TITLE: How to prove that the CDF of the standard normal distribution is log-concave?
QUESTION [5 upvotes]: How can I prove $\log(\Phi(x))$ is concave, where $\Phi(x)$ is CDF of N(0, 1) ?
Since
$\frac{d^2}{dx^2} \log \Phi(x) = \cfrac{\phi(x)[-x \Phi(x) - \phi(x)]}{(\Phi(x))^2}$, it is enough to show that $-x \Phi(x) - \phi(x) < 0$ for all $x \in \mathbb{R}$.
Here's a plot of $\log \Phi$:

P.S.
This may be inaccurate but I consider like below:

$-x \Phi(x) - \phi(x)$ is decreasing since

\begin{align}
\frac{d}{dx}( -x \Phi(x) - \phi(x)) &= -\Phi(x) - x \phi(x) + x \phi(x) \\
&= -\Phi(x) < 0
\end{align}

$\lim_{x \to -\infty} -x \Phi(x) - \phi(x) = 0$, since $\phi(-\infty)=0$ and by l'Hospital's rule

\begin{align}
\lim_{x \to -\infty} -x \Phi(x) &= - \lim_{x \to -\infty} \cfrac{\Phi(x)}{\frac{1}{x}} \\
&= \lim_{x \to -\infty} x^2 \phi(x) \\
&= \frac{1}{\sqrt{2\pi}} \lim_{x \to -\infty} \frac{x^2}{\exp(\frac{x^2}{2})}\\
&= \frac{1}{\sqrt{2\pi}} \lim_{x \to -\infty} \frac{2}{\exp(\frac{x^2}{2})} \\
&= 0
\end{align}
Therefore, $-x \Phi(x) - \phi(x) < 0$ for all $x \in \mathbb{R}$

REPLY [3 votes]: Alternatively to the (not easy to prove) result on preservation of log-concavity under integration one can take a straightforward approach. Here is the exercise from Boyd, Vandenberghe, Convex optimization with solution steps