More complications for Dutch Book results

Think of a wager as a sequence of event-payoff pairs:

• W = ((e₁, u₁),...,(e_n, u_n)).

There are then two different ways to calculate the expected value of the wager. First, directly:

1. E^D(W)=u₁P(e₁)+...+u_nP(e_n).

Second, indirectly by letting U_W be the utility function defined by W, i.e., U_W = u₁ ⋅ 1_e1 + ... + u_n ⋅ 1_en (where 1_e is the function that is 1 if e happens and 0 otherwise) and then calculating the expected utility of the function U_W:

2. E^I(W)=E(U_W).

If the credence function P is additive, then the two ways are equivalent. But without additivity, they come apart. Moreover, there is more than one way of calculating E(U) if the credences are inconsistent, but for now I will assume the standard Lebesgue sum way where, assuming U has only finitely many values, E(U)=∑_yyP(U = y).

The most common de Finetti Dutch Book Theorem, which says that inconsistent probabilities give rise to a Dutch Book, makes use of the direct way of calculating the values of wagers. Specifically, it considers wagers where you pay an amount x for a chance to win amount y if event E eventuates, and it calculates the value of such a wager as yP(E)−x. However, if instead one uses the indirect method of calculation, the value of such a wager becomes (y − x)P(E)−xP(E^c), where E^c is the complement of E.

This actually makes a real difference to Dutch Book theorems. Consider this inconsistent credence for a coin toss:

• P(H)=1/4

• P(T)=1/4

• P(H&T)=0

• P(H ∨ T)=1.

Then for any credence function U, it turns out that E^I(U)>0 if and only if the expected value of U is positive given the standard consistent fair-toss measure. The reason is this. Either U has the same value at heads and tails or it does not. If it has the same value at heads and tails, then E^I(U) has the same value as the expectation using the fair measure, since P agrees with the fair measure regarding H ∨ T. On the other hand, if U has different values at heads and tails, then E^I(U)=(1/4)U(H)+(1/4)U(T) which is exactly half of the fair measure’s expectation for U, and hence, again, E^I(U)>0 if and only if the fair measure says the expectation is positive. It seems to follow that E^I recommends exactly the same wagers as the standard consistent fair-toss measure.

Except that this isn’t quite true, either. For in addition to two ways of calculating expected values, there are two ways of making decisions on their basis in the case where a sequence of wagers is offered:

3. Accept a wager whose individual expected utility is positive.

4. Accept a wager when the expected utility of the already-accepted wagers combined with the currently offered wager exceeds the expected value of the combination of the already-accepted wagers.

Here, the combination of two wagers is concatenation. For instance ((e₁, u₁),(e₂, u₂)) combiness with ((e₃, u₃)) to form the wager ((e₁, u₁),(e₂, u₂),(e₃, u₃)). Given consistent credences, we have, E(W₁ + W₂)=E(W₁)+E(W₂), and (3) and (4) are equivalent. But, again, for inconsistent credences this additivity property can fail, and so a choice needs to be made between (3) and (4).

Note that (4) is itself an oversimplification. For theoretically, what wagers one accepts earlier on may depend on one’s best estimate as to what wagers will be offered later.

All in all, I know of five utility maximization decision procedures for sequences of wagers, generated by the answers to these questions:

• Direct or indirect utility calculation for a wager? (D or I)

• If indirect, Lebesgue sum or level set integral for calculating expectations? (LSum or LSet)

• If indirect, is the presently offered wager combined with previously accepted wagers in calculating expectations? (Indiv or Combo)

For consistent probabilities, these are all equivalent.

Moreover, there are two kinds of Dutch Books. There are Simple Dutch Books, where from the original position the agent accepts a Dutch Book, and Incremental Dutch Books, where after accepting some wagers, the agent goes on to accept a Dutch Book.

