1
$\begingroup$

Statement: Given a sequence $(\mathcal{F}_n)_n$ of $\sigma$-Algebras with $\mathcal{F}_{n+1} \subset \mathcal{F}_n$ (i.e. we lose informationen as time progresses) and defining $\mathcal{F}_\infty := \bigcap_n \mathcal{F}_n$.

If $Y_n \to Y_\infty$ a.s. and $|Y_n| \leq Z \in L^1$ then $$ E(Y_n \mid \mathcal{F}_n) \to E(Y_\infty \mid \mathcal{F}_\infty) \text{ a.s.} \tag{*}$$


My approach: I thought the above statement was obvious until I tried to came up with a proof for it, by the "regular" dominated convergence theorem for conditional expectation I can obtain two statements:

$$ E(Y_n \mid \mathcal{F}_\infty) \to E(Y_\infty \mid \mathcal{F}_\infty) \text{ a.s.} \tag{1} $$ and for a arbitrary but fixed $k \in \mathbb{N}$ also $$ E(Y_n \mid \mathcal{F}_k) \to E(Y_\infty \mid \mathcal{F}_k) \text{ a.s.} \tag{2}$$ Since $\mathcal{F}_\infty, \mathcal{F}_k$ are $\sigma$-Algebras. Especially in (2) the $k$ can coincide with $n \in \mathbb{N}$ (I doubt that this helps). I haven't made use of the fact yet that we lose information over time, i.e. we have the condition that $\mathcal{F}_{n+1} \subset \mathcal{F}_n$, the only idea I have concerning this statement is to integrate it into the tower property.

But I don't see how I could possibly relate 1,2 (and the tower property) to obtain (*)

Any hints?

$\endgroup$
1
$\begingroup$

I assume that $\{Y_n\}$ is a reversed MTG, i.e. $\forall n, \mathbb{E}[Y_{n-1}\mid\mathcal{F}_n]=Y_n$. First,

$$ |\mathbb{E}[Y_n\mid \mathcal{F}_n]-\mathbb{E}[Y_\infty\mid \mathcal{F}_\infty]|\\ \le |\mathbb{E}[Y_n\mid \mathcal{F}_n]-\mathbb{E}[Y_\infty\mid \mathcal{F}_n]|+|\mathbb{E}[Y_\infty \mid \mathcal{F}_n]-\mathbb{E}[Y_\infty\mid \mathcal{F}_\infty]| \\ \le \mathbb{E}[|Y_n-Y_\infty|\mid \mathcal{F}_n]+|\mathbb{E}[Y_\infty \mid \mathcal{F}_n]-\mathbb{E}[Y_\infty\mid \mathcal{F}_\infty]| \quad\text{a.s.} $$

For fixed $m\in\mathbb{N}$,

$$ \limsup_n\mathbb{E}[|Y_n-Y_\infty|\mid \mathcal{F}_n]\le \lim_n\mathbb{E}[\sup_{l,k\ge m}|Y_l-Y_k|\mid \mathcal{F}_n] \\ \overset{(*)}{=}\mathbb{E}[\sup_{l,k\ge m}|Y_l-Y_k|\mid \mathcal{F}_\infty] \quad \text{a.s.} $$

and the last term converges to $0$ a.s. as $m\to \infty$ (the supremum is dominated by $2Z\in L^1$). As for the second term, $\{\mathbb{E}[Y_{\infty}\mid \mathcal{F}_n]\}$ is a reversed MTG so it converges a.s. (and in $L^1$) to $\mathbb{E}[Y_{\infty}\mid \mathcal{F}_{\infty}]$ (the equality $(*)$ holds by the same reason).

$\endgroup$
2
  • $\begingroup$ Thanks a lot for this. One question, is it not possible to say that we always have $$ E(Y_n \mid \mathcal{F}_n) \to E( Y_\infty \mid \mathcal{F}_n) \text{ a.s.}$$ by dominated convergence? (I suppose you prove that in fact in your second paragraph). The idea with the backwards martingale indeed takes care of the second term. $\endgroup$
    – Spaced
    Nov 5 '16 at 13:17
  • $\begingroup$ @Spaced It would work if $\mathcal{F}_n$ was a constant $\sigma$-field. In your case it changes with $n$... $\endgroup$
    – d.k.o.
    Nov 5 '16 at 18:10

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.