20 | April | 2012 | Eventually Almost Everywhere

I continue the theme of explaining important bookwork from the Part III course, Advanced Probability, as succinctly as possible. In this post, we consider the convergence properties of discrete time martingales.

1) Theorem: Assume $X$ is a martingale bounded in $L^1$ . Then

$X_n\rightarrow X_\infty\in L^1(\mathcal{F}_\infty)$ a.s.

Remark: The theorem and proof works equally well for $X$ a supermartingale.

Proof: Essentially, we want to reduce convergence of the random variables which make up the martingale to a countable collection of events. We do this by considering upcrossings, which counts the number of times the process alternates from less than a given interval to greater than a given interval. The formal definition will be too wide for this format, so we summarise as

$N_n([a,b],X)=$ the number of disjoint time intervals up to time n in which X goes from $[-\infty a)$ to $(b,\infty]$ . Define $N([a,b],X)$ to be the limit as n increases to infinity.

It is a genuinely easy check that a sequence converges (possibly to $\infty$ ) iff this number of upcrossings of any interval with rational bounds is finite. We will show that the martingale almost surely has this property. The key lemma is a bound due to Doob:

Lemma: $(b-a)\mathbb{E}[N_n([a,b],X)]\leq \mathbb{E}[(X_n-a)^-]$

Proof: Say $S_1<T_1<S_2<T_2,\ldots$ are the successive hitting times of $[-\infty,a),(b,\infty]$ respectively. So $N_n=\inf\{k:T_k\leq n\}$ . We decompose, abbreviating the number of upcrossings as N.

$\sum_{k=1}^n(X_{T_k\wedge n}-X_{S_k\wedge n})=\sum_{k=1}^N(X_{T_k}-X_{S_k})+(X_n-X_{S_{N+1}})1_{\{S_{N+1}\leq n\}}$

Now take an expectation of both sides, applying the Optional Stopping Theorem to the bounded stopping times on the LHS. (If we are working with a supermartingale, formally we need to take $\mathbb{E}[\cdot|\mathcal{F}_{S_k}]$ of each summand on LHS to show that they are non-negative, and so taking a further expectation over $\mathcal{F}_{S_k}$ gives the required result.) We obtain:

$0\geq (b-a)\mathbb{E}N-\mathbb{E}(X_n-X_{S_{N+1}})1_{\{S_{N+1}\leq n\}}$

If $S_{N+1}>n$ then both $1_{\{S_{N+1}\leq n\}}=(X_n-a)^-=0$ . Otherwise $(X_n-a)^-\geq X_{S_{N+1}}-X_n$ . This complete the proof of the lemma.

Since $\mathbb{E}[(X_n-a)^-]\leq \mathbb{E}|X_n|+a<\infty$ , where this last bound is uniform in n by assumption, applying monotone convergence, we get that $N([a,b],X)$ is almost surely finite for every pair $a<b\in\mathbb{Q}$ . Because this set is countable, we can deduce that this holds almost surely for every pair simultaneously. We therefore define $X_\infty(w)=\lim X_n(w)$ when this limit exists, and 0 otherwise. With probability one the limit exists. Fatou’s lemma confirms that $X_\infty\in L^1(\mathcal{F}_\infty)$ .

2) We often want to have convergence in $L^1$ as well. Recall for Part II Probability and Measure (or elsewhere) that

UI + Convergence almost surely is necessary and sufficient for convergence in $L^1$ .

This applies equally well to this situation. Note that for a martingale, this condition is often convenient, because, for example, we know that the set $\{\mathbb{E}[X_\infty|\mathcal{F}_n],n\}$ is UI for any integrable $X_\infty$ .

3) Convergence in $L^p$ is easier to guarantee.

Theorem: i) X a martingale bounded in $L^p$ iff ii) $X_n\rightarrow X_\infty$ in $L^p$ and almost surely iff iii) $\exists Z\in L^p$ s.t. $X_n=\mathbb{E}[Z|\mathcal{F}_n]$ a.s.

Remark: As part of the proof, we will show, as expected, that $X_\infty,Z$ are the same.

Proof: i)->ii) Almost sure convergence follows from the above result applied to the p-th power process. We apply Doob’s inequality about running maxima in a martingale process:

$||X_n^*||_p:=||\sup_{m\leq n}X_m||_p\leq \frac{p}{p-1}||X_n||_p$

Using this, we see that $X_n^*\uparrow X_\infty^*:=\sup|X_k|$ . Now consider $|X_n-X_\infty|\leq 2X_\infty^*\in L^p$ and use Dominated Convergence to confirm convergence in $L^p$ .

Note that Doob’s $L^p$ inequality can be proven using the same author’s Maximal inequality and Holder.

ii)->iii) As we suspected, we show $Z=X_\infty$ is suitable.

$||X_n-\mathbb{E}[X_\infty|\mathcal{F}_n]||_p\stackrel{\text{large }m}{=}||\mathbb{E}[X_m-X_\infty|\mathcal{F}_n]||_p\stackrel{\text{Jensen}}{\leq}||X_m-X_\infty||_p\rightarrow 0$

iii)->i) is easy. Z bounded in $L^p$ implies X bounded by a simple application of the triangle inequality in the definition of conditional expectation.

Eventually Almost Everywhere

A blog about probability and olympiads by Dominic Yeo

Daily Archives: April 20, 2012

Martingale Convergence