4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I do not see why assuming filtration will help, but a general technique for proving uniform convergence is called "generic chaining" and the reference is Talagrand's book with the same title. See also an introductory paper link.springer.com/chapter/10.1007%2F978-1-4684-0559-0_9#page-1 $\endgroup$ – Henry.L Mar 13 '17 at 15:42
4
+100
$\begingroup$

(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

$\endgroup$
  • $\begingroup$ Thanks for this. Is [1, Sec39,2] meant to be [1, Thm39.2]? $\endgroup$ – aduh Mar 13 '17 at 19:24
  • $\begingroup$ Sorry for the confusion, I mean Sec 39 in [1] and the post in [2]. For general remark, i would still recommend Talagrand's masterpiece commented above. It is generally hopeless to discuss uniform convergence even if we are dealing with convex processes without further restrictions on $Y_n$ is my primary feeling. $\endgroup$ – Henry.L Mar 13 '17 at 19:28
  • $\begingroup$ Great, thanks again. Regarding the Talagrand book, perhaps in light of the author's comments here(!), I ought to check out his more recent "Upper and Lower Bounds for Stochastic Processes". $\endgroup$ – aduh Mar 13 '17 at 19:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.