# Prove that $L^2$ martingales with bounded increments that converge almost surely to a finite limit has converging quadratic variation

If $$X_n$$ is a sequence of $$L^2$$ martingales with bounded increments (i.e. $$|X_n-X_{n-1}| for some $$K>0$$) such that $$X_n$$ converge almost surely to a finite limit, prove that the quadratic variation $$\langle X_n\rangle$$ also converges almost surely to a finite limit.

This question is a partial converse to the famous result that the convergence of quadratic variation of a Martingale implies the convergence of the sequence. The hint asked us to apply optional stopping theorem to $$M_n:= X_n^2-\langle X_n\rangle$$. The first trouble I encountered is showing the uniform integrability of the stopped martingale $$M_{n\wedge\tau}$$, where the stopping time $$\tau:=\min_n |X_n|>A$$ for a fixed $$A>0$$. The stopped martingale $$X^2_{n\wedge \tau}$$ is uniformly bounded by $$A^2$$, but how about the quadratic variation? We have to use the bounded increments assumption, since otherwise the result is not true. But I don't have much idea about how to use it. There are similar assumption assuming the stopping time has finite expectation, but I don't think this is the case here since we are assuming $$X_n$$ converges to some finite limit almost surely.

Assuming we have shown the uniform integrability, how do we use the optional stopping theorem though? I would like to have some intuition for this question. Thanks.

• What is the definition of $L^2$ martingale you're using here? Is it that $\mathbb{E}[X_n^2] < \infty$ for all $n$, or $\sup_n \mathbb{E}[X_n^2] < \infty$? Also, is the assumption that the increments $|X_n-X_{n-1}|$ converge to a finite limit, or that $X_n$ converges to a finite limit? Oct 25, 2023 at 16:10
• 1. No, we don't assume the $L^2$ norms are bounded. 2. Sorry, the assumption is the X_n converges to a finite limit, let me edit it. Oct 26, 2023 at 0:08

It suffices to show $$P(\sup_n\langle X,X\rangle_n <\infty)=1$$. Note $$\langle X,X\rangle_n=\sum_{\ell \leq n}E[(X_\ell-X_{\ell-1})^2|\mathscr{F}_{\ell-1}]\leq nK^2$$ a.s. Consider $$\tau_C:=\inf\{n:|X_n|>C\}$$. Wlog let $$X_0=0$$. Then by martingality of $$X^2-\langle X,X\rangle$$, OST on the bounded stopping time $$n\wedge \tau_C$$, monotone convergence on the lhs and the bounded increments (indeed: $$|X_{n\wedge \tau_C}|\leq C+K$$ implies $$X_{n\wedge \tau_C}^2\leq (C+K)^2$$) we get $$E[\sup_n\langle X,X\rangle_{n\wedge \tau_C}]=\sup_nE[X_{n\wedge \tau_C}^2]<\infty$$ so $$P(\sup_n\langle X,X\rangle_{n\wedge \tau_C}<\infty)=1$$. Now, $$P(\tau_C=\infty)\to 1$$ as $$C\uparrow \infty$$. Indeed: $$P(\tau_C<\infty)\leq P(\sup_n|X_n|>C)\stackrel{C\uparrow \infty}\to P(\sup_n|X_n|=\infty)=0$$ since $$|X_n|$$ converges to a finite rv a.s. by continuous mapping. So we conclude: $$P(\sup_n \langle X,X\rangle_n =\infty)\leq P(\{\sup_n\langle X,X\rangle_{n\wedge \tau_C}=\infty\}\cup \{\tau_C<\infty\})\leq P(\tau_C<\infty)\stackrel{C\uparrow \infty}\to 0$$ where the first inequality follows from $$P(\{\sup_n\langle X,X\rangle_{n\wedge \tau_C}<\infty\}\cap \{\tau_C=\infty\})\leq P(\sup_n \langle X,X\rangle_n <\infty)$$.