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Casella & Berger (2001) write:

the Axiom of Countable Additivity, is not universally accepted among statisticians. ... [It] is rejected by a school of statisticians led by deFinetti (1972), who chooses to replace this axiom with the Axiom of Finite Additivity.

What might possibly be wrong or objectionable about the Axiom of Countable Additivity?

(Simple examples would be helpful!)

The Axiom of Countable Additivity is the 3rd condition below:

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1 Answers1

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De Finetti was a subjectivist, which means that he interpreted probabilities as judgments of a certain kind. On his view, countable additivity is too strong to be a reasonable constraint on judgment. Basically, de Finetti thought that countable additivity amounts to a prohibition on judgments that ought to be allowed.

The classic example is the following. A natural number is going to be chosen according to some distribution, you know not which. In your state of ignorance, surely it is permissible for you to judge that the selection will be fair. That is, it doesn't seem like you're making an error in judgment by taking the selection to be fair (though, of course, you're not required to think it's fair; this is just one permissible judgment among perhaps many others). In order for the selection to be fair, each natural number must have the same probability of being chosen. But that means the distribution of the selection is not countably additive. To see this, consider two cases: (1) if each number has a positive probability of being chosen, the sum of the probabilities diverges; (2) if each number has probability zero of being chosen, the sum of the probabilities is likewise zero. In both cases, countable additivity is violated.

On the other hand, there are finitely, but not countably, additive distributions that assign every natural number probability zero while still assigning the set of all natural numbers probability 1. De Finetti objected to requiring countable additivity in this case because it precludes the seemingly permissible judgment that the selection is fair. This judgment is consistent, however, with a probability theory that requires finite but not countable additivity, and it is precisely this kind of probability theory that de Finetti endorsed.

aduh
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    "[T]here are finitely, but not countably, additive distributions that assign every natural number probability zero while still assigning the set of all natural numbers probability $1.$" Can you give an example or a link to one, please? Off the top of my head, I don't understand how this is possible. – saulspatz Nov 30 '19 at 04:48
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    @saulspatz Examples are a bit difficult because the existence of such measures cannot be shown in ZF set theory. You need something at least as strong as the Hahn-Banach theorem. One way of obtaining the desired measure is by using HB to extend asymptotic density, for instance. Here's a paper with more discussion. You can also consult Rao and Rao's seminal text on finitely additive measure theory, if you have access. – aduh Nov 30 '19 at 04:53
  • In order for the selection to be fair, each natural number must have the same probability of being chosen. >> I don't quite see though why this implies each number must have a positive probability of being chosen. Can't we judge that the selection is fair AND assign probability zero to each number? –  Nov 30 '19 at 05:03
  • @JerryS1988 Yes, in fact we must. I never said that positive probability is implied. What I said was: if the numbers have the same positive probability, then countable additivity is violated. I've reworded the argument a bit to clarify. – aduh Nov 30 '19 at 05:05
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    Thanks very much. Rao and Rao looks too heavy for me; $40$ or $50$ years ago I'd have found it very interesting, I've no doubt. – saulspatz Nov 30 '19 at 05:20
  • @BrianMoehring Yes, the domain is the issue. Not every subset has a density. In fact the sets with densities don’t even form an algebra. – aduh Nov 30 '19 at 08:56
  • Ahhh. I thought about it more after reading your final comment, and sure enough, the sets with density are closed under disjoint union and complement, but not intersection, so my recall [of that durrett problem] was wrong after all. Thanks. – Brian Moehring Nov 30 '19 at 09:38
  • Related, I suppose: Uniform Probability and Riemann Sum. And, BTW, I still would like to have an answer to this question. – Han de Bruijn Dec 06 '19 at 14:50