Standard normal distribution hazard rate

Question

Is the hazard rate of the standard normal distribution convex? Can you give a reference?

Answer 1

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:

$$g'(x) = \Phi(x)-1 < 0 .$$