<p><em><strong>Note:</strong> &quot;risk averse&quot; implies <em>strict</em> concavity, I would not therefore allow for linear segments or saturation.</em></p>
<p><span class="math-container">$$u(40)=\frac{1}{2}[u(0)+u(100)] \implies u(0) = 2u(40) - u(100).$$</span></p>
<p>Insert into the expression for <span class="math-container">$m$</span>,</p>
<p><span class="math-container">$$u(m)=\frac{1}{2}\left[2u(40) - u(100)+u(180)\right] = u(40) + \frac{u(180)-u(100)}{2}.$$</span></p>
<p>Because we do not allow for saturation, we have that actually <span class="math-container">$m$</span> will be strictly higher than <span class="math-container">$40$</span>. But because we do not specify otherwise the utility function, the distance of <span class="math-container">$m$</span> from <span class="math-container">$40$</span> can be made arbitrarily small (note that strict concavity implies continuity). So <span class="math-container">$40$</span> is indeed the infimum (and see a comment below on how one could go about proving it formally). We already have that <span class="math-container">$m&lt;90$</span>, so the range <span class="math-container">$(40,90)$</span> is a correct although loose from above open range.</p>
<p>Next, by strict concavity,
<span class="math-container">$$u(180)-u(100) &lt; u(80),$$</span>
so
<span class="math-container">$$u(m)&lt;u(40) + \frac{u(80)}{2}.$$</span></p>
<p>This <em>appears</em> to say that <span class="math-container">$m$</span> is allowed to take the value <span class="math-container">$80$</span>, since <span class="math-container">$m=80$</span> satisfies the above inequality, due to the strict concavity of <span class="math-container">$u$</span>.</p>
<p>But as an insgihtful comment pointed out if we put <span class="math-container">$m=80$</span> we will get</p>
<p><span class="math-container">$$u(80) = u(40) + \frac{u(180)-u(100)}{2} \implies u(80) - u(40) = \frac{u(180)-u(100)}{2},$$</span></p>
<p>which is not compatible with strict concavity. And any higher value for <span class="math-container">$m$</span> is likewise rejected.</p>
<p>So, from this path, we get <span class="math-container">$m &lt;80.$</span></p>

CC BY-SA 4.0
2022-12-21T15:44:44.283Z
$m=80$ is not possible if $u$ is strictly concave. Reason is that if we put $m= 80$ in the equality $u(m) = u(40)+\frac{u(180)-u(100)}{2}$, we get $u(80) - u(40) = \frac{u(180)-u(100)}{2}$ which contradicts strict concavity. For strictly concave $u$, $m < 80$.
Amit
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2022-12-21T19:37:18.043Z
@Amit Correct, I will edit accordingly.
Alecos Papadopoulos
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2022-12-22T13:45:35.493Z
The infimum is $40$; that is not just "a loose lower bound."
Michael Greinecker
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2022-12-22T18:47:52.467Z
@MichaelGreinecker Hmm... so I guess we can formally prove that an infimum exists?
Alecos Papadopoulos
CC BY-SA 4.0
2022-12-22T19:01:13.680Z
@AlecosPapadopoulos That an infimum exists is trivial, to show that it is indeed 40 can be done by taking an example in which $u$ is only weakly increasing and weakly concave and approximating it by strictly increasing and concave functions. An explicit construction might be messy, but if you draw the picture, it is clear that it can be done.
Michael Greinecker
CC BY-SA 4.0
2022-12-25T15:37:39.663Z
Sry but I am considering when $u(0) \neq 0$, could m be larger than 80? It looks like u suppose u(0)=0 in your first inequality.
Foracustle
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2022-12-25T20:45:13.437Z
@Foracustle. No. the first 80 units provide utility higher than the 80 units after 100.
Alecos Papadopoulos
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2022-12-26T10:58:42.773Z
@Foracustle $u(0)=0$ is without loss of generality. In fact, you can consider any affine transformation ($v = au + b$). In other words, result does not depend on this assumption.
Amit
CC BY-SA 4.0
2022-12-21T13:32:14.217Z
2022-12-23T18:13:56.230Z
2022-12-23T18:13:56.230Z
Alecos Papadopoulos
<p>WLOG let <span class="math-container">$u(0)=0$</span> and <span class="math-container">$u(100)=60$</span>. Therefore, <span class="math-container">$u(40) = \dfrac{60}{2} = 30$</span>. By concavity of <span class="math-container">$u$</span>,
<span class="math-container">$\dfrac{u(180)-u(100)}{80} \leq \dfrac{u(100)-u(40)}{60}=\dfrac{1}{2}$</span>. This gives an upper-bound on <span class="math-container">$u(180)\leq 100$</span>. Consequently, we get an upper bound on <span class="math-container">$u(m) = \dfrac{u(180)}{2} \leq 50$</span>. This bound is achieved by the concave utility function <span class="math-container">$u(x)=\min\left(\frac{3}{4}x, 10+\frac{1}{2}x\right)$</span>. Therefore, <span class="math-container">$m\leq 80$</span>.</p>
<p><a href="https://i.sstatic.net/GELSa.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/GELSa.png" alt="enter image description here" /></a></p>
<hr />

