guide:D50f15a52d

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Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. We start with a useful remark. If [math](X_n)_{n\geq 0}[/math] is a submartingale, we get in particular that [math]X_n\in L^1(\Omega,\F,(\F_n)_{n\geq0},\p)[/math] for all [math]n\geq 0[/math]. Moreover, we know that we can write

[[math]] \E[X_n]=\E[X_n^+]-\E[X_n^-] [[/math]]

and hence

[[math]] \E[X_n^-]=\E[X_n^+]-\E[X_n]. [[/math]]

The submartingale property implies that [math]\E[X_0]\leq \E[X_n][/math] and thus

[[math]] \E[X_n^-]\leq \E[X_n^+]-\E[X_0]. [[/math]]

Therefore, if [math]\sup_{n\geq 0}\E[X_n^+] \lt \infty[/math], then [math]\E[X_n^-]\leq \sup_{n\geq 0}\E[X_n^+]-\E[\vert X_0\vert] \lt \infty[/math]. Since [math]\vert X_n\vert =X_n^++X_n^-[/math], we have that [math]\sup_{n\geq 0}\E[X_n^+] \lt \infty[/math] if and only if [math]\sup_{n\geq 0}\E[\vert X_n\vert] \lt \infty[/math].

Lemma (Doob's upcrossing inequality)

Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. Let [math]X=(X_n)_{n\geq 0}[/math] be a supermartingale and [math]a \lt b[/math] two real numbers. Then for all [math]n\geq 0[/math] we get

[[math]] (b-a)\E[N_n([a,b],X)]\leq \E[(X_n-a)^-], [[/math]]

where [math]N_n([a,b],x)=\sup\{ k\geq 0\mid T_k(x)\leq n\}[/math], i.e. the number of uncrossings of the interval [math][a,b][/math] by the sequence [math]x=(x_n)_n[/math] by time [math]n[/math], and [math](T_k)_{k\geq 0}[/math] is a sequence of stoppping times. Moreover, as [math]n\to\infty[/math] we have

[[math]] N_n([a,b],x)\uparrow N([a,b],x)=\sup\{ k\geq 0\mid T_k(x) \lt \infty\}, [[/math]]

i.e., the total number of up crossings of the interval [math][a,b][/math].

Lemma

A sequence of real numbers [math]x=(x_n)_n[/math] converges in [math]\bar \R=\R\cup\{\pm\infty\}[/math] if and only if [math]N([a,b],x) \lt \infty[/math] for all rationals [math]a \lt b[/math].

Show Proof

Suppose that [math]x[/math] converges. Then if for some [math]a \lt b[/math] we had that [math]N([a,b],x)=\infty[/math], that would imply that [math]\liminf_n x_n\leq a \lt b\leq \limsup_nx_n[/math], which is a contradiction. Next suppose that [math]x[/math] does not converge. Then [math]\liminf_nx_n \lt \limsup_nx_n[/math] and so taking [math]a \lt b[/math] rationals between these two numbers gives that [math]N([a,b],x)=\infty[/math].

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\begin{proof}[Proof of Lemma] We will omit the dependence on [math]X[/math] from [math]T_k[/math] and [math]S_k[/math] and we will write [math]N=N_n([a,b],X)[/math] to simplify notation. By the definition of the times [math](T_k)_{k\geq 0}[/math] and [math](S_k)_{k\geq 0}[/math], it is clear that for all [math]k[/math]

[[math]] \begin{equation} X_{T_k}-X_{S_k}\geq b-a. \end{equation} [[/math]]

We have

[[math]] \begin{align*} \sum_{k=1}^n(X_{T_k\land n}-X_{S_k\land n})&=\sum_{k=1}^N(X_{T_k}-X_{S_k})+\sum_{k=N+1}^n(X_n-X_{S_k\land n})\one_{\{N \lt n\}}\\ &=\sum_{k=1}^N(X_{T_k}-X_{S_k})+(X_n-X_{S_{N+1}})\one_{\{S_{N+1}\leq n\}}, \end{align*} [[/math]]

since the only term contributing in the second sum appearing in the middle of the last equation chain is [math]N+1[/math], by the definition of [math]N[/math]. Indeed, if [math]S_{N+2}\leq n[/math], then that would imply that [math]T_{N+1}\geq n[/math], which would contradict the definition of [math]N[/math]. using induction on [math]k\geq 0[/math], it is easy to see that [math](T_k)_{k\geq 0}[/math] and [math](S_k)_{k\geq 0}[/math] are stopping times. Hence for all [math]n\geq 0[/math], we have that [math]S_k\land n\leq T_k\land n[/math] are bounded stopping times and thus we get that [math]\E[X_{S_k\land n}]\geq \E[X_{T_k\land n}][/math], for all [math]k\geq 0[/math]. Therefore, taking expectations in the equations above and using the inequality (9) we get

[[math]] 0\geq \E\left[\sum_{k=1}^n(X_{T_k\land n}-X_{S_k\land n})\right]\geq (b-a)\E[N]-\E[(X_n-a)^-], [[/math]]

since [math](X_n-X_{S_{N+1}})\one_{\{S_{N+1}\leq n\}}\geq -(X_n-a)^-[/math]. Rearranging gives the desired inequality.

\end{proof}

Theorem (Almost sure martingale convergence theorem)

Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. Let [math]X=(X_n)_{n\geq 0}[/math] be a submartingale such that [math]\sup_{n\geq 0}\E[\vert X_n\vert] \lt \infty[/math]. Then the sequence [math](X_n)_{n\geq 0}[/math] converges a.s. to a r.v. [math]X_\infty\in L^1(\Omega,\F_\infty,(\F_n)_{n\geq 0},\p)[/math] as [math]n\to\infty[/math], where [math]\F_\infty=\sigma\left(\bigcup_{n\geq 0}\F_n\right)[/math].

