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I am a bit confused in my reading of martingales. I am using Breiman's book and there is an example that doesn't quite make sense to me. Let us define a sequence of random variables in this way:

Let $Y_0=0$ and $Y_1,Y_2,\ldots$ i.i.d. with distribution $P[Y_i=1]=p$ and $P[Y_i=-1]=1-p$ for some $p\in(0,1)$ . Define $X_n:=\sum_{i=0}^n Y_i$. Then he says: If $p=\frac{1}{2}$ , then $(X_n)_{n\in\mathbb{N}}$ is a martingale with respect to its natural filtration $\mathcal{F}_n^X:=\sigma(X_0,\ldots,X_n)$. I dont really understand this statement, because I have to show that $E[X_{n}\mid\mathcal{F}_{s}]=X_{s}$ whenever $n\geq s$ , and I have no idea why this has to be true. By definition of conditional expectation, for any $A\in\mathcal{F}_{s}$ , we have $\int_{A}E[X_{n}\mid\mathcal{F}_{s}]=\int_{A}X_{n}\, dP=\int_{A}\, X_{s}\, dP$ . For some reason, $p=\frac{1}{2}$ should play a role here. I don't really see it.

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    To be more concrete $s=1, n=2$ ... Let $A = \{X_1=1\}$ and show that $\int_A X_2\,dP = \int_A X_1\,dP$.2011-12-14
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    [By definition of conditional expectation, for any $A\in\mathcal{F}_{s}$ , we have $\int_{A}E[X_{n}\mid\mathcal{F}_{s}]=\int_{A}X_{n}\, dP=\int_{A}\, X_{s}\, dP$]: in fact, the first equal sign is the definition of conditional expectation $E[X_{n}\mid\mathcal{F}_{s}]$, the second is the relation you want to show.2011-12-14
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    @Didier Does my answer below make sense?2011-12-14
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    @r.g. Since you ask... a crucial argument is missing when you assert that $\mathrm E(Y_{n+1}:A)=0$ if $p=\frac12$, namely, the fact that $Y_{n+1}$ and $A$ are independent (do you see why?), hence $\mathrm E(Y_{n+1}\mathbf 1_A)=\mathrm E(Y_{n+1})\mathrm P(A)=(2p-1)\mathrm P(A)$. This shows that $\mathrm E(X_{n+1}\mid\mathcal F_n)=X_n+2p-1$ and you are done.2011-12-14
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    yes. that's an important point that I forgot to state in my proof. Of course, $Y_n+1$ and $A$ are independent, because $A$ is in the sigma algebra that is not generated by $Y_n+1$.2011-12-15

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