1
$\begingroup$

Consider EU on monetary outcomes. Say we have a utility function $u:\mathbb R\to\mathbb R$

The common axioms of EU are continuity, independence and weak order.

These axioms do not imply that the utility function is continuous.

My question: how to reach a continuous utility?

Edit: the usual continuity conditions for EU are mixture continuity and Archimedean.

Improve this question
$\endgroup$
4
  • $\begingroup$ Expected Utility is linear is probabilities and therefore, it is continuous. $\endgroup$
    – Amit
    Commented Jul 12 at 9:03
  • $\begingroup$ This is a fairly technical issue. There is a nice discussion in the "Notes on Theory of Choice" by Kreps. $\endgroup$
    – Michael Greinecker
    Commented Jul 12 at 11:25
  • 1
    $\begingroup$ @Amit We are talking about the continuity of small u function. $\endgroup$
    – dodo
    Commented Jul 12 at 16:33
  • 1
    $\begingroup$ @dodo Thanks. Yes, I misinterpreted. $\endgroup$
    – Amit
    Commented Jul 12 at 22:02

1 Answer 1

Reset to default
1
$\begingroup$

I believe we get it from the axioms you have provided with a few technical conditions on the spaces we are working with.

Let $X$ be a separable, metrizable space (like your $\mathbb{R}$) and let $\Delta(X)$ be the set of countably additive probability distributions, endowed with the weak convergence topology.

The complete preorder $\succeq$ on $\Delta(X)$ satisfies Continuity and Independence if and only if it has an expected utility representation:

$$p \succeq q \quad \text{ if and only if } \quad \int_X u(x) dp \geq \int_X u(x)dq$$

where $u : X \rightarrow \mathbb{R}$ is continuous and bounded.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Hi, the usual continuity conditions for EU are mixture continuity and Archimedean. Could you please explain how is your continuity stronger than the mixture continuity? $\endgroup$
    – dodo
    Commented Jul 12 at 16:38
  • $\begingroup$ Its stronger in the sense that we require the preference to be continuous with respect to the weak convergence topology. i.e. $(p_n) \rightarrow p$ if for every bounded and continuous $f$, $\int f(x) dp_n \rightarrow \int f(x) dp$. $\endgroup$
    – Walrasian Auctioneer
    Commented Jul 12 at 17:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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