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 simplexΔN; define ≺, ∼, ≽ and ≻ in the obvious ways (e.g., x∼y iff both x≼y and y≼x, and x≺y iff x≼y but noty≼x). Write ei for the i’th unit vector in RN.
In the following, I will always assume that ≼satisfies the independence axiom: that is, for all x,y,z∈ΔN and p∈(0,1], we have x≺y if and only if px(1−p)z≺py(1−p)z. Note that the analogous statement with weak preferences follows from this: x≼y holds iff y⊀x, which by independence is equivalent to py(1−p)z⊀px(1−p)z, which is just px(1−p)z≼py(1−p)z.
Lemma 1 (more of a good thing is always better).If x≺y and 0≤p<q≤1, then (1−p)xpy≺(1−q)xqy.
Proof. Let r:=q−p. Then, (1−p)xpy=((1−q)xpy)rx and (1−q)xqy=((1−q)xpy)ry. Thus, the result follows from independence applied to x, y, 11−r((1−q)xpy), and r.□
Lemma 2.If x≼y≼z and x≺z, then there is a unique p∈[0,1] such that (1−q)xqz≺y for q∈[0,p) and y≺(1−q)xqz for q∈(p,1].
Proof. Let p be the supremum of all r∈[0,1] such that (1−r)xrz≼y (note that by assumption, this condition holds for r=0). Suppose that 0≤q<p. Then there is an r∈(q,p] such that (1−r)xrz≼y. By Lemma 1, we have (1−q)xqz≺(1−r)xrz, and the first assertion follows.
Suppose now that p<q≤1. Then by definition of p, we do not have (1−q)xqz≼y, which means that we have (1−q)xqz≻y, which was the second assertion.
Finally, uniqueness is obvious, because if both p and p′ satisfied the condition, we would have y≺(1−pp′2)xpp′2z≺y.□
Definition 3.x is much better than y, notation x≻∗y or y≺∗x, if there are neighbourhoods U of x and V of y (in the relative topology of ΔN) such that we have x′≻y′ for all x′∈U and y′∈V. (In other words, the graph of ≻∗ is the interior of the graph of ≻.) Write x≼∗y or y≽∗x when x⊁∗y (x is not much better than y), and x∼∗y (x is about as good as y) when both x≼∗y and x≽∗y.
Theorem 4 (existence of a utility function). There is a u∈RN such that for all x,y∈ΔN,
∑ixiui<∑iyiui⟺x≺∗y⟹x≺y.
Unless x∼y for all x and y, there are i,j∈{1,…,N} such that ui≠uj.
Proof. Let i be a worst and j a best outcome, i.e. let i,j∈{1,…,N} be such that ei≼ek≼ej for all k∈{1,…,N}. If ei∼ej, then ei∼ek for all k, and by repeated applications of independence we get x∼ei∼y for all x,y∈ΔN, and therefore x∼∗y again for all x,y∈ΔN, and we can simply choose u=0.
Thus, suppose that ei≺ej. In this case, let u be such that for every k∈{1,…,N}, uk equals the unique p provided by Lemma 2 applied to ei≼ek≼ej and ei≺ej. Because of Lemma 1, ui=0≠1=uj. Let f(r):=(1−r)eirej.
We first show that p:=∑kxkuk<∑kykuk=:q implies x≺y. For every k, we either have uk<1, in which case by Lemma 2 we have ek≺f(ukϵk) for arbitrarily small ϵk>0, or we have uk=1, in which case we set ϵk:=0 and find ek≼ej=f(ukϵk). Set ϵ:=∑kxkϵk. Now, by independence applied N−1 times, we have x=∑kxkek≼∑kxkf(ukϵk)=f(pϵ); analogously, we obtain y≽f(q−δ) for arbitrarily small δ>0. Thus, using p<q and Lemma 1, x≼f(pϵ)≺f(q−δ)≼y and therefore x≺y as claimed. Now note that if ∑kxkuk<∑kykuk, then this continues to hold for x′ and y′ in a sufficiently small neighbourhood of x and y, and therefore we have x≺∗y.
Now suppose that ∑kxkuk≥∑kykuk. Since we have ui=0 and uj=1, we can find points x′ and y′ arbitrarily close to x and y 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, x′≻y′ by the preceding paragraph. But this implies that x⊀∗y, 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.u is unique up to affine transformations.
Proof. Since u is a VNM utility function for ≼∗, this follows from the analogous result for that case.□
Corollary 7.Unless x∼y for all x,y∈ΔN, for all r∈R the set {x∈ΔN:∑ixiui=r} has lower dimension than ΔN (i.e., it is the intersection of ΔN with a lower-dimensional subspace of RN).
Proof. First, note that the assumption implies that N≥2. Let v∈RN be given by vi=1, ∀i, and note that ΔN is the intersection of the hyperplane A:={x∈RN:x⋅v=1} with the closed positive orthant
. By the theorem, u is not parallel to v, so the hyperplane Br:={x∈RN:x⋅u=r} is not parallel to A. It follows that A∩Br has dimension N−2, and therefore
can have at most this dimension. (It can have smaller dimension or be the empty set if A∩Br only touches or lies entirely outside the positive orthant.)□
Math appendix for: “Why you must maximize expected utility”
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.
