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.

  • $\begingroup$ 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). $\endgroup$ Mar 10, 2017 at 4:19

1 Answer 1


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


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