Let $X_n, n\geq 0$ be a martingale. We know that $E[X_n]=E[X_m]$ for all $m, n \geq 0$. Moreover suppose that $X_n \rightarrow X$ P-a.s.
What do we know about $E[X]$? Is it clear that $E[X]=E[X_0]$?
What if $X_n$ is a submartingale? Is it clear that $E[X] \geq E[X_0]$? And the analogous result for a supermartingale?
What if the convergence is not P-a.s. but in $L^p$ for some $p \geq 1$?
Best Answer
$X := \lim X_n$ right?
Well we want to know if $E[\lim X_n] = \lim E[X_n] (= \lim E[X_0] = E[X_0])$.
Depending on the $X_n$, we may or may not have $E[\lim X_n] = \lim E[X_n]$.
We could use Fatou's Lemmas to say that
$$E[\liminf X_n] \le \liminf E[X_n] \le \limsup E[X_n] \le E[\limsup X_n] \to E[\lim X_n] = \lim E[X_n]$$
but we need the $X_n$'s to satisfy the assumptions of Fatou's Lemmas
For a super/sub mart, we have:
$$E[X] := E[\lim X_n] \stackrel{if}{=} \lim E[X_n] \le / \ge \lim E[X_0] = E[X_0]$$