hazard increase

Question

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?

Answer 1

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 $x$.

Answer 2

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.