# Certainty equivalence when the utility is semi-continuous instead of continuous

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?

Clarification:

$$U(x_1,x_2)$$ 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"?

• @HerrK. I updated my answer based on your comment. I hope it is now clearer. Many thanks. Nov 24, 2020 at 0:09
• Yes it does make sense. Nov 24, 2020 at 4:43
• 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? Nov 24, 2020 at 5:22
• @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. Nov 24, 2020 at 5:50
• Your counterexample works; the answer is, indeed, no. Nov 24, 2020 at 23:01