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I have been wondering whether it is possible to extend the $L^2$-martingale convergence theorem to submartingales that are not necessarily non-negative (the case of non-negative submartingales is treated here). Thus, if $\{X_n, F_n\}_{n=0}^\infty$ denotes a submartingale with filtration $F_n$ such that $\sup_{n} E(X_n^2)<\infty$, can we say that $X_n$ converges in $L^2$?

A good reference suffices. Thank you in advance.

JohnK
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1 Answers1

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Let $R_i,\,i\ge1,$ be independent, nonnegative r.v. with mean $1$. Then $M_n:=\prod_{i=1}^nR_i,\,n\ge0,$ defines a nonnegative martingale w.r.t. its natural filtration $F_n=\sigma(R_1,\ldots,R_n),\,n\ge0$. By the martingale convergence theorem, it converges almost surely as $n\to\infty$ towards a nonnegative r.v. $M_\infty$, and further $\mathbb E[M_\infty]\le1$ (by Fatou's lemma).

Next, let $X_n:=-\sqrt{M_n}$. It is clear that $X_n,\,n\ge0,$ is a $\{F_n\}_{n=0}^\infty$-adapted process bounded in $\mathrm L^2(\mathbb P)$ (we have $\mathbb E[X_n^2]=\mathbb E[M_n]=1$ for every $n\in\mathbb N$). Further, by convexity of the function $-\sqrt{\cdot}$ and (conditional) Jensen's inequality, $$\mathbb E[X_{n+1}\mid F_n]=\mathbb E\!\left[-\sqrt{M_{n+1}}\mid F_n\right]\ge-\sqrt{\mathbb E[M_{n+1}\mid F_n]}=-\sqrt{M_n}=X_n$$ for every $n\in\mathbb N$. Thus $(X_n)_{n\ge0}$ is a submartingale bounded in $\mathrm L^2(\mathbb P)$.

Now $X_n$ converges almost surely as $n\to\infty$ towards $X_\infty:=-\sqrt{M_\infty}$. Using the Riesz-Scheffé lemma, the following assertions are therefore equivalent:

  1. $X_n\to X_\infty$ in $\mathrm L^2(\mathbb P)$,
  2. $\mathbb E[X_n^2]\to\mathbb E[X_\infty^2]$,
  3. $\mathbb E[M_n]\to\mathbb E[M_\infty]$,
  4. $M_n\to M_\infty$ in $\mathrm L^1(\mathbb P)$,
  5. $\mathbb E[M_\infty]=1$.

By Kakutani's martingale theorem, these assertions are also equivalent to $$\prod_{i=1}^\infty\mathbb E\!\left[\sqrt{R_i}\right]>0,$$ or equivalently $$\sum_{i=1}^\infty\left(1-\mathbb E\!\left[\sqrt{R_i}\right]\right)<\infty.$$

For instance, choose the independent $R_i$'s such that $$R_i=\begin{cases}\frac{(i+1)^2}{i^2},&\text{with probability $\frac{i^2}{(i+1)^2}$,}\\0,&\text{with remaining probability.}\end{cases}$$ In this case $(X_n)_{n\ge0}$ is a (negative) submartingale, bounded but not converging in $\mathrm L^2(\mathbb P)$.

nejimban
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