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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.

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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, $m\leq 80$.

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  • $\begingroup$ 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. $\endgroup$
    – Foracustle
    Dec 20, 2022 at 15:21
  • $\begingroup$ Edited the proof. This one is clearer and shorter. $\endgroup$
    – Amit
    Dec 21, 2022 at 6:59
  • $\begingroup$ Could you, please, describe your answer in more detail? How do you obtain each intermediate step? $\endgroup$
    – Athaeneus
    Dec 22, 2022 at 12:45
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    $\begingroup$ 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. $\endgroup$
    – Amit
    Dec 22, 2022 at 13:48
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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 $$u(m)<u(40) + \frac{u(80)}{2}.$$

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 $m <80.$

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    $\begingroup$ $m=80$ 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 $u$, $m < 80$. $\endgroup$
    – Amit
    Dec 21, 2022 at 15:44
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    $\begingroup$ @Amit Correct, I will edit accordingly. $\endgroup$
    – Alecos Papadopoulos
    Dec 21, 2022 at 19:37
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    $\begingroup$ The infimum is $40$; that is not just "a loose lower bound." $\endgroup$
    – Michael Greinecker
    Dec 22, 2022 at 13:45
  • $\begingroup$ @MichaelGreinecker Hmm... so I guess we can formally prove that an infimum exists? $\endgroup$
    – Alecos Papadopoulos
    Dec 22, 2022 at 18:47
  • $\begingroup$ @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. $\endgroup$
    – Michael Greinecker
    Dec 22, 2022 at 19:01

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