Update: I've cross posted at MO here.

Let $(\Omega, \mathcal{F})$ be a measurable space equipped with a filtration $\{\mathcal{F}_n\}_{n \in \mathbb{N}}$ such that $\mathcal{F}_n \uparrow \mathcal{F}$.

Let $\mathcal{C}$ be convex set of mutually absolutely continuous probabilities on $(\Omega, \mathcal{F})$ generated by finitely many extreme points $P_1,...,P_n$.

Suppose that $\{R_n \}_{n \in \mathbb{N}}$ is a sequence of probability measures defined, respectively, on $(\Omega, \mathcal{F}_n)$, and suppose that for all $Q \in \mathcal{C}$, $R_n \ll Q|_{\mathcal{F}_n}$ for all $n$. Let $Y^Q_n = dR_n/dQ|_{\mathcal{F}_n}$ be the corresponding Radon-Nikodym derivative. Let us assume that, for all $Q \in \mathcal{C}$, $\{Y_n^Q \}_{n \in \mathbb{N}}$ is a martingale in $\{\mathcal{F}_n\}_{n \in \mathbb{N}}$ with respect to $Q$.

Since the $Y_n^Q$ are non-negative, the martingale convergence theorem guarantees that $Y_n^Q \to Y^Q_\infty$ almost surely (with respect to any $Q \in \mathcal{C}$, by mutual absolute continuity).

Question. Does it follow from our convexity assumptions that the martingale convergence mentioned above is uniform in $Q \in \mathcal{C}$? That is, is it true that $\sup_Q |Y^Q_n - Y^Q_\infty| \to 0 \ $ almost surely as $n \to \infty$?

If it helps, we can assume that the filtration is very simple. For instance, we can assume that each $\mathcal{F}_n$ is generated by a finite measurable partition. Also, if it helps, we can assume that for all $Q \in \mathcal{C}$ the sequence $\{ Y_n^Q\}$ is uniformly integrable and so $Y_n^Q \to Y^Q_\infty$ in $L^1$ as well as almost surely.

  • $\begingroup$ 1. As I understand, $N=N(\epsilon,\omega)$ in your question. In other words, $\sup_{Q} |Y_n^Q - Y_\infty^Q| \to 0$, $n\to\infty$, a.s. Correct? 2. Why are you interested in this question at all? $\endgroup$
    – zhoraster
    Feb 18, 2017 at 10:20
  • $\begingroup$ @zhoraster 1. Yes, you are right. I will edit for clarity. 2. This came up as I was working on a bigger problem, but I'm afraid it would take up too much space to write all the details. I hope the question is self-contained as it is, but if it's missing something crucial, then I will add it. $\endgroup$
    – aduh
    Feb 18, 2017 at 15:28
  • $\begingroup$ It is unlikely that the statement holds in this generality. Probably, there is some additional information on $R_n$ that may be useful (I suspect so because $Y_n^Q$ is a martingale). $\endgroup$
    – zhoraster
    Feb 18, 2017 at 18:00
  • $\begingroup$ We can assume that for all $n$ and all $F \in \mathcal{F}_{n-1}$, $R_{n-1}(F) = R_n(F)$. Hence, the martingale. We can also assume that $\{ R_n\}$ is uniformly absolutely continuous with respect to $Q \in \mathcal{C}$, hence $\{ Y_n^Q\}$ is uniformly integrable. $\endgroup$
    – aduh
    Feb 18, 2017 at 18:06
  • $\begingroup$ Doesn't this first condition imply that $R_n = R|_{\mathcal F_n}$ for some $R$? $\endgroup$
    – zhoraster
    Feb 18, 2017 at 18:09

1 Answer 1


We can understand $\mathcal C$ as a subset of $\mathbb{R}^{n-1}$ ($Q\in\mathcal C\leftrightarrow$ the coefficients before $P_1,\dots,P_{n-1}$ in a convex combination).

I claim that $\sup_{Q\in \mathcal C'}| Y_n^Q - Y^Q|\to 0$, $n\to\infty$, a.s. for any $\mathcal C'$ such that its closure is in the interior of $\mathcal C$. This would follow from almost sure pointwise convergence on a countable dense subset of $\mathcal C$ and the fact that $Y_n^Q$ is convex. So let us prove the latter.