CC BY-SA 4.0
2022-12-20T15:21:57.760Z
Imo 60 should be the lower bound for u(100). And I'm a little confused about why m cannot be greater than 80? Hope for more details since an example is not enough in math.
Foracustle
CC BY-SA 4.0
2022-12-21T06:59:13.627Z
Edited the proof. This one is clearer and shorter.
Amit
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2022-12-22T12:45:43.170Z
Could you, please, describe your answer in more detail? How do you obtain each intermediate step?
Athaeneus
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2022-12-22T13:48:34.093Z
Just think of the problem in this way: Find a concave function $u$ that passes through these three points $(0,0)$, $(40,30)$ and $(100, 60)$ in a such way that the value of $m$ satisfying $u(m)=u(180)/2$ is maximum.
Amit
CC BY-SA 4.0
2022-12-20T13:10:45.007Z
2022-12-21T07:04:30.143Z
2022-12-21T07:04:30.143Z
Amit
<p>The body of question is:</p>
<p>Assume the decision maker is risk averse, <span class="math-container">$u(40)=\frac{1}{2}(u(0)+u(100))$</span>, <span class="math-container">$u(m)=\frac{1}{2}(u(0)+u(180))$</span>, try to estimate the range of m.</p>
<p>It is easy to get the infimum of m: let u(180)=u(100) and <span class="math-container">$m \geq 40 $</span>. However, getting the supremum of m is a little complex since I don't know the exact utility function. I only know <span class="math-container">$90 \geq m $</span> since it is concave.</p>
<p>Here is my idea: in order to get the maximal value of m, we need to make the slope of u(x) keep constant when <span class="math-container">$x \geq m$</span>(in fact I don't know whether this speculation is true or not). Concavity guarantees the righ-hand and left-hand derivates of u(x) exist. Then I don't know how to continue.</p>
<p>If somewhere is not clear, plz tell me. Any advance on it would be highly appreciated.</p>

<microeconomics><mathematical-economics>
estimation of certainty equivalent without given utility function
Foracustle
# estimation of certainty equivalent without given utility function

(Asked by: username_0 on 2022-12-19)

The body of question is:

Assume the decision maker is risk averse, $u(40)=\frac{1}{2}(u(0)+u(100))$, $u(m)=\frac{1}{2}(u(0)+u(180))$, try to estimate the range of m.

It is easy to get the infimum of m: let u(180)=u(100) and $m \geq 40 $. However, getting the supremum of m is a little complex since I don't know the exact utility function. I only know $90 \geq m $ since it is concave.

Here is my idea: in order to get the maximal value of m, we need to make the slope of u(x) keep constant when $x \geq m$(in fact I don't know whether this speculation is true or not). Concavity guarantees the righ-hand and left-hand derivates of u(x) exist. Then I don't know how to continue.

If somewhere is not clear, plz tell me. Any advance on it would be highly appreciated.

