# (PTE) 4.4. Martingale inequalities and convergence in $L^p$

$\newcommand{\argmin}{\mathop{\mathrm{argmin}}\limits}$ $\newcommand{\argmax}{\mathop{\mathrm{argmax}}\limits}$

In this section we look into the condition that makes a martingale converges in $L^p$, $p>1$ in detail. We start by proving Doob’s inequality. By using this result we prove martingale inequalities which will then be used to prove Doob’s $L^p$ maximal inequality. $L^p$ convergence is direct from them. Lastly, as an extension of Doob’s inequality, I will briefly state a version of optional stopping.

## Martingale inequalities

Let $X_n$ be a submartingale, $N$ be a stopping time such that $N \le k$ a.s. Then $$EX_0 \le EX_N \le EX_k.$$

(i) Observe that $X_{n\wedge N}$ is also a submartingale. Thus $EX_{0\wedge N} \le EX_{k\wedge N}$ and we get the first inequality.
(ii) Let $K_n = \mathbf{1}_{N\le n-1}$ be a non-negative bounded predictable sequence then $(K\cdot X)_n = X_n - X_{n\wedge N}$ is a submatringale. Thus $0 = E(K\cdot X)_0 \le E(K\cdot X)_k$ which leads to the second inequality.

This natural result will be the foundation of numerous theorems that will be introduced from now on. For simplicity, I will call stopping times with almost sure upper bound bounded stopping times.

Let $X_n$ be a submartingale. Define $\bar X_n = \max_{0\le m\le n}X_m.$ For $\lambda > 0,$ $$\lambda P(\bar X_n \ge \lambda) \le EX_n\mathbf{1}_{\bar X_n \ge \lambda}.$$

Let $A=\{\bar X_n \ge \lambda\}.$ Let $N=\inf\{m: X_m\ge\lambda\}\wedge n$ be a bounded stopping time. Since $\lambda \mathbf{1}_A \le X_N \mathbf{1}_A,$ $\lambda P(A) \le EX_N \mathbf{1}_A.$
On $A,$ $EX_N \le EX_n$ by Doob's inequality. On $A^c,$ $N=n$ a.s. Thus in either case $EX_N \mathbf{1}_A \le EX_n \mathbf{1}_A$ and we get the result.

A more comprehensive form might be $P(\bar X_n \ge \lambda) \le \frac{1}{\lambda} EX_n\mathbf{1}_{\bar X_n \ge \lambda},$ which can be viewed as a version of inequality that resembles Chebyshev’s inequality.

Similarly, we can also derive supermartingale inequality.

Let $X_n$ be a supermartingale. For $\lambda>0,$ $$\lambda P(\bar X_n \ge \lambda) \le EX_0 - EX_n\mathbf{1}_{\bar X_n < \lambda}.$$

Let $A$ and $N$ as in the proof of submartingale inequality. The result is direct from $$EX_0 \ge EX_N = EX_N\mathbf{1}_A + EX_N\mathbf{1}_{A^c}.$$

## $L^p$ convergence theorem

With the help of submartingale inequality, we get the following theorem.

Let $X_n$ be a non-negative submartingale. For $1<p<\infty,$ $$E\bar X_n^p \le \left(\frac{p}{p-1}\right)^p EX_n^p.$$
expand proof

Let $M>0.$ By properly using Foubini's theorem \begin{aligned} E(\bar X_n \wedge M)^p &= \int_0^\infty P((\bar X_n \wedge M)^p \ge t) dt \\ &= \int_0^\infty P(\bar X_n \wedge M \ge \lambda)p\lambda^{p-1}d\lambda \\ &= \int_0^M P(\bar X_n \ge \lambda) p\lambda^{p-1}d\lambda \\ &\le \int_0^M \frac{1}{\lambda} EX_n\mathbf{1}_{\bar X_n \ge \lambda} p\lambda^{p-1}d\lambda \\ &= \int_0^M \int_\Omega X_n\mathbf{1}_{\bar X_n \ge \lambda}dP p\lambda^{p-2}d\lambda \\ &= \frac{p}{p-1} EX_n(\bar X_n \wedge M)^{p-1} \\ &\le \frac{p}{p-1} (EX_n^p)^{1/p} (E(\bar X_n \wedge M)^p)^{1/q} \end{aligned} The first inequality is follows submartingale inequality and the second one is from Holder's inequality. Transposition and applying MCT ($M\uparrow\infty$) leads to the result.

It is often called $L^p$ maximal inequality. Note that we used $\bar X_n\wedge M$ in order to prove that the inequality holds even if $E\bar X_n$ is not finite. $L^p$ convergence of a martingale is derived from this.

