<p>I take it that <span class="math-container">$u: \mathbb{R} \to \mathbb{R}$</span> and not <span class="math-container">$u: \mathbb{R}^N \to \mathbb{R}$</span> (as in the question). Otherwise <span class="math-container">$u(c)$</span> for <span class="math-container">$c \in \mathbb{R}$</span> does not make sense.</p>
<p>tldr:</p>
<ul>
<li>if <span class="math-container">$u$</span> is continuous, a certainty equivalence exists (if the mean utility is well defined)</li>
<li>if <span class="math-container">$u$</span> is strictly monotone, then the certainty equivalence (if it exists) must be unique.</li>
</ul>
<p>Let <span class="math-container">$P(x)$</span> be a probability distribution over <span class="math-container">$\mathbb{R}$</span>. Then the expected utility is given by
<span class="math-container">$$
\overline{u} = \int u(x) dP(x).
$$</span>
Let us assume that this is well defined. A certainty equivalence <span class="math-container">$c$</span> of the lottery satisfies <span class="math-container">$u(c) = \overline{u}$</span>.</p>
<p>Notice that at least for some <span class="math-container">$\underline{x} \in \mathbb{R}$</span> for which <span class="math-container">$u(\underline{x}) \le \overline{u}$</span>. If not then for all <span class="math-container">$x \in \mathbb{R}$</span>, <span class="math-container">$u(x) &gt; \overline{u}$</span> and
<span class="math-container">$$
\overline{u} = \int u(x) d P(x) &gt; \int \overline{u} d P(x) = \overline{u},
$$</span>
a contradiction. Similarly, one can show that there is an <span class="math-container">$\overline{x}$</span> such that <span class="math-container">$u(\overline{x}) \ge \overline{u}$</span>.</p>
<p>Define
<span class="math-container">$$
f(t) = u((1-t)\underline{x} + t \overline{x}) - \overline{u}.
$$</span>
Then if <span class="math-container">$u$</span> is continuous, <span class="math-container">$f$</span> is also continuous and we have that <span class="math-container">$f(0) \le 0$</span> and <span class="math-container">$f(1) \ge 0$</span>. By the intermediate value theorem, there exists a <span class="math-container">$t^\ast \in [0,1]$</span> for which <span class="math-container">$f(t^\ast) = 0$</span>. Then,
<span class="math-container">$$
u(c) = \overline{u}
$$</span>
where we defined <span class="math-container">$c = (1-t^\ast) \underline{x} + t^\ast \overline{x}$</span>.</p>
<p>This shows the existence of the certainty equivalence. So far, we did not use strict monotonicity.</p>
<p>Next, using strict monotonicity, one can easily show that the certainty equivalence is unique.</p>
<p>If not, there are at least two, say <span class="math-container">$c_1$</span> and <span class="math-container">$c_2$</span>. Assume without loss of generality that <span class="math-container">$c_1 &lt; c_2$</span>. As they are certainty equivalences, both should satisfy <span class="math-container">$u(c_1) = u(c_2) = \overline{u}$</span>. This is impossible as <span class="math-container">$u(c_1) &lt; u(c_2)$</span> by strict monotonicity.</p>

CC BY-SA 4.0
2023-09-06T13:52:36.827Z
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.
High GPA
CC BY-SA 4.0
2023-09-06T12:47:51.347Z
2023-09-06T12:47:51.347Z
tdm
<p>Here is an example that shows that certainty equivalents need not exist: Let <span class="math-container">$f:\mathbb{R}\to (0,1)$</span> be an increasing bijection. Let <span class="math-container">$0&lt;\alpha&lt; f(1)-f(0)$</span> Define <span class="math-container">$u:\mathbb{R}^\mathbb{N}\to\mathbb{R}$</span> by
<span class="math-container">$$u(x)=\alpha\sum 2^{-n} f(x_n)+\limsup_n f\big(n^{-1} x_n\big).$$</span>
This function is strictly increasing. The second term is <span class="math-container">$f(0)$</span> for every constant act but is <span class="math-container">$f(1)$</span> for the act <span class="math-container">$x$</span> given by <span class="math-container">$x_n=n$</span>. Since the first term is always less than <span class="math-container">$\alpha$</span>, <span class="math-container">$x$</span> 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.</p>
<p>An obvious sufficient condition for the existence of a certainty equivalent is that <span class="math-container">$u$</span> restricted to constant acts is continuous and neither bounded below nor above. Then, all possible utility values are taken by constant acts.</p>

CC BY-SA 4.0
2023-09-07T11:48:53.300Z
2023-09-07T12:01:34.070Z
2023-09-07T12:01:34.070Z
Michael Greinecker
<p>Let <span class="math-container">$R$</span> be the set of real number. Let <span class="math-container">$N$</span> be an infinite set. Let utility <span class="math-container">$u:R^N\to R$</span>. The utility function is strictly monotonic.</p>
<p>My question is, does the certainty equivalence <span class="math-container">$CE$</span> exist? Do we need additional assumptions?</p>
<p>For <span class="math-container">$x\in R^N$</span>, The definition of <span class="math-container">$CE$</span> is follow: if <span class="math-container">$u(x)=u(c)$</span> where <span class="math-container">$c\in R$</span>, then <span class="math-container">$c=CE(x)$</span>.</p>
<p>Note that <span class="math-container">$c$</span> also means constant act by the usual abusing of notations.</p>
<p>One thing I am sure is that, if we further assume <span class="math-container">$u$</span> is continuous on <span class="math-container">$R^N$</span>, then the <span class="math-container">$CE$</span> does exist. My question is, can we weaken the assumption? For example, let's assume <span class="math-container">$u$</span> is continuous <span class="math-container">$R$</span>.</p>

<microeconomics><decision-theory><choice-theory><uncertainty>
Minimal assumption for a “certainty equivalence” exists
High GPA
# Minimal assumption for a “certainty equivalence” exists

(Asked by: username_0 on 2023-09-06)

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 $R$.

**Tags:** microeconomics, decision-theory, choice-theory, uncertainty


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## 2 answers

### Answer by username_1 on 2023-09-06. Score: 3

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.

tldr:

* 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.

> **Comments:**
> **username_0**: 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.  

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### Answer by username_2 on 2023-09-07. Score: 3

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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