# (PTE) 4.2. Martingales and a.s. convergence

Remaining sections in chapter 4 is about martingales and convergence of it. Regarding martingales, our first topic will be convergence in almost sure sense. Next we will look into convergence in $L^p,$ with $p>1$ and $p=1$ separately. In the meantime the theory of optional stopping will be covered.

## Martingales

We say $X_n$ is adapted to $\mathcal{F}_n$ if $X_n \in \mathcal{F}_n$ for all $n.$ For simplicity instead of denoting $\mathcal{F}_n$ together, we could just say $X_n$ is a (sub/super)martingale if the adapted $\sigma$-fields are clear. If $X_n$ is a martingale, $\int_A X_{n+1} dP = \int_A X_n dP$ for all $A \in \mathcal{F}n,$ so trivially $EX_{n+1} = EX_n$ for all $n.$ $X_n$ is a martingale if and only if $X_n$ is both a submartingale and a supermartingale. In addition, if $X_n$ is a submartingale, then $-X_n$ is a supermartingale.

The easiest but important examples are random walks and square martingales.

(i) $X_n := \xi_1 + \cdots \xi_n$ is a martingale.

(ii) $X_n := (\xi_1 + \cdots + \xi_n)^2 - n\sigma^2$ is a martingale.

Though we cannot guarantee that functions of martingales are also martingales, we can say for sure that a function of martingale is a submartingale if the function is convex.

The proof is direct by conditional Jensen’s inequality. The obvious corollary is for submartingales.

The following two examples will be useful in the section comes later.

(ii) If $X_n$ is a supermartingale, then $X_n \wedge a$ is a supermartingale.

## Martingale convergence theorems

For martingale convergence theorems, we need to define and prove predictable sequences, stopping times, upcrossing inequality and related properties.

### Upcrossing inequality

Intuitively, consider $n$ as time index. The term “predictable” is from the fact that we knows every information about the behavior of $H_{n+1}$ in the time point $n.$

We get the result that the sum of submartingale increments weighted by a bounded predictable sequence is also a submartingale.

(i) $E|(H\cdot X)_n| \le \sum_{m=1}^n M_nE(|X_m|+|X_{m-1}|) < \infty$ for all $n.$

(ii) Clearly, $(H\cdot X)_n \in \mathcal{F}_n$ for all $n.$

(iii) $$\begin{aligned} E((H\cdot X)_{n+1} | \mathcal{F}_n) &= (H\cdot X)_n + E(H_{n+1}(X_{n+1}-X_n)|\mathcal{F}_n) \\ &= (H\cdot X)_n + H_{n+1}\cancel{\{E(X_{n+1}|\mathcal{F}_n)-X_n\}}. \end{aligned}$$

We already get a glimpse of stopping times while studying coupon collector’s problem and renewal theory. They were random variables that specify the time that an event occurs. Here, we define it formally.

It is highly useful to define a predictable sequence as an indicator function related to stopping times. With such sequence, we can easily derive the following theorem.

Let $H_m = \mathbf{1}_{m\le N}$ then it is a non-negative bounded predictable sequence since $\{m\le N\} = \{N \le m-1\}^c \in \mathcal{F}_{m-1}.$ By theorem 4.2.8 $X_{n\wedge N}-X_0$ is a submartingale, so $X_{n\wedge N}$ is also a submartingale.

As an example and a lemma for our main theorem - martingale convergence - I will state and prove the upcrossing inequality.

## expand proof

First we show that $N_m$'s are stopping times. For given $n,$ $$\{N_1=n\} = \{X_0>a, \cdots, X_{n-1}>a,X_n\le a\} \in \mathcal{F}_n. \\ \{N_1=n\} = \bigcup_{\ell=1}^{n-1}\{N_1=\ell, X_{\ell+1}<b,\cdots, X_{n-1}<b,X_n\ge b\} \in \mathcal{F}_n. \\ \cdots$$ Thus $N_m$'s are stopping times. Next, we define $Y_n = a + (X_n-a)^+$ so that $Y_{N_{2k}} \ge b$ and $Y_{N_{2k-1}} = a$ for all $k.$ Since $x\mapsto a + (x-a)^+$ is increasing and convex, $Y_n$ is also a submartingale.

