# Geometric probability independence

I have a basic question about the independence of two events.

Let $$x$$ be some fixed point in the interior of some set (say it's a convex bounded set $$C$$ in $$\mathbb{R}^2$$ or something). Choose $$a, b, y$$ uniformly at random and independently from the interior of $$C$$. Are the events that $$||x - a|| < ||x - b||$$ and $$||y - a|| < ||y-b||$$ independent? That is, is it true that $$\Pr(||x-a|| < ||x-b||\text{ and } ||y-a|| < ||y-b||) \\ = \Pr(||x-a|| < ||x-b||)\cdot\Pr(||y-a|| < ||y-b||) = 1/2?$$ I would really think so, but I'm not sure how to rigorously show this. Maybe it's enough to say that $$||x-a||, ||y-a||$$ are independent random variables?

What if now $$a, b$$ are chosen uniformly at random from some subset $$C'\subset C$$ (again, some "nice" subset), and $$y$$ is as before chosen uniformly at random from $$C$$. Are the events independent?

• The random variables $||x-a||, ||x-b||, ||y-a||, ||y-b||$ are obviously not independent since for example $||x-a||$ and $||x-b||$ are not. – A. Bailleul Jan 31 at 14:17
• Edited. Wouldn't ||x-a|| and ||y-a|| be independent? – user114743 Jan 31 at 14:20
• Well I guess they are not. If you know where $x$ is and know what $||x-a||$ is, I gues that tells you something about $||y-a||$? But intuition seems to strongly suggest that the probability I wrote should be 1/2, no? – user114743 Jan 31 at 14:21
• have you tried the case of a circle – phdmba7of12 Jan 31 at 14:26
• Well that seems like it would be messy, and I want to try to find an argument that doesn't really rely on any properties of the set itself. – user114743 Jan 31 at 14:33