<p>Let <span class="math-container">$\succeq$</span>  be a complete and transitive binary relation on the set of all lotteries on finite sets of prizes. Then <span class="math-container">$\succeq$</span> is mixture continuous if and only if it satisfies the Archimedean axiom and a condition called local mixture dominance.</p>
<p>They also have two examples in there.</p>
<p><a href="http://www.econ2.jhu.edu/people/karni/archimedean.pdf" rel="nofollow noreferrer">http://www.econ2.jhu.edu/people/karni/archimedean.pdf</a></p>

CC BY-SA 4.0
2023-10-19T06:24:29.460Z
2023-10-19T06:35:00.687Z
2023-10-19T06:35:00.687Z
Paul H.Y. Cheung
<p>In the context of preferences on a set of lotteries on a finite set <span class="math-container">$X$</span>, what is an example of a preference that is independent, Archimedean but <strong>not</strong> mixture continuous?</p>

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

<hr>

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

<ul>
<li><strong>Independent</strong>: if <span class="math-container">$\forall$</span> lotteries <span class="math-container">$\pi,\rho,\sigma$</span> and <span class="math-container">$\forall \alpha\in(0,1)$</span>:
<span class="math-container">$$
\pi\succ\rho\iff\alpha\pi+(1-\alpha)\sigma\succ \alpha\rho+(1-\alpha)\sigma,\\
\pi\sim\rho\iff\alpha\pi+(1-\alpha)\sigma\sim \alpha\rho+(1-\alpha)\sigma;
$$</span></li>
<li><strong>Archimedean</strong>: if <span class="math-container">$\forall$</span> lotteries <span class="math-container">$\pi,\rho,\sigma$</span>,
<span class="math-container">$$
\pi\succ\rho\succ\sigma\implies\exists\alpha,\beta\in(0,1): \alpha \pi+(1-\alpha)\sigma\succ\rho, \beta \pi+(1-\beta)\sigma\prec\rho
$$</span></li>
<li><strong>Mixture continuous</strong>: if for all <span class="math-container">$\pi,\rho,\sigma$</span>, the sets
<span class="math-container">$$
\{\alpha\in[0,1]:\alpha\pi+(1-\alpha)\rho\succeq\sigma\}\quad\text{and}\quad \{\alpha\in[0,1]:\alpha\pi+(1-\alpha)\rho\preceq\sigma\}
$$</span>
are closed in <span class="math-container">$[0,1]$</span></li>
</ul>

CC BY-SA 4.0
2018-11-16T02:47:49.393Z
Independence and Archimedean together imply (mixture) continuity. So you can't have the first two without the third. See Kreps (1988) for a proof.
Herr K.
CC BY-SA 4.0
2018-11-16T03:24:19.900Z
@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.
yurnero
CC BY-SA 4.0
2018-11-16T05:09:16.583Z
It's Lemma 5.6(b)
Herr K.
<preferences><uncertainty><choice-theory>
Archimedean but not mixture continuous
yurnero
# Archimedean but not mixture continuous

(Asked by: username_0 on 2018-11-15)

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 $[0,1]$

**Tags:** preferences, uncertainty, choice-theory
> **Comments:**
> **username_1**: Independence and Archimedean together imply (mixture) continuity. So you can't have the first two without the third. See Kreps (1988) for a proof.  
> **username_0**: @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.  
> **username_1**: It's Lemma 5.6(b)  

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## 1 answers

### Answer by username_2 on 2023-10-19. Score: 2

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.

<http://www.econ2.jhu.edu/people/karni/archimedean.pdf>


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