In the context of preferences on a set of lotteries on a finite set $X$, what is an example of a preference that is independent, Archimedean but not mixture continuous?

I know the mixture continuous implies Archimedean but cannot think of an example where the reverse implication fails.

For completeness, I provide here the definitions of the terms above. Suppose $X=\{x_1,\ldots,x_n\}$ is finite. Let $\Delta X=\{\pi\in\mathbb{R}^n:\sum_{i=1}^n\pi_i=1, \pi_i\geq 0\}$. A preference $\succeq$ on the elements of $\Delta X$ is:

  • Independent: if $\forall$ lotteries $\pi,\rho,\sigma$ and $\forall \alpha\in(0,1)$: $$ \pi\succ\rho\iff\alpha\pi+(1-\alpha)\sigma\succ \alpha\rho+(1-\alpha)\sigma,\\ \pi\sim\rho\iff\alpha\pi+(1-\alpha)\sigma\sim \alpha\rho+(1-\alpha)\sigma; $$
  • Archimedean: if $\forall$ lotteries $\pi,\rho,\sigma$, $$ \pi\succ\rho\succ\sigma\implies\exists\alpha,\beta\in(0,1): \alpha \pi+(1-\alpha)\sigma\succ\rho, \beta \pi+(1-\beta)\sigma\prec\rho $$
  • Mixture continuous: if for all $\pi,\rho,\sigma$, the sets $$ \{\alpha\in[0,1]:\alpha\pi+(1-\alpha)\rho\succeq\sigma\}\quad\text{and}\quad \{\alpha\in[0,1]:\alpha\pi+(1-\alpha)\rho\preceq\sigma\} $$ are closed in $[0,1]$
  • $\begingroup$ Independence and Archimedean together imply (mixture) continuity. So you can't have the first two without the third. See Kreps (1988) for a proof. $\endgroup$
    – Herr K.
    Nov 16, 2018 at 2:47
  • $\begingroup$ @HerrK. I haven't used that book in the past but on your comment, I've looked at its Chapter 5. But I cannot find the result you refer to. $\endgroup$
    – yurnero
    Nov 16, 2018 at 3:24
  • 1
    $\begingroup$ It's Lemma 5.6(b) $\endgroup$
    – Herr K.
    Nov 16, 2018 at 5:09

1 Answer 1


Let $\succeq$ be a complete and transitive binary relation on the set of all lotteries on finite sets of prizes. Then $\succeq$ is mixture continuous if and only if it satisfies the Archimedean axiom and a condition called local mixture dominance.

They also have two examples in there.



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.