Show Proof

Let [math]a \lt b\in \Q[/math]. By Doob's upcrossing inequality, we get that

[[math]] \E[N_n([a,b],X)]\leq (b-a)^{-1}\E[(X_n-a)^-]\leq (b-a)^{-1}\E[\vert X_n\vert +a]. [[/math]]

By monotone convergence, since [math]N_n([a,b],X)\uparrow N([a,b],X)[/math] as [math]n\to\infty[/math], we get that

[[math]] \E[N([a,b],X)]\leq (b-a)^{-1}\left(\sup_n\E[\vert X_n\vert]+a\right) \lt \infty, [[/math]]

by the assumption on [math]X[/math] being bounded in [math]L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)[/math]. Therefore, we get that [math]N([a,b],X) \lt \infty[/math] a.s. for every [math]a \lt b\in \Q[/math]. Hence,

[[math]] \p\left[\bigcap_{a \lt b\in \Q}\{ N([a,b),X) \lt \infty\}\right]=1. [[/math]]

Writing [math]\Omega_0=\bigcap_{a \lt b\in\Q}\{ N([a,b),X) \lt \infty\}[/math], we have that [math]\p[\Omega_0]=1[/math] and by lemma on [math]\Omega_0[/math] we have that [math]X[/math] converges to a possible infinite limit [math]X_\infty[/math]. So we can define

[[math]] X_\infty=\begin{cases}\lim_{n\to\infty}X_n,& \text{ on $\Omega_0$,}\\ 0,&\text{ on $\Omega\setminus\Omega_0$,}\end{cases} [[/math]]

Then [math]X_\infty[/math] is [math]\F_\infty[/math]-measurable and by Fatou and the assumption on [math]X[/math] being in [math]L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)[/math] we get

[[math]] \E[\vert X_\infty\vert]=\E\left[\liminf_{n\to\infty} \vert X_n\vert\right]\leq \liminf_{n\to\infty}\E[\vert X_n\vert] \lt \infty. [[/math]]

Hence [math]X_\infty\in L^1(\Omega,\F_\infty,(\F_n)_{n\geq 0},\p)[/math].

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Corollary

Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. Let [math](X_n)_{n\geq 0}[/math] be a nonnegative supermartingale. Then [math](X_n)_{n\geq 0}[/math] converges a.s. to a limit [math]X_\infty\in L^1(\Omega,\F_\infty,(\F_n)_{\infty},\p)[/math] and which satisfies

[[math]] X_n\geq \E[X_\infty\mid \F_n]a.s. [[/math]]

Show Proof

Note that [math](-X_n)_{n\geq 0}[/math] is a submartingale, thus [math](-X_n)^+=0[/math] for all [math]n\geq 0[/math], which implies that [math]\sup_{n\geq 0}\E[-X_n^+]=0 \lt \infty[/math]. Hence

[[math]] X_n\xrightarrow{n\to\infty\atop \text{a.s. and $L^1$}}X_\infty\in L^1(\Omega,\F_\infty,(\F_n)_{n\geq 0},\p). [[/math]]

Moreover, for all [math]m\geq n[/math] we have

[[math]] X_n\geq \E[X_m\mid\F_n]. [[/math]]

By Fatou we get

[[math]] X_n\geq \liminf_{m\to\infty}\E[X_m\mid \F_n]\geq \E\left[\liminf_{m\to\infty}X_m\mid \F_n\right]=\E[X_\infty\mid \F_n]. [[/math]]

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Example

[Simple random walk on [math]\mathbb{Z}[/math]] Let [math]Y_n=1+Z_1+\dotsm +Z_n[/math], for [math]Z_j[/math] iid with [math]\p\left[Z_j=\pm1\right]=\frac{1}{2}[/math], [math]Y_0=1[/math], [math]\F_0=\{\varnothing,\Omega\}[/math] and [math]\F_n=\sigma(Z_1,...,Z_n)[/math]. Then we have already seen that that [math](Y_n)_{n\geq 0}[/math] is a martingale. Let [math]T=\inf\{n\geq 0\mid Y_n=0\}[/math]. We need to show that [math]T \lt \infty[/math] a.s. Let [math]X_n=Y_{n\land T}[/math]. Then [math](X_n)_{n\geq 0}[/math] is also a martingale. Moreover, [math]X_n\geq 0[/math] for all [math]n\geq 0[/math] and [math](X_n)_{n\geq 0}[/math] converges a.s. to a r.v. [math]X_\infty\in L^1(\Omega,\F_\infty,(\F_n)_{n\geq 0},\p)[/math]. Since [math]X_n=Y_{n\land T}[/math], we get [math]X_\infty=Y_T[/math]. The convergence of [math](X_n)_{n\geq 0}[/math] implies that [math]T \lt \infty[/math] a.s., indeed on the set [math]\{T=\infty\}[/math] we get [math]\vert X_{n+1}-X_n\vert=1[/math]. Consequently, on [math]\{T=\infty\}[/math], we get that [math](X_n)_{n\geq 0}[/math] is not a Cauchy sequence and therefore cannot converge. This implies that [math]\p[T=\infty]=0[/math] and thus [math]T \lt \infty[/math]. Hence

[[math]] \lim_{n\to\infty}X_n=Y_T=0. [[/math]]

We also note that [math]\E[X_n]=1 \gt \E[X_\infty]=0[/math] for all [math]n\geq 0[/math] and so [math]X_n[/math] does not converge to [math]X_\infty[/math] in [math]L^1[/math].

General references

Moshayedi, Nima (2020). "Lectures on Probability Theory". arXiv:2010.16280 [math.PR].