then

$$E({\eta }^2)=2{\int }_0^{\mathrm{\infty }}tP(\eta \ge t)dt.$$

Using this, Doob’s maximal inequality, Fubini’s theorem, and the Cauchy-Schwarz inequality,

$E({\zeta }_n^2)$ $=2{\displaystyle {\int }_0^{\mathrm{\infty }}}tP({\zeta }_n>t)dt$

$\le 2{\displaystyle {\int }_0^{\mathrm{\infty }}}E({\xi }_n{1}_{\{{\zeta }_n\ge t\}}dt$

$=2{\displaystyle {\int }_0^{\mathrm{\infty }}}\left({\displaystyle {\int }_{\{{\zeta }_n\ge t\}}}{\xi }_n(\omega )𝑑P(\omega )\right)𝑑t$

$=2{\displaystyle {\int }_{\mathrm{\Omega }}}\left({\displaystyle {\int }_0^{{\zeta }_n(\omega )}}𝑑t\right){\xi }_n(\omega )𝑑P(\omega )$

$=2{\displaystyle {\int }_{\mathrm{\Omega }}}{\zeta }_n(\omega ){\xi }_n(\omega )𝑑P(\omega )$

$\le 2{(E({\zeta }_n^2))}^{1/2}{(E({\xi }_n^2))}^{1/2}.$

If $E({\zeta }_n^2)=0$ the claim is immediate. Otherwise, we divide this inequality by ${(E({\zeta }_n^2))}^{1/2}$ and obtain

$${(E({\zeta }_n^2))}^{1/2}\le 2{(E({\xi }_n^2))}^{1/2},$$

and so

$$E({\zeta }_n^2)\le 4E({\xi }_n^2),$$

proving the claim. ∎

7 Upcrossings

Suppose that ${\xi }_n$ is a sequence of random variables that is adapted to a filtration ${ℱ}_n$ and let be real numbers. Define

$${\tau }_0=0,$$

and by induction for $m\ge 1$,

$${\sigma }_m(\omega )=inf\{k\ge {\tau }_{m-1}(\omega ):{\xi }_k(\omega )\le a\}$$

and

$${\tau }_m(\omega )=inf\{k\ge {\sigma }_m(\omega ):{\xi }_k(\omega )\ge b\},$$

where $inf\mathrm{\varnothing }=\mathrm{\infty }$. For each $m$, ${\tau }_m$ and ${\sigma }_m$ are each stopping times with respect to the filtration ${ℱ}_k$. For $n\ge 0$ we define

$$U_n[a,b](\omega )=sup\{m\ge 0:{\tau }_m(\omega )\le n\}.$$

For $x\in ℝ$, we write

$$x^{-}=\max \{0,-x\}=-\min \{0,x\}.$$

namely, the negative part of $x$.

We now prove the upcrossings inequality.

Theorem 7 (Upcrossings inequality).

If ${\xi }_n$, $n\mathrm{\ge }\mathrm{1}$, is a supermartingale with respect to a filtration ${\mathrm{F}}_n$ and , then for each $n\mathrm{\ge }\mathrm{1}$,

$$(b-a)E(U_n[a,b])\le E({({\xi }_n-a)}^{-}).$$

Proof.

For $n\ge 1$ and $\omega \in \mathrm{\Omega }$, and writing $N=U_n[a,b](\omega )$, for which $N\le n$, we have

Because ${\xi }_{{\sigma }_{N+1}}(\omega )\le a$, we have

$$(b-a)N\le 1_{\{{\sigma }_{N+1}\le n\}}(\omega )(a-{\xi }_n(\omega ))+\sum _{m=1}^n({\xi }_{{\tau }_m\wedge n}(\omega )-{\xi }_{{\sigma }_m\wedge n}(\omega )).$$

One proves that

$$1_{\{{\sigma }_{N+1}\le n\}}(\omega )(a-{\xi }_n(\omega ))\le \min \{0,a-{\xi }_n(\omega )\}={({\xi }_n(\omega )-a)}^{-}.$$

Thus

$$(b-a)E(U_n[a,b])\le E({({\xi }_n-a)}^{-})+\sum _{m=1}^{n}E({\xi }_{{\tau }_m\wedge n}-{\xi }_{{\sigma }_m\wedge n}).$$

Using that ${\xi }_n$ is a supermartingale, for each $1\le m\le n$,

$E({\xi }_{{\tau }_m\wedge n}-{\xi }_{{\sigma }_m\wedge n})$ $\le 0.$

Therefore

$$(b-a)E(U_n[a,b])\le E({({\xi }_n-a)}^{-}).$$

∎

8 Doob’s martingale convergence theorem

We now use the uprossings inequality to prove Doob’s martingale convergence theorem.

