A Normal distribution inequality

Question

Denote by $\phi(z)$ and $\Phi(z)$ the density and distribution of a standard Normal variable (mean=0, sd=1), respectively. It appears that for $z > 0$, $$ \frac{\phi(z)}{1 - \Phi(z)} < z + \frac{1}{z} $$ Is there a proof (or disproof) of this?

Answer 1

A similar question has been answered Here

Anyway, this is an alternative proof:

Let's rewrite your inequality in the following way

$$\frac{1-\Phi(z)}{\phi(z)}>\frac{z}{1+z^2}$$

$$1-\Phi(z) > \phi(z)\frac{z}{1+z^2}$$

Let's define the following function:

$g(z)=1-\Phi(z)-\phi(z)\frac{z}{1+z^2}$

As:

• $g(0)=\frac{1}{2}>0$

• $g'(z)=-2 \frac{\phi(z)}{(1+z^2)^2}<0$ for every z

• $g(\infty) \rightarrow 0$

thus $g(z)$ is always positive... and this immediately proves what you are requested to do