<p>After a while, I have not found conditions or proof the function is quasi-concave at <span class="math-container">$\alpha \in [0,1]$</span>.</p>

<p>However, the ultimate goal was to show that there is an intermediate solution to the equation. It was indeed possible to condition on the second derivatives w.r.t. to <span class="math-container">$\alpha$</span> and <span class="math-container">$h$</span> under which: 
- the function equals to zero and increases at <span class="math-container">$0$</span> 
- the function equals to 0 and decreases at <span class="math-container">$1$</span>.  </p>

<p>Those conditions grant that there is at least one inflection point somewhere in <span class="math-container">$[0,1]$</span></p>

<p>I can provide more details if somebody is interested.</p>

CC BY-SA 4.0
2019-12-08T18:27:03.443Z
2019-12-08T18:27:03.443Z
Oleh
<p>Your conjecture seems to be contradicted, at least for small values of <span class="math-container">$\sigma$</span>. 
You can draw the function with the following R-code:  </p>

<pre><code>qq_f = function(x,k,h,sig){   
-pnorm(-k, sd=sig)*( (dnorm(h*(1-x), sd=sig))^2 ) - 0.5*dnorm(-k^2, sd=sig)*( 2*pnorm(h*(1-x), sd=sig) -1 )^2  
}  
curve(qq_f(x,k=0,h=1,sig=0.5),col='blue',xlim=c(-1,3),type='l',main="A candidate quasi-concave function")
</code></pre>

<p><a href="https://i.sstatic.net/Bxa7A.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Bxa7A.png" alt="enter image description here"></a></p>

CC BY-SA 4.0
2019-11-23T23:15:48.333Z
The function seems to be quasi-concave only for sufficiently high variance. However, I am still looking for analytical form of constituons under witch it is always true.
Oleh
CC BY-SA 4.0
2019-11-26T14:18:59.927Z
Added to the question body: I should prove quasi-concavity on the interval $\alpha \in [0,1]$
Oleh
CC BY-SA 4.0
2019-11-27T12:32:39.387Z
Your function and objective are interesting. Both $f$ and $F$ are quasi-concave in $-\alpha$, but I am unable to find a general result regarding their nonlinear transformation, given that even a linear combination of two quasi-concave functions is not necessarily quasi-concave.
Bertrand
CC BY-SA 4.0
2019-11-23T20:10:22.533Z
2019-11-23T21:14:07.253Z
2019-11-23T21:14:07.253Z
Bertrand
<p>I am looking for a method to prove that the following function is quasi-concave in <span class="math-container">$\alpha$</span>  (or find conditions under which it is true):</p>

<p><span class="math-container">$ \pi=F(-k)(f(0)^2-f(h(1-\alpha))^2)+ \frac{1}{2}-\frac{1}{2}f(0)^2-\frac{1}{2}f(-k^2)(2F(h(1-\alpha))-1)^2 $</span>,</p>

<p>where:</p>

<p><span class="math-container">$F$</span> - is a CDF function of Normal Distribution, <span class="math-container">$N[0,\sigma^2]$</span>;</p>

<p><span class="math-container">$f$</span> - is a PDF function of Normal Distribution, <span class="math-container">$N[0,\sigma^2]$</span>;</p>

<p><span class="math-container">$k$</span> - exogenous patameter, k <span class="math-container">$\in [0,+∞)$</span>;</p>

<p><span class="math-container">$h$</span> - exogenous patameter, h <span class="math-container">$\in [0,+∞)$</span>.</p>

<p>Any help would be appreciated.</p>

<p><strong>UPD</strong></p>

<p><strong>I have forgotten to mention that I am trying to prove quasi-concavity on the interval <span class="math-container">$\alpha \in [0,1]$</span></strong></p>

