Let $U:\mathbb R^2\to\mathbb R$ be a utility function.

If $U$ is strictly increasing and continuous, then it is well known that for any $(x_1,x_2)$ there exists a certainty $(c,c)$ such that $$U(x_1,x_2)=U(c,c).$$

If we assume that $U$ is strictly increasing and upper semi continuous, can we still find a certainty equivalence $(c,c)$ or not?


$U(x_1,x_2)$ is not necessarily expected utility (EU). It could be other utility models under uncertainty, such as the max-min expected utility, choquet expected utility, and others.

For example, see: http://www.columbia.edu/~md3405/BE_Risk_4_17.pdf

My try:

Of course one important example of $U$ is the additive representation (subjective expected utility): $U=\sum_ip_iu(x_i)$.

Consider a special case: let $p_1=p_2=0.5$; let $u$ be upper-semi continuous, for example $u(a)=a$ when $a<1$ and $u(a)=a+3$ when $a\geq 1$.

In this case, $U(0,1)=2$.

$U(c,c)=c$ when $c<1$; $U(c,c)=c+3$ when $c\geq 1$. None of the certainty has a utility of 2.

So the answer seems to be "NO"?

  • 1
    $\begingroup$ @HerrK. I updated my answer based on your comment. I hope it is now clearer. Many thanks. $\endgroup$ – High GPA Nov 24 '20 at 0:09
  • 1
    $\begingroup$ Yes it does make sense. $\endgroup$ – Herr K. Nov 24 '20 at 4:43
  • 1
    $\begingroup$ I am sorry why are you defining certainty like this. It seems like $x_1,x_2$ are different goods in the utility function. Shouldn't it be $c$ such that $E(u(x))=u(c)$, where $c, x$ are lotteries? $\endgroup$ – Dayne Nov 24 '20 at 5:22
  • 1
    $\begingroup$ @Dayne: That threw me off at first, but I realized after OP's clarification that $U(x_1,x_2)$ is an expected utility where $x_i$ is the payoff in state $i$. So $U(c,c)$ is the EU where one gets the same payoff $c$ in both states. I guess maybe OP wanted to allow for the possibility that the probability of a state somehow depends on the amount of payoff one gets in it, e.g. $U(x_1,x_2)=p(x_1)u(x_1)+(1-p(x_1))u(x_2)$, where $p(\cdot)$ is a payoff-dependent probability function. But ultimately, it's an incarnation of $E(u(x))=u(c)$ as you pointed out. $\endgroup$ – Herr K. Nov 24 '20 at 5:50
  • 2
    $\begingroup$ Your counterexample works; the answer is, indeed, no. $\endgroup$ – Michael Greinecker Nov 24 '20 at 23:01

Your Answer

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

Browse other questions tagged or ask your own question.