Find probability distribution of $X+Y$

Question

By knowing that \begin{equation} f_{(X,Y)}(x,y) = \begin{cases} \frac{1}{2} (x+y)e^{-(x+y)} & x,y>0 \\ 0 & \text{otherwise}\end{cases} \end{equation} I need to prove that $X+Y \sim \Gamma(3,1)$. I tried to apply Jacobian transformation formula, $(T,U) = (X+Y,Y)$, then I find $f_{(T,U)}(t,u)$, but I'm only interested to $T=X+Y$, so $f_T(t)=\int_\mathbb{R_+} f_{(T,U)}(t,u)du$. Obviously, by doing that, I get $f_{(T,U)}(t,u)=\frac{1}{2}te^{-t}$, and I don't understand how to proceed. Probably the path to the answer is easier than I think.

Answer 1

This is the probability (distribution function) what we have to evaluate:

$$F_{X+Y}(u)=P(X+Y<u)=\iint_{A_u}\frac{1}{2} (x+y)e^{-(x+y)}dxdy$$ where

That is, we have

$$\frac{1}{2}\int_0^u\left[\int_0^{-x+u} (x+y)e^{-(x+y)}dy\right]dx.$$

Regarding the inner integral

$$\int_0^{-x+u} (x+y)e^{-(x+y)}dy=$$ $$=u\sinh(u)+\sinh(u)-(u+1)\cosh(u)-x\sinh(x)-\sinh(x)+(x+1)\cosh(x).$$

Finally, this function (times $\frac12$) has to be integrated from $0$ to $u$ with respect to $x$.

$$\frac12\int_0^u u\sinh(u)+\sinh(u)-$$ $$-(u+1)\cosh(u)-x\sinh(x)-\sinh(x)+(x+1)\cosh(x)dx=$$ $$=\frac12\left[(u^2+2u+2)\sinh(u)-(u^2+2u+2)\cosh(u)+2\right].$$

This is the cdf of $X+Y$. The pdf is the derivative of the above with respect to $u$:

$$f_{X+Y}(u)=\frac12u^2(\cosh(u)-\sinh(u))=\frac12u^2e^{-u}$$ for $u\geq 0$ and zero otherwise.

This is what I managed to compute. And yes, this is the $\Gamma(3,1)$ distribution (see https://en.wikipedia.org/wiki/Gamma_distribution#Characterization_using_shape_α_and_rate_β). The definite integral of this function from $0$ to $\infty$ is one and its shape is shown below