# Is it possible to construct an $L^1$ dominating random variable for a non-negative uniformly integrable martingale?

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.

• 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. Jan 17, 2020 at 0:03
• To what end are you trying to find a dominating random variable? Jan 17, 2020 at 1:17
• @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. Jan 17, 2020 at 2:52
• My question is whether you are trying to use a dominating random variable to prove a further result. Jan 17, 2020 at 3:37
• 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 $M_n$. Jan 17, 2020 at 6:34

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 $$\mathbb{E}(\sup_{n \in \mathbb{N}} |M_n|) \leq \mathbb{E}Y<\infty$$).
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.