If $f_{X,Y}(x,y)=4e^{-2(x+y)}$, find the pdf of $X+Y$

Question

I need to find the pdf of $X+Y$.

I tried Jacobian method and Cdf method but two outputs are different.

Can someone plz help me solve this problem? These are my solution process.

jacobian-> $Z=X+Y$ and $X=W$. So $f(Z,W)=|J|f(x,y)=f(x,y)$ as $|J|=1$ $\int^z _0 4e^{-2z}dw=4ze^{-2z}$

CDF->$P(X+Y\le Z)=\int^z_0\int^{z-x}_0 e^{-2(x+y)}dy\,dx=-2ze^{-2z}+e^{-2z}-1$

$d/dz(F(Z))=f(z)=4(z-1)e^{-2z}$

Answer 1

I suggest you a third method.

Observe that

$$f_{XY}(x,y)=2e^{-2x}\cdot2e^{-2y}=f_X(x)\cdot f_Y(y)$$

Thus you have two iid rv which are both exponential with mean 0.5.

That means $Z=X+Y$ is a Gamma

$$Z\sim \text{Gamma}(2;2)$$

$$f_Z(z)=4z e^{-2z}$$

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As per the CDF method there are some errors in your calculations; your $F(+\infty)=-1$ and it is not possible.

My calculations are the following

$$F_Z(z)=\int_0^z 2 e^{-2x}\left[\int_0^{z-x}2 e^{-2y}dy \right]dx=\dots=1-2^{-2z}-2z e^{-2z}$$

and thus

$$\frac{d}{dz}F=4z e^{-2z}$$

as expected