Let $X_n$ be a martingale with $\sup_n E|X_n|^p < \infty.$ For $p>1,$ there exists $X$ such that $X_n \to X$ a.s. and in $L^p.$

By submartingale convergence, there exists $X\in L^1$ such that $X_n\to X$ a.s. By MCT and $L^p$ maximal inequality, \begin{aligned} E\sup_n|X_n|^p &= \lim_n E\max_{0\le m\le n} |X_m|^p \\ &\le \lim_n \left(\frac{p}{p-1}\right)^p E|X_n|^p \\ &\le \left(\frac{p}{p-1}\right)^p \sup_n E|X_n|^p < \infty. \end{aligned} Thus $|X_n - X|^p \le (2\sup_n|X_n|^p)$ is integrable and by DCT, the result follows.

## Bounded optional stopping

As a sidenote, I would like to cover the fact that bounded stopping times preserve submartingale properties.

For a stopping time $\tau,$ $$\mathcal{F}_\tau := \{A\in\mathcal{F}: A\cap(\tau=n)\in\mathcal{F}_n,\forall n\}$$

It is not difficult to check that $\mathcal{F}_\tau$ is a $sigma$-field with $\tau \in \mathcal{F}_\tau.$

Let $X_n$ be a submartingale, $\sigma,\tau$ be stopping times that $0\le\sigma\le\tau\le k$ a.s. Then $E(X_\tau|\mathcal{F}_\sigma) \ge X_\sigma$ a.s.

The proof can be done in two different ways. The first proof uses Doob’s inequality.

Since $Y_{n\wedge \tau}$ is a submartingale, by Doob's inequality $EY_\sigma \le EY_\tau.$ For given $A\in\mathcal{F}_\sigma,$ let $$N = \begin{cases} \sigma &~\text{on } A\\ \tau &~\text{on } A^c \end{cases}$$ Then $N$ is a stopping time since $$(N=n) = ((\sigma=n)\cap A)\cup((\tau=n)\cap(\sigma \le n)\cap A^c) \in \mathcal{F}_n.$$ Hence $$\begin{gathered} EY_N = EY_\sigma\mathbf{1}_A + EY_\tau\mathbf{1}_A^c \le EY_\tau.\\ \int_A Y_\sigma dP \le \int_A Y_\tau dP = \int_A E(Y_\tau|\mathcal{F}_\sigma) dP. \end{gathered}$$

The second approach uses the lemma and inductive process:

$$E(X_\tau|\mathcal{F}_\sigma)\mathbf{1}_{\sigma=n} = E(X_\tau|\mathcal{F}_n)\mathbf{1}_{\sigma=n} \text{ a.s.}$$
expand proof

We first show that the right hand side is $\mathcal{F}_\sigma$-measurable. Given $a\in\mathbb{R}$ and $k\ge0,$ \begin{aligned} &(E(X_\tau|\mathcal{F}_n)\mathbf{1}_{\sigma=n}\le a) \cap (\sigma=k) \\ &= \begin{cases} (E(X_\tau|\mathcal{F}n) \le a)\cap (\sigma=k) \in \mathcal{F}_k &,~ k=n \\ (0\le a)\cap(\sigma=k) \in \mathcal{F}_k &,~ \text{otherwise} \end{cases} \end{aligned} Next for given $A \in \mathcal{F}_\sigma,$ \begin{aligned} &\int_A E(X_\tau|\mathcal{F}_\sigma)\mathbf{1}_{\sigma=n} dP \\ &= \int_{A\cap(\sigma=n)} E(X_\tau|\mathcal{F}_\sigma) dP \\ &= \int_{A\cap(\sigma=n)} X_\tau dP \\ &= \int_{A\cap(\sigma=n)} E(X_\tau|\mathcal{F}_n) dP \\ &= \int_A E(X_\tau|\mathcal{F}_n)\mathbf{1}_{\sigma=n} dP. \end{aligned}
it sufficies to show that for all $A \in \mathcal{F}_n$ $$\int_A E(X_\tau|\mathcal{F}_\sigma)\mathbf{1}_{\sigma=n} dP \ge E(X_\tau|\mathcal{F}_n)\mathbf{1}_{\sigma=n}.$$ Given $A \in \mathcal{F}_n,$ \begin{aligned} &\int_A E(X_\tau|\mathcal{F}_\sigma)\mathbf{1}_{\sigma=n} dP - E(X_\tau|\mathcal{F}_n)\mathbf{1}_{\sigma=n} \\ &= \int_{A\cap(\sigma=n)} E(X_\tau|\mathcal{F}_n)-X_n dP \\ &= \int_{A\cap(\sigma=n)} X_\tau - X_n dP \\ &= \int_{A\cap(\sigma=n)\cap(\tau\ge n+1)} X_\tau - X_n dP \\ &\ge \int_{A\cap(\sigma=n)\cap(\tau\ge n+1)} X_\tau - X_{n+1} dP \\ &= \int_{A\cap(\sigma=n)\cap(\tau\ge n+2)} X_\tau - X_{n+1} dP \\ &\;\;\vdots \\ &\ge \int_{A\cap(\sigma=n)\cap(\tau=k)} X_\tau - X_k dP = 0. \end{aligned}

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).