X ∼ Exp(1) and Y ∼ uniform(0, 1) are independent. Compute $E(e^{-XY^2})$.

Question

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!

Answer 1

Alternatively, use moment-generating functions. We have $$\operatorname{E}[e^{-XY^2}] = \operatorname{E}[\operatorname{E}[e^{-XY^2} \mid Y]] = \operatorname{E}[M_X(-Y^2)] = \operatorname{E}\left[\frac{1}{1-(-Y^2)}\right] = \int_{y=0}^1 \frac{dy}{1+y^2} = \tan^{-1} 1 = \frac{\pi}{4},$$ where the MGF of $X$ is $M_X(t) = \operatorname{E}[e^{tX}] = \frac{1}{1-t}$.

Answer 2

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?

Answer 3

We have \begin{align*} \mathbb E[e^{-XY^2}] &=\int_0^1\int_0^\infty e^{-xu^2} e^{-x} dx du\\ & = \int_0^1 \Big(-\frac{1}{1+u^2} e^{-(1+u^2)x} \vert_{x=0}^\infty \Big) du\\ &=\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: $$ f_{X,Y}(x,u)=e^{-x}1_{x>0}1_{u\in(0,1)}. $$