<p>Let <span class="math-container">$U:\mathbb R^2\to\mathbb R$</span> be a utility function.</p>
<p>If <span class="math-container">$U$</span> is strictly increasing and continuous, then it is well known that for any <span class="math-container">$(x_1,x_2)$</span> <strong>there exists a certainty</strong> <span class="math-container">$(c,c)$</span> such that <span class="math-container">$$U(x_1,x_2)=U(c,c).$$</span></p>
<p>If we assume that <span class="math-container">$U$</span> is strictly increasing and upper semi continuous, <strong>can we still find a certainty equivalence</strong> <span class="math-container">$(c,c)$</span> or not?</p>
<hr />
<p><strong>Clarification</strong>:</p>
<p><span class="math-container">$U(x_1,x_2)$</span> 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.</p>
<p>For example, see: <a href="http://www.columbia.edu/%7Emd3405/BE_Risk_4_17.pdf" rel="nofollow noreferrer">http://www.columbia.edu/~md3405/BE_Risk_4_17.pdf</a></p>
<hr />
<p><strong>My try:</strong></p>
<p>Of course one important example of <span class="math-container">$U$</span> is the additive representation (subjective expected utility): <span class="math-container">$U=\sum_ip_iu(x_i)$</span>.</p>
<p>Consider a special case: let <span class="math-container">$p_1=p_2=0.5$</span>; let <span class="math-container">$u$</span> be upper-semi continuous, for example <span class="math-container">$u(a)=a$</span> when <span class="math-container">$a&lt;1$</span> and <span class="math-container">$u(a)=a+3$</span> when <span class="math-container">$a\geq 1$</span>.</p>
<p>In this case, <span class="math-container">$U(0,1)=2$</span>.</p>
<p><span class="math-container">$U(c,c)=c$</span> when <span class="math-container">$c&lt;1$</span>; <span class="math-container">$U(c,c)=c+3$</span> when <span class="math-container">$c\geq 1$</span>. None of the certainty has a utility of 2.</p>
<p>So the answer seems to be &quot;NO&quot;?</p>

CC BY-SA 4.0
2020-11-23T17:18:44.527Z
Did you mean certainty equivalent, as in a $c\in\mathbb R$ such that $u(c)=\sum_ip_ix_i$?
Herr K.
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2020-11-24T00:09:22.760Z
@HerrK. I updated my answer based on your comment. I hope it is now clearer. Many thanks.
High GPA
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2020-11-24T01:13:19.363Z
Looks like you've answered your own question. I guess you're also implicitly requiring that $U$ is *not* simultaneously lower semicontinuous.
Herr K.
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2020-11-24T03:21:59.410Z
@HerrK. Does my answer make sense?
High GPA
CC BY-SA 4.0
2020-11-24T04:43:46.473Z
Yes it does make sense.
Herr K.
CC BY-SA 4.0
2020-11-24T05:22:30.370Z
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?
Dayne
CC BY-SA 4.0
2020-11-24T05:50:38.577Z
@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.
Herr K.
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2020-11-24T07:22:28.147Z
@HerrK.: Yeah! Perhaps we can request the OP to edit the question to further clarify this?
Dayne
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2020-11-24T08:19:48.743Z
@Dayne $U$ is not necessarily expected utility. It could be other utility functions, as clarified in the question.
High GPA
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2020-11-24T08:43:58.970Z
Okay. So just for further clarification, $x_i$ represents a particular good $i$ or payoff from state $i$?
Dayne
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2020-11-24T13:02:39.417Z
@Dayne Those two make no difference mathematically I guess; for example, Arrow Debreu use the same framework for those two cases: consumer choice and Arrow-Debreu state-contingent claims
High GPA
CC BY-SA 4.0
2020-11-24T15:22:23.613Z
I guess utility function is allowed much more freedom in choice of functions than E(U) as a function of states. In any case, mathematical equivalence, I think, is not enough. Unless certainty equivalence is defined for utility functions (as opposed to expected utility), this is confusing.
Dayne
CC BY-SA 4.0
2020-11-24T17:45:54.350Z
Like Dayne says, there is a significant difference. An EU utility is defined as a linear function on $U: \Delta(X) \rightarrow \mathbb{R}$. What does strictly increasing mean in such a context? If $X = \{0,1\}$, then each lottery is of the form $(p,1-p)$. How can we even compare them with respect to a set order?

