## Wednesday, January 22, 2020

### Lebesgue sums previsions don't always lead to Dutch Books for inconsistent credences

Suppose E is the Lebesgue-sum prevision. Namely, if W is a wager on a finite space Ω with a credence (perhaps inconsistent P) and UW is the utility function corresponding to W, then EW = ∑yP({ω : UW(ω)=y}).

Suppose your decision procedure for repeated wagers is to accept a wager if and only if the wager’s value is non-negative (independently of whatever other wagers you might have accepted). Suppose, further, that Ω has exactly two points and the credence of each point is non-negative and of at least one it is positive.

Proposition: Then, no finite sequence of wagers forms a Dutch Book.

Proof: Consider the sequence of utility functions U1, ..., Un that corresponds to a Dutch Book sequence of wagers W1, ..., Wn. Then U1 + ... + Un < 0 everywhere on Ω and yet EWi ≥ 0 for all i. Let a and b be the two points of Ω. Reordering the wagers if necessary (the order doesn’t matter on this decision procedure), we can assume that the wagers W1, ..., Wm are such that Ui(a)≠Ui(b) for i ≤ m, and that Wm + 1, ..., Wn are such that Ui(a)=Ui(b) for i > m. Then EWi = Ui(a)=Ui(b) for i > m. Hence, Ui is positive everywhere on Ω for i > m. So, if W1, ..., Wn form a Dutch Book, so do W1, ..., Wm. Now, EWi = αUi(a)+βUi(b) where α and β are the probabilities of a and b respectively. It follows that EW1 + ... + EWn = α(U1(a)+...+Um(a)) + β(U1(b)+...+Um(b)). Since this is a Dutch Book, it follows that the two sums on the right hand side are both negative. Since α and β are non-negative and at least one is positive, it follows that EW1 + ... + EWn < 0, and hence this isn’t a Dutch Book.

IanS said...

It is good to avoid Dutch books, but it is not enough. You also want to avoid choosing dominated options. With inconsistent credences, the value assigned by the Lebesgue sum prevision is discontinuous in the neighbourhood of wagers with equal payoffs on two or more outcomes. This leads to cases in which the Lebesgue sum prevision will assign a higher value to a wager with lower payoffs than another, whatever the outcome. (Note that level set method avoids these issues.)

Here is an example. Suppose you have inconsistent credences Cr(a) = Cr(b) = 0.2, Cr(a or b) = 1. Wager#1 pays 1 on a and o.9 on b. This is given value 0.38. Wager#2 pays 0.5 on a or b. This has value 0.5. So, based on Lebesgue sum prevision, you would choose Wager#2. But Wager#2 pays less than Wager#1 whatever the outcome. This is clearly the wrong choice.

As you show, there are no Dutch books if Ω has two points. But if Ω has three or more points, Dutch books are possible. An example follows.

Suppose Ω has three points. Call them a, b and c. Suppose your credences are Cr(a) = Cr(b) = Cr(c) = 1/3 and (inconsistently) Cr(a or b) = Cr(b or c) =Cr(c or a) = 1/2. Then you would accept the wager with payoffs 1.8 on a and -1 on b or c. (Expectation is 1.8/3 – 1/2 = +0.1) You would also accept each of the two similar wagers with the payoffs cyclically permuted. The three wagers together make a Dutch book. (Whatever the outcome, you win one bet and lose two with net payoff 1.8 – 2*1 = -0.2.)

Alexander R Pruss said...

That all looks correct.

In fact, Lebesgue sum prevision always gives failures of domination in cases of inconsistency.

I was just surprised that it doesn't always give a dutch book.

IanS said...

Here is another way of looking at it.

For Dutch Books as applied in the post, the only relevant feature of a prevision (for given credences) is its “acceptance set”, i.e. the set of wagers assigned non-negative value. Previsions with the same acceptance set will be subject (or not) to the same Dutch Books, however else they may differ.

In the 2-point case, the acceptance set for credences α and β (your notation) is the same as for the consistent credences α/(α+β) and β/(α+β). Hence no Dutch Books.