Abstract: Classical countably additive real-valued probabilities come at a philosophical cost: in many infinite situations, they assign the same probability value---namely, zero---to cases that are impossible as well as to cases that are possible. There are three non-classical approaches to probability that can avoid this drawback: full conditional probabilities, qualitative probabilities and hyperreal probabilities. These approaches have been criticized for failing to preserve intuitive symmetries that can easily be preserved by the classical probability framework, but there has not been a systematic study of the conditions under which these symmetries can and cannot be preserved. This paper fills that gap by giving complete characterizations under which symmetries understood in a certain "strong" way can be preserved by these non-classical probabilities, as well as by offering some results to make it plausible that the strong notion of symmetry here may be the right one. Philosophical implications are briefly discussed, but the main purpose of the paper is to offer technical results to inform more sophisticated further philosophical discussion.
Preprint here.
Great work! And a handy summary. I won’t pretend to have studied it.:-)
ReplyDeleteOn the philosophy: A possible move for defenders of non-classical probability would be to give up completeness (i.e. not require full conditional probabilities etc.). Classical probability does this; surely non-classical approaches could do likewise.
Granted, classically non-measurable sets are necessarily weird and non-constructible. By contrast, applying this approach to hyperreal probabilities on a label-independent fair countably infinite lottery would make any infinite co-infinite set non-measurable. But is this really a problem? You could see it as properly reflecting the fact that the natural concept of fairness (i.e. fairness between singletons) implies nothing non-trivial about the probability of any particular infinite co-infinite set.
Here is an interesting case: circle/spinner under rational rotation. (‘rational’ = rational multiple of 2π)
ReplyDeleteThe action of rational rotations is locally finite. So, by your Theorem 2, regular hyperreal and regular qualitative probabilities exist. By contrast, for unrestricted real rotations they don’t. This difference is not surprising: the standard counterexample uses images under an iterated irrational rotation. But it does raise the question of whether the difference between rational and real rotations should matter philosophically.
Ian:
ReplyDeleteOn completeness, note that classical probability does measure all countable sets, and it is the countable ones that cause the problems.
The rational rotation thing is really interesting.
Another case: countable lottery with finite permutations (i.e. those that permute only a finite number of tickets). Full conditional, regular hyperreal and regular qualitative probabilities all exist. For unrestricted permutations, they don’t.
ReplyDeleteMathematically, this is no surprise – restricting to finite permutations blocks paradoxical decompositions. But the philosophy is interesting. Both full and finite permutation symmetry reflect the intuition of label independence. Both permit any label to be swapped with any other. They differ on whether you require invariance under an infinite number of swaps at once.
Very nice case, too. I should include them if I am asked to revise the paper.
ReplyDeleteI wonder what happens if you combine finite permutations and shifts (in a bidirectional setup). Do you have invariant conditional probabilities?
I think so. Here's an attempted proof sketch. Suppose that a set A of integers has a paradoxical decomposition under finite permutations and shifts. Clear, A must be infinite. Let P be a shift-invariant probability on A (which exists because the group of shifts is abelian). Note that P assigns zero to every finite subset of A by finite additivity and shift-invariance.
That paradoxical decomposition involves a partition B1,...,Bn,C1,...,Cm such
that there are symmetries g1,...,gn,h1,...,hm with g1B1,...,gnBn being a partition of A, and h1C1,...,hmCm being a partition of A as well. For any symmetry g, let g* be the symmetry obtained by dropping all the finite permutations in g. More precisely, g* is a shift by the limit of g(n)-n as n goes to infinity (or minus infinity). Then, for any set U, g*U and gU differ in only finitely many places. Then, g1*B1,...,gn*Bn is a partition of A modulo finite sets, and h1*C1,...,hn*Cn is a partition of A modulo finite sets. Finite sets have zero measure with respect to P. So, these are partitions modulo sets that have zero P-measure. But that will violate the shift-invariance and finite additivity of P.
This is interesting, because in the bi-infinite coin sequence case, if you combine finite reversals and shifts, you don't have invariant conditional probabilities. Moreover, the lottery shift and finite-permutation group is not supramenable, because it contains a subgroup isomorphic to the finite-reversals-and-shifts group from the coin case.
The method of proof here might help to generate some other cases. But I need to run off for a Philosophy Kayaking event now.
The proof seems to require that A has non-zero P-measure. Is it obvious that it must?
ReplyDeleteThe proof requires that there be a shift-invariant probability measure *on A*. It doesn't require that that measure be a restriction of a shift-invariant probability on the integers.
ReplyDeleteThere is such a measure on A, because the group of shifts (which has a partial action on A) is supramenable.
I now see. Very neat.
ReplyDeleteMore generally, the proof shows that if G is a group that has a partial action on Omega that admits strongly G-invariant full conditional probabilities and that has the property that for any x and y in Omega there is a g in G such that gx=y, and G* is the group generated by G and by the finite permutations of Omega, then there are strongly G*-invariant probabilities on Omega.
ReplyDelete