On the other hand if we think of the consumption space as $\mathbb{R}^2$, then a utility function $U: \mathbb{R}^2 \rightarrow \mathbb{R}$ makes sense. Please clarify your question to the exact setting you are interested in.
Walrasian Auctioneer
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2020-11-24T23:01:15.317Z
Your counterexample works; the answer is, indeed, no.
Michael Greinecker
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2020-11-25T01:28:50.103Z
@WalrasianAuctioneer There is no mathematical difference between: "$x_i$  represents a particular good i or payoff from state i", as I replied to the "Okay" comment. Of course there is a difference between SEU and other utility functions. Dayne's "Okay" comment did not mentioned SEU or EU.
High GPA
<microeconomics><mathematical-economics><consumer-theory><decision-theory><theory>
Certainty equivalence when the utility is semi-continuous instead of continuous
High GPA
# Certainty equivalence when the utility is semi-continuous instead of continuous

(Asked by: username_0 on 2020-11-23)

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](http://www.columbia.edu/%7Emd3405/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"?

**Tags:** microeconomics, mathematical-economics, consumer-theory, decision-theory, theory

> **Comments:**
> **username_1**: Did you mean certainty equivalent, as in a $c\in\mathbb R$ such that $u(c)=\sum_ip_ix_i$?  
> **username_0**: @HerrK. I updated my answer based on your comment. I hope it is now clearer. Many thanks.  
> **username_1**: Looks like you've answered your own question. I guess you're also implicitly requiring that $U$ is *not* simultaneously lower semicontinuous.  
> **username_0**: @HerrK. Does my answer make sense?  
> **username_1**: Yes it does make sense.  
> **username_2**: 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?  
> **username_1**: @username_2: 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.  
> **username_2**: @HerrK.: Yeah! Perhaps we can request the OP to edit the question to further clarify this?  
> **username_0**: @username_2 $U$ is not necessarily expected utility. It could be other utility functions, as clarified in the question.  
> **username_2**: Okay. So just for further clarification, $x_i$ represents a particular good $i$ or payoff from state $i$?  
> **username_0**: @username_2 Those two make no difference mathematically I guess; for example, Arrow Debreu use the same framework for those two cases: consumer choice and Arrow-Debreu state-contingent claims  
> **username_2**: I guess utility function is allowed much more freedom in choice of functions than E(U) as a function of states. In any case, mathematical equivalence, I think, is not enough. Unless certainty equivalence is defined for utility functions (as opposed to expected utility), this is confusing.  
> **username_3**: Like username_2 says, there is a significant difference. An EU utility is defined as a linear function on $U: \Delta(X) \rightarrow \mathbb{R}$. What does strictly increasing mean in such a context? If $X = \{0,1\}$, then each lottery is of the form $(p,1-p)$. How can we even compare them with respect to a set order?  On the other hand if we think of the consumption space as $\mathbb{R}^2$, then a utility function $U: \mathbb{R}^2 \rightarrow \mathbb{R}$ makes sense. Please clarify your question to the exact setting you are interested in.  
> **username_4**: Your counterexample works; the answer is, indeed, no.  
> **username_0**: @WalrasianAuctioneer There is no mathematical difference between: "$x_i$  represents a particular good i or payoff from state i", as I replied to the "Okay" comment. Of course there is a difference between SEU and other utility functions. username_2's "Okay" comment did not mentioned SEU or EU.  

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## 0 answers

