A strict propriety argument for probabilism without any continuity assumptions

Here’s an accuracy-theoretic argument for probabilism (the thesis that only probabilities are rationally admissible credences) on finite spaces that does not make any continuity assumptions on the scoring rule. I will assume all credence functions take values on [0,1].

1. All probabilities are rationally admissible credences.

2. If any non-probabilities are rationally admissible, then all non-probabilities satisfying Normalization (whole space has credence 1) and Subadditivity (P(A) + P(B) ≤ P(A∪B) when A and B are disjoint) are rationally admissible with the appropriate prevision being given by a level set integral [correction: actually, I need LSI^↑, not the version of LSI in the earlier blog post].

3. A rationally appropriate scoring rule s satisfies strict propriety for all rationally admissible credences with an appropriate prevision: if V is an appropriate prevision then V_us(u) is better than V_us(v) whenever u and v are different rationally admissible credences.

4. There is a rationally appropriate scoring rule.

But now we have a cute theorem:

• On any finite space Ω with at least two points, no scoring rule satisfies strict propriety for the credences with Normalization and Subadditivity and level set integral prevision.

It follows no non-probabilities are rationally admissible.

Is this a good argument? I find (2) somewhat plausible—it’s hard to think of a less problematic weakening of the axioms of probability than from Additivity to Subadditivity, and I have not been able to find a better prevision than the level set integral one. Standard arguments for probabilism assume strict propriety for all probabilities. But it seems to me that a non-probabilist will find strict propriety for all probabilities plausible only insofar as they find strict propriety for all admissible credences plausible. Thus (3) is dialectically as good as the usual strict propriety assumption.

I think the non-probabilist’s best way out is to deny strict propriety or to deny that there is a rationally appropriate scoring rule. Both of these ways out work just as well against more standard arguments for probabilism, and I think both are good ways out.

Technically speaking, the advantage of this argument over standard arguments for probabilism is that it makes no assumptions of continuity.

The accuracy argument for probabilism

A standard scoring rule argument for probabilism—the doctrine that credence assignments should satisfy the axioms of probability—goes as follows. If s is a scoring rule on a finite probability space Ω, so that s(c)(ω) is the epistemic utility of credence assignment c at ω in Ω, and (a) s is strictly proper and (b) s is continuous, then for any credence c that does not satisfy the axioms of probability, there is a credence p that does satisfy them such that s(p)(ω) is better than s(c)(ω) for all ω. This means that it’s stupid to have a non-probabilistic credence c, since you could instead replace it with p, and do better, no matter what.

Here is a problem with the dialectics behind this argument. Let P be the set of all credence assignments that satisfy the axioms of probability. But suppose that I think that there is some nonempty set M of credence assignments that do not satisfy the axioms of probability but are rationally just as good as those in P. Then I will think there is some way of making decisions using credences in M, just as good as the way of making decisions using credences in P. The best candidate in the literature for this is to use a level set integral, which allows one to assign an expected value E_cU to any utility assignment U even if c is not a probability. Note that E_pU is the standard mathematical expectation with respect to p if p is a probability.

The argument for probabilism assumed two things about the scoring rule: strict propriety and continuity. Strict propriety is the claim that:

1. E_ps(p) > E_ps(c) whenever c is a credence other than p

for any probability p. In words, by the lights of a probability p, then we get the best expected epistemic utility if we make p be our credence.

Now, if I am not convinced by the argument that (1) should hold for any probability p and any credence c other than p, then I will be unmoved by the scoring rule argument for probabilism. So suppose that I am convinced. But recall that I think that credences in M are just as rationally good as the probabilities in P. Because of this, if I find (1) convincing for all probabilities p, I will also find it convincing for all credences p in M, where E_p is my preferred way of calculating expected utilities—say, a level set integral.

Thus, if I am convinced by the argument for strict propriety, I will just as much accept (1) for p in M as for p in P. But now we have:

Theorem 1. If E_p is strongly monotonic for all p ∈ P ∪ M and coincides with mathematical expectation for p ∈ P, and (1) holds for all p in P ∪ M, where M is non-empty, then s is not continuous on P.

