Uniform martingale convergence of Radon-Nikodym derivatives of a convex set of probabilities

Question

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.

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.