*
I represent preference relations as total preorders ≼ on a simplex ΔN; define ≺, ∼, ≽ and ≻ in the obvious ways (e.g., x∼y iff both x≼y and y≼x, and x≺y iff x≼y but not y≼x). Write ei for the i’th unit vector in RN.
In the following, I will always assume that ≼ satisfies the independence axiom: that is, for all x,y,z∈ΔN and p∈(0,1], we have x≺y if and only if px(1−p)z≺py(1−p)z. Note that the analogous statement with weak preferences follows from this: x≼y holds iff y⊀x, which by independence is equivalent to py(1−p)z⊀px(1−p)z, which is just px(1−p)z≼py(1−p)z.
Lemma 1 (more of a good thing is always better). If x≺y and 0≤p<q≤1, then (1−p)xpy≺(1−q)xqy.
Proof. Let r:=q−p. Then, (1−p)xpy=((1−q)xpy)rx and (1−q)xqy=((1−q)xpy)ry. Thus, the result follows from independence applied to x, y, 11−r((1−q)xpy), and r.□
Lemma 2. If x≼y≼z and x≺z, then there is a unique p∈[0,1] such that (1−q)xqz≺y for q∈[0,p) and y≺(1−q)xqz for q∈(p,1].
Proof. Let p be the supremum of all r∈[0,1] such that (1−r)xrz≼y (note that by assumption, this condition holds for r=0). Suppose that 0≤q<p. Then there is an r∈(q,p] such that (1−r)xrz≼y. By Lemma 1, we have (1−q)xqz≺(1−r)xrz, and the first assertion follows.
Suppose now that p<q≤1. Then by definition of p, we do not have (1−q)xqz≼y, which means that we have (1−q)xqz≻y, which was the second assertion.
Finally, uniqueness is obvious, because if both p and p′ satisfied the condition, we would have y≺(1−pp′2)xpp′2z≺y.□
Definition 3.x is much better than y, notation x≻∗y or y≺∗x, if there are neighbourhoods U of x and V of y (in the relative topology of ΔN) such that we have x′≻y′ for all x′∈U and y′∈V. (In other words, the graph of ≻∗ is the interior of the graph of ≻.) Write x≼∗y or y≽∗x when x⊁∗y (x is not much better than y), and x∼∗y (x is about as good as y) when both x≼∗y and x≽∗y.
Theorem 4 (existence of a utility function). There is a u∈RN such that for all x,y∈ΔN,
∑ixiui<∑iyiui⟺x≺∗y⟹x≺y.
Unless x∼y for all x and y, there are i,j∈{1,…,N} such that ui≠uj.
Proof. Let i be a worst and j a best outcome, i.e. let i,j∈{1,…,N} be such that ei≼ek≼ej for all k∈{1,…,N}. If ei∼ej, then ei∼ek for all k, and by repeated applications of independence we get x∼ei∼y for all x,y∈ΔN, and therefore x∼∗y again for all x,y∈ΔN, and we can simply choose u=0.
Thus, suppose that ei≺ej. In this case, let u be such that for every k∈{1,…,N}, uk equals the unique p provided by Lemma 2 applied to ei≼ek≼ej and ei≺ej. Because of Lemma 1, ui=0≠1=uj. Let f(r):=(1−r)eirej.
We first show that p:=∑kxkuk<∑kykuk=:q implies x≺y. For every k, we either have uk<1, in which case by Lemma 2 we have ek≺f(ukϵk) for arbitrarily small ϵk>0, or we have uk=1, in which case we set ϵk:=0 and find ek≼ej=f(ukϵk). Set ϵ:=∑kxkϵk. Now, by independence applied N−1 times, we have x=∑kxkek≼∑kxkf(ukϵk)=f(pϵ); analogously, we obtain y≽f(q−δ) for arbitrarily small δ>0. Thus, using p<q and Lemma 1, x≼f(pϵ)≺f(q−δ)≼y and therefore x≺y as claimed. Now note that if ∑kxkuk<∑kykuk, then this continues to hold for x′ and y′ in a sufficiently small neighbourhood of x and y, and therefore we have x≺∗y.
Now suppose that ∑kxkuk≥∑kykuk. Since we have ui=0 and uj=1, we can find points x′ and y′ arbitrarily close to x and y 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, x′≻y′ by the preceding paragraph. But this implies that x⊀∗y, 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.u is unique up to affine transformations.
Proof. Since u is a VNM utility function for ≼∗, this follows from the analogous result for that case.□
Corollary 7. Unless x∼y for all x,y∈ΔN, for all r∈R the set {x∈ΔN:∑ixiui=r} has lower dimension than ΔN (i.e., it is the intersection of ΔN with a lower-dimensional subspace of RN).
Proof. First, note that the assumption implies that N≥2. Let v∈RN be given by vi=1, ∀i, and note that ΔN is the intersection of the hyperplane A:={x∈RN:x⋅v=1} with the closed positive orthant