Question on interchanging of random variables with the same distribution inside expectation

Question

Let $T\in\mathbb{N}$ and let $(S_t)_{0\leq t\leq T}$ be such that the increments $S_1-S_0,\dots,S_T-S_{T-1}$ are independent and identically distributed, and let $(\mathcal{F}_t)_{0\leq t\leq T}$ be the filtration generated by $S$. Let $f:\mathbb{R}\to\mathbb{R}$ be a measurable function. Suppose that $f(S_t)$ is integrable for all $t$. I am attempting to show that $$\max\{f(S_{T-1}),\mathbb{E}[f(S_T)|\mathcal{F}_{T-1}]\}$$ can be written as $V(T-1,S_{T-1})$ for some deterministic function $V$. Unless I am missing something implied by the maximum, it is then necessary to show that $\mathbb{E}[f(S_T)|\mathcal{F}_{T-1}]$ is an explicit function of $S_{T-1}$. I am unsure of how to do this. My only thoughts was something along the lines of $$\begin{align}\mathbb{E}[f(S_T)|\mathcal{F}_{T-1}]&=\mathbb{E}[f(S_{T-1}+\{S_T-S_{T-1}\})|\mathcal{F}_{T-1}]\\&=\mathbb{E}[f(S_{T-1}+\{S_{T-1}-S_{T-2}\})|\mathcal{F}_{T-1}]\\&=f(2S_{T-1}-S_{T-2})\end{align}$$ because of increments being identically distributed. I don't think this is valid though because of the function $f$ (and also I am not expecting to use any value of $S$ other than at the time $T-1$ in the expression). If someone could give some advice it would be greatly appreciated!

Answer 1

The substitution you made is wrong. Here is a general useful statement which helps in many situations including yours.

Let $\mathcal{S}$ be any sub-$\sigma$-field. When you have an integrable random variable, which is expressed as a function of two random variables, $g(X,Y)$ say, where $X$ is $\mathcal{S}$-measurable and $Y$ is independent of $\mathcal{S}$, the conditional expectation $E[g(X,Y)|\mathcal{S}]$ is a function of $X$ only, $\psi(X)$ say, where $\psi(x) := E[g(x,Y)]$.

Be careful, you cannot simply replace $x$ by $X$ in $E[g(x,Y)]$!

Indeed, for every $\omega$ in the probability space $(\Omega,\mathcal{A},P)$, $$\psi(X)(\omega) = \psi(X(\omega)) = E[g(x(\omega),Y)] = \int_\Omega E[g(x(\omega),Y(\omega'))] \mathrm{d}P(\omega'),$$ which is different (in general) from $$E[g(X,Y)]= \int_\Omega E[g(x(\omega),Y(\omega))] \mathrm{d}P(\omega).$$