Wednesday, July 31, 2013

More fun with conditional probabilities

Let X and Y be independent random variables uniformly distributed over [0,1), and suppose our setup is symmetric between X and Y (i.e., any probabilities, conditional or not, are symmetric under interchange of X and Y). Let Z be the point in the plane with polar coordinates (r,θ)=(√X,2πY). It is easy to see that Z is uniformly distributed over the unit disc D (not including the boundary).

Let A be the horizontal line segment from the center O of the disc to the right edge.

Question: What is P(Z=O|ZA)?

The obvious answer is zero or infinitesimal. After all, Z is uniformly distributed over the disc D, and O is just one of the infinitely many points on A.

But the obvious answer seems to be mistaken. Here's why. We have Z on the line segment A if and only if either X=0 (in which case, no matter what Y is, Z=O) or Y=0. We have Z=O if and only if X=0 (it doesn't matter what Y is). Let E be the event that X=0 or Y=0, i.e., that ZA. Then P(Z=O|ZA)=P(X=0|E). But surely P(X=0|E)=P(Y=0|E) by symmetry. So 1=P(X=0 or Y=0|E)≤P(X=0|E)+P(Y=0|E)≤2P(X=0|E), and so P(X=0|E)≥1/2. Thus, P(Z=O|ZA)≥1/2.

Many of these posts on conditional probabilities, infinitesimals and uniform distribution should be going into a paper which may be entitled "In search of true uniformity."