Theorem 8 (Doob’s martingale convergence theorem).

Suppose that ${\xi }_n$, $n\mathrm{\ge }\mathrm{1}$, is a supermartingale with respect to a filtration ${\mathrm{F}}_n$ and that

Then there is some $\xi \mathrm{\in }{L}^{\mathrm{1}}\mathit{}\mathrm{(}\mathrm{\Omega }\mathrm{,}\mathrm{A}\mathrm{,}P\mathrm{)}$ such that for almost all $\omega \mathrm{\in }\mathrm{\Omega }$,

$$\underset{n\to \mathrm{\infty }}{lim}{\xi }_n(\omega )=\xi (\omega )$$

and with $E\mathit{}\mathrm{(}\mathrm{|}\xi \mathrm{|}\mathrm{)}\mathrm{\le }M$.

Proof.

For any and $n\ge 1$, the upcrossings inequality tells us that

$$E(U_n[a,b])\le \frac{E{({\xi }_n-a)}^{-})}{b-a}\le \frac{E(|{\xi }_n-a|)}{b-a}\le \frac{E(|{\xi }_n|+|a|)}{b-a}\le \frac{M+|a|}{b-a}.$$

For each $\omega \in \mathrm{\Omega }$, the sequence $U_n[a,b](\omega )\in [0,\mathrm{\infty })$ is nondecreasing, so by the monotone convergence theorem,

$$E\left(\underset{n\to \mathrm{\infty }}{lim}{U}_n[a,b]\right)=\underset{n\to \mathrm{\infty }}{lim}E(U_n[a,b])\le \frac{M+|a|}{b-a}.$$

This implies that

Let

This is an intersection of countably many sets each with measure $1$, so $P(A)=1$.

Let

If $\omega \in B$, then there are $a,b\in \mathbb{Q}$, , such that

It follows from this ${lim}_{n\to \mathrm{\infty }}{U}_n[a,b](\omega )=\mathrm{\infty }$. Thus $\omega \notin A$, so $B\cap A=\mathrm{\varnothing }$, and because $P(A)=1$ we get $P(B)=0$.

We define $\xi :\mathrm{\Omega }\to ℝ$ by

which is Borel measurable. Furthermore, since $|\xi |={lim\; inf}_n|{\xi }_n|$ almost everywhere, by Fatou’s lemma we obtain

$E(|\xi |)$ $=E(\underset{n}{lim\; inf}|{\xi }_n|)$

$\le \underset{n}{lim\; inf}E(|{\xi }_n|)$

$\le \underset{n}{sup}E(|{\xi }_n|)$

$=M.$

∎

9 Uniform integrability

Let $\xi :(\mathrm{\Omega },𝒜,P)\to ℝ$ be a random variable. It is a fact that $\xi \in L^1$ if and only if for each $ϵ>0$ there is some $M$ such that

(One’s instinct might be to try to use the Cauchy-Schwarz inequality to prove this. This doesn’t work.) Thus, if ${\xi }_n$ is a sequence in $L^1(\mathrm{\Omega },𝒜,P)$ then for each $ϵ>0$ there are $M_n$ such that, for each $n$,

A sequence of random variables ${\xi }_n$ is said to be uniformly integrable if for each $ϵ>0$ there is some $M$ such that, for each $n$,

If a sequence ${\xi }_n$ is uniformly integrable, then there is some $M$ such that for each $n$,

and so

The following lemma states that the conditional expectations of an integrable random variable with respect to a filtration is a uniformly integrable martingale with respect to that filtration.

Lemma 9.

Suppose that $\xi \mathrm{\in }{L}^{\mathrm{1}}\mathit{}\mathrm{(}\mathrm{\Omega }\mathrm{,}\mathrm{A}\mathrm{,}P\mathrm{)}$ and that ${\mathrm{F}}_n$ is a filtration of $\mathrm{A}$. Then $E\mathit{}\mathrm{(}\xi \mathrm{|}{\mathrm{F}}_n\mathrm{)}$ is a martingale with respect to ${\mathrm{F}}_n$ and is uniformly integrable.

