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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 $\alpha \in [0,1]$

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    $\begingroup$ Are $a$ and $ \alpha$.the same? And is this really a utility function? Have you tried some simplifying monoton transformations? $\endgroup$
    – Giskard
    Commented Nov 23, 2019 at 6:29
  • $\begingroup$ 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? $\endgroup$
    – Oleh
    Commented Nov 23, 2019 at 8:59
  • $\begingroup$ 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. $\endgroup$
    – Giskard
    Commented Nov 23, 2019 at 9:33
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    $\begingroup$ Seems to me that in the end you are left with $$ F(-k)(-f(h(1-\alpha))^2)-\frac{1}{2}f(-k^2)(2F(h(1-\alpha))-1)^2 $$ $\endgroup$
    – Giskard
    Commented Nov 23, 2019 at 9:34
  • $\begingroup$ That’s true. I am stuck somewhere there $\endgroup$
    – Oleh
    Commented Nov 23, 2019 at 9:35

2 Answers 2

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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")

enter image description here

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  • $\begingroup$ 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. $\endgroup$
    – Oleh
    Commented Nov 23, 2019 at 23:15
  • $\begingroup$ Added to the question body: I should prove quasi-concavity on the interval $\alpha \in [0,1]$ $\endgroup$
    – Oleh
    Commented Nov 26, 2019 at 14:18
  • $\begingroup$ 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. $\endgroup$
    – Bertrand
    Commented Nov 27, 2019 at 12:32
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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.

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