Let $R$ be the set of real number. Let $N$ be an infinite set. Let utility $u:R^N\to R$. The utility function is strictly monotonic.

My question is, does the certainty equivalence $CE$ exist? Do we need additional assumptions?

For $x\in R^N$, The definition of $CE$ is follow: if $u(x)=u(c)$ where $c\in R$, then $c=CE(x)$.

Note that $c$ also means constant act by the usual abusing of notations.

One thing I am sure is that, if we further assume $u$ is continuous on $R^N$, then the $CE$ does exist. My question is, can we weaken the assumption? For example, let's assume $u$ is continuous $R$.


2 Answers 2


I take it that $u: \mathbb{R} \to \mathbb{R}$ and not $u: \mathbb{R}^N \to \mathbb{R}$ (as in the question). Otherwise $u(c)$ for $c \in \mathbb{R}$ does not make sense.


  • if $u$ is continuous, a certainty equivalence exists (if the mean utility is well defined)
  • if $u$ is strictly monotone, then the certainty equivalence (if it exists) must be unique.

Let $P(x)$ be a probability distribution over $\mathbb{R}$. Then the expected utility is given by $$ \overline{u} = \int u(x) dP(x). $$ Let us assume that this is well defined. A certainty equivalence $c$ of the lottery satisfies $u(c) = \overline{u}$.

Notice that at least for some $\underline{x} \in \mathbb{R}$ for which $u(\underline{x}) \le \overline{u}$. If not then for all $x \in \mathbb{R}$, $u(x) > \overline{u}$ and $$ \overline{u} = \int u(x) d P(x) > \int \overline{u} d P(x) = \overline{u}, $$ a contradiction. Similarly, one can show that there is an $\overline{x}$ such that $u(\overline{x}) \ge \overline{u}$.

Define $$ f(t) = u((1-t)\underline{x} + t \overline{x}) - \overline{u}. $$ Then if $u$ is continuous, $f$ is also continuous and we have that $f(0) \le 0$ and $f(1) \ge 0$. By the intermediate value theorem, there exists a $t^\ast \in [0,1]$ for which $f(t^\ast) = 0$. Then, $$ u(c) = \overline{u} $$ where we defined $c = (1-t^\ast) \underline{x} + t^\ast \overline{x}$.

This shows the existence of the certainty equivalence. So far, we did not use strict monotonicity.

Next, using strict monotonicity, one can easily show that the certainty equivalence is unique.

If not, there are at least two, say $c_1$ and $c_2$. Assume without loss of generality that $c_1 < c_2$. As they are certainty equivalences, both should satisfy $u(c_1) = u(c_2) = \overline{u}$. This is impossible as $u(c_1) < u(c_2)$ by strict monotonicity.

  • $\begingroup$ It is usual that $c\in R$ also mean the constant act $(c,c,...)$. So $u(c)$ still make sense. But still super appreciate your work. $\endgroup$
    – High GPA
    Sep 6, 2023 at 13:52

Here is an example that shows that certainty equivalents need not exist: Let $f:\mathbb{R}\to (0,1)$ be an increasing bijection. Let $0<\alpha< f(1)-f(0)$ Define $u:\mathbb{R}^\mathbb{N}\to\mathbb{R}$ by $$u(x)=\alpha\sum 2^{-n} f(x_n)+\limsup_n f\big(n^{-1} x_n\big).$$ This function is strictly increasing. The second term is $f(0)$ for every constant act but is $f(1)$ for the act $x$ given by $x_n=n$. Since the first term is always less than $\alpha$, $x$ has no certainty equivalent. The utility function is not continuous in the product topology, but the restriction to constant acts (or even to all bounded acts) is continuous.

An obvious sufficient condition for the existence of a certainty equivalent is that $u$ restricted to constant acts is continuous and neither bounded below nor above. Then, all possible utility values are taken by constant acts.


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