I've cross-posted this question at Mathoverflow with the hope of generating more ideas.

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?

  • $\begingroup$ We know that for any $F\in\mathcal F$, the quantity $\delta_n\left(F\right):=\inf_{G\in\mathcal F_n}\mathbb P\left(F\Delta G\right)$ goes to $0$ as $n$ goes to infinity. Probably a fast rate or a uniform convergence can help. $\endgroup$ Jun 28, 2017 at 11:42

1 Answer 1


I believe there are, theoretically, several ways to quantify the rate of convergence of $\mathcal{F}_n$. Here is my suggestion. Disclaimer: It is a very crude, probably naive, approach, but maybe someone may find creative ideas by reading this.

Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space. The idea is to construct a sequence of non-negative numbers $(a_n)_{n \in \mathbb{N}}$ that goes to zero when $\mathcal{F}_n \rightarrow \mathcal{F}$, as $n \rightarrow +\infty$. Then, the rate of convergence $R(n) = R_n(\mathcal{F}_n \rightarrow \mathcal{F})$ of the filtration will be understood as the speed of convergence of $(a_n)$, which is, at least in principle, an easier problem. We establish the convention that if $a_n=0$ for all $n$, then $R(n) = +\infty$ (instant convergence).

Following the idea of Davide Giraudo, define, for any $F \in \mathcal{F}$, $$ M_{n}(F) = \inf_{G \in \mathcal{F}_n} \mathbb{P}(F \Delta G) $$ This acts as the baseline. Some observations: $M_n(\Omega) = M_n(\emptyset) = 0$ (we must take care of this), if $\mathcal{F}_n = \mathcal{F}$ for all $n$, then $M_n(F)=0$ for all $F \in \mathcal{F}$. Now, let $L_n$ and $U_n$ be defined as $$ L_n = \inf_{ F \in \mathcal{F}, \mathbb{P}(F) \neq 0 } \{ M_n(F) \}, \quad U_n = \sup_{ F \in \mathcal{F} } \{ M_n(F) \} $$ which is like considering the "worst-case" (minimal difference, $L_n$) scenario and the "best-case" (maximal difference, $U_n$) scenario, where we intentionally excluded the trivial cases , e.g. $\emptyset$. Observe that, if a particular set $\bar{F} \in \mathcal{F}$ attains $U_n$, then all the other sets $F \in \mathcal{F}$ satisfy $ 0 \leq M_n(F) \leq M_n(\bar{F})$. Finally, let $$ a_n = \frac{L_n + U_n}{2} $$ which acts like an "average difference". Let us check that it works when $\mathcal{F}_n = \mathcal{F}$. So, $M_n(F)=0$, $L_n = U_n = 0$ and also $a_n = 0$. Let us now check that if $\mathcal{F}_n$ doesn't actually converge to $\mathcal{F}$, $a_n$ doesn't go to zero. It is enough to consider the limit $n \rightarrow +\infty$, so $L_n \rightarrow L$ and $U_n \rightarrow U$. Reason by contradiction. $a_n = 0$ implies $L=U=0$. But $U=0$ would imply that $M_{+\infty}(\bar{F})=0$ for the set $\bar{F}$ which maximizes the difference, so all the other sets also have $M_{+\infty}(F)=0$ (by previous property), which implies $\mathcal{F}_{+\infty} = \mathcal{F}$ ("almost surely"), which is a contradiction, so $U \neq 0$ and $a_n$ can't go to zero.

I am sure my answer could be improved and probably I was not mathematically precise enough and there may be some oversights, this is just to generate ideas for a very interesting question.

EDIT: To be clear, the general procedure I am suggesting is the following $$ \mathcal{F}_n \color{red}{\rightarrow} a_n \rightarrow R(n) \color{red}{\rightarrow} E(X|\mathcal{F}_n) $$ Start with $\mathcal{F}_n$. From that, you generate a sequence $a_n$. You compute $R(n)$ from $a_n$ using the concepts given in the Wikipedia page, which go from the simple $R(n) = a_{n+1}/a_n$ up to more elaborate things. Or just ask a mathematician. After that, ideally, you need a general formula linking $R(n)$ to $E(X|\mathcal{F}_n) \rightarrow X$. This is an independent problem, which can be solved just assuming to start with $R(n)$. The red arrows denote the (hard) parts where great care must be taken (in order to avoid trivial results).

  • $\begingroup$ Thanks for the interesting answer. I'll get back to you with more comments after I've had time to think about your suggestions some more. $\endgroup$
    – aduh
    Jul 6, 2017 at 15:11
  • $\begingroup$ It seems to me that $L_n$ is not so useful. As you've defined it, it's always $0$ because you haven't ruled out the trivial case $\Omega$. Even if you rule that out with more modifications, since $\mathcal{F}_n$ is a sub-algebra of $\mathcal{F}$ we will still always have $L_n=0$. $U_n$, on the other hand, seems a more promising thing to look at. As a proof of concept, I was hoping for a result linking the convergence rates of conditional expectations with quantities like $a_n$ or $U_n$. $\endgroup$
    – aduh
    Jul 6, 2017 at 15:57
  • $\begingroup$ You are right, $a_n = U_n$ is probably enough. The second question, i.e., linking the rate of convergence of filtrations to the convergence rate of conditional expectations, is beyond me. Suppose someone gives you $R(n)$ (filtration convergence rate, as a function of $n$). How would you link it to $E(X|\mathcal{F}_n) \rightarrow X$ ? If you manage to find a general formula for that link, then finding $R(n)$ will be the easy part, as rate of convergence of a sequence can be quantified: en.wikipedia.org/wiki/Rate_of_convergence $\endgroup$ Jul 7, 2017 at 8:18
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    $\begingroup$ Thanks. I will award you the bounty because I appreciate your input and because I don't want it to go to waste. But I am leaving the question open as I think there is much more to be said about it. $\endgroup$
    – aduh
    Jul 7, 2017 at 15:53
  • $\begingroup$ Thanks, @aduh. If you don't mind, could you have a look at my question: math.stackexchange.com/questions/2341151/… I found this formula in an old document, without any proof. Since I would like to know if it is true or not, proving/disproving it is essential and I would really appreciate it. $\endgroup$ Jul 7, 2017 at 20:20

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