I'm given the joint density $$f_{X,Y}(x,y)= \left\{\begin{matrix}e^{-y}, \mbox{ } 0\leq x \leq y \le \infty \\ 0, \mbox{ otherwise}\end{matrix}\right.$$ and I'm told to the find the probability density of the sum $Z = X+Y$.
The domains are what are throwing me off. Any comments on the domain or other incorrect parts of the solution are appreciated! Thanks.
My solution:
$$F_{Z}(z) = P(Z \leq z) = P(X+Y \leq z) = \int\int_{x+y \leq z}f_{X,Y}(x,y)dxdy$$
$$=\int_{-\infty}^{\infty}dx\int_{-\infty}^{z-x}f_{X,Y}(x,y)dxdy.$$
$$ $$ $$f_{Z}(z) = \frac{d}{dz}F_{Z}(z)=\int_{-\infty}^{\infty}f_{X,Y}(x,z-x)dx.$$
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NOTE: $0\leq x \leq y \le \infty$. So, $x \geq 0$ and $x \leq \frac{z}{2}$.
So, $$f_{Z}(z) =\int_{0}^{z/2}f_{X,Y}(x,z-x)dx $$
$$= \int_{0}^{z/2}e^{x-z}dx$$ $$= e^{-z}\int_{0}^{z/2}e^{x}dx = e^{-z}\left [ e^{x} \right]\Big|_0^\frac{z}{2} $$
$$ = e^{-y}[e^{\frac{z}{2}}-1]. $$ $$
Thus, $$f_{Z}(z)= \left\{\begin{matrix}e^{-z}[e^{\frac{z}{2}}-1], \mbox{ } z \geq 0 \\ 0, \mbox{ z < 0}\end{matrix}\right.$$
We know $0 \leq X \leq Y$ for sure from the definition of $f_{X, Y}(x, y)$. In order to have $X + Y \leq z$ we can start by thinking about the range of $Y$ values, i.e. $0 \leq Y \leq z$. Then given $Y$, $X$ mustlie in the range $0 \leq X \leq z - Y$. Combining this with the very first condition we have $0 \leq X \leq \min(Y, z - Y)$. Therefore $$F_Z(z) = \mathbb{P}(Z \leq z) = \mathbb{P}(X + Y \leq z) = \int_{y = 0}^{y = z}\int_{x = 0}^{x = \min(y, z - y)}f_{X, Y}(x, y) \thinspace dx \thinspace dy.$$ Integrating this gives $$F_Z(z) = \int_{y = 0}^{y = z}\int_{x = 0}^{x = \min(y, z - y)}e^{-y} \thinspace dx \thinspace dy = \int_{y = 0}^{y = z} e^{-y}\min(y, z - y) dy.$$ When $\min(y, z - y) = y \Leftrightarrow y < z - y \Leftrightarrow y < z/2$, so we break the integral into regions: $$F_Z(z) = \int_{0}^{z/2} ye^{-y} dy + \int_{z/2}^{z} (z - y)e^{-y} dy = \int_{0}^{z/2} ye^{-y} dy + z\int_{z/2}^{z} e^{-y} dy - \int_{z/2}^{z} ye^{-y} dy$$ And solving: $$F_Z(z) = \left[-(y + 1)e^{-y} \right]_{y = 0}^{y = z/2} + z[-e^{-y}]_{y = z/2}^{y = z} - \left[-(y + 1)e^{-y} \right]_{y = z/2}^{y = z}$$ $$F_Z(z) = -\frac{z}{2}e^{-z/2} - e^{-z/2} + e^{-0} - ze^{-z} + ze^{-z/2} + ze^{-z} + e^{-z} - \frac{z}{2}e^{-z/2} - e^{-z/2}$$ $$F_Z(z) = 1 + e^{-z} - 2e^{-z/2} \thinspace (\text{for } 0 \leq z, 0 \text{ otherwise})$$ (we can check that when $F_Z(0) = 0$ and as $z \rightarrow \infty$, $F_Z(z) \rightarrow 1$. Differentiating gives $$f_Z(z) = -e^{-z} + e^{-z/2} \thinspace (\text{for } 0 \leq z, 0 \text{ otherwise})$$