We now prove that a uniformly integrable supermartingale converges in $L^1$.

Theorem 10.

Suppose that ${\xi }_n$ is a supermartingale with respect to a filtration ${\mathrm{F}}_n$, and that the sequence ${\xi }_n$ is uniformly integrable. Then there is some $\xi \mathrm{\in }{L}^{\mathrm{1}}\mathit{}\mathrm{(}\mathrm{\Omega }\mathrm{,}\mathrm{A}\mathrm{,}P\mathrm{)}$ such that ${\xi }_n\mathrm{\to }\xi$ in $L^{\mathrm{1}}$.

Proof.

Because the sequence ${\xi }_n$ is uniformly integrable, there is some $M$ such that for each $n\ge 1$,

$$E(|{\xi }_n|)\le M+1.$$

Thus, because ${\xi }_n$ is a supermartingale, Doob’s martingale convergence theorem tells us that there is some $\xi \in L^1(\mathrm{\Omega },𝒜,P)$ such that for almost all $\omega \in \mathrm{\Omega }$,

$$\underset{n\to \mathrm{\infty }}{lim}{\xi }_n(\omega )=\xi (\omega ).$$

Because ${\xi }_n$ is uniformly integrable and converges almost surely to $\xi$, the Vitali convergence theorem tells us that ${\xi }_n\to \xi$ in $L^1$. ∎

The above theorem shows in particular that a uniformly integrable martingale converges to some limit in $L^1$. The following theorem shows that the terms of the sequence are equal to the conditional expectations of this limit with respect to the natural filtration.

Theorem 11.

Suppose that a sequence of random variables ${\xi }_n$ is uniformly integrable and is a martingale with respect to its natural filtration

$${ℱ}_n=\sigma ({\xi }_1,\mathrm{\dots },{\xi }_n).$$

Then there is some $\xi \mathrm{\in }{L}^{\mathrm{1}}\mathit{}\mathrm{(}\mathrm{\Omega }\mathrm{,}\mathrm{A}\mathrm{,}P\mathrm{)}$ such that ${\xi }_n\mathrm{\to }\xi$ in $L^{\mathrm{1}}$ and such that for each $n\mathrm{\ge }\mathrm{1}$, for almost all $\omega \mathrm{\in }\mathrm{\Omega }$,

$${\xi }_n(\omega )=E(\xi |{ℱ}_n)(\omega ).$$

Proof.

By Theorem 10, there is some $\xi \in L^1(\mathrm{\Omega },𝒜,P)$ such that ${\xi }_n\to \xi$ in $L^1$. The hypothesis that the sequence ${\xi }_n$ is a martingale with respect to ${ℱ}_n$ tells us that for that for $n\ge 1$ and for any $m\ge n$,

$$E({\xi }_m|{ℱ}_n)={\xi }_n,$$

and so for $A\in {ℱ}_n$,

$${\int }_A{\xi }_m𝑑P={\int }_{A}E({\xi }_m|{ℱ}_n)𝑑P={\int }_A{\xi }_n𝑑P.$$

Thus

$\left|{\displaystyle {\int }_A}({\xi }_n-\xi )𝑑P\right|$ $=\left|{\displaystyle {\int }_A}({\xi }_m-\xi )𝑑P\right|$

$\le {\displaystyle {\int }_A}|{\xi }_m-\xi |𝑑P$

$\le E(|{\xi }_m-\xi |).$

But $E(|{\xi }_m-\xi |)\to 0$ as $m\to \mathrm{\infty }$. Since $m$ does not appear in the left-hand side, we have

$$\left|{\int }_A({\xi }_n-\xi )𝑑P\right|=0,$$

and thus

$${\int }_A{\xi }_n𝑑P={\int }_A\xi 𝑑P.$$

But $E(f|{ℱ}_n)$ is the unique element of $L^1(\mathrm{\Omega },{ℱ}_n,P)$ such that for each $A\in {ℱ}_n$,

$${\int }_{A}E(f|{ℱ}_n)𝑑P={\int }_{A}f𝑑P,$$

and because ${\xi }_n$ satisfies this, we get that ${\xi }_n=E(f|{ℱ}_n)$ in $L^1$, i.e., for almost all $\omega \in \mathrm{\Omega }$,

