Dominance and countably infinite fair lotteries

Suppose we have a finitely-additive probability assignment p (perhaps real, perhaps hyperreal) for a countably infinite lottery with tickets 1, 2, ... in such a way that each ticket has infinitesimal probability (where zero counts as an infinitesimal). Now suppose we want to calculate the expected value or previsio E_pU of any bounded wager U on the outcome of the lottery, where we think of the wager as assigning a value to each ticket, and the wager is bounded if there is a finite M such that |U(n)| < M for all n.

Here are plausible conditions on the expected value:

1. Dominance: If U₁ < U₂ everywhere, then E_pU₁ < E_pU₂.

2. Binary Wagers: If U is 0 outside A and c on A, then E_pU = cP(A).

3. Disjoint Additivity: If U₁ and U₂ are wagers supported on disjoint events (i.e., there is no n such
   that U₁(n) and U₂(n) are both non-zero), then E_p(U₁+U₂) = E_pU₁ + E_pU₂.

But we can’t. For suppose we have it. Let U(n) = 1/(2n). Fix a positive integer m. Let U₁(n) be 2 for n ≤ m + 1 and 0 otherwise. Let U₂(n) be 1/m for n > m + 1 and 0 for n ≤ m + 1. Then by Binary Wagers and by the fact that each ticket has infinitesimal probability, E_pU₁ is an infinitesimal α (since the probability of any finite set will be infinitesimal). By Binary Wagers and Dominance, E_pU₂ ≤ 1/(m+1). Thus by Disjoint Additivity, E_p(U₁+U₂) ≤ α + 1/(m+1) < 1/m. But U < U₁ + U₂ everywhere, so by Dominance we have E_pU < 1/m. Since 0 < U everywhere, by Dominance and Binary Wagers we have 0 < E_pU.

Thus, E_pU is a non-zero infinitesimal β. But then β < U(n) for all n, and so by Binary Wagers and Dominance, β < E_pU, a contradiction.

I think we should reject Dominance.

Must we accept free stuff?

Suppose someone offers you, at no cost whatsoever, something of specified positive value. However small that value, it seems irrational to refuse it.

But what if someone offers you a random amount of positive value for free. Strict dominance principles say it’s irrational to refuse it. But I am not completely sure.

Imagine a lottery where some positive integer n is picked at random, with all numbers equally likely, and if n is picked, then you get 1/n units of value. Should you play this lottery for free?

The expected value of the lottery is zero with respect to any finitely-additive real-valued probability measure that fits the description (i.e., assign equal probablity to each number). And for any positive number x, the probability that you will get less than x is one. It’s not clear to me that it’s worth going for this.

If you like infinitesimals, you might say that the expected value of the lottery is infinitesimal and the probability of getting less than some positive number x is 1 − α for an infinitesimal α. That makes it sound like a better deal, but it’s not all that clear.

Of course, infinite fair lotteries are dubious. So I don’t set much store by this example.

An argument for probabilism without assuming strict propriety

Suppose that s is a proper scoring rule on a finite space Ω continuous on probabilities and suppose that for no probability p is the expectation E_ps(p) infinitely bad (i.e., no probability is infinitely bad by its own lights). Suppose that s is probability distinguishing: there isn’t a non-probability c and probability p such that s(c) = s(p) everywhere. Then any non-probability credence c is weakly s-dominated by some probability p: i.e., s(p)(ω) is at least as good as s(c)(ω) for all ω, and strictly better for at least one ω. (This follows from the fact that Lemma 1 of this short piece holds with the same proof when q is a non-probability.)

If one thinks that one should always switch to a weakly dominating option, then this conclusion provides an argument for probabilism.

One might, however, reasonably think that it is only required to switch to a weakly dominating option when one assigns non-zero probability of the weakly dominating option being better. If so, then we get a weaker conclusion: your credences should either be irregular (i.e., assign zero to some non-empty set) or probabilistic. But a view that permits violations of the axioms of probability but only when one has irregular credences seems really implausible. So your credences should be probabilistic.

The big question is whether probability distinguishing is any more plausible as a condition on a scoring rule than strictness of propriety. I think it has some plausibility, but I am not quite sure how to argue for it.

Some possible progress on continuous scoring rules and dominance in an infinite case

On finite sample spaces, we have the Pettigrew-Nielsen-Pruss domination theorem for strictly proper scoring rules that are continuous when restricted to the probabilities that shows that the score of any non-probability is dominated by the score of a probability. Last year, I showed that for a reasonable sense of “continuous”, this is not true on countably infinite sample spaces (when we take probabilities to be countably additive; for if we take probabilities to be finitely additive, there are no strictly proper scoring rules).

In the comments, Ian then suggested that we want our scoring rule to be continuous on all credences, not just the probabilities.

Here are two preliminary responses, though not all the details of the proof of the second have yet been checked, so I could just be wrong.