$$\begin{aligned} (b-a)EU_n &\le \sum_{k=1}^{U_n}(Y_{N_{2k}}-Y_{N_{2k-1}}) \\ &= \sum_{k=1}^{U_n} \sum_{i\in J_k}(Y_i-Y_{i-1}),\\ &\;\;\;\;\text{where } J_k=\{N_{2k-1}+1,\cdots,N_{2k}\} \\ &= \sum_{m\in J} (Y_m - Y_{m-1}),\\ &\;\;\;\;\text{where } J=\cup_{k=1}^{U_n} J_k \\ &\le \sum_{m=1}^n \mathbf{1}_{m\in J}(Y_m - Y_{m-1}). \end{aligned}$$ Let $H_m = \mathbf{1}_{m\in J},$ then since $$\{m\in J\} = \{N_{2k-1} < m \le N_{2k} \text{ for some } k\}$$ $H_m$ is a bounded, non-negative predictable sequence. Thus $$(b-a)U_n \le (H\cdot Y)_n$$ and the right hand side is a submartingale. Let $K_m = 1-H_m$ then similarly $(K\cdot Y)_n$ is a submartingale and $E(K\cdot Y)_n \ge 0.$ Hence $$\begin{aligned} E(Y_n-Y_0) &= E(H\cdot Y)_n + E(K\cdot Y)_n \\ &\ge E(H\cdot Y)_n \ge (b-a)EU_n. \end{aligned}$$

We call $U_n$ the number of upcrossings. An important fact directly follows from the theorem is $EU_n \le \frac{1}{b-a}(EX_n^+ + |a|).$ This will be the key to prove the martingale convergence.

### Martingale convergence theorems

We get our first convergence theorem for dependent sequence.

## expand proof

Given $a<b,$ let $U_n[a,b]$ be the number of upcrossings of $X_1,\cdots,X_n$ over $[a,b].$ By the upcrossing inequality, $EU_n[a,b] \le \frac{EX_n^++|a|}{b-a}.$ Let $U[a,b] = \lim_n U_n[a,b]$ then $$EU[a,b] = \lim_n EU_n[a,b] \le \sup_n \frac{EX_n^++|a|}{b-a} < \infty.$$ Thus by Markov's inequality, $0\le U[a,b]\le \infty$ a.s.

Now suppose $\liminf_n X_n < \limsup_n X_n.$ Then for some $a<b,$ $X_n<a$ and $X_n>b$ infinitely often. Thus $$\begin{aligned} P(\liminf_n X_n < \limsup_n X_n) &= P(\liminf_n X_n < a < b< \limsup_n X_n \text{ for some } a,b\in\mathbb{Q}) \\ &\le \sum_{a,b\in\mathbb{Q}}P(\liminf_n X_n <a<b< \limsup_n X_n) \\ &= \sum_{a,b\in\mathbb{Q}}P(U[a,b]=\infty) = 0 \end{aligned}$$ so there exists $X$ such that $X_n \to X$ a.s. We now need to show that such $X$ is integrable. By Fatou's lemma, $$\begin{aligned} EX^+ &\le \liminf_nEX_n^+ \le \sup_n EX_n^+ < \infty.\\ EX^- &\le \liminf_nEX_n^- = \liminf_nE(X_n^+-X_n) \\ &\le \sup_nEX_n^+ - EX_0 < \infty. \end{aligned}$$

As corollaries, we get supermartingale convergence and closability of negative submartingales.

The next example shows that even if a martingale converges almost surely, we cannot guarantee $L^p$ convergence to the same limit. The following sections will be about the condition that a martingale converges a.s. *and* in $L^p$.

By supermartingale convergence, $X_n \to X$ for some $X\in L^1.$ Note that on $(N=\infty),$ $X_n = S_n.$ By the law of iterated logarithm, $P(\liminf_n S_n = -\infty, \limsup_n S_n = \infty)=1.$ It follows that $$\begin{aligned} P(N=\infty) &= P(N=\infty, \liminf_n S_n = -\infty, \limsup_n S_n = \infty) \\ &= P(N=\infty, \liminf_n X_n = -\infty, \limsup_n X_n = \infty) \\ &\le P(\liminf_n X_n = -\infty, \limsup_nX_n = \infty) = 0. \end{aligned} $$ and $N<\infty$ a.s. Hence $X=\lim_nS_{n\wedge N} = S_N = 0$ a.s.

However, $E|X_n| = ES_{n\wedge N} = ES_0 = 1$ for all $n$ since $X_n$ is a martingale.

*Acknowledgement*

This post series is based on the textbook *Probability: Theory and Examples*, 5th edition (Durrett, 2019) and the lecture at Seoul National University, Republic of Korea (instructor: Prof. Sangyeol Lee).