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How can one prove that $\log{\Phi(x)}$ is a concave function in x?

I tried taking second derivative, but so far it isn't helpful. I read a hint on my textbook that says it is easy to show its first derivative $\frac{\phi(x)}{\Phi(x)}$ is decreasing, but I have no clue on how I should work that out either.

Can anybody give me some hint? Thanks very much!

EDIT: Sorry if I didn't make it clear. $\Phi(.)$ refers to the cumulative distribution function of the standard normal distribution, while $\phi(.)$ is its derivative - the density function.

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  • $\begingroup$ I don't agree that it is easy, except for $x > 0$, and the inequality you need is for $x > 0 , 1 - \Phi(x) < \frac {\phi(x)} x$, which follows from $1 - \Phi(x) = \int_x^{\infty} \phi(y) dy < \int_x^{\infty} \frac yx \phi(y) dy$ $\endgroup$ – mike May 7 '13 at 16:36
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The second derivative of $\log\Phi$ is $-\psi/\Phi^2$ where $\psi(x)=x\Phi(x)+\phi(x)$ hence $\psi'=\Phi\gt0$. Since $\psi(-\infty)=0$, $\psi\gt0$ everywhere, QED.

This uses only that $\phi'(x)=-x\phi(x)$ for every $x$ and that $x\Phi(x)\to0$ when $x\to-\infty$.

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