1
$\begingroup$

Setting: We have two choices of goods $(x_1,y_1)$ and $(x_2,y_2)$ from the set of choices $[-1,1]^2$. Moreover, we have the following preference relation $$(x_1,y_1)\mathcal{R}(x_2,y_2)\iff |x_1|\geq|x_2|\>\>\text{or}\>\> |y_1|\geq|y_2|$$

Question: We have to check if there exists a utility function reprensation of this preference relation.

My attempt: So from what I have learned, we know that a preference relation admits a utility function representation if it is rational (reflexive, complete, transitive) and continuous. I have found that this preference relation is not transitive, but this does not mean that there does not exist a utility function representation, because the aforementioned statement is not an if and only if statement.

Moreover, I thought we could try to derive a contradiction from the fact that if there exists a utility function $u$ representation of the preference relation, then we have $$(x_1,y_1)\mathcal{R}(x_2,y_2)\iff u(x_1,y_1)\geq u(x_2,y_2)$$ I tried to use the fact that the relation is not transitive to derive a contradiction by using the statement above, but was unsuccessful.

Sadly, these are the two main theorems/propositions that I've learned to solve these problems.

Any help is appreciated!

$\endgroup$
2
$\begingroup$

You just need to use the violation of transitivity and proceed by contradiction.

Suppose you have that $(x_1,y_1)R(x_2,y_2)$ and $(x_2,y_2)R(x_3,y_3)$ and a utility function, $u:[-1,1]^2\rightarrow \mathbb{R}$, exists, then $u(x_1,y_1)\geq u(x_2,y_2)$ (these are two reals) and $u(x_2,y_2)\geq u(x_3,y_3)$. (another two reals). Since the reals are transitive, we conclude $u(x_1,y_1)\geq u(x_3,y_3)$ which in turn implies that $(x_1,y_1)R(x_3,y_3)$. However this is a contradiction (if you choose carefully your three bunddles).

$\endgroup$
2
$\begingroup$

Transitivity and completeness are actually necessary for the existence of a utility representation. Whenever you prove that preferences fail to be complete or transitive you can conclude that they do not admit a utility function. For finite choice sets $X$, transitivity and completeness are necessary and sufficient, see Theorem 5 here.

Contrary to what you suggest, continuity is not itself necessary, but it is almost necessary. What you need is a condition called separability. Say that $\succcurlyeq$ is separable if there exists a countable set $Z \subseteq X$ such that for every $x,y\in X$ there exists some $z\in Z$ such that $x\succcurlyeq z \succcurlyeq y$.

Theorem A preference relation $\succcurlyeq$ on $X$ admits a utility representation if and only if it is complete, transitive, and separable.

This is actually a very old result by Cantor that precedes the well-known result by Debreu—which assumes continuity. Cantor's result was brought into Economics by Kreps. You can find a proof here, Theorem 9.

$\endgroup$
  • 1
    $\begingroup$ "Whenever you prove that preferences fail to be continuous or transitive..." Did you mean "complete or transitive"? $\endgroup$ – Herr K. Aug 6 at 17:12
  • 1
    $\begingroup$ Let me suggest the edit above. $\endgroup$ – Bayesian Aug 6 at 18:20

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.