A surprisingly tricky question regarding conditional expectation

Question

Let $S_n := X_1 + \dots + X_n$ for some i.i.d. r.v. $X_i$ (specifics not so important) and let $f$ be a function satisfying

\begin{align} Ef(x+X_1) = f(x). \end{align} Then, as always the following would hold \begin{align} E\left(f(S_n + X_{n+1}) \vert X_1, \dots X_n\right) = f(S_n) \quad \text{(correct?)} \end{align} Now, what if we additionally have some r.v. $Y_1, \dots, Y_n$ depending on $X_1, \dots, X_n$ in some way but independent of $X_{n+1}$ and consider \begin{align} E\left( f(S_n + X_{n+1}) \vert X_1,\dots, X_n, Y_1, \dots, Y_n\right), \end{align} is this still equal to $f(S_n)$?

It is weird that I have never thought about this before... Also, this question could probably be asked in a more general form, but I thought giving a hands-on example might make it more palpable.

Edit: As was correctly pointed out, I should've been clearer: The $Y_1, \dots, Y_n$ are not stochastically independent of the $X_1, \dots, X_n$. They're not measurebale functions of the $X_1, \dots, X_n$.

Answer 1

In general, if $\mathcal{G}$ is a $\sigma$-field s.t. $S$ is $\mathcal{G}$-measurable and $X$ is independent of $\mathcal{G}$, then for any integrable function $\varphi$, $\mathsf{E}[\varphi(S,X)\mid \mathcal{G}]=g(S)$ a.s., where $g(s):=\mathsf{E}\varphi(s,X)$. Apply this result to your case with $\mathcal{G}=\sigma\{X_1,\ldots,X_n,Y_1,\ldots,Y_n\}$, assuming that $X_{n+1}$ is independent of $\mathcal{G}$ (see @Michael's comments and examples 1, 2, and 3).

Answer 2

If $Y_1,\ldots,Y_n$ are measurable functions of $X_1,\ldots,X_n$, then $\sigma(X_1,\ldots,X_n,Y_1,\ldots,Y_n) = \sigma(X_1,\ldots,X_n)$. Hence $$\mathbb E[f(S_n + X_{n+1})| X_1,\ldots,X_n,Y_1,\ldots,Y_n] = \mathbb E[f(S_n + X_{n+1}) | X_1,\ldots,X_n ] = \big.\int f(x+y) dQ_{X_{n+1}}(y)\big|_{y=S_n} = f(S_n) $$