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Let $X_k$ be iid with $P(X_k=1)=p$, $P(X_k=-1)=q$, where $p\ne q$. Now define $S_n=\sum\limits^n_{k=1}X_k$ with $S_0=0$ and $Y'_n=S^2_n$, and $Y_n=Y'_n-\alpha_n =S^2_n-\alpha_n$, we want to find $\alpha_n$ such that $Y_n$ is a martingale.

I found $\alpha_n=1+2S_n(p-q)$ but according to Doob's decomposition the compensator should be $F_{n-1}$ measurable, but we don't have that here.

Thanks.

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The process $(S_n^2-A_n)_{n\geqslant0}$ is a martingale if $A_0=0$ and $A_{n+1}=A_n+2(p-q)S_n+1$, for example $$ A_n=\sum\limits_{k=0}^{n-1}(2(p-q)S_k+1)=n+2(p-q)\sum\limits_{k=1}^{n-1}S_k. $$ For every $n\geqslant1$, $A_n$ is $F_{n-1}$ measurable hence $(A_n)_{n\geqslant1}$ is a compensator of $(S_n^2)_{n\geqslant0}$.

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So $\alpha_n n$ is actually $F_{n-1}$ measurable? –  John Klein Nov 12 '11 at 22:19
    
It seems your $\alpha_nn$ should actually be $\alpha_n$. Anyway, as I said, $A_n$ (which is $\alpha_n$ if the typo in your post is actually as I described) is $F_{n-1}$ measurable. –  Did Nov 12 '11 at 22:22
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@John, did you read my answer? Please do (carefully), and then we'll talk. –  Did Nov 12 '11 at 22:25
    
Sorry, Didier, I am not sure how you calculated the $A_{n+1}$, my answer is similar to yours at the top, but I Don't have it in a recursive definition. thanks –  John Klein Nov 12 '11 at 22:32
    
Thanks so much, I figured it out now! –  John Klein Nov 12 '11 at 22:36
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