# Quasi concavity of utility function

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 $$\alpha \in [0,1]$$

• Are $a$ and $\alpha$.the same? And is this really a utility function? Have you tried some simplifying monoton transformations? Nov 23 '19 at 6:29
• 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
Nov 23 '19 at 8:59
• 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. Nov 23 '19 at 9:33
• Seems to me that in the end you are left with $$F(-k)(-f(h(1-\alpha))^2)-\frac{1}{2}f(-k^2)(2F(h(1-\alpha))-1)^2$$ Nov 23 '19 at 9:34
• That’s true. I am stuck somewhere there
– Oleh
Nov 23 '19 at 9:35

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")


• 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
Nov 23 '19 at 23:15
• Added to the question body: I should prove quasi-concavity on the interval $\alpha \in [0,1]$
– Oleh
Nov 26 '19 at 14:18
• 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. Nov 27 '19 at 12:32

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.