Math appendix for: "Why you must maximize expected utility"

by Benya 7y13th Dec 20123 min read4 comments

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This is a mathematical appendix to my post "Why you must maximize expected utility", giving precise statements and proofs of some results about von Neumann-Morgenstern utility theory without the Axiom of Continuity. I wish I had the time to make this post more easily readable, giving more intuition; the ideas are rather straight-forward and I hope they won't get lost in the line noise!

The work here is my own (though closely based on the standard proof of the VNM theorem), but I don't expect the results to be new.

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I represent preference relations as total preorders on a simplex ; define , , and in the obvious ways (e.g., iff both and , and iff but not ). Write for the 'th unit vector in .

In the following, I will always assume that satisfies the independence axiom: that is, for all and , we have  if and only if . Note that the analogous statement with weak preferences follows from this: holds iff , which by independence is equivalent to , which is just .

Lemma 1 (more of a good thing is always better). If and , then .

Proof. Let . Then, and . Thus, the result follows from independence applied to , , and .

Lemma 2. If  and , then there is a unique such that for and for .

Proof. Let be the supremum of all such that (note that by assumption, this condition holds for ). Suppose that . Then there is an such that . By Lemma 1, we have , and the first assertion follows.

Suppose now that . Then by definition of , we do not have , which means that we have , which was the second assertion.

Finally, uniqueness is obvious, because if both and satisfied the condition, we would have .

Definition 3. is much better than , notation or , if there are neighbourhoods of and of (in the relative topology of ) such that we have for all and . (In other words, the graph of is the interior of the graph of .) Write  or when ( is not much better than ), and ( is about as good as ) when both and .

Theorem 4 (existence of a utility function). There is a such that for all ,

Unless for all and , there are  such that .

Proof. Let be a worst and a best outcome, i.e. let be such that for all . If , then  for all , and by repeated applications of independence we get for all , and therefore again for all , and we can simply choose .

Thus, suppose that . In this case, let be such that for every , equals the unique provided by Lemma 2 applied to and . Because of Lemma 1, . Let .

We first show that implies . For every , we either have , in which case by Lemma 2 we have for arbitrarily small , or we have , in which case we set  and find . Set . Now, by independence applied times, we have ; analogously, we obtain for arbitrarily small . Thus, using and Lemma 1, and therefore  as claimed. Now note that if , then this continues to hold for and in a sufficiently small neighbourhood of and , and therefore we have .

Now suppose that . Since we have  and , we can find points and arbitrarily close to and such that the inequality becomes strict (either the left-hand side is smaller than one and we can increase it, or the right-hand side is greater than zero and we can decrease it, or else the inequality is already strict). Then, by the preceding paragraph. But this implies that , which completes the proof.

Corollary 5. is a preference relation (i.e., a total preorder) that satisfies independence and the von Neumann-Morgenstern continuity axiom.

Proof. It is well-known (and straightforward to check) that this follows from the assertion of the theorem.

Corollary 6. is unique up to affine transformations.

Proof. Since  is a VNM utility function for , this follows from the analogous result for that case.

Corollary 7. Unless for all , for all the set has lower dimension than (i.e., it is the intersection of with a lower-dimensional subspace of ).

Proof. First, note that the assumption implies that . Let be given by , , and note that is the intersection of the hyperplane with the closed positive orthant . By the theorem, is not parallel to , so the hyperplane is not parallel to . It follows that has dimension , and therefore can have at most this dimension. (It can have smaller dimension or be the empty set if only touches or lies entirely outside the positive orthant.)

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