X ∼ Exp(1) and Y ∼ uniform(0, 1) are independent. Compute $E(e^{-XY^2})$. X ∼ Exp(1) and Y ∼ uniform(0, 1) are independent. Compute $E(e^{-XY^2})$.
I am having trouble answering this question, I am not sure if I should take advantage of moment generating functions or multivariable LOTUS. I think multivariable law of the unconscious statistician would be easier, but I am unsure on how to derive the joint pdf.Any help would be greatly appreciated!
We have
\begin{align*}
\mathbb E[e^{-XY^2}] &amp;=\int_0^1\int_0^\infty e^{-xu^2} e^{-x} dx du\\
&amp; = \int_0^1 \Big(-\frac{1}{1+u^2} e^{-(1+u^2)x} \vert_{x=0}^\infty \Big) du\\
&amp;=\int_0^1\frac{1}{1+u^2} du = {\rm arctan}(1) - {\rm arctan}(0)= \frac{\pi}{4},
\end{align*}
since ${\rm arctan}’(x) = 1/(1+x^2).$ Here in the first line I am using the fact that the joint density of $X$ and $Y$ is:
$$
f_{X,Y}(x,u)=e^{-x}1_{x&gt;0}1_{u\in(0,1)}.
$$
If $X$ and $Y$ are independent with respective densities $f$ and $g$, then for any reasonable function $F$,
$$E[F(X,Y)] = \iint F(x,y)f(x)g(y) dxdy.$$
Do you know the densities of $X$ and $Y$ in this case?