Another confusion: from the definition, if $M_n$ is a martingale $~E(M_{n+1}|\mathscr{F}_n)=M_n~$ could it implies $~E(M_{n+1}|\mathscr{F}_k)=M_k~,~~ k\le n~$.
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1For your "Another confusion" comment: Can you simplify $$E\bigg(E(X|\mathscr{F}{k+1})\bigg|\mathscr{F}{k}\bigg)?$$ If you know how to do that, try inducting on $n-k$ to evaluate $E(M_{n+1}|\mathscr{F}_k)$ – Brian Moehring Mar 02 '20 at 04:06
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Sorry, could you tell me how to simplify it? and Mk and Mn are independent or not? – AlieZ777777 Mar 02 '20 at 22:25