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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.

Foracustle
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2 Answers2

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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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Amit
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  • 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 Dec 20 '22 at 15:21
  • Edited the proof. This one is clearer and shorter. – Amit Dec 21 '22 at 06:59
  • Could you, please, describe your answer in more detail? How do you obtain each intermediate step? – Athaeneus Dec 22 '22 at 12:45
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    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 Dec 22 '22 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 $m <80.$

Alecos Papadopoulos
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    $m=80$ 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 Dec 21 '22 at 15:44
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    @Amit Correct, I will edit accordingly. – Alecos Papadopoulos Dec 21 '22 at 19:37
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    The infimum is $40$; that is not just "a loose lower bound." – Michael Greinecker Dec 22 '22 at 13:45
  • @MichaelGreinecker Hmm... so I guess we can formally prove that an infimum exists? – Alecos Papadopoulos Dec 22 '22 at 18:47
  • @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 Dec 22 '22 at 19:01
  • 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 Dec 25 '22 at 15:37
  • @Foracustle. No. the first 80 units provide utility higher than the 80 units after 100. – Alecos Papadopoulos Dec 25 '22 at 20:45
  • @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 Dec 26 '22 at 10:58