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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!

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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

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    I used the implicitplot function in Maple.2012-05-14