Strict dominance and symmetry

Abstract

The strict dominance principle that a wager always paying better than another is rationally preferable is one of the least controversial principles in decision theory. I shall show that (given the Axiom of Choice) there is a contradiction between strict dominance and plausible isomorphism or symmetry conditions, by showing how in several natural cases one can construct isomorphic wagers one of which strictly dominates the other. In particular, I will show that there is a pair of wagers on the outcomes of a uniform spinner which differ simply in where the zero degrees point of the spinner is defined to be but where one wager dominates the other. I shall also argue that someone who accepts Williamson’s famous argument that the probability of an infinite sequence of heads is zero should accept the symmetry conditions, and thus has reason to weaken the strict dominance principle, and I shall propose a restriction of the principle to “implementable” wagers. Our main result also has implications for social choice principles.

Appendix: Some proofs

Proof of Proposition 1

We will show (i)$\rightarrow$(iii)$\rightarrow$(iv)$\rightarrow$(ii)$\rightarrow$(i).

That (i) implies (iii) is clear: if we had $V^g$ strictly dominating V and yet $\precapprox$ was a G-invariant preference structure satisfying strict dominance, then by strict dominance we would have $V\prec V^g$, contrary to G-invariance.

Now we show that (iii) implies (iv). Assume (iv) is false. Thus there is a symmetry g such that $g^n \omega \ne \omega$ whenever $n>0$ and $\omega \in \Omega$. More generally, it follows that if $g^n \omega = g^m \omega$, then $n=m$. For, otherwise, we could suppose without loss of generality that $n>m$ and we would have $g^{n-m} \omega = \omega$.

Let $a\sim b$ if and only if $a=g^n b$ for some integer n. Then $\sim$ is an equivalence relation, and we can let E contain exactly one element from each equivalence class by Choice. Given $a\in \Omega$, let $b\in E$ be such that $a=g^n b$ for some integer n. The integer n is unique, since otherwise we would have $g^n b = g^m b$ for distinct n and m. Let $U(a)=n$, observe that $U^g(a)=U(a)+1$ and, as before, let $V(a)=\phi (U(a))$ for a bounded strictly increasing $\phi$. Then $V^g$ strictly dominates V, and we have not-(iii).

Now we show (iv) implies (ii). Assume (iv). Say that $V \precapprox W$ just in case there is a $g\in G$ such that $V(\omega ) \le W^g(\omega )$ for all $\omega \in \Omega$. Then $\precapprox$ is a G-invariant partial preorder, and it clearly satisfies the Principle of Non-Strict Dominance. We now show that it satisfies Strict Dominance as well. For suppose it does not, so that there are $W_1$ and $W_2$ such that $W_1 < W_2$ everywhere but not $W_1 \prec W_2$. By Non-Strict Dominance, we have $W_1 \precapprox W_2$. Thus, for $W_1 \prec W_2$ to fail, we must also have $W_2 \precapprox W_1$. Hence there is a $g\in G$ such that $W_2 \le W_1^g$ everywhere. Therefore, $W_2 < W_2^g$ everywhere, since $W_1 < W_2$ everywhere. By (iv), there is a positive n and an $\omega$ such that $g^n\omega = \omega$. We then have $W_2(\omega )< W_2(g\omega )< W_2(g^2\omega )< \dots < W_2(g^n\omega )$, which contradicts $g^n\omega =\omega$.

However, $\precapprox$ may be only a partial preorder, and we need a total one. To that end, let [W] be the equivalence class of the wager W under the relation $\approx$ (where, recall, $W_1\approx W_2$ if and only if $W_1\precapprox W_2$ and $W_2\precapprox W_1$). Define the partial order $\preceq$ on these equivalence classes by stipulating that $[W]\preceq [V]$ if and only if $W\precapprox V$ (this is well defined because $\precapprox$ is transitive). By the Szpilrajn order extension theorem (Szpilrajn, 1930) (this uses the Axiom of Choice), we can extend $\preceq$ to a total order $\preceq ^*$. Now define $W_1\precapprox ^* W_2$ if and only if $[W_1]\preceq ^* [W_2]$. Then $\precapprox ^*$ is easily seen to be a total preorder that extends $\precapprox$. It is strongly G-invariant because it extends the strongly G-invariant preorder $\precapprox$. It remains to check that $\precapprox ^*$ satisfies the Principles of Non-Strict and Strict Dominance. Non-Strict Dominance follows from the fact that $\precapprox ^*$ extends $\precapprox$ and the latter satisfies Non-Strict Dominance. That leaves the Strict case. Suppose $W_2$ strictly dominates $W_1$. By Strict Dominance for $\precapprox$, we have $W_1 \prec W_2$. Therefore, $[W_1] \preceq [W_2]$ and not $[W_2] \preceq [W_1]$. Hence $[W_1] \preceq ^* [W_2]$, since $\preceq ^*$ extends $\preceq$. Since we do not have $[W_2] \preceq [W_1]$, we have $[W_2]\ne [W_1]$, and since $\preceq ^*$ is an order, and not merely a preorder, it follows that we do not have $[W_2] \preceq ^* [W_1]$. Thus, we have $W_1 \precapprox ^* W_2$ but not $W_2 \precapprox ^* W_1$.

Thus, (iv) implies (ii).

Finally, (ii) trivially implies (i), which completes the proof. $\square$