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?

  • $\begingroup$ 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. $\endgroup$ – A. Bailleul Jan 31 at 14:17
  • $\begingroup$ Edited. Wouldn't ||x-a|| and ||y-a|| be independent? $\endgroup$ – user114743 Jan 31 at 14:20
  • $\begingroup$ 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? $\endgroup$ – user114743 Jan 31 at 14:21
  • $\begingroup$ have you tried the case of a circle $\endgroup$ – phdmba7of12 Jan 31 at 14:26
  • $\begingroup$ 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. $\endgroup$ – user114743 Jan 31 at 14:33

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