4
$\begingroup$

This question is cross-posted at MSE with a soon to expire bounty that hasn't generated much discussion.

Let $(\Omega, \mathcal{F},P)$ be a probability space and $(\mathcal{F}_n)_n$ a filtration that increases to $\mathcal{F}$.

Is there a way to quantify the "rate of convergence" of $\mathcal{F}_n \uparrow \mathcal{F}?$

I'll now try to clarify the question by explaining its motivation. I was wondering what can be said about the rate of convergence in Levy's martingale convergence theorem. By that result, for an integrable random variable $X$ we have $$E(X \mid \mathcal{F}_n) \to X$$ almost surely. I was wondering if we can make a statement like $$P(|E(X \mid \mathcal{F}_n) - X| > \epsilon) = O(n^{-a})$$ under fairly general assumptions.

As pointed out to me by Bananach in this question, it's hopeless to expect the rate to be independent of the particular filtration because we can replace $\mathcal{F}_n$ with $$\bar{\mathcal{F}_n} = \mathcal{F}_{\sqrt{n}}$$ (rounding to the nearest integer), and obtain a new filtration for which conditional expectations converge more slowly.

But perhaps if we knew "how quickly" $\mathcal{F}_n$ increases to $\mathcal{F}$, we could find a rate of convergence of conditional expectations that depends on the "rate of convergence" of the filtration.

An example is given in Michael's answer to the same question that I linked to above. Let $X$ be uniformly distributed on $[-1,1]$, assume $\mathcal{F} = \sigma(X)$, and define $$Z_n = X \mathbf{1}_{|X|>2^{-n}} \ \ \text{and} \ \ \mathcal{F}_n = \sigma(Z_n,...,Z_1).$$ Then, $$E(X \mid \mathcal{F}_n) = X \mathbf{1}_{|X|>2^{-n}} \to X,$$ and $$P(|E(X \mid \mathcal{F}_n) - X)| > \epsilon) = 2^{-n}.$$

Another, trivial, example is $\mathcal{F} = \sigma(X)$ and $\mathcal{F}_n = \sigma(X)$ for all $n$, and then $$E(X \mid \mathcal{F}_n) = X.$$ The filtration converges "instantly" and so do the conditional expectations.

The examples suggest that the rate of convergence of the conditional expectations is the same as the "rate of convergence" of the filtration. Can this idea be made precise in general?

$\endgroup$
2
$\begingroup$

Rather than measuring the rate of convergence of a sequence, it is more widely accepted that we measure the rate of convergence of a stochastic process that converges to a stationary measure.For a general stochastic process that generates the data $x_1,x_2,\dots x_n$ there is not a uniform rate available since we can always construct a (normal) ergodic process that breaks down the rate. The common measure of convergence rate mentioned in ergodic theory is the so-called rate function $r_k:\mathbb{R}^+\times\mathbb{Z}^+\mapsto\mathbb{R}^+$ of frequencies for a process with joint meausre $\mu$ if $$\mu_n(\{ x_1^n: |\mu_k-p_k(\bullet\mid x_1^n)| \})\leq r_k(\epsilon,n)\text{ for fixed k,}\epsilon>0 \text{ as }n\rightarrow\infty$$ where $\mu_k$ is the finite dimensional distribution (of a sample of size $k$) and $p_k$ is the empirical measure defined by a sample of size $k$. Note that we also require $r_k(\epsilon,n)\rightarrow 0$ as $n\rightarrow \infty$. If the log-rate function $$\frac{1}{n}logr_k(\epsilon,n)>0$$ is bounded from zero then we usually want to choose an optimal value for $\frac{1}{n}logr_k(\epsilon,n)$ which is studies by large deviation theory.

But if you want to measure the convergence rate between two filtrations then a more appropriate (yet not equivalent) notion is the Rosenblatt's mixing coefficient. This notion measures the similarity between two $\sigma$-algebras directly, when two algebras are independent their mixing coefficient is zero.

This is more like an entrophy costraint on convergence rate as following. The measure of rate in terms of entropy we be specified similarly as $$\mu_n(\{ x_1^n: 2^{-n(h+\epsilon)}\le\mu_n(x_1^n)\le2^{-n(h-\epsilon)}\})\geq 1-r_k(\epsilon,n)\text{ for fixed }\epsilon>0 \text{ as }n\rightarrow\infty$$ where $h$ is the entropy of the joint measure $\mu$.

Since a martingale adapted to a filtration sequence can as well be regarded as a stochastic process, its convergence rate in terms of filtration can also be measured by the rate function defined above.

[Shiedls]Shields, Paul C. "The ergodic theory of discrete sample paths." Graduate Studies in Mathematics, American Mathematics Society (1996).

$\endgroup$
  • $\begingroup$ Thanks, this looks very helpful. I'll have to spend some time studying some of the details. $\endgroup$ – aduh Jul 24 '17 at 15:07

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.