(Strong monotonicity means that if U < V everywhere then E_pU < E_pV. The Theorem follows immediately from the Pettigrew-Nielsen-Pruss domination theorem.)

Suppose then that I am convinced that a scoring rule s should be continuous (either on P or on all of P ∪ M). Then the conclusion I am apt to draw is that there just is no scoring rule that satisfies all the desiderata I want: continuity as well as (1) holding for all p ∈ P ∪ M.

In other words, the only way the argument for probabilism will be convincing to me is if my reason to think (1) is true for all p in P is significantly stronger than my reason to think (1) is true for all p in M, and I have a sufficiently strong reason to think that there is a scoring rule that satisfies all the true rational desiderata on a scoring rule to conclude that (1) holding for all p in M is not among the true rational desiderata even though its holding for all p in P is.

And once I additionally learn about the difficulties in defining sensible scoring rules on infinite spaces, I will be less confident in thinking there is a scoring rule that satisfies all the true rational desiderata on a scoring rule.

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.

Truth-directedness and propriety of scoring rules does not imply strict propriety

A scoring rule assigns a score to a credence assignment (which can but need not satisfy the axioms of probability), where a score is a random variable measuring how close the credence assignment is to the truth.

A scoring rule is strictly truth-directed provided that if c′ is a credence assignment that is closer to the truth than c is at ω, then c′ gets a better a score at ω. A scoring rule is proper provided that for all probabilities p, the p-expected value of the score of a probability p is at least as good as the p-expected value of the score of any other credence, and is strictly proper.

Propriety for a scoring rule is a pretty plausible condition, but it’s a bit harder to argue philosophically for strict propriety. But scoring-rule based philosophical arguments for probabilism—the doctrine that credences ought to be probabilities—require strict propriety.

In a clever move, Campbell-Moore and Levinstein showed that propriety plus strict truth-directedness and additivity (the idea that the score can be decomposed into a sum of single-event scores) implies strict propriety.

Here’s an interesting fact I will show: propriety plus strict truth-directedness do not imply strict propriety in the absence of additivity. Further, my counterexample will be bounded, infinitely differentiable and strictly proper on the probabilities. Personally don’t find additivity all that plausible, so I conclude the Campbell-Moore and Levinstein move does not move the discussion of strict propriety and probabilism ahead much.

Let Ω = {0, 1}. Given a credence function c (with values in [0,1]) on the powerset of Ω, define the credence function c^* which has the same value as c on the empty set and on Ω, but where c^*({0}) is the number z in [0,1] that minimizes (c({0})−z)² + (c({1})−(1−z))², and where c^*({1}) = 1 − c^*({0}). In other words, c^* is the credence function closest to c in the Euclidean metric such that c^*({0}) + c^*({1}) = 1.

Now let b^*(c) = b(c^*). Then b^* agrees with b score on the probabilities, and hence is strictly proper on them. Further, every value of b^* is a Brier score of some credence, and hence b^* is proper.

We now check that it is strictly truth-directed. Brier scores are strictly truth-directed. Thus, replacing a credence function with one that is closer to the truth on Ω or on the empty set will improve the b^* score. Moreover, it is easy to check that c^*({0}) = (1+c({0})−c({1}))/2. It’s easy to check that if we tweak c({0}) to move us closer to the truth at some fixed ω ∈ {0, 1}, then c^* will be closer to the truth at ω as well, and similarly if we tweak c({1}) to be closer to the truth at ω, and in both cases we will improve the score by the strict truth-directedness of Brier scores.

Finally, however, note that b^* is not strictly proper and does not have a domination theorem of the sort used in arguments for probabilism, since the b^*-score of any credence c that fails to be a probability due to its being the case c({0}) + c({1}) ≠ 1 but that gets the right values on the empty set and Ω (zero and one, respectively) is equal to the b^*-score of c^*, and c^* will be a probability in that case.

Note that in the example above we don't have quasi-strict propriety either.

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.

