If $X|Y=y \sim U[0,y^2]$, where $Y \sim exp(1/2)$. Find the distribution of $\frac{X}{Y^2}$ and $Cov(\frac{X}{Y^2},Y)$

Question

Given $X|Y=y \sim U[0,y^2]$, where $Y \sim exp(1/2)$, I want to find the distribution $f_Z(z)$ where $Z = \frac{X}{Y^2}$ and $Cov(\frac{X}{Y^2},Y)$. The thing is that the excercise asks to do it without using a change of variables nor calculating integrals. I think the idea is to use some kind of "trick" involving conditional expectation.

I was able to do this: $Cov(\frac{X}{Y^2},Y) = E(\frac{X}{Y^2}Y)-E(\frac{X}{Y^2})E(Y) = E(\frac{X}{Y})-E(\frac{X}{Y^2})E(Y)$.

Now $E(\frac{X}{Y}) = E(E(\frac{X}{Y}|Y)) = E(\frac{1}{Y}E(X|Y)) = E(\frac{1}{Y}\frac{Y^2}{2}) = \frac{1}{2}E(Y) = 1$ and we can find $E(\frac{X}{Y^2})$ in the same way.

How can I find $f_Z$ withouth making a change of variables though?

Answer 1

Since $X\mid Y$ is uniform on $[0,Y^2]$, then $X/Y^2$ is uniform on $[0/Y^2,Y^2/Y^2]=[0,1]$. The distribution of $Y$ itself is irrelevant as long as it is a non-zero random variable with probability $1$.

Answer 2

$$ \mathbb{P}(Z \le z) = \mathbb{E}[\mathbb{P}(X \le zY^2 \vert Y)] $$ Since $X \vert Y = y \sim \mathcal{U}[0, y^2]$, $$ \mathbb{P}(X \le zY^2 \vert Y) = \dfrac{zY^2 - 0}{Y^2 - 0} = z \ \forall z \in [0, 1] $$ Thus, $$ \mathbb{P}(Z \le z) = \mathbb{E}(z) = z \ \forall z \in [0, 1] $$ or, $Z \sim \mathcal{U}[0, 1]$