Let $Q',Q''\in \mathcal C$ and $\alpha\in(0,1)$. Then for any $A\in \mathcal F_n$ $$ \int_A \left(\alpha \frac{dR_n}{dQ'|_{\mathcal F_n}} + (1-\alpha) \frac{dR_n}{dQ''|_{\mathcal F_n}}\right)d(\alpha Q' + (1-\alpha)Q'') \\ = \int_A \frac{dR_n}{dQ'|_{\mathcal F_n}} \left(\alpha + (1-\alpha) \frac{dQ'|_{\mathcal F_n}}{dQ''|_{\mathcal F_n}}\right)\left(\alpha + (1-\alpha) \frac{dQ''|_{\mathcal F_n}}{dQ'|_{\mathcal F_n}}\right)dQ' \\ = \int_A \frac{dR_n}{dQ'|_{\mathcal F_n}} \left(\alpha^2 + (1-\alpha)^2 + \alpha(1-\alpha)\left(\frac{dQ'|_{\mathcal F_n}}{dQ''|_{\mathcal F_n}}+\frac{dQ''|_{\mathcal F_n}}{dQ'|_{\mathcal F_n}}\right)\right)dQ' \\ \ge \int_A \frac{dR_n}{dQ'|_{\mathcal F_n}} \left(\alpha^2 + (1-\alpha)^2 + 2\alpha(1-\alpha)\right)dQ' = R_n(A). $$ Consequently, $$ \alpha \frac{dR_n}{dQ'|_{\mathcal F_n}} + (1-\alpha) \frac{dR_n}{dQ''|_{\mathcal F_n}}\ge \frac{dR_n}{d\bigl(\alpha Q' + (1-\alpha)dQ''\bigr)|_{\mathcal F_n}}, $$ which is exactly our claim.

In order to get the convergence on the whole $\mathcal C$, this may be useful. Another potentially useful fact is the uniform convergence of reciprocals (but $R$ should be equivalent to each $P_i$) $1/Y_n^Q \to 1/Y_n^Q$, $n\to\infty$, which follows from linearity.

  • $\begingroup$ @aduh, I thought $\mathcal C$ is closed by your definition. The closedness won't help though (see the suggested link). tldr: the uniform convergence on closed subsets of the interior of $\mathcal C$ follows from the convexity of $Y_n^Q$ in $Q$. $\endgroup$
    – zhoraster
    Feb 18, 2017 at 19:56
  • $\begingroup$ Sorry, you're right. I didn't understand what was going on at first. Having thought about it more and having looked at your link, I think I see your idea. This is very nice! $\endgroup$
    – aduh
    Feb 18, 2017 at 22:41
  • $\begingroup$ Besides in the first paragraph, the hypothesis that $\mathcal{C}$ is finitely generated does not seem to be used (perhaps this initial identification of $\mathcal{C}$ and $\mathbb{R}^{n-1}$ is more significant than I realize, though). Do you think that we can use this somehow to extend the above result? We know that the convergence is uniform on the set of extremal points, after all, since there are only finitely many of them. $\endgroup$
    – aduh
    Feb 19, 2017 at 18:39
  • 1
    $\begingroup$ The fact that $\mathcal C$ is finitely generated is used in my argument: if it were not, one would have to replace "$\mathcal C'$ with closure in the interior of $\mathcal C$" by "compact subset $\mathcal C'$ lying in the interior of $\mathcal C$". $\endgroup$
    – zhoraster
    Feb 19, 2017 at 20:00
  • 1
    $\begingroup$ @aduh, well, if we could "extend", it would mean that there is some $\epsilon>0$, $(1+\varepsilon)P_i - \varepsilon P_j$ must be a probability measure for all $i,j$. The density of the latter w.r.t. $P_i$ is $(1+\varepsilon) - \varepsilon dP_j/dP_i$. It is positive iff $dP_j/dP_i<1/\varepsilon -1$. $\endgroup$
    – zhoraster
    Mar 12, 2017 at 19:38

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