hazard increase I had a question regarding this proof. The hazard function is being defined as $\frac{\phi(x)}{1-F(x)}$ with $\phi(x)$ as the density function and $F(x)$ the CDF.
How can I prove that this is increasing in $x$ if we’re dealing with a normal distribution?
 A: You can just use differentiation:
Assuming that $\phi(x) $ is specified by the normal distribution with mean $\mu$ and variance $\sigma^2$, we let:
$$ h(x) = \frac{\phi (x)}{ 1- F(x)} = \frac{\phi (x)}{ 1- \int_{-\infty} ^x \phi(t)dt}$$
Then:
\begin{align}
\frac{d h(x)}{d x} &= \frac{-\frac{(x-\mu)}{\sigma^2} \phi (x)}{ 1- \int_{-\infty} ^x \phi(t)dt} + \frac{(\phi(x))^2}{(1- \int_{-\infty} ^x \phi(t)dt)^2} \\
&=\Bigg[\frac{\phi (x)}{( 1- F(x))^2} \Bigg] \Big(\phi(x) - \frac{x-\mu}{\sigma^2} (1-F(x)) \Big)
\end{align}
Since the term in square brackets is always positive, We just need to focus on the term in the normal brackets (and show that it is always $>0$).
When $x \leq \mu$, it is clear that:
\begin{align}
\phi(x) - \frac{x-\mu}{\sigma^2} (1-F(x)) > 0
\end{align}
When $x> \mu$, it is not so straightforward. So let's do it step by step. We begin by looking at the second term in the expression:
\begin{align}
\frac{x-\mu}{\sigma^2} (1-F(x)) 
&= \frac{x-\mu}{\sigma^2}\int_x^\infty \phi(t)dt \\
&= \frac{x-\mu}{\sigma^2} \int_x^\infty\frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(t-\mu)^2}{2\sigma^2}} dt\\
&< \frac{x-\mu}{\sigma^2} \int_x^\infty \frac{(t-\mu)/\sigma^2}{(x-\mu)/\sigma^2} \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(t-\mu)^2}{2\sigma^2}}dt \\
&= \frac{1}{\sqrt{2\pi}\sigma} \int_x^\infty \frac{t-\mu}{\sigma^2}e^{-\frac{(t-\mu)^2}{2\sigma^2}} dt \\
&= \frac{1}{\sqrt{2\pi}\sigma} \Big[ -e^{-\frac{(t-\mu)^2}{2\sigma^2}}\Big]_x^\infty \\
&= \phi(x)
\end{align}
The inequality follows from the fact that the ratio $\frac{t-\mu}{x-\mu} > 1$.
Thus, from this, we see that when $x > \mu$ we have
\begin{align}
\phi(x) - \frac{x-\mu}{\sigma^2} (1-F(x)) > \phi(x) - \phi(x) = 0
\end{align}
So we can conclude that $$\frac{dh(x)}{dx} > 0$$ Which shows that $h(x)$ is an increasing function of $x$.
A: First, for $x>0$,
$$
\int_x^\infty e^{-t^2/2}\,dt < \frac1x\int_x^\infty te^{-t^2/2}\,dt
= \frac1x e^{-x^2/2}.
$$
So, for a standard normal,
$$
\frac d{dx}\bigg(\frac{1-F(x)}{\phi(x)}\bigg)
= \frac d{dx}\bigg(e^{x^2/2}\int_x^\infty e^{-t^2/2}\,dt\bigg)
= xe^{x^2/2}\int_x^\infty e^{-t^2/2}\,dt - 1 < 0,
$$
for all $x$. (If $x\le 0$, the left-hand side is less than or equal to $-1$. If $x>0$, it follows from the initial inequality.)
Thus, $(1-F(x))/\phi(x)$ is decreasing, so the hazard function, $h(x)$, is increasing. Since the hazard function of a $N(\mu,\sigma^2)$ is $\sigma^{-1} h((x-\mu)/\sigma)$, the same is true for a general normal distribution.