$${\xi }_n(\omega )=E(f|{ℱ}_n)(\omega ),$$

proving the claim. ∎

10 Lévy’s continuity theorem

For a metrizable topological space $X$, we denote by $𝒫(X)$ the set of Borel probability measures on $X$. The narrow topology on $\mathrm{P}\mathbf{}\mathrm{(}X\mathrm{)}$ is the coarsest topology such that for each $f\in C_b(X)$, the map

$$\mu \mapsto {\int }_{X}f𝑑\mu$$

is continuous $𝒫(X)\to ℂ$.

A subset $\mathscr{H}$ of $𝒫(X)$ is called tight if for each $ϵ>0$ there is a compact subset $K_{ϵ}$ of $X$ such that if $\mu \in \mathscr{H}$ then , i.e. $\mu (K_{ϵ})>1-ϵ$. (An element $\mu$ of $𝒫(X)$ is called tight when $\{\mu \}$ is a tight subset of $𝒫(X)$.)

For a Borel probability measure $\mu$ on ${ℝ}^d$, we define its characteristic function $\tilde{\mu }:{ℝ}^d\to ℂ$ by

$$\tilde{\mu }(u)={\int }_{{ℝ}^d}{e}^{ix\cdot u}𝑑\mu (x),u\in {ℝ}^d.$$

$\tilde{\mu }$ is bounded by $1$ and is uniformly continuous. Because $\mu ({ℝ}^d)=1$,

$$\tilde{\mu }(0)=1.$$

Lemma 12.

Let $\mu \mathrm{\in }\mathrm{P}\mathit{}\mathrm{(}\mathrm{R}\mathrm{)}$. For $\delta \mathrm{>}\mathrm{0}$,

$$\mu \left(\{x\in ℝ:|x|\ge \frac{2}{\delta }\}\right)\le \frac{1}{\delta }{\int }_{-\delta }^{\delta }(1-\tilde{\mu }(u))𝑑u;$$

in particular, the right-hand side of this inequality is real.

Proof.

Using Fubini’s theorem and the fact that all real $t$, $1-\frac{\sin t}{t}\ge 0$,

${\displaystyle {\int }_{-\delta }^{\delta }}(1-\tilde{\mu }(u))𝑑u$ $={\displaystyle {\int }_{-\delta }^{\delta }}\left({\displaystyle {\int }_{ℝ}}(1-e^{ixu})𝑑\mu (x)\right)𝑑u$

$={\displaystyle {\int }_{ℝ}}\left({\displaystyle {\int }_{-\delta }^{\delta }}1-e^{iux}du\right)𝑑\mu (x)$

$={\displaystyle {\int }_{ℝ}}{\left(u-{\displaystyle \frac{e^{iux}}{ix}}\right)}_{-\delta }^{\delta }𝑑\mu (x)$

$={\displaystyle {\int }_{ℝ}}\left(2\delta -{\displaystyle \frac{e^{i\delta x}}{ix}}+{\displaystyle \frac{e^{-i\delta x}}{ix}}\right)𝑑\mu (x)$

$=2\delta {\displaystyle {\int }_{ℝ}}\left(1-{\displaystyle \frac{\sin (\delta x)}{\delta x}}\right)𝑑\mu (x)$

$\ge 2\delta {\displaystyle {\int }_{|\delta x|\ge 2}}\left(1-{\displaystyle \frac{\sin (\delta x)}{\delta x}}\right)𝑑\mu (x)$

$\ge 2\delta {\displaystyle {\int }_{|\delta x|\ge 2}}\left(1-{\displaystyle \frac{1}{|\delta x|}}\right)𝑑\mu (x)$

$\ge 2\delta {\displaystyle {\int }_{|\delta x|\ge 2}}{\displaystyle \frac{1}{2}}𝑑\mu (x)$

$=\delta \mu (\{x\in ℝ:|\delta x|\ge 2\}).$

∎

The following lemma gives a condition on the characteristic functions of a sequence of Borel probability measures on $ℝ$ under which the sequence is tight.

Lemma 13.

