Tuesday, July 21, 2020

Do Popper functions carry enough information?

Let P be a Popper function on some algebra F of events on Ω. There is a natural way to define a probability comparison given P, namely A ⪅ B iff P(A|A ∪ B)≤P(B|A ∪ B), which I think I’ve used before. Unfortunately, this can violate a common axiom of probability comparisons, namely that A ⪅ B iff Ω − B ⪅ Ω − A.

For instance, consider the Popper function generated by a regular hyperreal probability on [0, 1] whose standard part agrees with Lebesgue measure on intervals. Then [0, 1]⪅[0, 1), since both have conditional probability 1 on [0, 1]∪[0, 1)=[0, 1]. But their complements are ∅ and {1}, respectively, and {1}⪅∅ is false.

This little fact may have some significance. There is a theorem in the literature that shows a correlation between Popper function and hyperreal probabilities: every Popper function with all non-empty sets regular can be generated from a normal hyperreal probability using the conditional probability formula, and given a Popper function with all non-empty sets regular there is a normal hyperreal probability from which the Popper function can be generated. This has led to a debate whether it’s better to work with Popper functions or hyperreal probabilities. One argument for working with Popper functions—an argument I’ve approved of in print—is that the hyperreal probabilities carry more information than reality does. I still think that this problem is there for hyperreal probabilities. But the above argument suggests that going to a Popper function may have discarded too much information. Popper functions allow for fine-grained comparisons of “small” events, such as singletons, but not for the complements of these events.


IanS said...

Maybe you could define the probability ranking differently. Rank A and B according to the comparison of P((A-B) | (A-B) ∪ (B-A)) with P((B-A) | (A-B) ∪ (B-A)).

Alexander R Pruss said...

That's clever.

IanS said...
This comment has been removed by the author.
Alexander R Pruss said...


Alas, that ranking is not transitive. Consider A=(1/4,1/2], B=(1/2,3/4], C=(1/4,1/2) in the uniform case on [0,1]. Then P(A-B|AΔB)=1/2, P(B-A|AΔB)=1/2, P(C-B|BΔC)=1/2, P(B-C|BΔC)=1/2, P(A-C|AΔC)=1, P(C-A|AΔC)=0. So, A≤B, B≤C, C<A.

IanS said...

True. Pity.