Motivation:
It might appear intuitive that $E(f(X,Y)|Y=y)=E(f(X,y)|Y=y)$, i.e. we just substitute in what is known in the conditional expectation. However, I want to prove this rigorously using the definition and properties of conditional expectation defined in measure theory.
The general setting:
Let $(\Omega, \mathcal{F}, P)$ be a probability space, consider a sub-sigma algebra $\mathcal{G}\subseteq \mathcal{F}$ and random elements $X: \Omega\rightarrow S$, $Y: \Omega\rightarrow T$. Assume $Y$ is $\mathcal{G}$-measurable. Also, let $f$ be a real-valued measurable function $f: S\times T\rightarrow R$ such that $E(f(X,Y))<\infty$.
Recall: $E(f(X,Y)|\mathcal{G}) $ is defined as the unique (up to a.s. equivalence) $\mathcal{G}$-measurable function in $(\Omega, \mathcal{G}, P|_{\mathcal{G}})$ such that $\int_A f(X,Y) dP = \int_A E(f(X,Y)|\mathcal{G}) dP$ for all $A\in \mathcal{G}$. Any properties deduced from this definition are allowed to use.
Questions:
(1). Can we deduce that $E(f(X,Y)|\mathcal{G})(w) = E(f(X,Y(w))|\mathcal{G})(w)$ $P-$almost surely in $w\in \Omega$?
(2). If (1) is not true in general, then under what regularity conditions does (1) hold?
(Edit: by the independence lemma, (1) holds under the assumption of X is independent of Y, when X and Y are random vectors, but assuming independence seems too strong)
Best Answer
A sufficient condition for (1) to be true is that there exists a regular conditional distribution for $X$ given $\mathcal{G}$. This is true, for instance, if $S$ is a standard Borel space. A regular conditional distribution for $X$ given $\mathcal{G}$ is a function $\mu(A,\omega)$, where $\mu(\cdot,\omega)$ is a probability measure on $S$ for each $\omega$, $\mu(A,\cdot)$ is a random variable for each measurable $A$, and $P(X\in A\mid\mathcal{G})=\mu(A)$ a.s., for each measurable $A$. In this case, if $Z$ is a version of $E[f(X,Y)\mid\mathcal{G}]$, then we have $$ Z(\omega) = \int f(x,Y(\omega))\,\mu(dx, \omega) $$ for almost every $\omega$. In particular, if we define $$ Z(y) = \int f(x,y)\,\mu(dx), $$ then $Z(y)$ is a version of $E[f(X,y)\mid\mathcal{G}]$ for every $y$, and $Z(Y)$ is a version of $E[f(X,Y)\mid\mathcal{G}]$. See Chapter 5 of Foundations of Modern Probability by Olav Kallenberg for further details.