Joint probability of uniform random variables and absolute difference

Question

Let $X,Y$ be independent standard uniform random variables. How do we compute the joint density $$P(a\leq X\leq 1-a, a \leq Y \leq 1-a, |X-Y|\geq a) $$

for some positive value $a$. I can split the absolute value into two separate conditions $X-Y\geq a$ and $-(X-Y) \geq a$. Now, I have way too many conditions. Any leads is appreciated.

Many thanks.

Answer 1

By symmetry the required probability is $2P(a\leq X \leq 1-a,a\leq Y \leq 1-a, X \geq Y+a)$. This gives $\int_a^{1-2a} \int_{y+a}^{1-a} dxdy$ provided $a<1-2a$ or $a<\frac1 3$. [The inside integral requires $y+a <1-a$ or $y<1-2a$. That is why the outside integral is from $a$ to $1-2a$].

I will leave the case $a\geq \frac 1 3$ to you.