CC BY-SA 4.0
2019-11-23T06:29:11.543Z
Are $a$ and $ \alpha$.the same?  And is this really a utility function?  Have you tried some simplifying monoton transformations?
Giskard
CC BY-SA 4.0
2019-11-23T08:59:52.710Z
They are indeed the same, edited the question. Also, it is indeed a utility function of a multi-stage game already solved for optimal effort levels. I have tried dividing by constants, but have not done any particularly useful in this direction. Am I missing something obvious?
Oleh
CC BY-SA 4.0
2019-11-23T09:33:18.760Z
I don't know the final answer, but getting rid of all things that are constant in $\alpha$, such as $+ \frac{1}{2}-\frac{1}{2}f(0)^2$ would be a good start.
Giskard
CC BY-SA 4.0
2019-11-23T09:34:14.953Z
Seems to me that in the end you are left with
$$
F(-k)(-f(h(1-\alpha))^2)-\frac{1}{2}f(-k^2)(2F(h(1-\alpha))-1)^2
$$
Giskard
CC BY-SA 4.0
2019-11-23T09:35:49.250Z
That’s true. I am stuck somewhere there
Oleh
<microeconomics><game-theory><industrial-organisation>
Quasi concavity of utility function
Oleh
# Quasi concavity of utility function

(Asked by: username_0 on 2019-11-23)

I am looking for a method to prove that the following function is quasi-concave in $\alpha$ (or find conditions under which it is true):

$ \pi=F(-k)(f(0)^2-f(h(1-\alpha))^2)+ \frac{1}{2}-\frac{1}{2}f(0)^2-\frac{1}{2}f(-k^2)(2F(h(1-\alpha))-1)^2 $,

where:

$F$ - is a CDF function of Normal Distribution, $N[0,\sigma^2]$;

$f$ - is a PDF function of Normal Distribution, $N[0,\sigma^2]$;

$k$ - exogenous patameter, k $\in [0,+∞)$;

$h$ - exogenous patameter, h $\in [0,+∞)$.

Any help would be appreciated.

**UPD**

**I have forgotten to mention that I am trying to prove quasi-concavity on the interval $\alpha \in [0,1]$**

**Tags:** microeconomics, game-theory, industrial-organisation

> **Comments:**
> **username_1**: Are $a$ and $ \alpha$.the same?  And is this really a utility function?  Have you tried some simplifying monoton transformations?  
> **username_0**: They are indeed the same, edited the question. Also, it is indeed a utility function of a multi-stage game already solved for optimal effort levels. I have tried dividing by constants, but have not done any particularly useful in this direction. Am I missing something obvious?  
> **username_1**: I don't know the final answer, but getting rid of all things that are constant in $\alpha$, such as $+ \frac{1}{2}-\frac{1}{2}f(0)^2$ would be a good start.  
> **username_1**: Seems to me that in the end you are left with $$ F(-k)(-f(h(1-\alpha))^2)-\frac{1}{2}f(-k^2)(2F(h(1-\alpha))-1)^2 $$  
> **username_0**: That’s true. I am stuck somewhere there  

---

## 2 answers

### Answer by username_0 on 2019-12-08. Score: 0 | Accepted answer

After a while, I have not found conditions or proof the function is quasi-concave at $\alpha \in [0,1]$.

However, the ultimate goal was to show that there is an intermediate solution to the equation. It was indeed possible to condition on the second derivatives w.r.t. to $\alpha$ and $h$ under which:
- the function equals to zero and increases at $0$
- the function equals to 0 and decreases at $1$.

Those conditions grant that there is at least one inflection point somewhere in $[0,1]$

I can provide more details if somebody is interested.


---

### Answer by username_2 on 2019-11-23. Score: 3

Your conjecture seems to be contradicted, at least for small values of $\sigma$. 
You can draw the function with the following R-code:  
qq_f = function(x,k,h,sig){   
-pnorm(-k, sd=sig)*( (dnorm(h*(1-x), sd=sig))^2 ) - 0.5*dnorm(-k^2, sd=sig)*( 2*pnorm(h*(1-x), sd=sig) -1 )^2  
}  
curve(qq_f(x,k=0,h=1,sig=0.5),col='blue',xlim=c(-1,3),type='l',main="A candidate quasi-concave function")

> **Comments:**
> **username_0**: The function seems to be quasi-concave only for sufficiently high variance. However, I am still looking for analytical form of constituons under witch it is always true.  
> **username_0**: Added to the question body: I should prove quasi-concavity on the interval $\alpha \in [0,1]$  
> **username_2**: Your function and objective are interesting. Both $f$ and $F$ are quasi-concave in $-\alpha$, but I am unable to find a general result regarding their nonlinear transformation, given that even a linear combination of two quasi-concave functions is not necessarily quasi-concave.  

---

