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If $X_n$ is a sequence of $L^2$ martingales with bounded increments (i.e. $|X_n-X_{n-1}|<K$ 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.

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  • $\begingroup$ 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? $\endgroup$ Oct 25, 2023 at 16:10
  • $\begingroup$ 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. $\endgroup$ Oct 26, 2023 at 0:08

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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 $P(\{\sup_n\langle X,X\rangle_{n\wedge \tau_C}<\infty\}\cap \{\tau_C=\infty\})\leq P(\sup_n \langle X,X\rangle_n <\infty)$.

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  • $\begingroup$ I see so it similar to a truncation argument where you force the stopping time is bounded so that OST is applicable. That's a clever argument! $\endgroup$ Oct 26, 2023 at 2:26

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