What happens with Dutch Books varies between the different procedures, and I am still working out the details. Say that a credence P is monotonic provided that P(∅)=0, P(Ω)=1 and P(A)≤P(B) whenever A ⊆ B. Here is what I have:

• D: Simple Dutch Books whenever probabilities are inconsistent.

• I+LSum+Indiv: I conjecture Incremental Dutch Books for some but not all inconsistent monotonic credences.

• I+LSum+Combo: I conjecture Incremental Dutch Books for all non-additive credences.

• I+LSet+Indiv: I don’t know.

• I+LSet+Combo: No Dutch Books of either sort for any monotonic credences.

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.

Dutch Books and infinite sequences of coin tosses

Suppose we have an infinite sequence of independent and fair coins. A betting portfolio is a finite list of subsets of the space of outcomes (heads-tails sequences) together with a payoff for each subset. Assume:

1. Permutation: If a rational agent would be happy to pay x for a betting portfolio, and A is one of the subsets in the betting portfolio, then she would also be happy to pay x for a betting portfolio that is exactly the same but with A replaced by A*, where A* is isomorphic to A under a permutation of the coins.
2. Equivalence: A rational agent who is happy to pay x for one betting scenario, will be willing to accept an equivalent betting scenario---one that is certain to give the same payoff for each outcome---for the same price.
3. Great Deal: A rational agent will be happy to pay $1.00 for a betting scenario where she wins $1.25 as long as the outcome is not all-heads or all-tails.

Leveraging the Axiom of Choice and using the methods of the Banach-Tarski Paradox, one can then find two betting portfolios that the agent would be happy to accept for $1.00 each such that it is certain that if she accepts them both, she will win at most $1.25; hence she will accept a sure loss of at least $0.75. For details of a closely related result, see Chapter 6 in Infinity, Causation and Paradox (draft temporarily here).

So what to do? I think one should accept Causal Finitism, the doctrine that causal antecedents are always finite. Given Causal Finitism, one can't have a real betting scenario based an infinite number of coin tosses. Moreover, the only known way to operationalize the use of the Axiom of Choice in the proof in a betting scenario also involves infinite causal dependencies.

Truth and Dutch Books

Suppose I initially assigned probability 0.5 to p and 0.5 to ~p. Suppose p is in fact true, and my credence in p comes to be magically increased to 0.8 without my credence in ~p being changed. I thus have inconsistent probabilities: 0.8 for p and 0.5 for ~p. This is supposed to be bad: it lays me open to Dutch Books. For instance, I will accept the following pair of options:

1. Pay $0.75 to win $1.00 if p
2. Pay $0.45 to win $1.00 if ~p.

But if I do that, then I will pay $1.20 and get $1.00, for a net loss of $0.20.

Yes, that's an unhappy result. But note that I am actually better off than earlier when my credences were consistent. Earlier I would have rejected (1) since my credence in p was 0.5, but I would have accepted (2). So I would have paid $0.45 and got nothing to show for it. Thus my revision in the direction of truth made me be better off, even though it also led me to accept a Dutch Book.

This suggests that pragmatically and synchronically speaking what matters is truth, not probabilistic consistency. Better be inconsistent and closer to truth than consistent and further from truth.

Diachronically, of course, at least logical inconsistency could be dangerous, as it can lead to lots of absurd conclusions. But in practice we all have inconsistent beliefs and we manage to contain the inconsistency without much in the way of explosion.

So what's so bad about Dutch Books? It seems to be this: an opponent who knows (with certainty) your credences and doesn't know (at least with certainty) whether p is true can offer you a series of bets that you are guaranteed to lose money on. This is a big deal if you're playing an adversarial game against such an opponent. But such games are, I think, a special case, and while they do occur in war, business, sport and other competitive pursuits, we should not let competitive pursuits against fellow humans dictate the nature of rationality to us. And note a curious thing: consistency is not the only available strategy against such an opponent—hiding your credences will also help. If you revise your credence in the direction of truth but your opponent doesn't know about your revision, you will do at least as well as before, and quite possibly better.