**Tags:** microeconomics, mathematical-economics

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## 2 answers

### Answer by username_2 on 2022-12-20. Score: 8 | Accepted answer

WLOG let $u(0)=0$ and $u(100)=60$. Therefore, $u(40) = \dfrac{60}{2} = 30$. By concavity of $u$,
$\dfrac{u(180)-u(100)}{80} \leq \dfrac{u(100)-u(40)}{60}=\dfrac{1}{2}$. This gives an upper-bound on $u(180)\leq 100$. Consequently, we get an upper bound on $u(m) = \dfrac{u(180)}{2} \leq 50$. This bound is achieved by the concave utility function $u(x)=\min\left(\frac{3}{4}x, 10+\frac{1}{2}x\right)$. Therefore, $m\leq 80$.

[![enter image description here](https://i.sstatic.net/GELSa.png)](https://i.sstatic.net/GELSa.png)

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> **Comments:**
> **username_0**: Imo 60 should be the lower bound for u(100). And I'm a little confused about why m cannot be greater than 80? Hope for more details since an example is not enough in math.  
> **username_2**: Edited the proof. This one is clearer and shorter.  
> **username_4**: Could you, please, describe your answer in more detail? How do you obtain each intermediate step?  
> **username_2**: Just think of the problem in this way: Find a concave function $u$ that passes through these three points $(0,0)$, $(40,30)$ and $(100, 60)$ in a such way that the value of $m$ satisfying $u(m)=u(180)/2$ is maximum.  

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### Answer by username_1 on 2022-12-21. Score: 4

***Note:** "risk averse" implies *strict* concavity, I would not therefore allow for linear segments or saturation.*

$$u(40)=\frac{1}{2}[u(0)+u(100)] \implies u(0) = 2u(40) - u(100).$$

Insert into the expression for $m$,

$$u(m)=\frac{1}{2}\left[2u(40) - u(100)+u(180)\right] = u(40) + \frac{u(180)-u(100)}{2}.$$

Because we do not allow for saturation, we have that actually $m$ will be strictly higher than $40$. But because we do not specify otherwise the utility function, the distance of $m$ from $40$ can be made arbitrarily small (note that strict concavity implies continuity). So $40$ is indeed the infimum (and see a comment below on how one could go about proving it formally). We already have that $m<90$, so the range $(40,90)$ is a correct although loose from above open range.

Next, by strict concavity,
$$u(180)-u(100) < u(80),$$
so
$$u(m)

This *appears* to say that $m$ is allowed to take the value $80$, since $m=80$ satisfies the above inequality, due to the strict concavity of $u$.

But as an insgihtful comment pointed out if we put $m=80$ we will get

$$u(80) = u(40) + \frac{u(180)-u(100)}{2} \implies u(80) - u(40) = \frac{u(180)-u(100)}{2},$$

which is not compatible with strict concavity. And any higher value for $m$ is likewise rejected.

So, from this path, we get $m <80.$

> **Comments:**
> **username_2**: $m=80$ is not possible if $u$ is strictly concave. Reason is that if we put $m= 80$ in the equality $u(m) = u(40)+\frac{u(180)-u(100)}{2}$, we get $u(80) - u(40) = \frac{u(180)-u(100)}{2}$ which contradicts strict concavity. For strictly concave $u$, $m < 80$.  
> **username_1**: @Amit Correct, I will edit accordingly.  
> **username_3**: The infimum is $40$; that is not just "a loose lower bound."  
> **username_1**: @MichaelGreinecker Hmm... so I guess we can formally prove that an infimum exists?  
> **username_3**: @AlecosPapadopoulos That an infimum exists is trivial, to show that it is indeed 40 can be done by taking an example in which $u$ is only weakly increasing and weakly concave and approximating it by strictly increasing and concave functions. An explicit construction might be messy, but if you draw the picture, it is clear that it can be done.  
> **username_0**: Sry but I am considering when $u(0) \neq 0$, could m be larger than 80? It looks like u suppose u(0)=0 in your first inequality.  
> **username_1**: @Foracustle. No. the first 80 units provide utility higher than the 80 units after 100.  
> **username_2**: @Foracustle $u(0)=0$ is without loss of generality. In fact, you can consider any affine transformation ($v = au + b$). In other words, result does not depend on this assumption.  

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