Wednesday, August 15, 2012

Probabilistic comparison and nonmeasurable sets

One problem with epistemological use of probability theory is that, given the Axiom of Choice, there are nonmeasurable sets. In plausible setups, these nonmeasurable sets give rise to situations that cannot be assigned a probability that satisfied plausible invariance conditions. One might try to get out of this problem (and some infinity problems, while one is at it) by replacing probability values with probability comparisons. Instead of saying how probable a proposition is, we have as our basic relation: "p is more likely than q". Unfortunately, that doesn't get rid of the problem of nonmeasurable sets, as can be seen from the Hausdorff paradox.

Write "p<q for "p is less likely than q". Say that p has a chance (of truth) provided that F<p, where F some denial of a tautology. The following axioms (which are not meant to be complete, but which are enough to generate the problem) are very plausible:

  1. It is not the case that p<p.
  2. If neither p nor q has a chance, then their disjunction (p or q) has no chance.
  3. If p and q are incompatible, and q has a chance, then p<(p or q).
  4. If q and r are equivalent, then p<q if and only if p<r.

Now imagine a process that randomly picks out a point on (the surface of) a sphere, in such a way that (a) there is a chance that some point on the sphere is picked out and (b) if two regions are rotations of one another (about the center of the sphere) then (i) neither is more likely to contain the point than the other and (ii) if one region has a chance of containing the point, so does the other.

It turns out that what I just supposed about the process—plausible as it is that it can hold—cannot be satisfied together with (1)-(4), assuming the Axiom of Choice. I'll give the proof in a moment.

So what should way say philosophically here? The axioms (1)-(4) are very plausible, and (a) and (b) seem to be compossible. Perhaps we need to reject the Axiom of Choice. Or perhaps we need to reject the metaphysical possibility of there being spaces built on the continuum in the way that the sphere is. Or maybe one of the axioms (1)-(4) needs to be rejected. I think the best bet is (3), which is a weak form of finite additivity. But (3) is still pretty plausible.

Proof of incompatibility: By the Hausdorff Paradox, (the surface of) the sphere can be divided up into four disjoint subsets A, B, C and D, such that D is countable, and the four sets A, B, C, and the union of B and C are all congruent—i.e., each can be transformed into any other by rotations.

Observe that if a proposition is equivalent to a disjunction and has a chance, then
at least one of its disjuncts has a chance by (2) and (4). We will use this fact.

Let p(X) be the proposition that the point is in region X.

Next observe (this is an easy counting argument) that if D is countable, then there is a rotation r such that S is the union of SD and SrD. Thus, p(S) is equivalent to the disjunction of p(SD) and p(SrD). Thus the disjunction of p(SD) and p(SrD) has a chance. Thus, at least one of the disjuncts has a chance, by our earlier observation. But if p(SrD) has a chance, so does p(SD) by condition (b)(ii). But now p(SD) is equivalent to the disjunction p(A) or p(B) or p(C). Thus at least one of these disjuncts has a chance. Thus all these disjuncts have a chance by (b)(ii).

Now, there is a rotation r such that A is the union of rB and rC. Thus, p(A) is equivalent to the disjunction of p(rB) and p(rC). But since p(rC) has a chance by (b)(ii) as p(C) does, and since p(rB) and p(rC) are incompatible (as B and C are disjoint), it follows from (3) that p(rB)<(p(rB) or p(rC)). But by (4) it follows that p(rB)<p(A). But A is a rotation of rB, so this contradicts (b)(i).

3 comments:

Tim said...

I'm not sure that axiom 2 is so obvious. Certainly there are stronger interpretations one could lay down on the relation that would satisfy 2, but what considerations move you to accept it in its spare form?

Alexander R Pruss said...

Here's a consideration: If neither p nor q is likelier than a contradiction, then (p or q) isn't likelier than a contradiction.

Of course, that's just a restatement. :-)

You can also prove it from the (stronger) plausible pair of axioms:
2a. If p and q are incompatible, and p and r are incompatible, and q<r, then (p or q) < (p or r).
2b. If p and q are equivalent, then p<r iff q<r.

The argument could also be modified by replacing 2 with the weaker 2':

2'. If p and q are incompatible and their disjunction has a chance, then at least one of them has a chance.

and adding the axiom:

5. If p entails q, and p has a chance, then q has a chance.

Alexander R Pruss said...

Tim:

One could try this interpretation. Work with interval-valued probabilities, and say that:
p<q iff (LP(p)<LP(q) and UP(p)<UP(q) and at least one of these two inequalities is strict),
where LP and UP are the lower and upper probabilities (no subscripts in comment box last I checked).

Then 2' follows from the fact (axiom?) that UP(A union B) is no greater than UP(A)+UP(B).

And 5 will follow from an appropriate monotonicity.

But presumably 3 won't hold in this setting. For instance, if the probability of p is [0,1] and the same is true of the probability of q, then we might well have the probability of p or q also being [0,1]. However, I think that just shows that interval valued probabilities aren't letting us encode the notion of "more likely than".

On the other hand, we do get both 2 and 3 from the three axioms of a Comparative Probability Structure here.