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The following is an old exam problem:

Let $\{X_n\}$, $n\geq0$, be a process adapted to a filtration $F_n$. Prove that $(X_n,F_n)$ is a martingale, if and only if for all bounded $F_n$-stopping time $\tau$, $EX_{\tau}=EX_0$ holds.

I know if $X_n$ is a martingale, then $X_{\tau}$ is a martingale by optional stopping theorem. Hence $E(X_{\tau})=E(E(X_{\tau}|F_{0}))=E(X_0)$. However I have trouble connecting the expectation to the conditional expectation for the other direction. It seems like it needs some smart way to define $\tau$ while I haven't thought of one. Thanks for any help.

guest625
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  • The question is unclear. What precisely are you trying to prove? – jdods Jul 24 '15 at 01:44
  • Sorry for that. I changed it a little bit. Hope it's clear now.. – guest625 Jul 24 '15 at 01:48
  • Can you argue somehow that any fixed time (e.g. $\tau=n$) is a bounded stopping time, therefore $E(X_n)=E(X_0)$ for any $n$, thus $X$ is a martingale. – jdods Jul 24 '15 at 01:55
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    I get that $E(X_n)=E(X_0)$ for all n, but does that mean $X$ is a martingale? Shouldn't we prove $E(X_n|F_{n-1})=X_{n-1}$? – guest625 Jul 24 '15 at 01:59
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    I believe you cannot go from $E[X_n] = E[X_0]$ to the martingale property. A counterexample: Let $X_n = aX_{n-1}+\epsilon_n$ where ${ \epsilon_n}$ is a sequence of iid rv's with mean 0. Then, if $|a|<1$, $E[X_n]=0=E[X_0]$ but $E[X_{n+1}\mid X_n] = aX_n$ (this is a so-called AR(1) process, btw ) –  Jul 24 '15 at 03:44
  • any i.i.d. sequence e.g. standard normal distributions will be a counterexample as well – Conformal Jul 24 '15 at 03:45
  • Yes, I was mistaken. Thanks for the counterexamples! – jdods Jul 24 '15 at 03:48

1 Answers1

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Let $A \in F_{n-1}$ and define $\tau =n-1$ on $A$,$\tau =n$ on $A^c$. Check that this is a stopping time.

Now

$$ E(X_{n-1} 1_A) +E(X_n 1_{A^c}) =E X_\tau= EX_0 = E X_n = E(X_n 1_A) + E(X_n 1_{A^c}), $$ which shows what you want (why?).

PhoemueX
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