Integrating a Uniform Joint PDF with $P[|X-Y|<1/2]$

Question

This question came up, and I guess it's both a calculus review question and a continuous probability one, because I don't really get why in part b, the limits of integration for $x$ is $0$ to $1/2-y$, because if y then goes from $1/2$ to $1$, $1/2-1/2 = 0$. Could someone walk through why we chose those bounds? Also if we chose to not do the complement, what would the integral be?

Answer 1

We want to integrate the density over the region satisfying the simultaneous conditions

$$0 \le X \le 1 \\ 0 \le Y \le 1 \\ Y \ge X \\ |X - Y| \ge 1/2.$$ The first three conditions represent a triangle with vertices $(0,0), (0,1), (1,1)$ in the Cartesian coordinate plane. Within the unit square, the last condition represents two triangles: one with vertices are $(0, 1/2), (0, 1), (1/2, 1)$, and another one that is the first's reflection about $y = x$, with vertices at $(1/2, 0), (1,0), (1, 1/2)$. So the intersection of both of these regions is the single triangle with vertices $$(0, 1/2), (0, 1), (1/2, 1).$$ This gives us two ways to write the double integral: either we may write it as $$\int_{x=0}^{1/2} \int_{y=x+1/2}^1 f(x,y) \, dy \, dx,$$ or as $$\int_{y=1/2}^1 \int_{x=0}^{1/2-y} f(x,y) \, dx \, dy.$$ The provided solution corresponds to the second choice.

How do we determine these regions bounded by the aforementioned inequalities? This is something you should know how to do from high school algebra. You plot the corresponding equalities, which are lines in the plane, and then determine using test points the region(s) that satisfy the respective inequalities.