The pdf of X and Y is: $$f_{X,Y}(x,y) = \begin{cases} 1 & 0<x<1, 0<y<1 \\ 0 & elsewhere \end{cases} $$ I need to find the cdf of $Z = X+ Y$. The answer is given as the following: $$F_Z{z} = \begin{cases} 0 & z<0 \\ \int_{0}^{z} \int_{0}^{z-x} dydx = \frac {z^2} 2 & 0\le z \le 1 \\ 1-\int_{z-1}^{1} \int_{z-x}^{1} dydx=1- \frac{(2-z)^2} 2 & 1\le z <2 \\ 1 & 2 \le z \end{cases}$$ In finding the cdf of $Z$, I understand how to calculate the integrals, but do not understand how to set the ranges of the integrals while $0 \le z \le 1$ and $1 \le z <2$. Can anyone explain to me please? Thanks a lot.
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Have a look at https://math.stackexchange.com/questions/357672/density-of-sum-of-two-uniform-random-variables-0-1. – StubbornAtom Mar 01 '18 at 06:34
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Geometrically.
Draw the unit square, and lines representing $x+y=z$ for various values of $z$.
When $0\leq z< 1$ the integral $\int_0^z\int_0^{z-x} \mathsf d y~\mathsf d x$ is the area of the lower left triangle and thus the probability that $X+Y\leq z$.
When $1\leq z\leq 2$ the integral $\int_{z-1}^1\int_{z-x}^1\mathsf d y~\mathsf d x$ is the area of the upper right triangle and thus the probability that $X+Y> z$, the complement of the required probability.
Graham Kemp
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