Monday, January 9, 2012

Cardinality and Bayesian regularity

Regularity is the Bayesian thesis that an ideal agent assigns probability zero only to impossible propositions. This creates obvious problems in the case of contingent propositions—such as that an infinitely thin dart will hit such-and-such a location or that a coin tossed infinitely often will always come up heads—that according to the standard probability calculus have probability zero. But maybe the Bayesian can hope that assigning some sort of hyperreal infinitesimal probability will do the trick? Timothy Williamson has a very nice argument that that's not going to work in the case of the coin tossed infinitely often. Here is another argument in the same direction, this one based on cardinalities.

The basic result is a theorem that shows that, assuming the Axiom of Choice, for any totally ordered finitely additive probability measure, there is a cardinality K such that as long as there are at least K mutually exclusive options, at least one of these options will receive probability zero (in fact, all but K of them will receive probability zero). But for any cardinality K, one can find a set of more than K mutually exclusive contingent propositions, for instance the set of propositions that there are exactly n entities, or n spatiotemporally disconnected island universes, where n ranges from 1 to something high enough to guarantee that there will be more than K such propositions.

Now on to the formal setting for my no-go theorem. Ordinary probabilities take real numbers as values. Here we allow that to be generalized. The generalization is this. Our optimistic Bayesian regularist, let us suppose, has some set V with a total ordering <, an identity 0 and an operation + satisfying the following conditions:
  1. a+0=a for all a
  2. if b<c then a+b<a+c for all a, b and c.
Additionally, we have a finitely-additive probability space taking values in V. This is a space X together with an algebra F of "measurable" subsets of X, where F is non-empty and closed under complementation and binary unions, and there is a function P from F to V such that:
  1. P(A)≥0 for all A
  2. P(A  B)=P(A)+P(B) whenever A  B=  .
Now comes the main formal result:

Theorem. Suppose that X also has a total ordering < and that (a) every singleton {x} where x is in X is measurable (i.e., a member of F) and (b) every set of the form {x<y} for a y in X is measurable. Let R be the range of P, i.e., the set of all values P(A) as A ranges over the members of F. Suppose that |X|+1 > |R|. Then there is at least one value of x such that P({x}) is not 0.

Now, it follows from the Axiom of Choice that every set has a total ordering (a claim somewhat weaker than the Axiom of Choice, apparently). Hence:

Corollary. Assuming the axiom of choice, if |X|+1>|V| and every subset of X is measurable, then there is a non-empty subset A of X such that P(A)=0.

In particular, if our probabilities take real values, then as long as we have more than continuum many mutually exclusive alternatives, at least one of them will have probability zero. (This particular result can be improved: all that's needed is that one have more than countably many exclusive alternatives.)

The proof of the theorem is pretty easy. To get a contradiction, suppose P({x}) is never zero. Let Sx={z < x}. Let U be the set of all sets of the form Sx for x in X as well as the set X itself.  Then:

Lemma. If A and B are distinct members of U, then P(A) and P(B) are distinct members of R.

Given the Lemma, since U has |X|+1 members, it follows immediately that R must have at least |X|+1 members, which contradicts our assumptions. The proof of the Lemma is all that's remaining. Well, suppose first that A=Sx and B=Sy and x and y are distinct. By total ordering, x<y or y<x. Suppose x<y—the proof in the other case is the same. Then it is easy to show using the facts that P({x})>0 and that x is a member of Sy but not of Sx together with (1)-(4) that P(Sy)>P(Sx), and hence P(A) and P(B) are distinct. Next we need to show that P(Sx) and P(X) are distinct. But that can be shown in much the same way, since x is a member of X but not of Sx.


Alexander R Pruss said...

I can now get similar results without using the Axiom of Choice.

Alexander R Pruss said...

A formal statement of the no-AC result is this. For any cardinality K there is a cardinality K* such that if |X| >= K* and |V| <= K, then a V-probability assignment on the powerset on X is not regular.

We can take K* to be K x K+ where K+ is the Hartog's number of K. I got the idea for this from here.

Alexander R Pruss said...

Just realized that I only need V to be partially ordered, not totally ordered. Wow.

And I can take K* to be just K+.

Alexander R Pruss said...

The paper giving this result has just been accepted by Philosophy of Science.

Yay! Yet another blog post turned into a publication.