I got the following question:
$X$ and $Y$ are random variables upon the probability space $(S,F,P)$. $Y$ is discrete with range $W$. Let $g: \Bbb R\to \Bbb R$ be a function so that $g(Y)$ is a random variable.
Prove that for every $y\in W$ for which $g(y)\neq 0$, and for $X$ continuous given $Y=y$ we can obtain that: $$E[g(Y)X|Y=y]=g(y)E[X|Y=y].$$
I managed to show that $$f_{g(Y)X|Y=y}(t)=f_{X|Y=y}\left(\frac{t}{g(Y)}\right).$$
And now I am trying to use that in order to calculate $ E[g(Y)X|Y=y]$, but I find it difficult to understand the definition of this conditional mean:
Is it an integral (sum) of two variables $X$ and $g(Y)$ or an integral of one variable. If this is the case, what is the variable I'm integrating on? It made me confused, and it would be helpful if you can explain it to me.
It is enough to observe here that:$$\mathbb E[g(Y)X\mid Y=y]=\mathbb E[g(y)X\mid Y=y]=g(y)\mathbb E[X\mid Y=y]$$
This also holds if $g(y)=0$ and continuity of $X$ under $Y=y$ is redundant as well.
Essential for the first equality is that under condition $Y=y$ it is immediate that $g(Y)X=g(y)X$.
The second equality is an application of $\mathbb E[aZ]=a\mathbb E[Z]$ where $a$ denotes a constant.
edit:
Integration:
$\mathbb E[g(Y)X\mid Y=y]=\int g(y)x P_{X|Y=y}(dx)=g(y)\int xP_{X|Y=y}(dx)=g(y)\mathbb E[X\mid Y=y]$
It is enough if $g$ is a measurable function. So continuity is okay, but is not necessary.