Let $\{X_n\}_{n \geq 1}$ be a sequence of random variables adapted to the filtration $\{\mathcal{F}_n\}_{n \geq 1}$. If $X_n \to X$ a.s., does it follow that \begin{equation*} \mathbb{E}\left[X_n \, \big| \, \mathcal{F}_{n-1}\right] \to X \text{ a.s.?} \end{equation*}
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- 2$\begingroup$ I gave an example of this not holding true as stated below, but I wonder if this holds if it is assumed that $X_n \to X$ in $L^1$ in addition to almost surely. $\endgroup$Nate River– Nate River2025-11-08 08:32:40 +00:00Commented 10 hours ago
- $\begingroup$ If $X_n \to X$ in $L_1$, then, I think, the statement is true: $\| \mathbb{E}[X_n | \mathcal{F}_{n-1}] - X \| \leq \| \mathbb{E}[X_n | \mathcal{F}_{n-1}] - X_{n-1} \| + \| X_{n-1} - X\| = \| \mathbb{E}[X_n -X_{n-1} | \mathcal{F}_{n-1}] \| + \| X_{n-1} - X\|$, etc $\endgroup$Vassilis Papanicolaou– Vassilis Papanicolaou2025-11-08 16:18:07 +00:00Commented 2 hours ago
- $\begingroup$ My argument proves $L_1$ convergence, but not almost sure convergence. $\endgroup$Vassilis Papanicolaou– Vassilis Papanicolaou2025-11-08 16:41:11 +00:00Commented 2 hours ago
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1 Answer
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1 Not necessarily. Let $Y_n$ be independent $0$-$1$ valued Bernoulli random variables with $Y_n = 1$ with probability $\frac{1}{2^{n+1}}$, and set
$$X_n = \sum_{k =1}^n 2^{k} Y_k.$$
By Borel-Cantelli, almost surely only finitely many of the $Y_n$ take value $1$, so $X_n$ converges almost surely to some $X$ that takes values in the set of integers.
However$$\mathbb E[X_n | \mathcal F_{n-1}] = X_{n-1} + \mathbb E[2^n Y_n] = X_{n-1} + \frac{1}{2},$$
which is an integer plus $\frac{1}{2}$ almost surely. So for every $n$, $|\mathbb E[X_n| \mathcal F_{n-1}] - X| \geq \frac{1}{2}$, almost surely.
Thus we cannot have the desired a.s. convergence.
- $\begingroup$ Is there a $p > 0$ such that the expectation of $X^p$ is finite? $\endgroup$Vassilis Papanicolaou– Vassilis Papanicolaou2025-11-08 15:13:04 +00:00Commented 3 hours ago