Being subject to a Dutch Book

I’ve periodically wondered why doing poorly when faced with a Dutch Book is supposed to be a sign of irrationality, but it’s not a sign of irrationality that rational people do poorly when faced with someone who hits all and only rational people on the head with a baseball bat.

This occurred to me today:

1. One cannot get a rational person to act against their own interest except by force, luck or superior information.

2. Putting a Dutch Book over someone with inconsistent credences does not require force, luck or superior information.

This seems to get at some of the intuition as to why being subject to a Dutch Book is supposed to be a sign of irrationality.

But I don’t know how much confidence we should have in (1). The exception clause already admits three exceptions. This sounds ad hoc. Would we be very surprised if more exceptions had to be added?

Still, there is some plausibility to (1), at least for self-interested rationality.

Previsions for inconsistent credences and arguments for probabilism

Fix a sample space Ω and an algebra F events on Ω. A gamble is an F-measurable real-valued function on Ω. A credence function is a function from a F to the reals. A prevision or price function on a set of set G of gambles is just a function from G to the real numbers. A previsory method E on a set of gambles G and a set of credence functions C assigns to each credence function P ∈ C a prevision E_P on G.

A previsory method on G and C has the weak domination property provided that if f and g are two gambles such as that f ≤ g everywhere on Ω, then E_P(f)≤E_P(g) for every f and g in G and P in C. It has the strong domination property provided that it has the weak domination property and if f < g everywhere on Ω, then E_P(f)<E_P(g). It has the zero property provided that E_P(0)=0.

Mathematical expectation is a previsory method on the set of all bounded gambles and all probability functions. It has the zero and strong domination properties.

The level set integral is a previsory method on the set of all bounded gambles and all monotonic credence functions (P is monotonic iff P(⌀)=0, P(Ω)=1 and P(A)≤P(B) whenever A ⊆ B). It has the zero and weak domination properties.

The level set integral has the strong domination property on the set of weakly countably additive monotonic credence functions, where P is weakly countably additive provided that Ω cannot be written as a countable union of sets each of credence 0. If F (or Ω) is finite, we get weak countable additivity for free from monotonicity.

A previsory method E requires (permits) a gamble f given a credence P provided that E_P(f)>0 (E_P(f)≥0); it requires (permits) it over some set S of gambles provided that E_P(f)>E_P(g) (E_P(f)≥E_p(g)) for every g in S.

A previsory method with the zero and weak domination properties cannot be strongly Dutch-Booked in a single wager: i.e., there is no gamble U such that U < 0 everywhere that the method requires. If it also has the strong domination property, it cannot be weakly Dutch-Booked in a single wager: there is no U such that U < 0 everywhere that the method permits.

Suppose we combine a previsory method with the following method of choosing which gambles to adopt in a sequence of offered gambles: you are required (permitted) to accept gamble g provided that E_P(g₁ + ... + g_n + g)>E_P(g₁ + ... + g_n) (≥, respectively) where g₁ + ... + g_n are the gambles already accepted. Then given the zero and weak domination properties, we cannot be strongly Dutch-Booked by a sequence of wagers, and given additionally the strong domination property, we cannot be weakly Dutch-Booked, either.

Given that level set integrals provide a non-trivial and mathematically natural previsory method with the zero and strong domination properties on a set of credence functions strictly larger than the consistent ones, Dutch-Book arguments for consistency fail.

What about epistemic utility, i.e., scoring-rule, arguments? I think these also fail. A scoring-rule assigns a number s(p, q) to a credence function p and a truth function q (i.e., a probability function whose values are always 0 or 1). Let T be truth, i.e., a function from Ω to truth functions such that T(ω)(A) if and only if ω ∈ A. Thus, T(ω) is the truth function that says “we are at ω” and we can think of s(p, T) as a gamble that measures how far p is from truth.

If E is previsory method on a set of gambles G and a set of credence functions C, then we say that s is an E-proper scoring rule provided that s(p, T) is in G for every p in C and E_ps(p, T)≤E_ps(q, T) for every p and q in C. We say that it is strictly proper if additionally we have strict inequality whenever p and q are different.

