PDF of summation of two random variables (different than uniform) My question is different than the questions about uniform distribution. 
Question: X and Y have the following pdf:
f(x,y) = (1/4)*x*y if 0 < x < 2 and 0 < y < 2
f(x,y) = 0 otherwise
Find the pdf of z = x + y. 
 A: You are probably intended to use the usual convolution procedure. Note that $X$ and $Y$ are independent, with densities $f_X(x)=\frac{x}{2}$ and $f_Y(y)=\frac{y}{2}$, each on the interval $(0,2)$. 
But let us instead go back to fundamentals. In the usual coordinate plane, draw the square on which the joint density function of $X$ and $Y$ lives. Let $Z=X+Y$. We find the cumulative distribution function $F_Z(z)$. After we have done that, we can differentiate to get the density function of $Z$.
The cdf of $Z$ is $0$ if $z\le 0$, and $1$ if $z\ge 4$. We now deal with the interesting part, $0\lt z\lt 4$. 
It turns out that the calculation is different for $0\lt z\le 2$ than for $2\lt z\lt 4$.  
Case (i): We first deal with $0\lt z\le 2$. For fixed $z$ between $0$ and $2$, draw the line $x+y=z$. Then $F_Z(z)$ is the probability of landing below that line. The interesting part of the world below that line is a triangle with corners $(0,0)$, $(z,0)$, and $(0,z)$.  To find $\Pr(Z\le z)$, we integrate the joint density over this triangle. We get
$$F_Z(z)=\int_{x=0}^z \left(\int_{y=0}^{z-x}\frac{1}{4}xy \,dy\right)\,dx.$$
Integrate, and then differentiate with respect to $z$ to get the density $f_Z(z)$. Actually, once we have done the inner integration, we don't need to do the outer one: since we will be differentiating, we can use the Fundamental Theorem of Calculus. But it may be clearer if we first compute $F_Z(z)$. 
Case (ii): Finally, we deal with $2\lt z\lt 4$. Again, this time for a fixed $z$ between $2$ and $4$, draw the line $x+y=z$. Again, $F_Z(z)$ is the probability of ending up below that line. The region of integration is less nice than in Case (i). So it is easier to first find $1-F_Z(z)$, the probability of ending up above the line $x+y=z$. Note that $1-F_Z(z)$ is the integral of our joint density over the triangle with corners $(2,z-2)$, $(2,2)$, and $(z-2,2)$. Thus for $2\lt z\lt 4$,
$$1-F_Z(z)=\int_{z-2}^2 \left(\int_{y=z-x}^{2}\frac{1}{4}xy \,dy\right)\,dx.$$
