For martingale, we have $E[X_{n+1}-EX_n|X_1,\dots,X_n]=0$, can we say $X_{n+1}-X_n$ is independent of $X_1,\dots,X_n$?
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On this site, the traditional prefix for obviously wrong statements is "I know that...". In reality, $X_n$ being a martingale means $E[X_{n+1}-X_n|X_1,\dots,X_n]=0$. And no, that doesn't mean independence. – Oct 21 '20 at 14:16
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Could you please give an example of this? – Boommm Oct 22 '20 at 01:19
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$X_{n+1}=X_n,(1+Y_n)$, where $Y_n$ is independent of $X_1,\ldots,X_n$ and $P(Y_n=1)=P(Y_n=-1)=\frac12$. – Oct 23 '20 at 14:18