Some stuff about models of PA+~Con(PA)

Assume Peano Arithmetic (PA) is consistent. Then it can’t prove its own consistency. Thus, there is a model M of PA according to which PA is inconsistent, and hence, according M, there is a proof of a contradiction from a finite set of axioms of PA. This sounds very weird.

But it becomes less weird when we realize what these claims do and do not mean in M.

The model M will indeed contain an M-natural number a that according to M encodes a finite sequence of axioms of PA, and it will also contain an M-natural number p that according to M encodes a proof of a contradiction using the axioms encoded in A.

However, here are some crucial qualifications. Distinguish between the M-natural numbers that are standard, i.e., correspond to an actually natural number, one that from the point of view of the “actual” natural numbers is finite, and those that are not. The latter are infinite from the point of view of the actual natural numbers.

First, the M-natural number a is non-standard. For a standard natural number will only encode a finite number of axioms, and for any finite subtheory of PA, PA can prove its consistency (this is the “reflexivity of PA”, proved by Mostowski in the middle of the last century). Thus, if a were a standard natural number, according to M there would be no contradiction from the axioms in a.

Second, while every item encoded in a is according to M an axiom of PA, this is not actually true. This is because any M-finite sequence of M-natural numbers will either be a standardly finite length sequence of standard natural numbers, or will contain a non-standard number. For let n be the largest element in the sequence. If this is standard, then we have a standardly finite length sequence of standard natural numbers. If not, then the sequence contains a non-standard number. Thus, a contains something that is not axiom of PA.

In other words, according to our model M, there is a contradictory collection of axioms of PA, but when we query M as to what that collection is, we find out that some of the things that M included in the collection are not actually axioms of PA. (In fact, they won’t even be well-formed formulas, since they will be infinitely long.) So a crucial part of the reason why M disagrees with the “true” model of the naturals about the consistency of PA is because M disagrees with it about what PA actually says!

A puzzle about consistency

Let T₀ be ZFC. Let T_n be T_n − 1 plus the claim Con(T_n − 1) that T_n − 1 is consistent. Let T_ω be the union of all the T_n for finite n.

Here’s a fun puzzle. It seems that T_ω should be able to prove its own consistency by the following reasoning:

If T_ω is inconsistent, then for some finite n we have T_n inconsistent. But Con(T_n) is true for every finite n.

This sure sounds convincing! It took me a while to think through what’s wrong here. The problem is that although for every finite n, T_ω can prove Con(T_n), it does not follow that T_ω can prove that for every finite n we have Con(T_n).

To make this point perhaps more clear, assume T_n is consistent for all n. Then Con(T_n) cannot be proved from T_n. Thus any finite subset of T_ω is consistent with the claim that for some finite n the theory T_n is inconsistent. Hence by compactness there is a model of T_ω according to which for some finite n the theory T_n is inconsistent. This model will have a non-standard natural number sequence, and “finite” of course will be understood according to that sequence.

Here’s another way to make the point. The theory T_ω proves T_ω consistent if and only if T_ω is consistent according to every model M. But the sentence “T_ω is consistent according to M” is ambiguous between understanding “T_ω” internally and externally to M. If we understand it internally to M, we mean that the set that M thinks consists of the axioms of ZFC together with the ω-iteration of consistency claims is consistent. And this cannot be proved if T_ω is consistent. But if we understand “T_ω” externally to M, we mean that upon stipulating that S is the object in M’s universe whose members^M correspond naturally to the members^V of T_ω (where V is “our true set theory”), according to M, it will be provable that the set S is consistent. But there is a serious problems: there just may be no such object as S in the domain of M and the stipulation may fail. (E.g., in non-standard analysis, the set of finite naturals is never an internal set.)

(One may think a second option is possible: There is such an object as S in M’s universe, but it can’t be referred to in M, in the sense that there is no formula ϕ(x) such that ϕ is satisfied by S and only by S. This option is not actually possible, however, in this case.)

Or so it looks to me. But all this is immensely confusing to me.

More dwindling of the prospects for an accuracy-based argument for probabilism in infinite cases

Let P be the set of countably additive probabilities on a countable set Ω. A strictly proper accuracy scoring rule on some set of credences C that includes P is a function s from C to [ − ∞, M]^Ω for some finite M such that E_ps(p)>E_ps(q) for any p ∈ P and any q ∈ C: i.e., from the point of view of each probability in P, that probability has the highest expected accuracy. (It’s a little easier for my example to work with accuracy scoring rules.)

We can identify P with the non-negative functions of ℓ¹(Ω) that sum to 1. This identification induces a topology on P (based on the norm and weak topologies on ℓ¹(Ω) which are the same). On [ − ∞, M]^Ω we will take the product topology.

We say that a scoring rule is uniformly bounded provided there is some finite R such that |s(p)(ω)| < R for all p ∈ C and ω ∈ Ω. We say that one function strictly dominates another provided that the former is strictly less than the latter everywhere.

