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

Question

Cross posted at MSE here. I'm hoping someone here can help complete zhoraster's answer. Any hints or references are appreciated.

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

(1) If you assumed that the $Y_{n}^{Q},Y_{\infty}^{Q}$ are all convex(thus continuous) functions along $\mathcal{C}$ being convex, then the uniform convergence follows easily from a classical result, see [1 Sec39,2]. But as the SE post's comment said if you are mainly dealing with (bounded) convex processes then ([1,Thm 39.1]) you are mostly dealing with a linear transition kernel and the statement becomes trivial.

(2)A slight restriction on your assumptions may be tightness of the collection of generating measures $P_1,\cdots P_n$. If this collection is tight, then you can construct for every $\epsilon>0$ a compact set $K_\epsilon\subset\Omega$ to normalize the $P_i$'s into a new collection of probability measures $P_i^{*}:= \frac{P_i\mid_{K_{\epsilon}}(\bullet)}{P_i(K_\epsilon)}$ and the convex set generated by this new collection of p.m. $\bar{\mathcal{C}_\epsilon}=convex\,hull_{i}{P_i^{*}}$. The uniform convergence of $Y_n^{Q}\Rightarrow Y_{\infty}^{Q}$ sequence on any subset $D$ of $\bar{\mathcal{C}_\epsilon}$ whose closure is in $\bar{\mathcal{C}_\epsilon}$ is proved in the SE post[3]

Note that absolutely continuity is transitive and $P_i^{*}\equiv P_i$ so we can assume $Y_n$'s w.r.t. $P_i^{*}$'s are $C_i\cdot Y_n$. And since $n<\infty$ we can take $C=max_i{C_i}<\infty$ for convenience.

And the whole $\mathcal{C}$ can be approximated by sets in form of $\bar{\mathcal{C}_\epsilon}$ with a sequence of smaller and smaller $\epsilon$'s without loss of uniformity. i.e.

$$sup_{Q\in\mathcal{C}} |Y^Q_n - Y^Q_\infty|\leq C\cdot sup_{Q\in \bar{D}} |Y^Q_n - Y^Q_\infty|+C\cdot sup_{Q\in\bar{\mathcal{C}_\epsilon}-\bar{D}} |Y^Q_n - Y^Q_\infty| \to 0 $$

$D\subset \bar{\mathcal{C}_\epsilon}$ is a subset of $\bar{\mathcal{C}_\epsilon}$ whose closure is again in $\bar{\mathcal{C}_\epsilon}$. The first and second term converges to zero as $\epsilon\rightarrow 0$ because [3].(Tightness is used to ensure the [3] is applicable on second term because $\bar{\mathcal{C}_\epsilon}-\bar{D}$ is a closed set in $\bar{\mathcal{C}_\epsilon}$.) I think the tightness is not too restrictive since if $\Omega$ is Polish then it is true that all $P_i$ form a tight collection.

Reference

[1]Rockafellar, Ralph Tyrell. Convex analysis. Princeton university press, 2015.

[2]https://math.stackexchange.com/questions/126142/uniform-convergence-of-sequence-of-convex-functions

[3]https://math.stackexchange.com/questions/2148640/uniform-martingale-convergence-of-radon-nikodym-derivatives-of-a-convex-set-of-p/2150360?noredirect=1#comment4426243_2150360