Find the density function of X+Y, where X and Y are random variables

Question

If $X$ and $Y$ have joint density function $f(x,y)=\frac{1}{2}(x+y)e^{-x-y}$ when $x,y>0$ (and $0$ otherwise), find the density function of $X+Y$.

Let $Z=X+Y$. Then from my textbook, it says that $f_Z(z)=\int_{-\infty}^{\infty} f_{x,y}(u,z-u)du$, where $u=x, u>0$. When I plug the joint density function into this equation, I get $f_Z(z)=\int_{0}^{\infty} \frac{1}{2}ze^{-z}du$. This integral doesn't evaluate because you would have a $u$ being evaluated at infinity. The answer in the book says $f_{X+Y}(u)=\frac{1}{2}u^2e^{-u}$ for $u>0$. I don't see how this is the answer.

Answer 1

You made a small mistake: $$ f_Z(z) = \int_0^{\color{blue}{z}} \frac{1}{2} ze^{-z} \ du = \frac{1}{2} z^2 e^{-z}$$

EDIT: One way to obtain this formula is to consider new variables $U = X$ and $Z = X+Y$. The Jacobian for this transformation is 1 and so $$ f_Z(z) = \int_{-\infty}^\infty f_{U,Z}(u,z) \ du = \int_{-\infty}^\infty f_{X,Y}(u,z-u) \ du. $$

Answer 2

One way to solve this is to transform the variables from $X,Y$ to $Z,W$ where $Z = X+Y$ and $W = X-Y$. The Jacobian is $-2$. So

$$f_{W,Z}(w,z) = \frac{1}{2} f_{X,Y}\big(\frac{1}{2}(z+w), \frac{1}{2}(z-w)\big)$$ Plugging in, we get $$f_{W,Z}(w,z) = \frac{1}{2}\cdot \frac{1}{2}(z)e^{-z}$$ The limits of the domain are $z$ from $0$ to $\infty$, and for a given $z$, the limits of $w$ are ${-z}$ to ${z}$. So if we integrate over $w$ to find the marginal distribution for $z$, we get $$ \frac{1}{4}ze^{-z}\cdot 2z = \frac{1}{2}z^2e^{-z}$$