<p>First, consider the preference relation $\succeq$ defined by
\begin{equation*}
x \succeq y \text{ if } x_1 + \cdots + x_n \geq y_1+\cdots+y_n
\end{equation*}</p>

<p>This preference relation satisfies your assumptions. Indeed, suppose that $x \succ y \succ z$. You can check that $y \sim \lambda x + (1-\lambda) z$ for a unique $\lambda$ defined by
\begin{equation*}
\lambda = \dfrac{y_1+\cdots+y_n-(z_1+\cdots+z_n)}{x_1+\cdots+x_n-(z_1+\cdots+z_n)} 
\end{equation*}</p>

<p>And these preferences obviously admit the utility representation $u(x)=x_1+\cdots+x_n$.</p>

<p>Now, consider the lexicographic preference relation $\succeq$ defined (I take $n=2$ for simplicity) by
\begin{equation*}
x \succ y \text{ if } (x_1&gt;y_1) \text{ or } (x_1=y_1,x_2&gt;y_2)
\end{equation*}
Lexicographic preferences have two important properties:</p>

<ul>
<li>$x \sim y$ if and only if $x=y$</li>
<li>they have no utility representation</li>
</ul>

<p>If $x=(1,1)$, $y=(0,1)$ and $z=(0,0)$, we have $x \succ y \succ z$. In addition, any $\lambda&gt;0$ is such that $\lambda x + (1-\lambda) z \succ y$. Thus assumption 2. is not valid, and the conclusion of the theorem (existence of a utility representation) is not valid either.</p>

CC BY-SA 3.0
2017-03-27T18:50:51.147Z
@HerrK. The contrapositive of assumption 1 is not $y \succ x \Rightarrow y>x$ but $y \succ x \Rightarrow y \not\leq x$, which is different (we are in $\mathbb{R}^2)$. Assumption 1 is indeed satisfied for lexicographic preferences.
Oliv
CC BY-SA 3.0
2017-03-27T19:21:47.467Z
You're right. My mistake.
Herr K.
CC BY-SA 3.0
2017-03-27T18:17:08.210Z
2017-03-27T18:17:08.210Z
Oliv
<p>Let $\succeq $ be a weak order in $\mathbb{R}^n$. If</p>

<ol>
<li>$ x\geq y \Rightarrow x\succeq y$</li>
<li>for any $x\succ y\succ z$ there exists a unique $\lambda \in (0,1)$ such that $y\sim \lambda x+(1-\lambda)z$, </li>
</ol>

<p>then there exists a value representation of $\succeq$.</p>

<p>Could anyone give me an example where the Value representation theorem(with out uncertainity) in the $n$ dimensional Euclidean space can be applied and a counter example that cannot be applied?</p>

CC BY-SA 3.0
2017-03-27T16:59:49.830Z
Can you please clarify condition 1)? Should we read $x \geq y \Rightarrow x \succeq y$? I am asking because your condition is trivially wrong for $x=y$.
Oliv
CC BY-SA 3.0
2017-03-27T19:25:09.003Z
Do we know (or care) what are the preferences when neither $x \geq y$ holds, nor $x \leq y$?
Alecos Papadopoulos
<microeconomics><preferences>
Preference relation value representation without uncertainity
user123
# Preference relation value representation without uncertainity

(Asked by: username_0 on 2017-03-27)

Let $\succeq $ be a weak order in $\mathbb{R}^n$. If

1. $ x\geq y \Rightarrow x\succeq y$
2. for any $x\succ y\succ z$ there exists a unique $\lambda \in (0,1)$ such that $y\sim \lambda x+(1-\lambda)z$,

then there exists a value representation of $\succeq$.

Could anyone give me an example where the Value representation theorem(with out uncertainity) in the $n$ dimensional Euclidean space can be applied and a counter example that cannot be applied?

**Tags:** microeconomics, preferences
> **Comments:**
> **username_1**: Can you please clarify condition 1)? Should we read $x \geq y \Rightarrow x \succeq y$? I am asking because your condition is trivially wrong for $x=y$.  
> **username_2**: Do we know (or care) what are the preferences when neither $x \geq y$ holds, nor $x \leq y$?  

---

## 1 answers

### Answer by username_1 on 2017-03-27. Score: 3 | Accepted answer

First, consider the preference relation $\succeq$ defined by
\begin{equation\*}
x \succeq y \text{ if } x\_1 + \cdots + x\_n \geq y\_1+\cdots+y\_n
\end{equation\*}

This preference relation satisfies your assumptions. Indeed, suppose that $x \succ y \succ z$. You can check that $y \sim \lambda x + (1-\lambda) z$ for a unique $\lambda$ defined by
\begin{equation\*}
\lambda = \dfrac{y\_1+\cdots+y\_n-(z\_1+\cdots+z\_n)}{x\_1+\cdots+x\_n-(z\_1+\cdots+z\_n)}
\end{equation\*}

And these preferences obviously admit the utility representation $u(x)=x\_1+\cdots+x\_n$.

Now, consider the lexicographic preference relation $\succeq$ defined (I take $n=2$ for simplicity) by
\begin{equation\*}
x \succ y \text{ if } (x\_1>y\_1) \text{ or } (x\_1=y\_1,x\_2>y\_2)
\end{equation\*}
Lexicographic preferences have two important properties:

* $x \sim y$ if and only if $x=y$
* they have no utility representation

If $x=(1,1)$, $y=(0,1)$ and $z=(0,0)$, we have $x \succ y \succ z$. In addition, any $\lambda>0$ is such that $\lambda x + (1-\lambda) z \succ y$. Thus assumption 2. is not valid, and the conclusion of the theorem (existence of a utility representation) is not valid either.

> **Comments:**
> **username_1**: @HerrK. The contrapositive of assumption 1 is not $y \succ x \Rightarrow y>x$ but $y \succ x \Rightarrow y \not\leq x$, which is different (we are in $\mathbb{R}^2)$. Assumption 1 is indeed satisfied for lexicographic preferences.  
> **username_3**: You're right. My mistake.  

---

