Monday, October 14, 2024

Defining comparative probabilities in terms of conditional probabilities

Suppose we have a full conditional probability P(AB) defined for all pairs of events (stipulating that P(A∣⌀) = 1 if we wish). I've proposed two methods for defining a probability comparison using conditional probabilities:

  1. A ⪅ B iff P(AAB) ≤ P(BAB).

  2. A ⪅ B iff P(ABAΔB) ≤ P(BAAΔB), where AΔB = (AB) ∪ (BA) is the symmetric difference.

In a footnote in a paper, I wrote about the second ordering, which I incorrectly attributed to de Finetti: “This ordering has the advantage that if A is a proper subset of B, then A < B, but it is somewhat harder to prove transitivity”.

Well, that was an understatement! It’s not just harder to prove transitivity: it’s impossible.

Define:

  • Ω: all integers

  • E0: non-negative even integers

  • E: positive even integers

  • D: positive odd integers.

Let P be a full conditional probability such that:

  1. P(E0E0D) = 1/2 = P(DE0D)

  2. P(DDE) = 1/2 = P(E|DE).

It is easy to see from (3) and (4) that because E0 and D are disjoint, and so are D and E, then by definition (2) we have E0 ⪅ D and D ⪅ E. (For disjoint A and B, the definition (2) of A ⪅ B is equivalent to thedefinition (1).) However, E0 − E = {0}, E − E0 = ⌀, and E0ΔE = {0}, so P(EE0E0ΔE) = 0 while P(E0EE0ΔE) = 1, and thus we cannot have E0 ⪅ E.

The only question is whether there actually is a full conditional probability satisfying (3) and (4). If there is, then (2) is not transitive in our case.

There is such a full conditional probability. Let Qn(A) =  ∣ A ∩ [−n,n] ∣ /(2n+1), where $B$ is the cardinality of a set B. Then Qn is a probability. Let Q be a limit of the Qn along an ultrafilter. This is a finitely additive hyperreal probability which is non-zero for all non-empty sets. Define P(AB) as the standard part of Q(AB)/Q(B) for B non-empty. This is a full conditional probability. Moreover, P(AB) = lim Qn(AB)/Qn(B) whenever the latter limit is defined. That limit is defined in the cases of the events involved in (3) and (4), and it is easy to evaluate the limits and see that (3) and (4) are true.

I don’t know what I was thinking when I wrote that footnote. My guess is that I had in my mind a proof sketch that doesn’t work (I have some idea what that might have been).

Whew! I noticed this afternoon that Theorem 1 of this paper of mine was incompatible with the transitivity of comparison (2). This made me really worried that my Theorem 1 was false. But since the comparison (2) isn’t transitive, I can relax.

This raises an interesting potential research problem. The Pruss definition of ⪅ does not satisfy the additivity axiom for comparative probabilities, namely that if C is disjoint from A ∪ B, then A ≲ B if and only if A ∪ C ≲ B ∪ C (it only preserves the left-to-right implication). Definition (2) does satisfy the additivity axiom, which is what I liked about it.

I suspect there is no good definition of comparative probabilities in terms of full conditional probabilities that satisfies the additivity axiom. (One reason for this intuition has to do with the fact that in Figure 1 here there are entries with a Yes in column 3 and a No in column 5.)

So, I now wonder: Is there some good combination of a definition of comparative probabilities in terms of full conditional probabilities with some weakened version of the additivity axiom?

2 comments:

  1. My counterexample to transitivity of the de Finetti order was needlessly complex, using an ultraproduct (and hence the Axiom of Choice). There is a much simpler example. Let Omega be {0,1,2}. Let P be a classical probability with P({0})=0 and P({1})=P({2})=1/2. Then P0 defines a conditional probability P(A|B) in the classical way, except when B is empty or B is {0}. When B is empty, stipulate P(A|B)=1. When B is {0}, stipulate that P(A|B)=1 if 0 is a member of A, and otherwise stipulate that it's zero. The result is a full conditional probability. It's easy to see that on the de Finetti definition we have {1} ~ {2} ~ {0,1}, but we don't have {1} ~ {0,1}.

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  2. The original version of this post incorrectly ascribed (2) to de Finetti, and the previous comment thus miscalled it the "de Finetti order".

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