## Wednesday, April 18, 2012

One of the Kolmogorov axioms of probability says that if A1,A2,... is a countable sequence of disjoint sets, then P(A1A2∪...)=P(A1)+P(A2)+.... I once (when writing my Philosophia Christi paper on fine and coarse tuning) thought that while we had intuitive reason to accept unrestricted additivity (where we do not restrict to countably many sets) and we had intuitive reason to accept finite additivity, there was no in-between reason for accepting countable additivity. Since unrestricted additivity is unsupportable (if you pick a random number between 0 and 1, the probability of picking any particular number is zero, and the sum of the uncountably many probabilities of particular numbers will be zero, but the probability of the union of these singleton sets is one), I thought we should go for finite additivity.

When I thought this, I was wrong, because at the time I didn't know about the phenomenon of non-conglomerability which countable additivity seems to be needed for ruling out. Non-conglomerability is where you have a measure P of probabilities (maybe not technically a probability measure), a set E of events where each event in E has non-zero probability and it is certain that exactly one of the events in E will happen, and an event A such that P(A|B)>x for all B in E but P(A)<x. In such a case, your probability of A is less than x even though you know that no matter which event in B will happen, you will have P(A)>x. This is pathological.

It is well-known that countable additivity entails conglomerability. I like proving this with a two-step argument. The first step is an easy argument that if the set E of events is countably additive, then because P(A)=P((AB1)∪(AB2)∪...)=P(AB1)+P(AB2)+..., if we have P(A|B)>x for all B in E, then P(A)>x as well.

The second step in the proof is that if the members of a set E of disjoint events each have non-zero probabilities, then E has only countably many events in it. This step allows us to rule out non-conglomerability using only countable additivity. This step follows from the following fact about real numbers

1. If E is a set and f is a function that assigns to each member of E a non-negative number such that for any finite sequence x1,...,xn of distinct members of E we have f(x1)+...+f(xn)≤1, then all but countably many members of E are zero,
by letting f=P and using the finite additivity of P.

If we need countable additivity precisely to rule out non-conglomerability, then we have an explanation of why it only has to be countable additivity. The reason has to do with the property (1) of real numbers, which property in turn follows from the fact that the real numbers are Archimedean—for every pair of positive real numbers x and y, there is a finite natural number n such that nx>y.

In other words, we have countable additivity in the probability calculus precisely because the probability values have a countable-like, i.e., Archimedean, structure. (Another way of seeing the countable structure of the reals: they are the completion of the rationals.)

And if generalize the values of the probability function to a larger, non-Archimedean field, we will need to require something stronger than countable additivity in order to avoid non-conglomerability.

#### 1 comment:

Alexander R Pruss said...

Paul Pedersen and John Earman have both corrected my claim that countable additivity implies conglomerability.

To be precise, I should have said that countable additivity implies conglomerability with respect to partitions into non-null sets.