If E is mathematical expectation, then E-propriety and strict E-propriety are just propriety and strict propriety.

It is thought (Joyce and others) that one can make use of the concept of strictly propriety to argue for that credence functions should be consistent. This uses a domination theorem that says that if s is a strictly proper additive scoring rule, then for any inconsistent credence function p there is a consistent function q such that s(p, T(ω)) < s(q, T(ω)) for all ω. (Roughly, an additive scoring rule adds up scores point-by-point over Ω.)

However, I think the requirement of additivity is one that someone sceptical of the consistency requirement can reasonably reject. There are mathematical natural previsory methods E that apply to some inconsistent credences, such as the monotonic ones, and these can be used to define (at least under some conditions) strictly E-proper scoring rules. And the domination theory won’t apply to these rules because they won’t be additive. Indeed, that is one of the things the domination theorem shows: if C includes an inconsistent credence function and E has the strong domination property, then no strictly E-proper scoring rule is additive.

So, really, how helpful the domination theorem is for arguing for consistency depends on whether additivity is a reasonable condition to require of a scoring rule. It seems that someone who thinks that it is OK to reason with a broader set of credences than the consistent ones, and who has a natural previsory method E with the strong domination property for these credences, will just say: I think the relevant notion isn’t propriety but E-propriety, and there are no strongly E-proper scoring rules that are additive. So, additiveness is not a reasonable condition.

Are there any strongly E-proper scoring rules in such cases?

[The rest of the post is based on the mistake that E-propriety is additive and should be dismissed. See my discussion with Ian in the comments.]

Sometimes, yes.

Suppose E is previsory method with the weak domination condition on the set of all bounded gambles on Ω. Suppose that E has the scaling property that E_p(cf)=cE_p(f) for any real constant c. (Level Set Integrals have scaling.) Further, assume the separability property that there is a countable set of B of bounded gambles such that for any two distinct credences p and q, there is a bounded gamble f in B such that E_pf ≠ E_qf. (Level Set Integrals on a finite Ω—or on a finite field of events—have separability: just let B be all functions whose values are either 0 or 1, and note that E_p1_A = p(A) where 1_A is the function that is 1 on A and 0 outside it.) Finally, suppose normalization, namely that E_p1_Ω = 1. (Level Set Integrals clearly have that.)

Note that given separability, scaling and normalization, there is a countable set H of bounded gambles such that if p and q are distinct, there exist f and g in H such that E_p requires f over g (i.e., E_pf > E_pg) and E_q does not or vice versa. To see this, let H consist of B together with all constant rational-valued functions, and note that if E_pf < E_qf, then we can choose a rational number r such that r lies between E_pf and E_qf, and then E_p and E_q will disagree on whether f is required over r ⋅ 1_Ω.

Let H be the countable set in the above remark. By scaling, we may assume that all the gambles in H are bounded by 1. Let (f₁, g₁),(f₂, g₂),... be an enumeration of all pairs of members of H. Define s_n(p, T(ω)) for a credence function p in C as follows: if E_p requires f_n over g_n then s_n(p, T(ω)) = −f_n(ω), and otherwise s_n(p, T(ω)) = −g_n(ω).

Note that s_n is an E-proper scoring rule. For suppose that q is a different credence function from p and E_ps_n(p, T)>E_ps_n(q, T). Now there are four possibilities depending on whether E_p and E_q require f_n over g_n and it is easy to see that each possibility leads to a contradiction. So, we have E-propriety.

Now, let s(p, T) be Σ_n = 1^∞ 2⁻ⁿs_n(p, T). The sum of E-proper scoring rules is E-proper, so this is an E-proper scoring rule.

What about strict propriety? Suppose that p and q are credence functions in C and E_ps(p, T)≤E_ps(q, T). By the E-propriety of each of the s_n, we must have E_ps_n(p, T)=E_ps_n(q, T) for all n. Thus, for all pairs of members of H, the requirements of E_p and E_q must agree, and by choice of H, p and q cannot be different.