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?

  • 1
    $\begingroup$ 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 $x=y$. $\endgroup$ – Oliv Mar 27 '17 at 16:59
  • 1
    $\begingroup$ Do we know (or care) what are the preferences when neither $x \geq y$ holds, nor $x \leq y$? $\endgroup$ – Alecos Papadopoulos Mar 27 '17 at 19:25

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.

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    $\begingroup$ @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. $\endgroup$ – Oliv Mar 27 '17 at 18:50
  • $\begingroup$ You're right. My mistake. $\endgroup$ – Herr K. Mar 27 '17 at 19:21

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