7
$\begingroup$

It seems like lexicographic isn't that "special". Like yes it is special in that supposing it has a utility function gives you a bijection from the rationals to the reals, but I mean unique in some sense. I know that it satisfies a bunch of nice properties, but it still feels like there should be plenty of other (rational) preference relations that don't have utility representations.

$\endgroup$
4
$\begingroup$

Yes, there are many. Here are some examples :

Consider the weak preference relation $\succsim$ defined over $\mathbb{R}^2$ as:

  • Example 1

$(x_1, y_1) \succsim (x_2, y_2)$

if and only if

either ($x_1+y_1 > x_2 + y_2$) or ($x_1+y_1 = x_2 + y_2$ and $x_1 \geq x_2$)

Claim : $\succsim$ cannot be represented by a utility function.

Proof : Suppose by contradiction that there existed a utility function $u$ representing these preferences. For each $a > 0$, we have $(a, 0) \succ (0, a)$, and therefore, $u(a, 0) > u(0, a)$. We can therefore assign to $a$ a non-degenerate interval of values satisfying the above inequality $I(a) = [u(0, a), u(a, 0)]$. For any $a > b > 0$, all commodity bundles generating utilities in the interval $I(a)$ are strictly preferred to those in the disjoint interval $I(b)$ and should therefore be assigned a greater utility level. Then in each of these intervals we can pick a distinct rational number in increasing order to represent preferences. Since $a \in \mathbb{R_{++}}$, there are uncountably many such intervals, but set of rational numbers are countable. This results in a contradiction.

  • Example 2

$(x_1, y_1) \succsim (x_2, y_2)$

if and only if

either ($\min(x_1,y_1) > \min(x_2,y_2)$) or ($\min(x_1,y_1) = \min(x_2,y_2)$ and $x_1+y_1 \geq x_2 + y_2$)

Claim : $\succsim$ cannot be represented by a utility function.

Proof : Suppose by contradiction that there existed a utility function $u$ representing these preferences. For each $a > 0$, we have $(a+1, a) \succ (a, a)$, and therefore, $u(a+1, a) > u(a, a)$. We can therefore assign to $a$ a non-degenerate interval of values satisfying the above inequality $I(a) = [u(a, a), u(a+1, a)]$. For any $a > b > 0$, all commodity bundles generating utilities in the interval $I(a)$ are strictly preferred to those in the disjoint interval $I(b)$ and should therefore be assigned a greater utility level. Then in each of these interval we can pick a distinct rational number in increasing order to represent preferences. Since $a \in \mathbb{R_{++}}$, there are uncountably many such intervals, but set of rational numbers are countable. This results in a contradiction.

$\endgroup$
  • 1
    $\begingroup$ @Giskard, I've updated example 1 with the proof that there is no utility representation. $\endgroup$ – Amit Oct 23 '19 at 21:46
  • $\begingroup$ arent these just variations of lexiographic preference relations? $\endgroup$ – EconJohn Oct 24 '19 at 4:37
  • 1
    $\begingroup$ Yes, these can be thought of as variations, but they are not called Lexicographic. $\endgroup$ – Amit Oct 24 '19 at 4:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.