Probability density function of X + Y

Question

Q: The random variables $X$ and $Y$ have joint probability density function

$f_{X,Y}$$(x,y)$ = \begin{cases} 8xy, & \text{$0 < x < y < 1$} \\ 0 & \text{otherwise} \end{cases}

Find the probability density function of X + Y

I attempted this question and got the answer wrong; when I looked at the solutions, they were as follows:

"First we find the CDF of Z = X + Y. A sketch of the region 0 < x < y < 1 and the line x + y = z shows that the region we need to integrate the joint PDF over is fundamentally different for $z$ $\in$ $(0,1]$ and $z$ $\in$ $(1,2)$. For z $\in$ $(0,1]$ we have

$F_Z$$(z)$ = P($Z$ $\leq$ $z$) = $\int_{0}^{z/2}$ $\int_{x}^{z-x}$ $8xy$ $dy$ $dx$

The question then calculates this, then does the same for z $\in$ (1,2), then eventually differentiating to get the answer.

The question is; where on earth have they gotten z/2 from? Graphically I can see the other limits but I can't see where they have gotten z/2 from at all?

Answer 1

For $z\in (0,1)$, the region to integrate is between the lines $x=0$, $x+y=z$ and $x=y$. It's a triangle with corners at $(0,0),(z/2,z/2),(0,1)$. If you write the integral as $\int (.) dydx$, the outer integral is for $x$ dimension and its limits are $0$ and $z/2$.