Theorem: There is a uniformly bounded strictly proper scoring rule s on C that is continuous on P and a credence c ∈ C − P such that s(c) is not strictly dominated by s(p) for any p ∈ P.

In a recent post, I showed that it’s not possible to use strictly proper scoring rules to produce a strict domination argument for probabilism (the thesis that all credences should be probabilities) in infinite cases when we take probabilities to be finitely additive, because in the finitely additive case there is no strictly proper scoring rule. The above Theorem is a complement for the countably additive case: it shows that there are nice strictly proper scoring rules for the countably additive probabilities, but they don’t support the strict domination results that arguments for probabilism seem to require.

There is a remaining open question as to what happens when one further assumes additivity of the scoring rule. But I do not think additivity of a scoring rule is a reasonable constraint, because it seems to me that epistemic utilities will depend on some global features of a forecast.

Here is a sketch of the proof of the theorem. Assume Ω is the natural numbers. Identify P with the non-negative members of ℓ¹ that sum to 1. Let s(p)(n)=p(n)/∥p∥₂ for p ∈ P. Note that the ℓ²-norm is continuous on ℓ¹, and hence s is continuous on Ω. Observe that s(p)∈ℓ¹. By the Cauchy-Schwarz inequality (together with the condition for equality in it), s is strictly proper on P. Define s(c)(n)=1/2(n + 1) for any credence c that is not in P. Note that E_ps(p)=∥p∥₂ for all p. Observe that E_ps(c)<∥s(c)∥₂∥p∥₂ for every p by Cauchy-Schwarz again. But ∥s(c)∥₂<1. Thus, s is strictly proper. But s(c) for c not a probability is not a member of ℓ¹, and hence is not dominated by any score of a probability.

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.

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.

Do inconsistent credences lead to Dutch Books?

It is said that if an agent has inconsistent credences, she is Dutch Bookable. Whether this is true depends on how the agent calculates expected utilities. After all, expected utilities normally are Lebesgue integrals over a probability measure, but the inconsistent agent’s credences are not a probability measure, so strictly speaking there is no such thing as a Lebesgue integral over them.

Let’s think how a Lebesgue integral is defined. If P is a probability measure and U is a measurable function on the sample space, then the expected value of U is defined as:

1. E(U)=∫₀^∞P(U > y)dy − ∫_−∞⁰P(U < y)dy

where the latter two integrals are improper Riemann integrals and where P(U > y) is shorthand for P({ω : U(ω)>y}) and similarly for P(U < y).

Now suppose that P is not a probability measure, but an arbitrary function from the set of events to the real numbers. We can still define the expected value of U by means of (1) as long as the two Riemann integrals are defined and aren’t both ∞ or both −∞.

Now, here is an easy fact:

Proposition: Suppose that P is a function from a finite algebra of events to the non-negative real numbers such that P(∅)=0. Suppose that U is a measurable (with respect to the finite algebra) function such that (a) P(U > y)=0 for all y > 0 and (b) P(U < 0)>0. Then if E(U) is defined by (1), we have E(U)<0.

Proof: Since the algebra is finite and U is measurable, U takes on only finitely many values. If y₀ is the largest of its negative values, then P(U < 0)=P(U < y) for any negative y > y₀, and hence ∫_−∞⁰P(U < y)dy ≥ |y₀|P(U < 0)>0 by (b), while ∫₀^∞P(U > y)dy by (a). □

But then:

Corollary: If P is a function from a finite algebra of events on the samples space Ω to the non-negative real numbers with P(∅)=0 and P(Ω)>0, then an agent who maximizes expected utility with respect to the credence assignment P as computed via (1) and starts with a baseline betting portfolio for which the utility is zero no matter what happens will never be Dutch Boooked by a finite sequence of changes to her portfolio.

Proof: The agent starts off with a portfolio with a utility assignment U₀ where P(U₀ > y)=0 for all y > 0 and P(U₀ < y)=0 for all y < 0, and hence once where E(U₀)=0 by (1). If the agent is in a position where the expected utility based on her current portfolio is non-negative, she will never accept a change to the portfolio that turns the portfolio’s expected utility negative, as that would violated expected utility maximization. By mathematical induction, no finite sequence of changes to her portfolio will turn her expected utility negative. But if a portfolio is a Dutch Book then the associated utility function U is such that P(U < 0)=P(Ω)>0 and P(U > y)=0 for all y > 0. Hence by the Proposition, E(U)<0, and hence a Dutch Book will not be accepted at any finite stage. □

Note that the Corollary does assume a very weak consistency in the credence assignment: negative credences are forbidden, impossible events get zero credence, and necessary events get non-zero credence.

Additionally, the Corollary does allow for the possibility of what one might call a relative Dutch Book, i.e., a change between portfolios that loses the agent money no matter what. The final portfolio won’t be a Dutch Book relative to the initial baseline portfolio, of course.

