# Does $\lim\limits_{n\rightarrow \infty}E[X|F_n]$ exists a.s if $F_n$ is the $\sigma-$field generated by $Y_0,...,Y_n$ where $Y_n=X+W_n$.

Let $X$ be a random variable such that $|X|<K$ a.s for some $K>0$. Let $Y_n=X+W_n$ for $n\in \mathbb{N}$ where $(W_n)$ is some process. If we define $F_n$ be the $\sigma-$field generated by $Y_0,Y_1,...,Y_n$. Does $\lim\limits_{n\rightarrow \infty}E[X|F_n]$ exists a.s?

My answer is "Yes". I proved it by using Martingale convergence Theorem, but I am not pretty sure whether I did it correct or not. If there is any problems of my proof, please let me know. Thanks!

The detail is as follows:

Let $Z_n=E[X|F_n]$, and note that $E|Z_n|=E|E[X|F_n]|\leq E(E[|X| | F_n])\leq K$ for all $n\in\mathbb{N}.$ This implies that $Z_n$ is $L^1$; moreover, $\sup\limits_{n\in\mathbb{N}} E|Z_n|<\infty.$

On the other hand, for $m<n,$ we have $E[Z_n|F_m]=E[E[X|F_n]|F_m]=E[X|F_m]=Z_m$.

Thus, we obtain $(Z_n)$ is a $(F_n)-$martingale, and $\{Z_n\}$ is bounded in $L^1$, by Martingale convergence theorem, $Z_n=E[X|F_n]$ converges almost surely.

• Your proof looks correct. The answer below shows how the theorem can be strengthened (you probably already noticed that you didn't use any particulars about the filtration). Commented Mar 10, 2017 at 4:19

If $X$ is any integrable random variable and $(\mathscr{F}_n)$ a filtration, then $\mathbf{E}(X \mid \mathscr{F}_n)$ is a martingale. The martingale is uniformly integrable and converges a.e. and in the $\mathscr{L}^1$ sense to $\mathbf{E}(X \mid \mathscr{F}_\infty),$ where $\mathscr{F}_\infty$ is the sigma field generated by the union of the $\mathscr{F}_n.$
You proved correctly that $Y_n$ is a martingale and that is bounded in $\mathscr{L}^1.$
Uniform integrability. Notice that $|\mathbf{E}(X \mid \mathscr{F}_n)| \leq \mathbf{E}(|X| \mid \mathscr{F}_n)$. Now, write $Y_n = \mathbf{E}(X \mid \mathscr{F}_n)$ then the set $\{Y_n > c\}$ belongs to $\mathscr{F}_n,$ and the definition of the conditional expectation yields $$\int\limits_{\{|Y_n| > c\}} |Y_n|\ d\mathbf{P} \leq \int\limits_{\{|Y_n| > c\}} \mathbf{E}(|X| \mid \mathscr{F}_n)\ d\mathbf{P} = \int\limits_{\{|Y_n| > c\}} |X|\ d\mathbf{P};$$ apply now Markov's inequality to conclude $\mathbf{P}(|Y_n| > c) \leq \dfrac{\mathbf{E}(|X|)}{c},$ and this inequality is uniform in $n.$ So, the absolute continuity of the measure $\displaystyle \mu(A) = \int\limits_A |X|\ d\mathbf{P}$ allows deducing that the martingale $(Y_n)$ is uniformly integrable.
Conclusion. Since $(Y_n)$ is a $\mathscr{L}^1$ bounded (as you proved) uniform integrable martingale, it converges to some random variable $Y_\infty$ for almost every point before $\mathbf{P}.$ To show $Y_\infty = \mathbf{E}(X \mid \mathscr{F}_\infty)$ let $A$ be in any $\mathscr{F}_n,$ and so the $\mathscr{L}^1$ convergence implies $$\int\limits_A Y_\infty\ d\mathbf{P} = \int\limits_A \mathbf{E}(X \mid \mathscr{F}_\infty)\ d\mathbf{P}.$$ Hence, the two measures $\displaystyle \nu(A) = \int\limits_A Y_\infty\ d\mathbf{P}$ and $\displaystyle \rho(A) = \int\limits_A \mathbf{E}(X \mid \mathscr{F}_\infty)\ d\mathbf{P}$ coincide on the field $\displaystyle \bigcup_{n = 1}^\infty \mathscr{F}_n,$ by uniqueness, they coincide in the minimal sigma field generated by said field, namely, on $\mathscr{F}_\infty.$ QED