Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I would like to verify this function $f:\mathbb{R} \to \mathbb{R}$ to be log-concave or determine the region where it is: $$ f(x) = 2 - \Phi(a-x) - \Phi(a) $$ where $a > 0$ is a constant, and $\Phi(x) := \frac12\left[1 + \operatorname{erf}\left( \frac{x}{\sqrt{2}}\right)\right] $, i.e. the CDF of the standard normal distribution.

  1. My attempt is based on "every concave function that is nonnegative on its domain is log-concave". $\Phi$ is convex over $(-\infty, 0)$ (and concave over $(0,\infty)$), so $f$ is concave over $(a, \infty)$. Since $f$ is nonnegative, $f$ is log-concave over $(a, \infty)$. Am I right?
  2. But I hope to decide if $f$ is log-concave over $\mathbb{R}$, or at least over $(0, \infty)$. Also is it possible to determine the region where $f$ is log-concave?
  3. To simplify the question, if $f$ is log-concave, is $f + c$ for any constant $c$ also log-concave?
  4. I am also puzzled when trying to determine where g is log-concave, $$ g(x) := 1 - \Phi(a-x) + \Phi(-a-x) $$ and whether it is over $(0, \infty)$?

Thanks and regards!

share|improve this question

1 Answer 1

It certainly can't be log-concave over $\mathbb R$, because $f$ has a positive limit $1 - \Phi(a)$ as $x \to -\infty$, and an increasing concave function can't have a finite limit at $-\infty$.

For $a=1$, numerical calculations show $\log f$ has an inflection at approximately $x=0.2180016571$, so it's not log-concave on $(0,\infty)$.

EDIT: here is a picture of part of the $a-x$ plane: $\log f$ is concave in the blue region and convex in the red region.

enter image description here

share|improve this answer
    
Thanks! I wonder why "an increasing concave function can't have a limit at −∞"? I can visualize such a function that is a counterexample by converging to $-\infty$. Do you mean "log-concave" instead of "concave"? –  Tim May 14 '12 at 18:02
    
No, I mean concave. If $f'(x_0) > 0$ and the function is concave, $f(x) < f(x_0) + (x-x_0) f'(x_0) \to -\infty$ as $x \to -\infty$. –  Robert Israel May 14 '12 at 18:05
    
I meant a finite limit, of course. –  Robert Israel May 14 '12 at 18:12
    
Thanks! How did you generate the plot of the concave/convex regions? My way is quite stupid, but is the best I can think of: I can only plot the functions for several values of $a$. –  Tim May 14 '12 at 20:58
    
I used the implicitplot function in Maple. –  Robert Israel May 14 '12 at 21:37

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.