Suppose I have a sequence of random variables $X_t$ adapted to a filtration $\mathcal{F}$
when is it true that if $\lim_{n \rightarrow \infty} E(X_t| \mathcal{F}_{t-n}) = 0 $ then $E(X_t) = 0$ ?
Could I have a proof of this fact
2026-04-04 08:19:50.1775290790
If $\lim_{n \rightarrow \infty} E(X_t| \mathcal{F}_{t-n}) = 0 $ then $E(X_t) = 0$?
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This very much depends on your definition of $\mathcal{F}_{t-n}$ for $n > t$. Usually one uses the convention $\mathcal{F}_0 = \{\emptyset, \Omega\}$ and you could always extend this to $\mathcal{F}_t = \mathcal{F}_0$ for all $t < 0$. Then, however, this statement is somewhat trivial since $E(X_t | \mathcal{F}_0) = E(X_t)$ for all $t$...