A pref B just when des(A) > des(B). (3)

I have been intentionally ambiguous about the domain of pref. In the canonical example it’s a boolean algebra of sets, where incompatible means disjoint, but it need not be. I first studied averaging relations in hopes of proving theorems about functions defined on the lattice of subspaces of Hilbert space – a nondistributive analogue of a boolean algebra of sets. These algebras occur in axiomatizations of quantum mechanics. I never came close to proving such theorems. But the definition of an averaging relation still makes sense in that context. Hence my first question for decision theorists:

Challenge 1: Can you find anything useful or interesting in the decision theory in which the space of propositions is modeled by the lattice of subspaces of Hilbert space?

Only a small corner of that central philosophical puzzle concerns us here. In the canonical example the two functions prob and value determine des and hence pref. (I choose prob and value as my primitives, but any two of prob , value and des determine the others. Since u is the Radon-Nikodym derivative of value with respect to prob it too is known when they are.) What more beyond the averaging hypothesis must we assume about pref in order to guarantee the existence of prob and value? In order to discover such conditions we will look at them (when they do exist) in a new way. For each real number x and proposition A define

jeffrey(x,A) = x prob(A) – value(A) (4)

The eponymous jeffrey function so defined has several interesting properties (as does the man for whom it is named):

1. For each x,jeffrey(x,× ) is a (signed) measure: when A and B are incompatible

 

jeffrey(x,AÈ B) = jeffrey(x,A) + jeffrey(x,B)

 

2. For each A, jeffrey(x,A) is an increasing function of x with a unique root.

The relation between jeffrey and des in the canonical example is captured by the assertion

It is straightforward to show that when jeffrey satisfies (a) and (b) the relation pref determined by (5) and (3) is an averaging relation. In my thesis (Bolker 1966) I prove the converse – an existence theorem for jeffrey. For each averaging relation pref (satisfying some additional continuity assumptions) there is a function jeffrey satisfying (a) and (b) such that the function des defined by (5) induces pref.

Note that all you need of jeffrey to determine des is knowledge of its roots. Do its other values convey any useful information? That may be a subtle question, since pref does not determine jeffrey. If f is a positive function then newj(x,A) = f(x)jeffrey(x,A) will satisfy (a) above. If newj is an increasing function of x for each A then it also satisfies (b) and serves just as well as jeffrey to define pref.

Challenge 2: What is the decision theoretic meaning of the jeffrey function?

The existence theorem in the previous section is not satisfactory for decision theory. The averaging relations it represents are too general. In Bolker 1966 I prove that when pref is impartial (see Jeffrey 1990 or Bolker 1966 for a definition) then it is in fact defined by a jeffrey function of the form (4). Now I want to take a short look inside that proof. Consider the vector space M of signed measures on the set of propositions. Let V be the linear span of the set of measures jeffrey(x,× ). The heart of the proof is the assertion that

pref is impartial (hence comes from des and value)

if and only if V is two-dimensional.

The geometry of M sheds light on the uniqueness of prob and des when they exist. Let M+ be the cone of positive measures in M and V+ the intersection of V with M+. The dimension of V+ must be one or two. When it’s one, prob is unique and des is determined up to a linear transformation. When it’s two the fractional linear transformations discussed in Jeffrey 1990 enter the picture. They provide rescaling functions f for jeffrey of the sort discussed in the last section.

If we assume merely that pref averages we have too little structure. If we assume it is impartial as well then V is two-dimensional and pref comes from a des and prob in the standard way. But perhaps there is a decision theory with weaker assumptions.

Challenge 3. Find interesting examples of averaging relations for which V is three-dimensional. Find conditions on pref analogous to impartiality that imply that V is three dimensional, or finite dimensional.