Standard normal distribution hazard rate Is the hazard rate of the standard normal distribution convex? Can you give a reference?
 A: $\phi $ is the standard normal density, $\Phi$ is the cdf, $\Phi'=\phi$, $\bar{\Phi}=1-\Phi $ is the survivor
function, and $\lambda =\phi /\bar{\Phi}$ is the hazard rate. We know that $%
\lambda(s) >s$.
Lemma
$\lambda(s) >\frac{3s}{4}+\frac{\sqrt{s^{2}+8}}{4}.$(Baricz / J.
Math. Anal. Appl. 340 (2008) 1362--1370, Thm. 2.3)
Proposition The hazard rate $\lambda $ is convex, $\lambda ^{\prime \prime }>0$.
Proof
We prove this by differentiation. Note that $\lambda ^{\prime }(s)=\frac{-s\phi(s)
}{\bar{\Phi}(s)}+\frac{\phi ^{2}(s)}{\bar{\Phi}^{2}(s)}=\lambda(s) \left( \lambda(s)
-s\right) >0$ and then
\begin{eqnarray*}
\lambda ^{\prime \prime }(s) &=&\lambda(s) \left( \lambda(s) -s\right) \left( \lambda(s)
-s\right) +\lambda(s) \left( \lambda(s) \left( \lambda(s) -s\right) -1\right)  \\
&=&\lambda(s) \left( s^{2}-3s\lambda(s) +2\lambda ^{2}(s)-1\right)  \\
&=&\lambda(s) \left( \left( 2\lambda(s) -s\right) \left( \lambda(s) -s\right)
-1\right) .
\end{eqnarray*}
This is positive iff $\left( 2\lambda(s) -s\right) \left( \lambda(s) -s\right) -1>0
$. If $s\leq 0$, then
$$
\left( 2\lambda(s) -s\right) \left( \lambda(s) -s\right) -1\geq 2\lambda
^{2}\left( 0\right) -1=\frac{4}{\pi }-1>0.
$$
For $s>0$, we find the upper root of the polynomial $\lambda \rightarrow \left( 2\lambda -s\right) \left( \lambda -s\right) -1$ , which is $\frac{3s}{4}+\frac{\sqrt{s^{2}+8}}{4}$. We conclude by the lemma that $\lambda(s)$ is greater than the upper root. This implies that $\lambda ^{\prime \prime }(s)>0$ and hence that
$\lambda $ is convex.
A: Yes, the hazard rate
$$ h(x) = \frac{\varphi(x)}{1-\Phi(x)}$$
of the normal distribution is convex. The second derivative is given by
$$h''(x) = \varphi(x) \frac{\left((1-\Phi(x))\frac{x-\mu}{\sigma}-\frac{3}{2}\sigma\varphi(x)\right)^2 - (1-\Phi(x))^2-\frac{1}{4}\sigma^2\varphi(x)^2}{(1-\Phi(x))^3\sigma^2}$$
This gives $h''(\mu) = \frac{\sqrt{2}}{\pi^{\frac{3}{2}}\sigma^3}(4-\pi) > 0$ and it is also elsewhere positive.
A more interesting question might be whether the log of the survival function $log(1-\Phi)$ is concave or convex since this gives hints about the tail behavior of the distribution. The normal distribution is very light-tailed so we expect a concave log survival function. And indeed, it is concave which is equivalently to an increasing hazard rate. We have
$$ h'(x) = \varphi(x) \frac{\varphi(x)-x(1-\Phi(x))}{(1-\Phi(x))^2} \geq 0 \;.$$
The last inequality holds since it is sufficient to show $$g(x) = \varphi(x)-x(1-\Phi(x)) \geq 0 \;.$$
At $x=0$, $g(x)=\varphi(0)>0$ and at $x=\infty$, $g(x)=0$. In order to be negative the derivative of $g(x)$ would have to have roots which it doesn't have:
$$g'(x) = \Phi(x)-1 < 0 .$$
