# Proving quasi-concavity of a certain function

1. May 6, 2013

### quantum_2000

1. The problem statement, all variables and given/known data

I need to prove that the function

(1-$\Phi$(x))(x-k)

is quasi-concave (i.e. first increasing and then decreasing), where $\Phi$ is the cdf of the normal distribution and k is a positive constant

2. Relevant equations

3. The attempt at a solution

I plotted the function and even tried to disprove it by simulation and the function is always quasi-concave, but so far have not been able to prove it.

The main issue is the cdf of the normal distribution, which can only be expressed using the Erf function, and this is where I get stuck.
Any help will be appreciated!

Q

Last edited by a moderator: May 7, 2013
2. May 7, 2013

### vela

Staff Emeritus

3. May 7, 2013

### quantum_2000

The first derivative of the function is

-$\phi$(x)(x-k)+(1-$\Phi$(x))

the first term is always negative, the second always positive, so to prove quasi-concavity is would be sufficient to show that the first term increases in magnitude and the second decreases as x goes up (this would actually prove concavity). The latter is of course true as the cdf is always an increasing function, but the former is not. So this way leads nowhere.
Writing the first derivative in detail I get

-$\frac{1}{\sqrt{2\pi}\sigma}$e$^{-\frac{(x-\mu)}{2\sigma^{2}}}$(x-k)+$\frac{1}{2}$-$\frac{1}{\sqrt{\pi}}$$\int$$^{\frac{x}{\sqrt{2}}}_{0}$e$^{-t^{2}}$dt

and I am stuck here, I don't know how to attack this beast. Going to second derivative is not going to help since the function is not concave, of this I am sure. I should prove that, as x increases, the first derivative as a "turning point" so that it is positive before that point and once it becomes negative it never gets positive again. I have litterally no idea how to prove this. One way would be to take x$_{1}$<x$_{2}$, impose that the first derivative calculated at x$_{1}$ is negative, that the first derivative calculated at x$_{2}$ is positive, and then show that this is not possible by contradiction. But again, once I write down these conditions I don't know how to proceed. I can't take the difference of the equations and show that it is monotone, for example, since what matters is not monotonicity of the first derivative - again, I am not trying to prove concavity.
An equivalent problem to solve is to show that once

$\frac{1}{\sqrt{2\pi}\sigma}$e$^{-\frac{(x-\mu)}{2\sigma^{2}}}$(x-k)+$\frac{1}{\sqrt{\pi}}$$\int$$^{\frac{x}{\sqrt{2}}}_{0}$e$^{-t^{2}}$dt

goes above 1/2 it stays there. Easy, right? :)

Any help, even on how quasi-concavity is typically proven, would be very much appreciated. Thank you in advance

4. May 7, 2013

### Ray Vickson

Looking at properties and behavior of the second derivative helps a lot in understanding more about the behavior of the first derivative.

5. May 9, 2013

### quantum_2000

True in general, and even the third derivative can help, but I do not see how it can be of help here. Please bring your ideas if you have any in this direction.

---

I came up with a (sufficient) set of conditions to show q-concavity, but it's rather complex. I put it down in case it helps you come up with other, simpler ideas.

1. Showing that the second derivative is negative until a certain x, and then becomes positive for all higher x's (which is what I observe graphically); this could be done by showing that:
1a. The second derivative is continuous;
1b. There exists a unique $\widehat{x}$ such that the third derivative is zero;
1c. The second derivative is positive for x1 and negative for x2 where x1<x2;

2. Showing that the integral of the second derivative from $\widehat{x}$ to +∞ is less than the first derivative calculated at $\widehat{x}$.

6. May 9, 2013

### Ray Vickson

what you say you observe graphically is incorrect; I observe something different graphically.

Anyway, if $\phi(x) = \exp(-x^2/2)/\sqrt{2 \pi}$ and $\Phi^c(x) = 1 - \Phi(x)$ your function is $F(x) = (x-k) \Phi^c(x).$ We have
$$F'(x) = \Phi^c(x) - (x-k) \phi(x),\\ F''(x) = \phi(x) [x(x-k)-2].$$
Note that $F''(x) < 0$ for x between 0 and k and $F''(x) > 0$ for large |x|. Thus, there exist $x_1(k) < x_2(k)$ such that $F''(x) > 0$ for $x < x_1(k)$ and $x > x_2(k)$, while $F''(x) < 0$ for $x_1(k) < x < x_2(k)$. Therefore, F is convex-concave-convex. Of course, $x_1(k), x_2(k)$ are just roots of a simple quadratic.

If you do further manipulations you can get what you need, but I cannot give further hints at this point.

7. May 10, 2013

### quantum_2000

Thank you for your suggestions. I realized that I did not mention that x is constrained to be grater than k, this is why I observed a concave-convex function... my mistake.

I could finally prove it. What I did was to look at the first order condition and show that it has a unique solution if $\frac{\Phi^c-k}{\phi}$ decreases in x, and this is the case because I know that $\frac{\Phi^c}{\phi}$ is decreasing in x since the normal distribution has Increasing Failure Rate.

I am not sure I'd be able to show q-concavity without resorting to the IFR property though, i.e. by just using math manipulation. Btw, is it possible that there is a typo in your second derivative? I was working on the third derivative trying to show that it has a unique positive solution that equates it to zero - this would be sufficient too given that the limit for x→+∞ is zero, third derivative is continuous, and the function is concave-convex - but I was stuck again, and then I got the idea of the IFR property.

Thanks again for taking the time to reply!

8. May 10, 2013

### Ray Vickson

You could also do it directly: as I said, the points 0 and k are in the concavity region, so if you are only interested in x >= k, the function is concave-convex. The inflection point x* > k is the largest root of x(x-k) = 2. We must have F'(x*) < 0, because if F'(x*) were >= 0 the (strict) convexity of F for x > x* would imply that F(x) → ∞ for x → ∞, which contradicts the easily-proven fact that F(x) → 0 for large x > 0. We have $F'(k) = \Phi^c (k) > 0$ (see below). Thus, F' is positive at x = k and negative at x = x*, so has a has a zero between k and x*; and that means that F has a single maximum, which lies in the concave region.

And no, I don't think my derivatives have typos. Letting
$$F(x) = (x-k)\Phi^c(x), \; \text{ where } \Phi^c(x) = \int_x^{\infty} \phi(t) \, dt,$$ we have
$$F'(x) = \Phi^c(x) - (x-k) \phi(x)\\ F''(x) = -\phi(x) - \phi(x) -(x-k)(-x \phi(x)) = \phi(x) [x(x-k) - 2].$$
This uses the fact that $\phi^{\prime} (x) = -x \phi(x).$

9. May 12, 2013

### quantum_2000

Crisp and clear! Reading your arguments was really instructive.

Thank you again for sharing your thoughts

Q