<p>Let <span class="math-container">$X$</span> be a convex subset of <span class="math-container">$\mathbb{R}^n$</span>. We define</p>
<p>C1: <span class="math-container">$\forall x, y, z\in X$</span>, if <span class="math-container">$x\succeq y$</span> and <span class="math-container">$z$</span> is a convex combination of <span class="math-container">$x$</span> and <span class="math-container">$y$</span>, then <span class="math-container">$z\succeq y$</span>.</p>
<p>C2: <span class="math-container">$\forall x, y, z\in X$</span>, if <span class="math-container">$x\succ y$</span> and <span class="math-container">$z\neq y$</span> is a convex combination of <span class="math-container">$x$</span> and <span class="math-container">$y$</span>, then <span class="math-container">$z\succ y$</span>.</p>
<p><strong>Proposition</strong>. C1 does not imply C2.</p>
<p><strong>Proof</strong>. Consider <span class="math-container">$X=[0,2)$</span>, define <span class="math-container">$\succeq$</span> on <span class="math-container">$X$</span> as follows
<span class="math-container">$x\succeq y$</span> if and only if <span class="math-container">$\lfloor x\rfloor\geq \lfloor y\rfloor$</span>. In other words, <span class="math-container">$\succeq$</span> is represented by the utility function <span class="math-container">$u:[0,2)\rightarrow\mathbb{R}$</span> defined as <span class="math-container">$u(x)=\lfloor x\rfloor = \begin{cases} 0 &amp; \text{if } x \in [0,1) \\ 1 &amp; \text{if } x \in [1,2) \end{cases}$</span>.
Clearly, <span class="math-container">$\succeq$</span> satisfy C1 but not C2. To see that it does not satisfy C2, consider <span class="math-container">$x=1.5$</span> and <span class="math-container">$y=0$</span> such that <span class="math-container">$x\succ y$</span>, now <span class="math-container">$z=0.5$</span> is a convex combination of <span class="math-container">$x$</span> and <span class="math-container">$y$</span>, but it is not the case that <span class="math-container">$z \succ y$</span>. Therefore, C2 is not satisfied.</p>

CC BY-SA 4.0
2024-08-22T21:30:48.547Z
Thank you so much!
RHyp
CC BY-SA 4.0
2024-08-22T08:12:52.657Z
2024-08-22T10:26:49.807Z
2024-08-22T10:26:49.807Z
Amit
<p>I was looking into convexity while reading Varian's intermediate micro textbook, and an article by Richter and Rubinstein said, &quot;The canonical definition of convex preferences requires that if a is preferred to b, then any convex combination of a and b is also preferred to b.&quot; (I interpret &quot;preferred&quot; in the strong sense in this statement, although that might be wrong.) I looked at Kreps' textbook, and he defined it slightly differently: a preference relation is convex if for all <span class="math-container">$x$</span> and <span class="math-container">$y$</span>, if <span class="math-container">$x\succeq y$</span> and <span class="math-container">$z$</span> is a convex combination of <span class="math-container">$x$</span> and <span class="math-container">$y$</span>, then <span class="math-container">$z\succeq y$</span>. I am trying to prove that Kreps' definition implies Richter and Rubinstein's. (As I think about this, though, I'm not sure whether this is true given my understanding of the latter.)</p>
<p>For the forwards direction, let <span class="math-container">$x\succ y$</span> and <span class="math-container">$z$</span> be a convex combination of <span class="math-container">$x$</span> and <span class="math-container">$y$</span>. Because <span class="math-container">$x\succeq y$</span> and <span class="math-container">$\succeq$</span> is convex, <span class="math-container">$z\succeq y$</span>. Now, I need to prove that the statement <span class="math-container">$y\succeq z$</span> is false, which is where I am stuck. I tried a proof by contradiction by trying to get <span class="math-container">$x$</span> as a convex combination of <span class="math-container">$y$</span> and <span class="math-container">$z$</span>, but that didn't get me anywhere.</p>

<microeconomics><preferences>
Convexity and Strict Preferences
RHyp
# Convexity and Strict Preferences

(Asked by: username_0 on 2024-08-22)

I was looking into convexity while reading Varian's intermediate micro textbook, and an article by Richter and Rubinstein said, "The canonical definition of convex preferences requires that if a is preferred to b, then any convex combination of a and b is also preferred to b." (I interpret "preferred" in the strong sense in this statement, although that might be wrong.) I looked at Kreps' textbook, and he defined it slightly differently: a preference relation is convex if for all $x$ and $y$, if $x\succeq y$ and $z$ is a convex combination of $x$ and $y$, then $z\succeq y$. I am trying to prove that Kreps' definition implies Richter and Rubinstein's. (As I think about this, though, I'm not sure whether this is true given my understanding of the latter.)

For the forwards direction, let $x\succ y$ and $z$ be a convex combination of $x$ and $y$. Because $x\succeq y$ and $\succeq$ is convex, $z\succeq y$. Now, I need to prove that the statement $y\succeq z$ is false, which is where I am stuck. I tried a proof by contradiction by trying to get $x$ as a convex combination of $y$ and $z$, but that didn't get me anywhere.

**Tags:** microeconomics, preferences


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## 1 answers

### Answer by username_1 on 2024-08-22. Score: 1 | Accepted answer

Let $X$ be a convex subset of $\mathbb{R}^n$. We define

C1: $\forall x, y, z\in X$, if $x\succeq y$ and $z$ is a convex combination of $x$ and $y$, then $z\succeq y$.

C2: $\forall x, y, z\in X$, if $x\succ y$ and $z\neq y$ is a convex combination of $x$ and $y$, then $z\succ y$.

**Proposition**. C1 does not imply C2.

**Proof**. Consider $X=[0,2)$, define $\succeq$ on $X$ as follows
$x\succeq y$ if and only if $\lfloor x\rfloor\geq \lfloor y\rfloor$. In other words, $\succeq$ is represented by the utility function $u:[0,2)\rightarrow\mathbb{R}$ defined as $u(x)=\lfloor x\rfloor = \begin{cases} 0 & \text{if } x \in [0,1) \\ 1 & \text{if } x \in [1,2) \end{cases}$.
Clearly, $\succeq$ satisfy C1 but not C2. To see that it does not satisfy C2, consider $x=1.5$ and $y=0$ such that $x\succ y$, now $z=0.5$ is a convex combination of $x$ and $y$, but it is not the case that $z \succ y$. Therefore, C2 is not satisfied.

> **Comments:**
> **username_0**: Thank you so much!  

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