Note, however, that we don’t need consistency to get rid of relative Dutch Books. Adding the regularity assumption that P(A)>0 for all non-empty A and the monotonicity condition that if A ⊂ B then P(A)<P(B) is all we need to ensure the agent will never accept even a relative Dutch Book. For regularity plus monotonicity ensures that a relative Dutch Book always decreases expected utility as defined by (1). But these conditions are not enough to rule out all inconsistency. For instance, if in the case of the flip of a single coin I assign probability 1 to heads-or-tails, probability 0.8 to heads, probability 0.8 to tails, and probability 0 to the empty event, then my assignment is patently inconsistent, but satisfies all of the above assumptions and hence is neither absolutely nor relatively Dutch Bookable.

How does all this cohere with the famous theorems about inconsistent credence assignments being Dutch Bookable? Simple: Those theorems define expected utility for inconsistent credences differently. Specifically, they define expected utility as ∑_iU_iP(E_i) where the E_i partition the sample space such that on E_i the utility has the constant value U_i. But that’s not the obvious and direct generalization of the Lebesgue integral!

I vaguely recall hearing something that suggests to me that this might be in the literature.

Also, I slept rather poorly, so I could be just plain mistaken in the formal stuff.

Truth and probabilistic consistency

Suppose Alice has an inconsistent probabilistic assignment P^A. Then, famously, there is a series of bets on single propositions (call these binary bets) that is a Dutch Book against Alice: i.e., Alice by her lights will accept each bet, and is guaranteed to lose money.

But now suppose Bob has a probabilistic assignment P^B—perhaps a consistent one—that is strictly further from the truth than Alice’s inconsistent one in the sense that

1. for any p, if p is false, then P^B(p)≥P^A(p),

2. for any p, if p is true, then P^B(p)≤P^A(p), and

3. at least one of the inequalities is strict.

Then Alice will do at least as well as Bob on every portfolio of offers of binary bets, and on some portfolios she will do strictly better than Bob. In particular, even if Bob’s probabilistic assignment is consistent, and there is a binary bet Dutch Book against Alice, Bob will fare no better than Alice with respect to that book.

Thus, if we start with a consistent assignment and then by some process move towards truth, we will do better (against binary bet portfolios) even if we lose consistency.

So why is Alice’s probabilistic assignment supposed to be rationally bad in a way that Bob’s isn’t? Well, the difference is this. A bookie can fleece Alice simply on the basis of knowing Alice’s probability assignment. But simply knowing Bob’s probability assignment won’t be enough to know which portfolio will fleece him.

However, the more I think about this, the more I lose the intuition that all this shows there is something particularly rationally problematic about Alice’s assignments just because they are inconsistent. Why should game-theoretic performance against a competitor who knows one’s credences be particularly indicative of rationality or the lack thereof? When nature offers us betting portfolios (to pursue this trail or that trail after a wounded deer in the woods, say), these portfolios are normally independent of our credences. Of course, in business and war, we have to worry about mind-reading competitors. But much of our life, we don’t.

Suppose I find myself with inconsistent credences. What should I do? Should I force them to be consistent? If I am dealing with mind-reading competitors who have no more information about the external world than I do, then I should go for consistency. But going for consistency will force me to modify some of my probabilities, and for all I know, these probabilities may get modified away from truth. And that might be more harmful.

There may be interesting trade-offs. Maybe some intellectual strategies work better against mind-reading competitors and others work better with the portfolios set by nature. We should not take doing well with respect to one selection of portfolio to be particularly informative about the nature of rationality.

Epistemic scores and consistency

Scoring rules measure the distance between a credence and the truth value, where true=1 and false=0. You want this distance to be as low as possible.

Here’s a fun paradox. Consider this sentence:

1. At t₁, my credence for (1) is less than 0.1.

(If you want more rigor, use Goedel’s diagonalization lemma to remove the self-reference.) It’s now a moment before t₁, and I am trying to figure out what credence I should assign to (1) at t₁. If I assign a credence less than 0.1, then (1) will be true, and the epistemic distance between 0.1 and 1 will be large on any reasonable scoring rule. So, I should assign a credence greater than or equal to 0.1. In that case, (1) will be false, and I want to minimize the epistemic distance between the credence and 0. I do that by letting the credence be exactly 0.1.

So, I should set my credence to be exactly 0.1 to optimize epistemic score. Suppose, however, that at t₁ I will remember with near-certainty that I was setting my credence to 0.1. Thus, at t₁ I will be in a position to know with near-certainty that my credence for (1) is not less than 0.1, and hence I will have evidence showing with near-certainty that (1) is false. And yet my credence for (1) will be 0.1. Thus, my credential state at t₁ will be probabilistically inconsistent.

Hence, there are times when optimizing epistemic score leads to inconsistency.

There are, of course, theorems on the books that optimizing epistemic score requires consistency. But the theorems do not apply to cases where the truth of the matter depends on your credence, as in (1).