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 $x \in \mathbb{R}$
Best Answer
Instead of using the second derivative, it is easier to verify the log-concavity using the Prekopa-Leindler inequality.
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.