Suppose that ${\mu }_n\mathrm{\in }\mathrm{P}\mathit{}\mathrm{(}\mathrm{R}\mathrm{)}$ and that ${\stackrel{\mathrm{~}}{\mu }}_n$ converges pointwise to a function $\varphi \mathrm{:}\mathrm{R}\mathrm{\to }\mathrm{C}$ that is continuous at $\mathrm{0}$. Then the sequence ${\mu }_n$ is tight.

Proof.

Write ${\varphi }_n={\tilde{\mu }}_n$. Because $|{\varphi }_n|\le 1$, for each $\delta >0$, by the dominated convergence theorem we have

$$\frac{1}{\delta }{\int }_{-\delta }^{\delta }(1-{\varphi }_n(t))𝑑t\to \frac{1}{\delta }{\int }_{-\delta }^{\delta }(1-\varphi (t))𝑑t.$$

On the other hand, that $\varphi$ is continuous at $0$ implies that for any $ϵ>0$ there is some $\eta >0$ such that when , , and hence for ,

$$\frac{1}{\delta }{\int }_{-\delta }^{\delta }(1-\varphi (t))𝑑t\le 2\underset{|t|\le \delta }{sup}|1-\varphi (t)|\le 2ϵ,$$

thus

$$\frac{1}{\delta }{\int }_{-\delta }^{\delta }(1-\varphi (t))𝑑t\to 0,\delta \to 0.$$

Let $ϵ>0$. There is some $\delta >0$ for which

Then there is some $n_{\delta }$ such that when $n\ge n_{\delta }$,

whence

Lemma 12 then says

Furthermore, any Borel probability measure on a Polish space is tight (Ulam’s theorem). Thus, for each , there is a compact set $K_n$ for which . Let

$$K_{ϵ}=K_1\cup \mathrm{\cdots }\cup K_{n_{\delta }-1}\cup \{x\in ℝ:|x|\le \frac{2}{\delta }\},$$

which is a compact set, and for any $n\ge 1$,

showing that the sequence ${\mu }_n$ is tight. ∎

For metrizable spaces $X_1,\mathrm{\dots },X_d$, let $X={\prod }_{i=1}^d{X}_i$ and let ${\pi }_i:X\to X_i$ be the projection map. We establish that if $\mathscr{H}$ is a subset of $𝒫(X)$ such that for each $1\le i\le d$ the family of $i$th marginals of $\mathscr{H}$ is tight, then $\mathscr{H}$ itself is tight.

Lemma 14.

Let $X_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{X}_d$ be metrizable topological spaces, let $X\mathrm{=}{\mathrm{\prod }}_{i\mathrm{=}\mathrm{1}}^d{X}_i$, and let $\mathrm{H}\mathrm{\subset }\mathrm{P}\mathit{}\mathrm{(}X\mathrm{)}$. Suppose that for each $\mathrm{1}\mathrm{\le }i\mathrm{\le }d$,

$${\mathscr{H}}_i=\{\pi _i{}_{*}\mu :\mu \in \mathscr{H}\}$$

is a tight set in $\mathrm{P}\mathit{}\mathrm{(}{X}_i\mathrm{)}$. Then $\mathrm{H}$ is a tight set in $\mathrm{P}\mathit{}\mathrm{(}X\mathrm{)}$.

Proof.

For $\mu \in \mathscr{H}$, write ${\mu }_i=\pi _i{}_{*}\mu$. Let $ϵ>0$ and take $1\le i\le d$. Because ${\mathscr{H}}_i$ is tight, there is a compact subset $K_i$ of $X_i$ such that for all ${\mu }_i\in {\mathscr{H}}_i$,

Let

$$K=\prod _{i=1}^d{K}_i=\bigcap _{i=1}^d{\pi }_i^{-1}(K_i).$$

Then for any $\mu \in \mathscr{H}$,

$\mu (X\setminus K)$ $=\mu \left(X\setminus {\displaystyle \bigcap _{i=1}^d}{\pi }_i^{-1}(K_i)\right)$

$=\mu \left({\displaystyle \bigcup _{i=1}^d}{\pi }_i^{-1}{(K_i)}^c\right)$

$=\mu \left({\displaystyle \bigcup _{i=1}^d}{\pi }_i^{-1}(X_i\setminus K_i)\right)$

$\le {\displaystyle \sum _{i=1}^d}\mu ({\pi }_i^{-1}(X_i\setminus K_i))$

$={\displaystyle \sum _{i=1}^d}{\mu }_i(X_i\setminus K_i)$

$=ϵ,$

which shows that $\mathscr{H}$ is tight. ∎

We now prove Lévy’s continuity theorem, which we shall use to prove the martingale central limit theorem.

