When can we have exact symmetries of hyperreal probabilities?

In many interesting cases, there is no way to define a regular hyperreal-valued probability that is invariant under symmetries, where “regular” means that every non-empty set has non-zero probability. For instance, there is no such measure for all subsets of the circle with respect to rotations: the best we can do is approximate invariance, where P(A)−P(rA) is infinitesimal for every rotation. On the other hand, I have recently shown that there is such a measure for infinite sequences of fair coin tosses where the symmetries are reversals at a set of locations.

So, here’s an interesting question: Given a space Ω and a group G of symmetries acting on Ω, under what exact conditions is there a hyperreal finitely-additive probability measure P defined for all subsets of Ω that satisfies the regularity condition P(A)>0 for all non-empty A and yet is fully (and not merely approximately) invariant under G, so that P(gA)=P(A) for all g ∈ G and A ⊆ Ω?

Theorem: Such a measure exists if and only if the action of G on Ω is locally finite. (Assuming the Axiom of Choice.)

The action of G on Ω is locally finite iff for every x ∈ Ω and every finitely-generated subgroup H of G, the orbit Hx = {hx : h ∈ H} of x under H is finite. In other words, we have such a measure provided that applying the symmetries to any point of the space only generates finitely many points.

This mathematical fact leads to a philosophical question: Is there anything philosophically interesting about those symmetries whose action is locally finite? But I’ve spent so much of the day thinking about the mathematical question that I am too tired to think very hard about the philosophical question.

Sketch of Proof of Theorem: If some subset A of Ω is equidecomposable with a proper subset A′, then a G-invariant measure P will assign equal measure to both A and A′, and hence will assign zero measure to the non-empty set A − A′, violating the regularity condition. So, if the requisite measure exists, no subset is equidecomposable with a proper subset of itself, which by a theorem of Scarparo implies that the action of G is locally finite.

Now for the converse. If we could show the result for all finitely-generated groups G, by using ultraproduct along an ultrafilter on the partially ordered set of all finitely generated subgroups of G we could show this for a general G.

So, suppose that G is finitely generated and the orbit of x under G is finite for all x ∈ Ω. A subset A of G is said to be G-invariant provided that gA = A for all g ∈ G. The orbit of x under G is always G-invariant, and hence every finite subset of A is contained in a finite G-invariant subset, namely the union of the orbits of all the points in A.

Consider the set F of all finite G-invariant subsets of Ω. It’s worth noting that every finite subset of G is contained in a finite G-closed subset: just take the union of the orbits under G. For A ∈ F, let P_A be uniform measure on A. Let F^* = {{B ∈ F : A ⊆ B}:A ∈ F}. This is a non-empty set with the finite intersection property. Let U be an ultrafilter extending F^*. Let ^*R be the ultraproduct of the reals over F with respect to U, and let P(C) be the equivalence class of the function A ↦ P_A(A ∩ C) on F. Note that C ↦ P_A(A ∩ C) is G-invariant for any G-invariant set A, so P is G-invariant. Moreover, P(C)>0 if C ≠ ∅. For let C′ be the orbit of some element of C. Then {B ∈ F : C′⊆B} is in F^*, and P_A(A ∩ C′) > 0 for all A such that C′⊆A, so the set of all A such that P_A(A ∩ C′) > 0 is in U. It follows that P(C′) > 0. But C′ is the orbit of some element x of C, so every singleton subset of C′ has the same P-measure as {x} by the G-invariance of P. So P({x}) = P(C′)/|C′| > 0, and hence P(C)≥P({x}) > 0.

Another impossibility result for finitely additive probabilities and invariance

Consider a countably infinite sequence of fair and independent coin tosses. Given the Axiom of Choice, there is no finitely additive probability measure that satisfies these conditions:

1. It is defined for all sets of outcomes.
2. It agrees with the classical probabilities where these are defined.
3. It is invariant under permutations of coins.

(Sketch of proof: Index the coins with members of the free group of rank two. The members of the group then induce permutations of coins, and hence act on the space of outcomes. The set of non-trivial fixed points under that action has classical probability zero. Throwing that out, we can use a standard paradoxical decomposition of the free group of rank two to generate a paradoxical decomposition of the rest of our space of results--here Choice will be used--and that rules out the possibility of a finitely additive probability measure.)

Betting on paradoxical sets

Suppose that a point z will be uniformly randomly chosen on the surface of a sphere S and you are asked to place bets as to which set z is in. Then, plausibly:

1. If two sets A and B are equivalent under rotations about the center of the sphere, you should accept this offer: get three dollars if z is in A and pay two dollars if z is in B.

But now consider a paradoxical decomposition of the whole sphere S, by a version of the Banach-Tarski Paradox[note 1]. In this, the sphere is partitioned into two subsets C and D, each of which can be decomposed into a finite number of subsets that can be rotated to form the whole sphere. Applying (1) to each set in the decomposition of C and its rotation, you will accept a sequence of deals that adds up to:

2. If z is in C, you get three dollars and if z is in S you pay two dollars.

Repeating this with D's decomposition, you get a sequence of deals that adds up to:

3. If z is in D, you get three dollars and if z is in S you pay two dollars.

But of course if z is in C or D, it is also in S, and if it's in S, then it's in exactly one of C or D. It follows that the deal adds up to:

4. No matter what, you get three dollars and you pay four dollars.

So, repeated application of (1) yields an unacceptable conclusion.

One might say that this is an artifact of the fact that there is no finitely additive rotation-invariant probability measure on the sphere. But I think the above formulation is a little bit more telling. I make no reference to probabilities here. All I assume is (1), which is a very intuitive rationality judgment, namely that when one has two equivalent scenarios, one should accept an unequal bet between them that is in one's favor.

What to conclude? One conclusion might be that a single application of (1) is fine, but the sequence of applications needed to yield (4) is not.

My own conclusion, however, is that it is metaphysically impossible to have a betting scenario like the above. But why not? What's wrong with it? Well, one possibility is that space is necessarily discrete, but that doesn't seem very plausible to me.

My own preference, however, is to conclude that it is impossible to have anything causally depend on whether a random point (or a particle or the like) is in one of these weird sets that are found in the paradoxical decomposition of the sphere. Why is that? I think it's because it would in effect be a violation of causal finitism, the thesis that no event can causally depend on infinitely many things. But the full story here requires significant amounts of work to complete.