<ol>
<li>It's well known that if <span class="math-container">$\succsim$</span> satisfies independence, then it is also <strong>convex</strong>.</li>
</ol>
<p>Since <span class="math-container">$\succsim$</span> satisfies independence,
<span class="math-container">$L\succsim L^{'} \iff \alpha L+(1-\alpha)L^{''}\succsim \alpha L^{'}+(1-\alpha)L^{''}$</span> for all <span class="math-container">$\alpha \in [0,1]$</span> and <span class="math-container">$ L, L^{'}, L^{''}\in \mathfrak{L} $</span></p>
<p>Convexity requires:</p>
<p><span class="math-container">$L\succsim L^{''}$</span> and  <span class="math-container">$L^{'}\succsim L^{''} \Longrightarrow \alpha L$</span> + <span class="math-container">$(1-\alpha)L^{'} \succsim L^{''}$</span> for all <span class="math-container">$\alpha \in \left[0,1\right]$</span>  and <span class="math-container">$ L, L^{'}, L^{''}\in \mathfrak{L} $</span></p>
<p>Pick any <span class="math-container">$ L, L^{'}, L^{''}\in \mathfrak{L} $</span>  with <span class="math-container">$L \succsim L^{''}$</span>  and <span class="math-container">$L^{'} \succsim L^{''}$</span>. By completeness, <span class="math-container">$L \succsim L^{'}$</span> or <span class="math-container">$L^{'} \succsim L$</span> or both. Without loss of generality, assume <span class="math-container">$L \succsim L^{'}$</span>. Then, by independence, for all <span class="math-container">$\alpha \in [0,1]:$</span></p>
<p><span class="math-container">$ \alpha L$</span> + <span class="math-container">$( 1- \alpha) L^{'} \succsim \alpha L^{'} + (1-\alpha)L^{'}  = L^{'} \succsim L^{''} $</span>, which we wanted to show.</p>
<ol start="2">
<li>Is it true if <span class="math-container">$\succsim$</span> are convex then they satisfy independence?</li>
</ol>
<p>On the other side, convexity does not imply independence. To see this, take a look
at Figure (see below). Triangles are indifference curves and arrows show the direction in which
utility increases. It is convex, but not independent.</p>
<p><a href="https://i.sstatic.net/omNGD.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/omNGD.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2020-11-30T16:57:09.723Z
2020-12-01T12:42:38.160Z
2020-12-01T12:42:38.160Z
Lorenzo Castagno
<p>Let´s say there is an uncertain situation with <span class="math-container">$N$</span> possible consequences <span class="math-container">$C = \{C_1, . . . C_N\}$</span>. Assume that
there is a rational preference relation <span class="math-container">$\succsim$</span>  over simple lotteries.</p>
<p>I know that if <span class="math-container">$\succsim$</span> satisfies independence, then it is also <strong>convex</strong>, but is it true if <span class="math-container">$\succsim$</span> are convex then they satisfy independence? How can I show this (if the implication is true)</p>

<microeconomics><game-theory><preferences><decision-theory><discrete-preferences>
If a rational preference relation over simple lotteries $\succsim$ are convex then they satisfy independence?
crosscut22
# If a rational preference relation over simple lotteries $\succsim$ are convex then they satisfy independence?

(Asked by: username_0 on 2020-11-30)

Let´s say there is an uncertain situation with $N$ possible consequences $C = \{C\_1, . . . C\_N\}$. Assume that
there is a rational preference relation $\succsim$ over simple lotteries.

I know that if $\succsim$ satisfies independence, then it is also **convex**, but is it true if $\succsim$ are convex then they satisfy independence? How can I show this (if the implication is true)

**Tags:** microeconomics, game-theory, preferences, decision-theory, discrete-preferences


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## 1 answers

### Answer by username_1 on 2020-11-30. Score: 7 | Accepted answer

1. It's well known that if $\succsim$ satisfies independence, then it is also **convex**.

Since $\succsim$ satisfies independence,
$L\succsim L^{'} \iff \alpha L+(1-\alpha)L^{''}\succsim \alpha L^{'}+(1-\alpha)L^{''}$ for all $\alpha \in [0,1]$ and $ L, L^{'}, L^{''}\in \mathfrak{L} $

Convexity requires:

$L\succsim L^{''}$ and $L^{'}\succsim L^{''} \Longrightarrow \alpha L$ + $(1-\alpha)L^{'} \succsim L^{''}$ for all $\alpha \in \left[0,1\right]$ and $ L, L^{'}, L^{''}\in \mathfrak{L} $

Pick any $ L, L^{'}, L^{''}\in \mathfrak{L} $ with $L \succsim L^{''}$ and $L^{'} \succsim L^{''}$. By completeness, $L \succsim L^{'}$ or $L^{'} \succsim L$ or both. Without loss of generality, assume $L \succsim L^{'}$. Then, by independence, for all $\alpha \in [0,1]:$

$ \alpha L$ + $( 1- \alpha) L^{'} \succsim \alpha L^{'} + (1-\alpha)L^{'} = L^{'} \succsim L^{''} $, which we wanted to show.

2. Is it true if $\succsim$ are convex then they satisfy independence?

On the other side, convexity does not imply independence. To see this, take a look
at Figure (see below). Triangles are indifference curves and arrows show the direction in which
utility increases. It is convex, but not independent.

[![enter image description here](https://i.sstatic.net/omNGD.png)](https://i.sstatic.net/omNGD.png)


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