Theorem 15 (Lévy’s continuity theorem).

Suppose that ${\mu }_n\mathrm{\in }\mathrm{P}\mathit{}\mathrm{(}{\mathrm{R}}^d\mathrm{)}$, $n\mathrm{\ge }\mathrm{1}$.

1. If $\mu \in 𝒫({ℝ}^d)$ and ${\mu }_n\to \mu$ narrowly, then for any $u\in {ℝ}^d$,

   $${\tilde{\mu }}_n(u)\to \tilde{\mu }(u),n\to \mathrm{\infty }.$$

2. If there is some $\varphi :{ℝ}^d\to ℂ$ to which ${\tilde{\mu }}_n$ converges pointwise and $\varphi$ is continuous at $0$, then there is some $\mu \in 𝒫({ℝ}^d)$ such that $\varphi =\tilde{\mu }$ and such that ${\mu }_n\to \mu$ narrowly.

Proof.

Suppose that ${\mu }_n\to \mu$ narrowly. For each $u\in {ℝ}^d$, the function $x\mapsto e^{ix\cdot u}$ is continuous ${ℝ}^d\to ℂ$ and is bounded, so

$${\tilde{\mu }}_n(u)={\int }_{{ℝ}^d}{e}^{ix\cdot u}𝑑{\mu }_n(x)\to {\int }_{{ℝ}^d}{e}^{ix\cdot u}𝑑\mu (x)=\tilde{\mu }(u).$$

Suppose that ${\tilde{\mu }}_n$ converges pointwise to $\varphi$ and that $\varphi$ is continuous at $0$. For $1\le i\le d$, let ${\pi }_i:{ℝ}^d\to ℝ$ be the projection map and define ${\iota }_i:ℝ\to {ℝ}^d$ by taking the $i$th entry of ${\iota }_i(t)$ to be $t$ and the other entries to be $0$. Fix $1\le i\le d$ and write ${\nu }_n=\pi _i{}_{*}{\mu }_n\in 𝒫(ℝ)$, and for $t\in ℝ$ we calculate

${\tilde{\nu }}_n(t)$ $={\displaystyle {\int }_{ℝ}}{e}^{ist}𝑑{\nu }_n(s)$

$={\displaystyle {\int }_{{ℝ}^d}}{e}^{i{\pi }_i(x)t}𝑑{\mu }_n(x)$

$={\displaystyle {\int }_{{ℝ}^d}}{e}^{ix\cdot {\iota }_i(t)}𝑑{\mu }_n(x)$

$={\tilde{\mu }}_n({\iota }_i(t)),$

so ${\tilde{\nu }}_n={\tilde{\mu }}_n\circ {\iota }_i$. By hypothesis, ${\tilde{\nu }}_n$ converges pointwise to $\varphi \circ {\iota }_i$. Because $\varphi$ is continuous at $0\in {ℝ}^d$, the function $\varphi \circ {\iota }_i$ is continuous at $0\in ℝ$. Then Lemma 13 tells us that the sequence ${\nu }_n$ is tight. That is, for each $1\le i\le d$, the set

$$\{\pi _i{}_{*}{\mu }_n:n\ge 1\}$$

is tight in $𝒫(ℝ)$. Thus Lemma 14 tells us that the set

$$\{{\mu }_n:n\ge 1\}$$

is tight in $𝒫({ℝ}^d)$.

Prokhorov’s theorem states that if $X$ is a Polish space, then a subset $\mathscr{H}$ of $𝒫(X)$ is tight if and only if each sequence of elements of $\mathscr{H}$ has a subsequence that converges narrowly to some element of $𝒫(X)$. Thus, there is a subsequence ${\mu }_{a(n)}$ of ${\mu }_n$ and some $\mu \in 𝒫({ℝ}^d)$ such that ${\mu }_{a(n)}$ converges narrowly to $\mu$. By the first part of the theorem, we get that ${\tilde{\mu }}_{a(n)}$ converges pointwise to $\tilde{\mu }$. But by hypothesis ${\tilde{\mu }}_n$ converges pointwise to $\varphi$, so $\varphi =\tilde{\mu }$.

