 Solution

The conditional density givesusing either integration by parts or the expected value of an exponential distribution. HenceThis is a finite value for almost every realization of $X$, but not an integrable random variable: $\mathbb{E}[1/X]={\int }_0^{\infty }{e}^{-x}/xdx=\infty$. Since $Y\ge 0$, its conditional expectation remains well-defined in the nonnegative, extended-expectation sense; the tower property of conditional expectation then gives $\mathbb{E}Y=\infty$.