Density of $\sqrt{Z}=\sqrt{X+Y}$

Question

Let $(X,Y)$ be a random variable with density $f_{XY}(x,y)=\frac{1}{2}(x+y)e^{-(x+y)},x>0,y>0$.

1. Verify that it is indeed a density i.e :

$\rightarrow \frac{1}{2}\int_{0}^{+\infty}[\int_{0}^{+\infty}(xe^{-x}e^{-y}+ye^{-x}e^{-y})dy]dx=1$

2. Find the marginal densities and the expected value of $X$.

$\rightarrow f_X(x)=\frac{1}{2}e^{-x}(x+1)\mathbb{I}_{[0,+\infty)}(x);f_Y(y)=\frac{1}{2}e^{-y}(y+1)\mathbb{I}_{[0,+\infty)}(y);\mathbb{E}(X)=\frac{3}{2}$

3. Find the density of $Z=X+Y$.

$\rightarrow X+Y\sim \Gamma(3,1)$

4. Find the density of $\sqrt{Z}$.

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For point 4), I wrote $\left\{\begin{matrix} \sqrt{x+y}=u\\ x=v\end{matrix}\right.\Rightarrow \left\{\begin{matrix} y=u^2-v\\ x=v\end{matrix}\right.\rightarrow \mathbb{E}[g(\sqrt{Z})]=\int_{0}^{+\infty}[\int_{\sqrt{v}}^{+\infty}u^3e^{-u^2}du]dv=\frac{1}{2}\int_{0}^{+\infty}[\int_{v}^{+\infty}te^{-t}dt]dv=\frac{\Gamma(3)}{2}$.

Where am I wrong? Thanks in advance for any help.

Answer 1

point 4) is requesting the density of $\sqrt{Z}$

Without a lot of calculation, having already found

$f_Z(z)=\frac{z^2}{2}e^{-z}$

$z>0$

Applying the known formula

$f_Y(y)=f_X[g^{-1}(y)]|\frac{d}{dy}g^{-1}(y)|$

immediately you get

$f_W(w)=\frac{w^4}{2}e^{-w^2}2w=w^5e^{-w^2}$

$w>0$