# How can I prove the backwards analog for the dominated convergence theorem for conditional expectation

Namely:

Suppose $$Y_n \rightarrow Y_{-\infty}$$ a.s. as $$n \rightarrow \infty$$ and $$|Y_n|\leq Z$$ a.s. where $$EZ <\infty$$. If $$F_n \downarrow F_{-\infty}$$ then

$$E(Y_n|F_n) \rightarrow E(Y_{-\infty}|F_{-\infty})$$ a.s.

This is the dominated convergence theorem for conditional expectations (it's from Durrett) Also $$Y_\infty = E(X|F_{\infty})$$

• You have Levy's Downward Theorem (check, e.g., David Williams, 'Probability with Martingales'): for a $W\in \mathcal{L}_1\left(\Omega,\mathcal{F},\mathbb{P}\right)$, you have that $M_{n}:=E\left[W\left|\mathcal{G}_n\right.\right]$ converges almost surely to $M_{-\infty}\in\mathcal{L}_1(\Omega)$ with $M_{-\infty}=E\left(W\left|\mathcal{G}_{-\infty}\right.\right)$. In the proof that you presented, you just need to apply this result to the identity in the second equation of Theorem 4.6.10. Also, the this result needs to be applied in the last equation when you use Jensen's inequality. Mar 12, 2019 at 8:54
• In my previous comment, I am assuming $\mathcal{G}_n\downarrow \mathcal{G}_{-\infty}$, of course. Mar 12, 2019 at 8:59
• @Mr.X If you have solved this problem, it would be better to post an answer. Mar 12, 2019 at 9:02

You have a useful result -- the backwards analog of Theorem 4.6.8. invoked in the proof of Theorem 14.6.10.:

Levy's 'Downward' Theorem [e.g., David Williams, Probability with Martingales]

Let $$W\in \mathcal{L}_1(\Omega,\mathcal{F},\mathbb{P})$$; let $$\mathcal{G}_n\downarrow \mathcal{G}_{-\infty}$$ and define $$M_n\overset{\Delta}=E\left[W\left|\mathcal{G}_n\right.\right]$$. Then, $$M_n\rightarrow M_{-\infty}$$, almost surely, with $$M_{-\infty}=E\left[W\left|\mathcal{G}_{-\infty}\right.\right]$$.

Now, we just need to adjust the proof of the forward dominated convergence theorem for conditional expectations that you presented to obtain its backwards counter-part.

The first place we can apply this is in the first identity of the second equation of Theorem 4.6.10, i.e., $$\lim_{n\rightarrow\infty}E\left[W_N\left|\mathcal{F}_n\right.\right]=E\left[W_N\left|\right.\mathcal{F}_{-\infty}\right]$$ for all $$N$$. This holds in view of Levy's Downward Theorem.

The second place where this result should be applied is in the last part, where it is observed that $$E\left[Y\left|\mathcal{F}_n\right.\right]\rightarrow E\left[Y\left|\mathcal{F}_{-\infty}\right.\right]$$, more precisely

$$\left|E\left[Y_n\left|\mathcal{F}_n\right.\right]-E\left[Y\left|\mathcal{F}_n\right.\right]\right|= \left|E\left[Y_n\left|\mathcal{F}_n\right.\right]-E\left[Y\left|\mathcal{F}_n\right.\right]-E\left[Y\left|\mathcal{F}_{-\infty}\right.\right]+E\left[Y\left|\mathcal{F}_{-\infty}\right.\right]\right|$$

$$\geq \left|E\left[Y_n\left|\mathcal{F}_n\right.\right]-E\left[Y\left|\mathcal{F}_{-\infty}\right.\right]\right|-\left|E\left[Y\left|\mathcal{F}_{-\infty}\right.\right]-E\left[Y\left|\mathcal{F}_n\right.\right]\right|$$

and observe that the last term in the last inequality above converges to zero, almost surely, in light of Levy's Downward Theorem.

Everything else holds as in the original forward version.

Remark. You apply Levy's Downward Theorem to obtain the Downward (or backwards) version of Theorem 14.6.10 just as you apply Levy's Upward Theorem to prove Theorem 14.6.10. (which in the proof to Theorem 14.6.10 is referred to as Theorem 4.6.8). Levy's Downward Theorem replaces Theorem 4.6.8.