Expection from joint density distribution

Question

Given $f(x,y)=\frac{e^{-y}}{y}$, $0<x<y$ and $0<y$, find $E(X)$. I tried to find $E(X)$ by finding the marginal pdf of $x$, but I realized that I was unable to do so because $\frac{e^{-y}}{y}$ is not integrable with respect to $y$. Is there any other way to find it?

Answer 1

Note that the joint density is constant with respect to $x$ over its support $\{(x,y): 0\leqslant y<\infty, 0<x<y\}$. It follows that conditioned on $Y=y$, $X$ is uniformly distributed on $(0,y)$, and so $\mathbb E[X\mid Y=y] = \frac12 y$. Now, the density of $Y$ is given by $$ f_Y(y) = \int_0^y \frac{e^{-y}}y\ \mathsf dx = e^{-y},$$ and so by the law of total expectation, \begin{align} \mathbb E[X] &= \int_0^\infty \mathbb E[X\mid Y=y]f_Y(y)\ \mathsf dy\\ &= \int_0^\infty \frac12 y e^{-y}\ \mathsf dy\\ &= \frac12 \int_0^\infty y e^{-y}\ \mathsf dy\\ &= \frac12. \end{align}