# 8

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 $\preccurlyeq$ on a simplex $\Delta_N$; define $\prec$, $\sim$, $\succcurlyeq$ and $\succ$ in the obvious ways (e.g., $x\sim y$ iff both $x\preccurlyeq y$ and $y\preccurlyeq x$, and $x\prec y$ iff $x\preccurlyeq y$ but not $y\preccurlyeq x$). Write $e^i$ for the $i$'th unit vector in $\mathbb{R}^N$.

In the following, I will always assume that $\preccurlyeq$ satisfies the independence axiom: that is, for all $x,y,z\in\Delta_N$ and $p\in(0,1]$, we have $x\prec y$ if and only if $px + (1-p)z \prec py + (1-p)z$. Note that the analogous statement with weak preferences follows from this: $x\preccurlyeq y$ holds iff $y\not\prec x$, which by independence is equivalent to $py + (1-p)z \not\prec px + (1-p)z$, which is just $px + (1-p)z \preccurlyeq py + (1-p)z$.

Lemma 1 (more of a good thing is always better). If $x\prec y$ and $0\le p < q \le 1$, then $(1-p)x + py\prec (1-q)x + qy$.

Proof. Let $r := q-p$. Then, $(1-p)x + py = \big((1-q)x + py\big) + rx$ and $(1-q)x + qy = \big((1-q)x + py\big) + ry$. Thus, the result follows from independence applied to $x$$y$, $\textstyle\frac{1}{1-r}\big((1-q)x + py\big)$, and $r$.$\square$

Lemma 2. If $x\preccurlyeq y\preccurlyeq z$ and $x\prec z$, then there is a unique $p\in[0,1]$ such that $(1-q)x + qz \prec y$ for $q\in[0,p)$ and $y\prec (1-q)x + qz$ for $q\in(p,1]$.

Proof. Let $p$ be the supremum of all $r\in[0,1]$ such that $(1-r)x + rz\preccurlyeq y$ (note that by assumption, this condition holds for $r=0$). Suppose that $0\le q. Then there is an $r\in(q,p]$ such that $(1-r)x + rz\preccurlyeq y$. By Lemma 1, we have $(1-q)x + qz \prec (1-r)x + rz$, and the first assertion follows.

Suppose now that $p < q \le 1$. Then by definition of $p$, we do not have $(1-q)x + qz\preccurlyeq y$, which means that we have $(1-q)x + qz\succ y$, which was the second assertion.

Finally, uniqueness is obvious, because if both $p$ and $p'$ satisfied the condition, we would have $\textstyle y \prec \big(1 - \frac{p+p'}2\big)x + \frac{p+p'}2z \prec y$.$\square$

Definition 3. $x$ is much better than $y$, notation $x\succ_* y$ or $y\prec_* x$, if there are neighbourhoods $U$ of $x$ and $V$ of $y$ (in the relative topology of $\Delta_N$) such that we have $x' \succ y'$ for all $x'\in U$ and $y'\in V$. (In other words, the graph of $\succ_*$ is the interior of the graph of $\succ$.) Write $x\preccurlyeq_* y$ or $y\succcurlyeq_* x$ when $x\nsucc_* y$ ($x$ is not much better than $y$), and $x\sim_* y$ ($x$ is about as good as $y$) when both $x\preccurlyeq_* y$ and $x\succcurlyeq_* y$.

Theorem 4 (existence of a utility function). There is a $u\in\mathbb{R}^N$ such that for all $x,y\in\Delta_N$,

$\sum_i x_i\,u_i \;<\; \sum_i y_i\,u_i\;\;\iff\;\; x\prec_* y\;\;\implies\;\;x\prec y.$

Unless $x\sim y$ for all $x$ and $y$, there are $i,j\in\{1,\dotsc,N\}$ such that $u_i\neq u_j$.