First, what happens seems to depend on the topology on the space of credences. Credences can be thought of as functions from PΩ to [0,1]. One possibility is to take the space of credences to get the product topology on [0,1]^PΩ. In that case, there is no continuous strictly proper (or even quasi-strictly proper) scoring rule. This follows from the uncountability of PΩ which shows that any countable intersection of neighborhoods of a probability function will contain infinitely many non-probability functions, so that any continuous score will have the property that for every probability there is a non-probability that gets the same score.

But, second, another reasonable topology on [0,1]^PΩ is the ℓ^∞(PΩ) topology. This topology is easily seen to be equivalent on the probabilities to the ℓ¹(Ω) topology (where a probability p on PΩ corresponds to a function p^* ∈ ℓ¹(Ω) defined by p^*(x) = p({x})). The example in my earlier post was a score s that was equal to the spherical score on all the probabilities and s(c)(n) = 1/2(n+1) for any non-probability credence, where we identify Ω with the natural numbers.

Let Q be the space of probability functions on PΩ. Let d(c) = inf_p ∈ Q∥c − p∥_∞ be the distance from c to Q. We can prove that d(c) = 0 iff c is a probability, and d is continuous in our topology. Let ϕ(c) = 0 if d(c) ≥ 1/4 and ϕ(c) = 4d(c) if d(c) < 1/4. This will be a continuous function. Now define s(c)(n) = ϕ(c)/2(n+1) + (1−ϕ(c))c^*(n)/∥c^*∥₂, where c^*(n) = c({n}), and where the second summand is deemed to be zero if ϕ(c) = 1 (regardless of the denominator). I haven’t checked all the details yet, but this s looks continuous to me in the relevant norm, but the domination result is false for any non-probability c. The important point is that the function c ↦ ∥c^*∥₂ is continuous and non-zero for c such that d(c) < 1/4, and that’s one of the points I might yet have an error in.

Dominance and infinite lotteries

Suppose we have an infinite fair lottery with tickets 1,2,3,…. Now consider a wager W where you get 1/n units of utility if ticket n wins. How should you value that wager?

Any value less than zero is clearly a non-starter. How about zero? Well, that would violate the dominance principle: you would be indifferent between W and getting nothing, and yet W is guaranteed to give you something positive. What about something bigger than zero? Well, any real number y bigger than zero has the following problem as a price: You are nearly certain (i.e., within an infinitesimal of certainty) that the payoff of W will be less than y/2, and hence you’ve overpaid by at least y/2.

But what about some price that isn’t a real number? By the above argument, that price would have to be bigger than zero, but must be smaller than every positive real number. In other words, it must be infinitesimal. But any such price will violation dominance as well: you would be indifferent between W and getting that price, yet it is certain that W would give you something bigger—namely one of the real numbers 1, 1/2, 1/3, ....

So it seems that no price, real numbered or other, will do.

(This argument is adapted from one that Russell and Isaacs give in the case of the St Petersburg paradox.)

One way out will be familiar to readers of my work: Reject the possibility of infinite fair lotteries, and thereby get yet another argument for causal finitism.

But for those who don’t like controversial metaphysics solutions to decision theoretic problems, there is another way: Deny the dominance principle, price W at zero, and hold that sometimes it is rational to be indifferent between two outcomes, one of which is guaranteed to be better than the other no matter what.

This may sound crazy. But consider someone who assigns the price zero to a dart tossing game where you get a dollar if the dart hits the exact center and nothing otherwise, reasoning that the classical mathematical expected value of that game for any continuous distribution of dart tosses (such as a normal distribution around the center) is zero. I think this response to an offer to play is quite rational: “I am nearly certain to lose, so what’s the point of playing?” Now, that case doesn’t violate the same dominance principle as the lottery case—it violates a stronger dominance principle that says that if one option is guaranteed to be at least as good as the other and in some possible scenario is better, then it should be preferred. But I think the dart case may soften one up for thinking this:

1. If (a) some game never has an outcome that’s negative, and (b) for any positive real, it is nearly certain that the outcome of some game will be less than that, I should value it at zero or less.

And if we do that, then we have to value W at zero. Yes, if you reject W in favor of nothing, you’ve lost something. But probably very, very little.

Here is another weakish reason to be suspicious of dominance. Dominance is too similar to conglomerability, and conglomerability should be suspicious to anyone who likes exotic probabilistic cases. (By the way, this paper connects with this.)

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:

2. 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:

3. 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).

Nonadditive scoring rules and domination

I wrote a rough draft of a paper proving geometrically that any strictly proper scoring rule continuous on the probabilities has every score of a non-probability dominated by a score of a probability, without assuming additivity of score. My proof is very much geometric.

Notes: Richard Pettigrew first announced this result in a forthcoming paper, but his proof is flawed. Then Michael Nielsen found a proof in the special case of bounded scoring rules. Finally, Nielsen and I approximately simultaneously (within hours of each other) found quite different proofs without the assumption of boundedness (though there could still be problems in one or the other proof). Research continues regarding how far the condition of continuity can be weakened.