$\newcommand{\E}{\operatorname{E}}$ Finding density function of $\E[X\mid Y]$ and $\E[Y\mid X]$, where $\E[X\mid Y]=Y/2$, $\E[Y\mid X]=X+1,$ $f(x,y)=e^{-y}$, and $0 \leq x \leq y$.
I am unsure on how to find the density of an expected value, could anyone please help me with the formula or method to use to find it?
$\newcommand{\E}{\operatorname{E}}$ $$ \iint\limits_{\{\,(x,y)\,:\, 0\,\le\,x\,\le\,y\,\}} e^{-y} \, d(x,y) = \int_0^\infty \left( \int_x^\infty e^{-y} \, dy \right) \, dx = \int_0^\infty e^{-x} \, dx = 1 $$ and thus this is indeed a probability density.
The conditional density of $y$ given the event $X=x$ is $\big( \text{constant}\cdot e^{-y} \big)$ for $y\ge x.$ Since $$ \int_x^\infty e^{-y}\,dy = e^{-x}, $$ the "constant" must be $1/e^{-x} = e^x.$ Thus the density is $e^x e^{-y} = e^{x-y} \text{ for } y\ge x.$ The conditional expect value is therefore $$ \E(Y\mid X=x) = \int_x^\infty e^{x-y} \, dy = x+1, $$ so we can say $$ \E(Y\mid X) = X+1. $$ For the conditional density of $X$ given $Y=y,$ note that the joint density does not depend on $x$ beyond the fact that $0\le x\le y,$ so the density is constant on the interval from $0$ to $y;$ in other words, given $Y=y,$ the conditional distribution of $X$ is the uniform distribution on that interval.