## Wednesday, February 16, 2022

### Domination and uniform spinners

About a decade ago, I offered a counterexample to the following domination principle:

1. Given two wagers A and B, if in every state B is at least as good as A and in at least one state B is better than A, then one should choose B over A.

But perhaps (1) is not so compelling anyway. For it might be that it’s reasonable to completely ignore zero probability outcomes. If a uniform spinner is spun, and on A you get a dollar as long as the spinner doesn’t land at 90 and on B you get a dollar no matter what, then (1) requires you to go for B, but it doesn’t seem crazy to say “It’s almost surely not going to land at 90, so I’ll be indifferent between A and B.”

But now consider the following domination principle:

1. Given two wagers A and B, if in every state B is better than A, then one should choose B over A.

This seems way more reasonable. But here is a potential counterexaple. Consider a spinner which uniformly selects a point on the circumference of a circle. Assume x is any irrational number. Consider a function u such that u(z) is a real number for any z on the circumference of the circle. Imagine two wagers:

• A: After the spinner is spun and lands at z, you get u(z) units of utility

• B: After the spinner is spun, the spinner is moved exactly x degrees counterclockwise to yield a new landing point z′, and you get u(z′) units of utility.

Intuitively, it seems absurd to think that B could be preferable to A. But it turns out that given the Axiom of Choice, we can define a function u such that:

1. For any z on the circumference of the circle, if z is the result of rotating z by x degrees counterclockwise around the circle, then u(z′) > u(z).

And then if we take the states to be the initial landing points of the spinner, B always pays strictly better than A, and so by the domination principle (2), we should (seemingly absurdly) choose B.

Remarks:

• The proof of the existence of u requires the Axiom of Choice for collections of countable sets of reals). In my Infinity book, I argued that this version of the Axiom of Choice is true. However, arguments similar to those in the book’s Axiom of Choice chapter suggest that the causal finitist has a good way out of the paradox by denying the implementability of the function u.

• Some people don’t like unbounded utilities. But we can make sure that u is bounded if we want (if the original function u is not bounded, then replace u(z) by arctan u(z)).

• Of course the function u is Lebesgue non-measurable. To see this, replacing u by its arctangent if necessary, we may assume u is bounded. If u were measurable and bounded, it would be integrable, and its Lebesgue integral around the circle would be rotation invariant, which is impossible given (3).

It remains to prove the existence of u. Let be the relation for points on the (circumference of the circle) defined by z ∼ z if the angle between z and z is an integer multiple of x degrees. This is an equivalence relation, and hence it partitions the circle into equivalence classes. Let A be a choice set that contains exactly one element from each of the equivalence classes. For any z on the circle, let z0 be the point in A such that z0 ∼ z. Let u(z) be the (unique!) integer n such that rotating z0 counterclockwise around the circle by an angle of nx degrees yields z. Then for any z, if z is the result of rotating z by x degrees around the circle, then u(z′) = u(z) + 1 > u(z) and so we have (3).

Alexander R Pruss said...

One can also do a similar, but not quite the same, thing with infinite sequences of coin flips. In other words, there is (assuming AC) a game based on a bidirectionally infinite sequence of fair coin flips (arranged left-to-right) with the property that there is an event E such that:
1. P(E) = 0 (indeed, E is countable)
2. the game obtained by shifting the coins one space to the right always pays at least as well
3. the game obtained by shifting the coins one space to the right pays better outside of E.

Here, E is the event that the coin flip outcomes are an endlessly repeating pattern (e.g., ...HHTHHTHHT...). We prove the above by letting ~ be the equivalence relation that holds between two sequences of coin flips when one sequence is a shift of the other, and then running an argument much as in the post, though it only works outside E.

Alexander R Pruss said...

Let rz be z rotated by x degrees. Then let g(z) = u(rz)-u(z) be the amount gained by the rotation.

In the main construction, where u(z) is unbounded, g(z) = 1 for every z. Constant gain everywhere. But if we make u bounded (say by the arctan move), then interestingly I think I can prove that although g(z) is strictly positive, it is (a) non-measurable, and (b) there is no strictly positive measurable function less than g(z). This makes the bounded case, to me, a bit less impressive.

IanS said...

By one of your theorems (Non-classical Probabilities Invariant Under Symmetries, Th 2), there is no rotation-invariant regular hyperreal distribution on the circle. So it’s not so clear how a ‘spinner’ should be represented mathematically. You have to accept that if you force rotation-invariance on some sort of sets, you can’t have it on others. The example illustrates this.

Another way of looking at it: If X is the outcome of a fair infinite lottery on the integers, then X+1 is also fair, and is always greater than X. (Fair here means fair between singletons. This does not imply translation invariant for arbitrary sets.) The post could be taken as ‘constructing’ a fair infinite lottery.

Alexander R Pruss said...

"The post could be taken as ‘constructing’ a fair infinite lottery." I didn't notice that. Cool!

Alexander R Pruss said...

BTW, if you take a model of set theory with countable choice but where all sets of reals are Lebesgue measurable, then there will be a partial preference ordering on wagers on the outcomes of a spinner that (a) is strongly rotation invariant (W1 is at least as good as W2 iff rW1 is at least as good as W2 iff W1 is at least as good as rW2) and (b) satisfies condition (2) (but not condition (1)): wager W1 is at least as good as W2 iff the Lebesgue integral of W1-W2 is defined and non-negative. For every positive function will have a well-defined Lebesgue integral.

I don't know if there is a model of set theory where there is a total preference ordering on wagers on the outcomes of a spinner that satisfies (a) and (b).

Alexander R Pruss said...

The construction in this post is pretty much the same as John Norton's Vitali set lottery: https://sites.pitt.edu/~jdnorton/papers/Infinite_lottery_not_final.pdf