Question

Conditional expectation:

Show that \[E[g(X,Y)|Y=y]=E[g(X,y)|Y=y]\]

Answer 1

I tried applying the definition, in the continuous case...
\[E[g(X,Y)|Y=y]=\int_{-\infty}^{\infty}g(x,y)~f_{X|Y}(x|y)dx\]
I'm not sure what else to do. I tried breaking up \(f_{X|Y}(x|y)\) using its definition but It doesn't really help.

In fact, I'm not even sure how to write \(E[g(X,y)|Y=y]\) using the definition... \(\int_{-\infty}^{\infty}g(x,_~~?)~f_{X|Y}(x|y)dx??\)

Answer 2

I mean I feel like I'd be writing the same thing for the RHS:
\[E[g(X,y)|Y=y]=\int_{-\infty}^{\infty}g(x,y)~f_{X|Y}(x|y)dx\]... I mean would there be more to this to prove or is it really just that the two expressions write out to be the same when using the definition?