For a long time I thought it was established that (given a standard axiomatisation of the probability calculus) countable additivity and countable conglomerability were equivalent. But I’ve lost confidence in my belief. So I’m wondering if anyone can tell me exactly what the answers are to a few questions below.

Just to make sure we’re clear, I’m taking countable additivity to be the principle that if each of the E_{i} in {E_{1}, …, E_{n}, …} are disjoint, then Pr(E_{1} v … v E_{n} v …) = Pr(E_{1}) + … + Pr(E_{n}) + ….

And I’m taking countable conglomerability to be the following. Again, if each of the E_{i} in {E_{1}, …, E_{n}, …} are disjoint, then there is some E_{i} such that Pr(E | E_{i}) <= Pr(E).

*Question One*: Does a failure of countable additivity entail a failure of countable conglomerability?

I’m pretty sure that, as stated, the answer to that is **no**. Consider a standard finitely additive probability function. So there’s some random variable X, and for all natural x, Pr(X=x)=0, while the Pr(X is a natural number)=1. Now insist that Pr is only defined over propositions of the form *X is in S*, where S is a finite or cofinite set of natural numbers. (By a cofinite set, I mean a set whose complement, relative to the naturals, is finite.) I’m reasonably sure that there’s no way to generate a failure of countable conglomerability.

*Question Two*: Assume there is a random variable X such that Pr(X is in S1 | X is in S2) is defined for every S1, S2 that are non-empty subsets of the naturals. And assume that whenever S2 is infinite, and the intersection of S1 with S2 is finite, then Pr(X is in S1 | X is in S2) is 0. (So Pr violates countable additivity.) Does Pr fail to respect countable conglomerability?

I’m even more confident that the answer to this is **yes**. Here’s the proof. Any positive integer *x* can be uniquely represented in the form 2^{n}(2*m*+1), with *n* and *m* non-negative integers. For short, let a statement of the form *n=x* mean that X is one of the numbers such that when represented this way, *n=x*, and similarly for *m*. Then for any non-negative integer, Pr(X is odd | *m=x*) = 0, since for any given *m* there is one way to be odd, and infinitely many ways to be even. By conglomerability, that implies Pr(X is odd) = 0. But an exactly parallel argument can be used to argue that Pr(X+1 is odd) = 0. And this leads to a contradiction.

*Question Three*: Assume there is a random variable X such that for any x, Pr(X=x)=0, while Pr(X is a natural number)=1, and that Pr(X is in S1 | X is in S2) is defined for every S1, S2 that are non-empty subsets of the naturals. Does Pr fail to respect countable conglomerability?

This is what I don’t know the answer to. I think the answer is **yes**, but I can’t see any obvious proof. Nor can I come up with a counterexample. Does anyone know (a) what the answer to this question is, and (b) where I might find a nice proof of the answer?

Much thanks in advance for helpful replies!

Posted by Brian Weatherson in *Uncategorized*