Determining bounds for change sum of continuous r.v.'s I'm trying to understand how to determine the bounds when computing the sum of continuous random variables.
Here is a sample question:
X and Y have the following joint pdf:
$f_{X,Y}(x,y) = 4xy, 0 \leq x \leq 1, 0 \leq y \leq 1$,
If $Z = X + Y$, calculate $f_z(z)$.
So I know that there should be two components to $f_z(z)$:
$f_z(z) = \int_{-\infty}^{\infty}f_{x,y}(x,z-x) dx $
i)$\large = \int_{0}^z 4x(z-x)dx = \frac{2z^3}{3}$
ii)$\large = f_z(z) = \int_{z-1}^1 4x(z-x) dx = 4z - \frac{2x^3}{3} - 8/3$
My question is how to is one to go about finding the bounds of the integrals (i.e. $(0,z)$ and $(z-1, 1)$)?
Is there a way to do this without drawing out a plot?
Thanks
 A: You have been given that $$f_{X,Y}(x,y) = \begin{cases}4xy & : 0 \leq x \leq 1,\; 0 \leq y \leq 1\\ 0 & : \text{ elsewhere}\end{cases}$$
You wish to know where $f_{X,Y}(x, z-x)$ is supported (ie: not zero) with respect to $x$, for values of $z$ where $0\leq z\leq 2$.   (Since $z=x+y$, then $0+0\leq z\leq 1+1$.)
$$f_{X,Y}(x,z-x) = \begin{cases}4x(z-x) & : 0 \leq x \leq 1,\; 0 \leq (z-x) \leq 1,\quad 0\leq z\leq 2\\ 0 &: \text{ elsewhere}\end{cases}$$
So this is where: $0\leq x\leq 1, 0\leq z-x\leq 1$ 
With a little rearranging: $0\leq x\leq 1, z-1\leq x\leq z$ 
So we integrate with respect to $x$ over the interval $\max(0, z-1)\leq x \leq \min(1,z)$ where $0\leq z\leq 2$$
But functions of $\min$ and $\max$ are messy to deal with, so we partition this depending on $z$ so that:$$\Big[0\leq z\leq 1, 0\leq x\leq z\Big]\cup\Big[1< z\leq 2, z-1\leq x\leq 1\Big]$$
Hence: $$f_{X+Y}(z) = \begin{cases}\int_0^z f_{X,Y}(x,z-x)\operatorname d x & : 0\leq z\leq 1\\[1ex]\int_{z-1}^1 f_{X,Y}(x, z-x)\operatorname d x & : 1< z\leq 2 \\[1ex] 0 & :\text{elsewhere}\end{cases}$$