Finally we prove that ${\mu }_n\to \mu$ narrowly. Let ${\mu }_{b(n)}$ be a subsequence of ${\mu }_n$. Because $\{{\mu }_n:n\ge 1\}$ is tight, Prokhorov’s theorem tells us that there is a subsequence ${\mu }_{c(n)}$ of ${\mu }_{b(n)}$ that converges narrowly to some $\lambda \in 𝒫({ℝ}^d)$. By the first part of the theorem, ${\tilde{\mu }}_{c(n)}$ converges pointwise to $\tilde{\lambda }$. By hypothesis ${\tilde{\mu }}_{c(n)}$ converges pointwise to $\varphi$, so $\tilde{\lambda }=\varphi =\tilde{\mu }$. Then $\lambda =\mu$. That is, any subsequence of ${\mu }_n$ itself has a subsequence that converges narrowly to $\mu$, which implies that the sequence ${\mu }_n$ converges narrowly to $\mu$. (For a sequence $x_n$ in a topological space $X$ and $x\in X$, $x_n\to x$ if and only if each subsequence of $x_n$ has a subsequence that converges to $x$.) ∎

11 Martingale central limit theorem

Let ${\gamma }_d$ be the standard Gaussian measure on ${\mathrm{R}}^d$: ${\gamma }_d$ has density

$$\frac{1}{\sqrt{{(2\pi )}^d}}{e}^{-\frac{1}{2}{|x|}^2}$$

with respect to Lebesgue measure on ${ℝ}^d$.

We now prove the martingale central limit theorem.

Theorem 16 (Martingale central limit theorem).

Suppose $X_j$ is a sequence in $L^{\mathrm{3}}\mathit{}\mathrm{(}\mathrm{\Omega }\mathrm{,}\mathrm{A}\mathrm{,}P\mathrm{)}$ satisfying the following, with ${\mathrm{F}}_k\mathrm{=}\sigma \mathit{}\mathrm{(}{X}_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{X}_k\mathrm{)}$:

1. $E(X_j|{ℱ}_{j-1})=0$.
   

2. $E(X_j^2|{ℱ}_{j-1})=1$.
   

3. There is some $K$ for which $E({|X_j|}^3|{ℱ}_{j-1})\le K$.

Then $\frac{S_n}{\sqrt{n}}$ converges in distribution to some random variable $Z\mathrm{:}\mathrm{\Omega }\mathrm{\to }\mathrm{R}$ with $Z_{\mathrm{*}}\mathit{}P\mathrm{=}{\gamma }_{\mathrm{1}}$, where

$$S_n=\sum _{j=1}^n{X}_j.$$

Proof.

For positive integers $n$ and $j$, define

$${\varphi }_{n,j}(u)=E(e^{iu\frac{1}{\sqrt{n}}{X}_j}|{ℱ}_{j-1}).$$

For each $\omega \in \mathrm{\Omega }$, by Taylor’s theorem there is some ${\xi }_{n,j}(\omega )$ between $0$ and $X_j(\omega )$ such that

$$e^{iu\frac{1}{\sqrt{n}}{X}_j(\omega )}=1+iu\frac{1}{\sqrt{n}}{X}_j(\omega )-\frac{u^2}{2n}{X}_j{(\omega )}^2-\frac{i{u}^3}{6n^{3/2}}{\xi }_{n,j}{(\omega )}^3.$$

Because $f\mapsto E(f|{ℱ}_{j-1})$ is a positive operator and ${|{\xi }_{n,j}|}^3\le {|X_j|}^3$, we have, by the last hypothesis of the theorem,

$$E({|{\xi }_{n,j}|}^3|{ℱ}_{j-1})\le E({|X_j|}^3|{ℱ}_{j-1})\le K$$ (3)

we use this inequality later in the proof. Now, using that $E(X_j|{ℱ}_{j-1})=0$ and $E(X_j^2|{ℱ}_{j-1})=1$,

${\varphi }_{n,j}(u)$ $=1+iu{\displaystyle \frac{1}{\sqrt{n}}}E(X_j|{ℱ}_{j-1})-{\displaystyle \frac{u^2}{2n}}E(X_j^2|{ℱ}_{j-1})-{\displaystyle \frac{i{u}^3}{6n^{3/2}}}E({\xi }_{n,j}^3|{ℱ}_{j-1})$

