Let $X\sim \text{Exp}(\lambda), \lambda <0$. $Y$ conditional on $X=x$ is exponential distributed with parameter $\sqrt x$. Calculate $E(Y)$ and show $V(Y)=\infty$.
My attempt $E(Y)=E(E(Y\mid X=x))=\frac {1}{\sqrt x}$
But I do not see how to handle the second part now. I see that it is sufficient to show $E(Y^2)=\infty$ since $V(Y)=E(Y^2)-E(Y)^2=E(Y^2)-\frac 1x$, some help is welcome!
No, you start with : $$\begin{align}\mathsf E(Y) &=\mathsf E(\mathsf E(Y\mid X)) \\ &=\int_\Bbb R \mathsf E(Y\mid X=x) ~f_{X}(x)~\mathsf d x\\ & = \int_0^\infty \frac 1{\surd x}\lambda e^{-\lambda x}~\mathsf d x\end{align}$$
Approach through the Law for Total Variance: $$\mathsf{Var}(Y)=\mathsf E(\mathsf {Var}(Y\mid X))+\mathsf {Var}(\mathsf E(Y\mid X))$$