Suppose that ≼ is a total preorder on simple utility functions on some space Ω with an algebra of subsets. Define f ≺ g iff f ≼ g but not g ≼ f. Think of ≼ as a decision procedure: you are required (permitted) to choose g over f iff f ≺ g (g ≼ f).
Suppose ≼ doesn’t allow choosing a dominated wager:
- If f < g everywhere, then f ≺ g.
Let 1A be the function that is 1 on A and 0 outside A. Define Ef = sup{c : c ⋅ 1Ω ≼ f}. Here are some facts about E:
If c < Ef < c′, then c ⋅ 1Ω ≺ f ≺ c′⋅1Ω.
E(c ⋅ 1Ω)=c
If f ≤ g everywhere, then Ef ≤ Eg.
If Ef < Eg, then f ≺ g.
(But we can’t count on its being the case that f ≺ g if and only if Ef < Eg.)
Now consider what I’ve called the independent and cumulative decision procedures for sequences of choices. On an independent decision procedure, at each stage you must choose a wager that is ≼-maximal (and you may choose any of the maximal ones). On a cumulative decision procedure, at each stage you must choose a wager that when added to what you’ve already chosen yields a ≼-maximal wager (and you may choose any of the maximal ones).
I think (I haven’t written it all down) I can prove that the following conditions are equivalent:
E is an expected value with respect to a finitely-additive probability P on Ω.
The independent decision procedure applied to a sequence of binary choices never permits you to choose a sequence of wagers whose sum is strictly dominated by the sum of a different sequence of wagers you could have chosen.
The cumulative decision procedure applied to a sequence of binary choices never permits you to choose a sequence of wagers whose sum is strictly dominated by the sum of a different sequence of wagers you could have chosen.
The probability P is defined by P(A)=E(1A).
All that said, I think that when your credences are inconsistent, you may need to decide neither independently nor cumulatively, but holistically, taking into account what wagers you made and what wagers you expect you will make.
Take \Sigma=\{0,1\} and define f ≼ g iff f \leq g, so that Ef=min f.
ReplyDeleteThen 1=P(\{0,1\})=P\{0\}+P\{1\}=E1_{\{0\}}+E1_{{\1\}}=0, a contradiction.
Ben:
ReplyDeleteYour ≼ isn't a total preorder. For instance, define f(x)=x and g(x)=1-x. Then neither f≼g nor g≼f is true.
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ReplyDeleteOops : ( wasnt paying attention.
ReplyDelete--ben
Does ‘finitely-additive’ in the post imply that Ω may be infinite?
ReplyDeleteI have a sketch proof that works for finite Ω. (Like you, I have not actually written out all the details.) But for infinite Ω, St Petersburg-like possibilities make things tricky.
Ian:
ReplyDeleteI've written up a proof on Saturday. It works for an infinite spaces, but remember that I'm only interested in *simple* utility functions, i.e., ones that have a finite number of values. So no St Petersburg.