## Friday, October 11, 2019

### 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)=∫0P(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 ∫0P(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 ∑iUiP(Ei) where the Ei partition the sample space such that on Ei the utility has the constant value Ui. 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.