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One condition for a martingale $M_k$ with a general filtration $\mathcal{F}_k$ is that the involved random variables $M_k$ are $\mathcal{F}_k$-measurable.

Now I have $M_n=Y_1+\dots +Y_n$ and $Y_i=S(X_i, Z_i)- E(S(X_i,Z_i) \mid \mathcal{F}_i)$, $\mathcal{F}_i=\{X_1, \dotsc, X_i\}$, which satisfies the martingale property $E(M_{n+1} | \mathcal{F}_n) = M_n$.

$S_i:\Omega \to R, \omega \mapsto S(X_i(\omega), Z_i(\omega))$, $S:R \times R \rightarrow R$.

I know that the conditional expectation is $\mathcal{F}_i$-measurable by definition, so the question is when is $S_k$ $\mathcal{F}_k$-measurable? I would have for example that $S_k$ is $f(X_k+Z_k)/Z_k$ where f is differentiable and $Y_k \neq 0$ a.s. and iid. (This is a follow-up question to this solution) So maybe it just boils down to: Is $Z_i$ $X_i$-measurable? And I am afraid that the answer should be no, generally not. But if it is this way maybe still $M_n$ is $\mathcal{F}_n$-measurable?

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Without hypotheses on the measurability of $(Z_n)$ it seems difficult to decide whether $S(X_n,Z_n)$ is $F_n$-measurable or not. –  Did Oct 5 '11 at 7:30
    
@DidierPiau you are right.. It turned out that the solution I linked to needs to be changed again -- It seems that the martingale property for $M_n$ can only be shown with the natural filtration $\mathcal{F}^M_n$ or $\mathcal{F}^{(X,Z)}_k$ –  Johannes L Oct 5 '11 at 13:46
    
What solution? // Tell me if what I just posted on the other page is related to the problem you want to solve. –  Did Oct 5 '11 at 14:26
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1 Answer

up vote 3 down vote accepted

$S_k$ is $\mathcal{F}_k=\sigma(X_1,...,X_k)$-measurable if and only if there exists a Borel-measurable function $f$ such that $S_k=f(X_1,...,X_k)$.

This is known as Doob's lemma (or sometime Doob-Dynkin's Lemma).

Regards

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Thanks a lot, this is really nice to think of whenever one has expressions with measurability in it –  Johannes L Oct 5 '11 at 13:47
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