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?


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).

  • $\begingroup$ 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

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