Proof. Let $i$ be a worst and $j$ a best outcome, i.e. let $i,j\in\{1,\dotsc,N\}$ be such that $e^i\preccurlyeq e^k\preccurlyeq e^j$ for all $k\in\{1,\dotsc,N\}$. If $e^i\sim e^j$, then $e^i \sim e^k$ for all $k$, and by repeated applications of independence we get $x\sim e^i\sim y$ for all $x,y\in\Delta_N$, and therefore $x\sim_* y$ again for all $x,y\in\Delta_N$, and we can simply choose $u=0$.

Thus, suppose that $e^i\prec e^j$. In this case, let $u$ be such that for every $k\in\{1,\dotsc,N\}$, $u_k$ equals the unique $p$ provided by Lemma 2 applied to $e^i\preccurlyeq e^k\preccurlyeq e^j$ and $e^i\prec e^j$. Because of Lemma 1, $u_i = 0 \neq 1 = u_j$. Let $f(r) := (1-r)e^i + re^j$.

We first show that $\textstyle p := \sum_k x_k\,u_k < \sum_k y_k\,u_k =: q$ implies $x\prec y$. For every $k$, we either have $u_k < 1$, in which case by Lemma 2 we have $e^k \prec f(u_k + \epsilon_k)$ for arbitrarily small $\epsilon_k > 0$, or we have $u_k = 1$, in which case we set $\epsilon_k := 0$ and find $e^k\preccurlyeq e^j = f(u_k + \epsilon_k)$. Set $\textstyle \epsilon := \sum_k x_k\,\epsilon_k$. Now, by independence applied $N-1$ times, we have $\textstyle x = \sum_k x_k\,e^k \preccurlyeq \sum_k x_k f(u_k + \epsilon_k) = f(p+\epsilon)$; analogously, we obtain $y \succcurlyeq f(q-\delta)$ for arbitrarily small $\delta > 0$. Thus, using $p and Lemma 1, $x\preccurlyeq f(p+\epsilon)\prec f(q-\delta)\preccurlyeq y$ and therefore $x\prec y$ as claimed. Now note that if $\textstyle\sum_k x_k\,u_k < \sum_k y_k\,u_k$, then this continues to hold for $x'$ and $y'$ in a sufficiently small neighbourhood of $x$ and $y$, and therefore we have $x\prec_* y$.

Now suppose that $\textstyle \sum_k x_k\,u_k \ge \sum_k y_k\,u_k$. Since we have $u_i = 0$ and $u_j = 1$, 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, $x'\succ y'$ by the preceding paragraph. But this implies that $x\not\prec_* y$, which completes the proof.$\square$

Corollary 5. $\preccurlyeq_*$ 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.$\square$

Corollary 6. $u$ is unique up to affine transformations.

Proof. Since $u$ is a VNM utility function for $\preccurlyeq_*$, this follows from the analogous result for that case.$\square$

Corollary 7. Unless $x\sim y$ for all $x,y\in\Delta_N$, for all $r\in\mathbb{R}$ the set $\textstyle \{x\in\Delta_N : \sum_i x_i\,u_i = r\}$ has lower dimension than $\Delta_N$ (i.e., it is the intersection of $\Delta_N$ with a lower-dimensional subspace of $\mathbb{R}^N$).

Proof. First, note that the assumption implies that $N\ge 2$. Let $v\in\mathbb{R}^N$ be given by $v_i = 1$, $\forall i$, and note that $\Delta_N$ is the intersection of the hyperplane $A := \{x\in\mathbb{R}^N : x\cdot v = 1\}$ with the closed positive orthant $\mathbb{R}^N_+$. By the theorem, $u$ is not parallel to $v$, so the hyperplane $B_r := \{x\in\mathbb{R}^N : x\cdot u = r\}$ is not parallel to $A$. It follows that $A\cap B_r$ has dimension $N-2$, and therefore $\textstyle\{x\in\Delta_N : \sum_i x_i\,u_i = r\} \;=\; A\cap B_r\cap\mathbb{R}^N_+$ can have at most this dimension. (It can have smaller dimension or be the empty set if $A\cap B_r$ only touches or lies entirely outside the positive orthant.)$\square$