Given a filtration $(\mathcal{F}_{n})_{n}$ and a sequence of random variables $(X_n)_{n}$, is it true that, for $m\leq n$,
$$\mathbb{E}\{X_n\}=\mathbb{E}\{X_m\}\Rightarrow\mathbb{E}\{X_n|\mathcal{F}_m\}=\mathbb{E}\{X_m|\mathcal{F}_m\}=X_m$$?
If the answer is 'yes', how can one prove this?
2026-03-26 04:32:48.1774499568
Is it true that $\mathbb{E}\{X_n\}=\mathbb{E}\{X_m\}\Rightarrow\mathbb{E}\{X_n|\mathcal{F}_m\}=\mathbb{E}\{X_m|\mathcal{F}_m\}=X_m$?
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