# How to prove that the CDF of the standard normal distribution is log-concave?

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 $x \in \mathbb{R}$

• Your P.S. looks good to me.
– A.Γ.
Dec 20 '16 at 17:41

Let $q$ be a log-concave density on $\mathbb{R}$ with corresponding CDF $Q$. For arbitrary $a, b \in \mathbb{R}$ and $\lambda \in [0, 1]$, set $t = \lambda a + (1 - \lambda)b$ and define the following three functions: \begin{align*} h(x) &= q(x) I\{x \le t\} \\ f(x) &= q(x) I\{x \le a\} \\ g(x) &= q(x) I\{x \le b\} \end{align*}
It is easy to see that these functions satisfy the inequality $$f(x)^\lambda g(y)^{1 - \lambda} \le h(\lambda x + (1 - \lambda) y)$$ for all $x, y \in \mathbb{R}$. This means we may apply the Prekopa-Leindler inequality: $$\left(\int f(x) \, dx\right)^\lambda \left(\int g(x) \, dx\right)^{1 - \lambda} \le \int h(x) \, dx$$
Taking the logarithm on both sides and applying the definition of the CDF $Q$ we now get $$\lambda \log Q(a) + (1 - \lambda) \log Q(b) \le \log Q(t) = \log Q(\lambda a + (1 - \lambda) b),$$ i.e. $Q$ is log-concave.