Ordinal dominance and risk aversion

Abstract

We find that, for sufficiently risk-averse agents, strict dominance by pure or mixed actions coincides with dominance by pure actions in the sense of (Börgers in Econometrica 61(2):423–430, 1993), which, in turn, coincides with the classical notion of strict dominance by pure actions when preferences are asymmetric. Since risk aversion is a cardinal feature, all finite single-agent choice problems with ordinal preferences admit compatible utility functions which are sufficiently risk averse as to achieve equivalence between pure and mixed dominance. This result extends to some infinite environments.

A Proofs

Proof

(Proposition 1) Fix an action $a\in A$ and a state $x\in X$, and let $u$ and $v$ be compatible vNM utility functions such that

$$\begin{aligned} \{ \alpha \in \Delta (A) \,|\, u(\alpha ,x) \ge u(a,x)\} \subseteq \{ \alpha \in \Delta (A) \,|\, v(\alpha ,x) \ge v(a,x)\}. \end{aligned}$$

(11)

We want to show that $\tau _u(a,x)\ge \tau _v(a,x)$. If $a$ is either ${\mathbin \succcurlyeq }_x$-maximum or ${\mathbin \succcurlyeq }_x$-minimum, then $\tau _u(a,x)=+\infty$ and $\tau _v(a,x)=+\infty$ by definition, and the result is trivial. Hence, we assume for the rest of the proof that $\underline{u}(x)<u(a,x)<\bar{u}(x)$.

By Assumption 1, there exists an action $b\in A$ such that $u(b,x) = u^-(a,x)$, and, consequently, $v(b,x) = v^-(a,x)$. Let $(a_m)$ be a sequence of actions such that $a_m{\mathbin \succ }_x a$ for all $m$, $\lim _{m\rightarrow \infty } u(a_m,x) = \bar{u}(x)$, and $\lim _{m\rightarrow \infty } v(a_m,x) = \bar{v}(x)$. Also, for each $\theta \in [0,1]$ and each $m\in {\mathbb {N}}$, let $\alpha _{m,\theta }$ be the mixed action that plays $a_m$ with probability $\theta$, and $b$ with probability $1-\theta$. For all such $m$ we have that $u(\alpha _{m,1},x) > u(a,x) > u(\alpha _{m,0},x)$. Hence, since expected utility is continuous in the mixing probabilities, there exists $\theta (m)\in (0,1)$ such that $u(\alpha _{m,\theta (m)},x)= u(a,x)$. After some simple algebra this implies that:

$$\begin{aligned} \dfrac{u(a,x)-u^-(a,x)}{u(a_m,x) - u(a,x)} = \dfrac{\theta (m)}{1-\theta (m)}. \end{aligned}$$

(12)

By (11), we have that $v(\alpha _{m,\theta (m)},x) \ge v(a,x)$, which implies that:

$$\begin{aligned} \dfrac{v(a,x)-v^-(a,x)}{v(a_m,x) - v(a,x)} \le \dfrac{\theta (m)}{1-\theta (m)}. \end{aligned}$$

(13)

Using (12) and (13) and taking limits as $m$ goes to infinity thus yield the desired result

$$\begin{aligned} \tau _v(a,x) \!=\! \lim _{m\rightarrow \infty } \dfrac{v(a,x)-v^-(a,x)}{v(a_m,x) - v(a,x)} \!\le \! \lim _{m\rightarrow \infty } \dfrac{u(a,x)-u^-(a,x)}{u(a_m,x) - u(a,x)} \!=\! \tau _u(a,x).\qquad \end{aligned}$$

(14)

$\square$

Proof

(Lemma 1) Let $\beta = \alpha (W_x(a))$. Since $u(b,x)\le u^-(a,x)$ for $b\in W_x(a)$, and $u(b,x)\le \bar{u}(x)$ for $b\in B$, it follows that:

$$\begin{aligned} u(\alpha ,x) - u(a,x)&\le \beta \big (u^-(a,x) - u(a,x)\big ) + \big (1-\beta \big ) \big ( \bar{u}(x) - u(a,x)\big ) \nonumber \\&= -\beta \left( \dfrac{u^-(a,x) - u(a,x)}{\bar{u}(x) - u(a,x) } \right) \big (\bar{u}(x) - u(a,x)\big )\nonumber \\&+ \big (1-\beta \big ) \big (\bar{u}(x) - u(a,x)\big ) \nonumber \\&= \big (1-\beta \cdot (\tau _u(a,x)+1)\big ) \big ( \bar{u}(x) - u(a,x)\big ) \le 0. \end{aligned}$$

(15)

$\square$

Proof

(Proposition 2) Fix a set $B\subseteq A$, an action $a \in A\backslash P(B)$, and a mixture $\alpha$ with $\alpha (B\backslash \{a\})=1$. There exists some $Y\subseteq X$ conditional on which $a$ is not weakly dominated in $B$. Assume without loss of generality that for all $b \in B\setminus \{a\}$ there exists some $x\in Y$ such that $b\not \sim _x a$. This implies that for all $b \in B\backslash \{a\}$ there also exists some $x\in Y$ such that $a{\mathbin \succ }_x b$, i.e., $B\backslash \{a\} \subseteq \cup _{x\in Y} W_x(a)$. Since $K = \min \{\Vert A\Vert ,\Vert X\Vert \}<+\infty$, there exists a finite subset $Z = \{x_1,\ldots ,x_k\}\subseteq Y$ with cardinality $k\le K$, and such that $B\backslash \{a\} \subseteq \cup _{x\in Z} W_x(a)$. Therefore:

$$\begin{aligned} \sum _{x\in Z} \alpha (W_x(a)) \ge \alpha (B\backslash \{a\}) = 1 \ge \dfrac{k}{K} = \sum _{x\in Z} \dfrac{1}{K} \ge \sum _{x\in Z} \dfrac{1}{\tau _u(a,x) + 1}. \end{aligned}$$

(16)

This implies that there exists a state $x$ such that $(\tau _u(a,x) + 1)\alpha (W_x(a))\ge 1$, and the result thus follows from Lemma 1. $\square$

Proof

(Proposition 3) Let $a$, $B$, $\alpha ,$ and $Y$ be as in the proof of Proposition 2. As before, we know that $B\setminus \{a\} \subseteq \cup _{x\in Y} W_x(a,B)$, and thus:

$$\begin{aligned} \sum _{x\in Y} \alpha (W_x(a,B)) \ge 1 \ge \sum _{x\in X} \dfrac{1}{1+\tau _u(a,x)} \ge \sum _{x\in Y} \dfrac{1}{1+\tau _u(a,x)}. \end{aligned}$$

(17)

Hence, there exists $x\in Y$ such that $(\tau _u(a,x) + 1)\alpha (W_x(a))\ge 1$, and the result follows from Lemma 1. $\square$