# Measurability question for martingale (Is $S(X_i, Z_i)- E(S(X_i,Z_i) \mid \mathcal{F}_i)$ $\mathcal{F}_i=\{X_1, \dots, X_i \}$-measurable?)

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?

• 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, 2011 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 $\mathcal{F}^{(X,Z)}_k$ Oct 5, 2011 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, 2011 at 14:26

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