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Consider a non-negative martingale $(M_n)_{n \in \mathbb{N}}$ that is also uniformly integrable. Does there exist $Y \in L^1$ such that $$|M_n| \leq Y \ \ \text{for all } n \in \mathbb{N}?$$

I haven't been able to make much progress on this problem for the past few days, besides the fact that we can note the existence of an $L^1$ limit $M_{\infty}$ for the martingale (from the uniform integrability). I tried to construct a dominating sequence $Y_n$ based on a running maximum of $M_n$ and acquire a $Y$ via a convergence (to the martingale limit, so our candidate $Y = M_{\infty}$) in the spirit of the generalized DCT but it didn't seem to work.

Prima facie, it looks like a partial converse to what David Williams, in Probability with Martingales, calls Hunt's Lemma:

Suppose that $(X_n)$ is a sequence of random variables such that $X := \lim_n X_n$ exists, almost surely and that $(X_n)$ is dominated by non-negative $Y \in L^1$. For any filtration $\{\mathcal{F}_n\}$, we can show that $E(X_n| \mathcal{F}_n) \to E(X|\mathcal{F}_{\infty})$.

Note: NOT a homework problem.

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    $\begingroup$ My intuition is that this is false, just because "uniformly integrable" is in general a weaker notion than "dominated", and I don't see any good reason why the martingale property should help upgrade it. But I also can't think of an obvious counterexample. $\endgroup$ Jan 17, 2020 at 0:03
  • $\begingroup$ To what end are you trying to find a dominating random variable? $\endgroup$
    – Math1000
    Jan 17, 2020 at 1:17
  • $\begingroup$ @Math1000 I'm not sure I understand you. It's a problem I've come across and am simply trying to consolidate my knowledge of martingale theory. $\endgroup$ Jan 17, 2020 at 2:52
  • $\begingroup$ My question is whether you are trying to use a dominating random variable to prove a further result. $\endgroup$
    – Math1000
    Jan 17, 2020 at 3:37
  • $\begingroup$ A note that might be helpful: such $Y$ exists iff $E[\sup_n |M_n|] < \infty$. And by Burkholder-Davis-Gundy, this in turn holds iff $E[ [M]_n^{1/2}]$ has a finite limit, where $[M]_n$ is the quadratic variation of $M_n$. $\endgroup$ Jan 17, 2020 at 6:34

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No, in general there is no such dominating random variable $Y$.

There exist non-negative uniformly integrable martingales $(M_n)_{n \in \mathbb{N}}$ such that $\mathbb{E}(\sup_{n \in \mathbb{N}} |M_n|)=\infty$, see this question for an example. For such a martingale there cannot exist $Y \in L^1$ such that $|M_n| \leq Y$ for all $n \in \mathbb{N}$ (because this would imply $\mathbb{E}(\sup_{n \in \mathbb{N}} |M_n|) \leq \mathbb{E}Y<\infty$).

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In the dominated case, one has the a.s. convergence of $E[M_n\mid\mathcal B]$ to $E[M_\infty\mid \mathcal B]$ for each sub-$\sigma$-field of the probability space on which these random variables are defined; this result can be found in Doob's 1953 book on stochastic processes. (Here $M_\infty:=\lim_n M_n$.) The $L^1$ convergence of these conditional expectations follows easily because $\{M_n\}$ is UI. There is a sort of converse in an interesting paper of Blackwell and Dubins ("A converse to the dominated convergence theorem") https://projecteuclid.org/euclid.ijm/1255644957 , to the effect that (after enlarging the probability space if necessary) if $\sup_n M_n$ is not integrable, then there exists a sub-$\sigma$-field $\mathcal B$ such that the convergence of $E[M_n\mid\mathcal B]$ to $E[M_\infty\mid \mathcal B]$ fails a.s.

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