$E[g(X,Y)|Y=y] = E[g(X,y)|Y=y]$ - Exercise $3.77$ Ross Probability Models.

Question

How would you show that $E[g(X,Y)|Y=y] = E[g(X,y)|Y=y]$ in the continuous case? This problem is in Probability Models by Ross. No solution is provided, and no information is given about the random variables, altough I suppose we could assume the functions involved are absolute convergent or something.

The penultimate step in the discrete case is that all $p(X=i, Y=j|Y=y)$ are $0$ unless $j = y$, which I can't translate into the continuous case. My real analysis skills are non-existent, but you aren't supposed to know that for this book either way.

Answer 1

I tried to prove it in the simplified case g(X,Y) = XY. (Part a in Ross' book)

Let Z = XY. Then $$E[Z|Y=y] = \int z \cdot f_{Z|Y}(z|y)dz$$ $$= y \int x \cdot f_{Z|Y}(z|y)dx$$ through change of variable. And $$ f_{Z|Y}(z|y) = f_{XY|Y}(xy|y)$$ $$ = \frac{f(XY=xy, Y=y)}{f(Y=y)}$$ $$=\frac{f(X=x, Y=y)}{f(Y=y)}$$ which gives the desired conclusion in this case.

Is this any more rigorous than the comment made by Peter Morfe in the comment above? To convince ourselves about the last step, I suppose we could look at the cumulative distribution functions and take derivatives.

Answer 2

Let $\mu_{X,Y}$ be the joint law of $(X,Y)$. This is the Borel probability measure in $\mathbb{R}^{d} \times \mathbb{R}^{d}$ (assuming $X$ and $Y$ are both random vectors taking values in $\mathbb{R}^{d}$) such that, for any two Borel sets $A, B \subseteq \mathbb{R}^{d}$, \begin{equation*} \mu_{X,Y}(A \times B) = \mathbb{P}\{X \in A, \, \, Y \in B\}. \end{equation*} Similarly, let $\mu_{Y}$ denote the law of $Y$.

It is possible to show that there is a weakly $\mu_{Y}$-measurable, probability measure-valued function $y \mapsto \mu_{y}^{X}$ such that if $f$ is a bounded, continuous function on $\mathbb{R}^{d} \times \mathbb{R}^{d}$, then \begin{equation*} \int_{\mathbb{R}^{d} \times \mathbb{R}^{d}} f(x,y) \, \mu_{X,Y}(dx \otimes dy) = \int_{\mathbb{R}^{d}} \left[ \int_{\mathbb{R}^{d}} f(x,y) \, \mu_{y}^{X}(dx) \right] \, \mu_{Y}(dy). \end{equation*} I have seen the measures $\{\mu^{X}_{y}\}_{y \in \mathbb{R}^{d}}$ referred to as a regular family of conditional probabilities --- the term disintegration is also used to describe this process of decomposing $\mu$.

One now chooses a suitable $\mu_{Y}$-measurable set $E \subseteq \mathbb{R}^{d}$ with $\mu_{Y}(E) = 1$ and defines $\mathbb{E}(\cdot \, \mid \, Y = y)$ for $y \in E$ and bounded, continuous $f : \mathbb{R}^{d} \times \mathbb{R}^{d} \to \mathbb{R}$ by the formula \begin{equation*} \mathbb{E}(f(X,Y) \, \mid \, Y = y) = \int_{\mathbb{R}^{d}} f(x,y) \, \mu^{X}_{y}(dx). \end{equation*} With this formula in hand, it is now clear that $\mathbb{E}(g(X,Y) \, \mid \, Y = y) = \mathbb{E}(g(X,y) \, \mid \, Y = y)$ for $\mu_{Y}$-a.e. $y \in \mathbb{R}^{d}$ (in particular, for $y \in E$).