$=1-{\displaystyle \frac{u^2}{2n}}-{\displaystyle \frac{i{u}^3}{6n^{3/2}}}E({\xi }_{n,j}^3|{ℱ}_{j-1}).$

For $p\ge 1$,

$E(e^{iu\frac{1}{\sqrt{n}}{S}_p})$ $=E(e^{iu\frac{1}{\sqrt{n}}{S}_{p-1}}{e}^{iu\frac{1}{\sqrt{n}}{X}_p})$

$=E(E(e^{iu\frac{1}{\sqrt{n}}{S}_{p-1}}{e}^{iu\frac{1}{\sqrt{n}}{X}_p}|{ℱ}_{p-1}))$

$=E(e^{iu\frac{1}{\sqrt{n}}{S}_{p-1}}E(e^{iu\frac{1}{\sqrt{n}}{X}_p}|{ℱ}_{p-1}))$

$=E(e^{iu\frac{1}{\sqrt{n}}{S}_{p-1}}{\varphi }_{n,p}(u))$

$=E\left(e^{iu\frac{1}{\sqrt{n}}{S}_{p-1}}\left(1-{\displaystyle \frac{u^2}{2n}}-{\displaystyle \frac{i{u}^3}{6n^{3/2}}}E\left({\xi }_{n,p}^3|{ℱ}_{p-1}\right)\right)\right),$

which we write as

$$E\left(e^{i\frac{u}{\sqrt{n}}{S}_p}-\left(1-\frac{u^2}{2n}\right)e^{i\frac{u}{\sqrt{n}}{S}_{p-1}}\right)=-E\left(e^{i\frac{u}{\sqrt{n}}{S}_{p-1}}\frac{i{u}^3}{6n^{3/2}}E\left({\xi }_{n,p}^3|{ℱ}_{p-1}\right)\right).$$

Now using (3) we get

Let $u\in ℝ$ and let $n=n(u)$ be large enough so that $0\le 1-\frac{u^2}{2n}\le 1$. For $1\le p\le n$, multiplying the above inequality by ${\left(1-\frac{u^2}{2n}\right)}^{n-p}$ yields

$$|{(1-\frac{u^2}{2n})}^{n-p}E(e^{iu\frac{1}{\sqrt{n}}{S}_p})-{(1-\frac{u^2}{2n})}^{n-p+1}E(e^{iu\frac{1}{\sqrt{n}}{S}_{p-1}}|\le K\frac{{|u|}^3}{6n^{3/2}}.$$ (4)

Now, because ${\sum }_{p=1}^n(a_p-a_{p-1})=a_n-a_0$,

Using this with (4) gives

$$\left|E(e^{iu\frac{1}{\sqrt{n}}{S}_n})-{\left(1-\frac{u^2}{2n}\right)}^n\right|\le n\cdot K\frac{{|u|}^3}{6n^{3/2}}=K\frac{{|u|}^3}{6n^{1/2}}.$$

But if $|a_n-b_n|\le c_n$, $c_n\to 0$, and $b_n\to b$, then $a_n\to b$. As

$$\underset{n\to \mathrm{\infty }}{lim}{\left(1-\frac{u^2}{2n}\right)}^n=e^{-\frac{u^2}{2}}$$

and $K\frac{{|u|}^3}{6n^{1/2}}\to 0$, we therefore get that

$$E(e^{iu\frac{1}{\sqrt{n}}{S}_n})\to e^{-\frac{u^2}{2}}$$

as $n\to \mathrm{\infty }$.

Let ${\mu }_n={\left(\frac{S_n}{\sqrt{n}}\right)}_{*}P$ and let $\varphi (u)=e^{-\frac{u^2}{2}}$. We have just established that ${\tilde{\mu }}_n\to \varphi$ pointwise. The function $\varphi$ is continuous at $0$, so Lévy’s continuity theorem tells us that there is a Borel probability measure $\mu$ on $ℝ$ such that $\varphi =\tilde{\mu }$ and such that ${\mu }_n$ converges narrowly to $\mu$. But $\varphi (u)=e^{-\frac{u^2}{2}}$ is the characteristic function of ${\gamma }_1$, so we have that ${\mu }_n$ converges narrowly to ${\gamma }_1$.

∎