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Setup:
Let $(epsilon_n)_{ngeq 1}$ be IID Bernoulli taking values in ${-1,1}$ and $mathscr{F}_n=sigma(epsilon_1,dotsc,epsilon_n)$. Then let $(C_n)_{ngeq 1}$ be any previsible process (i.e. $C_n$ is $mathscr{F}_{n-1}$ measurable) and $Z_0>0$ such that we may define the process $(Z_n)$ by $Z_n=Z_{n-1}+epsilon_n C_n$ with $0<C_n<Z_{n-1}$ for $ngeq 1$.

Specific question:
How do I justify that if $Y_n=C_n/Z_{n-1}$ then
$mathbb{E}(log(1+epsilon_n Y_n) | mathscr{F}_{n-1})=g(Y_n)$ where $g(x)=plog(1+x)+qlog(1-x)$ wherever this is defined?

My thoughts:
Intuitively, $Y_n$ would be known from $mathscr{F}_{n-1}$ since $C_n$ is and so would $Z_{n-1}$ but I’ve been stumped from translating this into detail. I understand simpler calculations where you can “pull out what is known”, e.g. $mathbb{E}(epsilon_n C_n | mathscr{F}_{n-1} )=C_n(2p-1)$ but here I directly know $C_n$ is previsible. It seems very natural to think “oh this RV is known so I can just average out the other one even when its a function of both” but I want to be sure why, especially when a function of the RVs is involved, if that’s true.

Generalized a bit:
More generally (but perhaps not totally well-posed), given some Borel function $f(x,y)$, and process $X, Y$ and $mathscr{F}_n$ generated by $X$ when can one say
$mathbb{E}(f(X_n, Y_n) | mathscr{F}_{n-1})=g(Y_n)$ where $g(y)=mathbb{E}(f(.,y))$? Is all that is needed that $Y_n$ is previsible? Or more? Thanks.

For every sigma-algebra $mathcal G$, every random variables $X$ and $Y$ and every measurable function $h$ such that $h(X,Y)$ is integrable, if $X$ is independent of $mathcal G$ and $Y$ is $sigma(mathcal G)$-mesurable, then $$E(h(X,Y)midmathcal G)=g(Y)$$ where the function $g$ is defined as $$g(y)=E(h(X,y))$$

Proof: It suffices to show that, for every bounded and $sigma(mathcal G)$-mesurable random variable $Z$, $$E(h(X,Y)Z)=E(g(Y)Z)$$ By hypothesis, $X$ is independent of $(Y,Z)$ hence, using Fubini, $$E(h(X,Y)Z)=int!!!!!iint h(x,y)zdP_X(x)dP_{Y,Z}(y,z)=iintleft(int h(x,y)dP_X(x)right)zdP_{Y,Z}(y,z)$$ Each inner parenthesis is $$int h(x,y)dP_X(x)=E(h(X,y))=g(y)$$ hence the whole is $$E(h(X,Y)Z)=iint g(y)zdP_{Y,Z}(y,z)=E(g(Y)Z)$$ and the proof is complete.