How to solve distribution function integral

Question

Given $f_{X,Y}(x,y)=\begin{cases}\frac{1}{2}x^2e^{-x}e^{-y} \ &x,y >0 \\ 0 & \text{O.T.} \end{cases}$

and $Z=X+Y$

Find $f_Z(z)$

My idea is to first find the distribution function

$F_Z(z)=P(Z\leq z)=P(X+Y\leq z)=P((X,Y)\in A)$, where $A=\{(x,y)\mid x\leq z-y\}$

This leads to the double integral

$$\int_0^\infty \int_0^{z-y} \frac 1 2 x^2e^{-x} e^{-y} \ \ dx \ dy$$

From here I´m kinda stuck as i can´t get a useful result out of the integral.

Is the integral wrong?

Answer 1

You need this: $$ \int_0^z \int_0^{z-y} \frac 1 2 x^2e^{-x} e^{-y} \, dx \, dy $$ If $y$ goes above $z$, then $z-y$ is negative, and you've got $x$ going from $0$ to a negative number, and the density is in fact $0$ there.

Answer 2

You're on the right track. Hint: try substituting $u=y$ and $x=v-u$, and reason about switching the order of integration.