1
$\begingroup$

It is well-known that a rational, continuous and monotone preference relation $\succeq$ defined on $\mathbb{R}^L$ admits a utility representation.

I would like to understand why monotonicity is required. In other words, what is an example of a rational and continuous but non-monotone preference relation that does not have a utility representation?

$\endgroup$
  • 1
    $\begingroup$ I don't think monotonicity is required for utility representation, at least in Debreu’s representation theorem. $\endgroup$ – Herr K. Jan 14 at 20:05
  • $\begingroup$ Thank you, you are right. I had never seen this result without the monotonicity axiom, which is probably used for simplifying the proof. I'll accept this answer if you post it. $\endgroup$ – Oliv Jan 14 at 20:30
3
$\begingroup$

A necessary condition for a preference relation to be represented by a utility function is that the preference relation is rational (where a "utility function" is a real-valued function that assigns a higher or equal numerical value to bundle A than to bundle B, when A is weakly preferred to B).

Preferences are rational when they are complete (I can express my preferences for all conceivable bundles), and transitive (If I weakly prefer A to B and B to C, then I weakly prefer A to C).

Given rationality, a sufficient condition is that the rational preference relation is continuous, i.e. if it is preserved under limits.

Then, the related utility function is also continuous.

Monotonicity is a property where we go from quantities to preferences, after assuming desirability ("goods" rather than "bads"). Local nonsatiation is actually the weaker assumption that is required for most of the theory.

$\endgroup$
  • 1
    $\begingroup$ (+1) Two related results may be worth noting. First, if $\succsim$ is defined on a finite or countable set, then rationality of $\succsim$ alone is both necessary and sufficient for utility representation. Second, if $\succsim$ is defined on an uncountable set, then rationality and continuity together are necessary and sufficient for (continuous) utility representation. $\endgroup$ – Herr K. Jan 15 at 1:46

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.