## Friday, May 11, 2012

### Hausdorff Paradox and conditional probabilities

The Hausdorff Paradox shows that given the Axiom of Choice, there is no finitely additive probability measure defined for all subsets of the surface of a ball that is invariant under rotations—that assigns the same measure to a subset of the surface of the sphere and to a rotation about any axis through the center of that subset. Because of results like that, the standard Lebesgue measure on the surface of a ball is only defined for some subsets, i.e., the measurable ones.

Now, in classical probability, we can only define conditional probability when we condition on an event A with non-zero probability. We then use the formula P(B|A)=P(B&A)/P(A). Some have tried to come up with axiomatizations that allow for conditioning on all non-empty measurable sets, including zero-probability ones.

We should not hold our breath for success. Here's why. Let C be a solid ball of unit volume. Let P be Lebesgue measure on C: the measure of a subset of C is just the volume of the subset. We expect the Lebesgue measure on C to be invariant under rotations about all axes through the center of the ball. We would also expect that the conditional probability to be thus invariant. I.e., if r is a rotation about an axis through the center, we would expect it to be that P(r(B)|r(A))=P(B|A). Unfortunately this cannot be done, at least not if one assumes the axioms that P(−|A) is a finitely additive measure on the P-measurable subsets of C and P(A|A)=1. For let A be the surface of a ball concentric with C but of smaller radius. Then the volume of A is zero: A is a two-dimensional surface, after all. Moreover, Lebesgue measure has the property that every subset of a set with zero measure is also measurable (and has zero measure). So P(−|A) will be a finitely additive probability measure on all subsets of A. If we have our rotation invariance condition, then P(B|A)=P(r(B)|r(A))=P(r(B)|A) since r(A)=A (the sphere A is invariant under rotation through its center). So, P(−|A) will be a finitely additive rotation-invariant probability measure on all subsets of A, which violates the Hausdorff Paradox (assuming the Axiom of Choice).

Put it differently: Any conditional probability assignment that allows conditioning on all non-empty subsets will exhibit an unacceptable rotational bias.