Bounds on the mills ratio

Question

How do I show the following bounds on the mills ratio :

$\frac{1}{x}- \frac{1}{x^3} < \frac{1-\Phi(x)}{\phi(x)} < \frac{1}{x}- \frac{1}{x^3} +\frac{3}{x^5} \ \ \ \ \ \ \ $ for $ \ \ \ x>0$ where $\Phi()$ is the CDF of the Normal distribution , and $\phi()$ is the density function of the Normal distribution ?

Also , is there a similar bound when $x < 0$ ?

I am aware of the proof of the fact that the mills ratio is bounded below by $\frac{x}{1+x^2}$ and above by $\frac{1}{x}$ , but I am unable to prove this inequality .

Answer 1

Here's a sketch and a link for how I prove it. Let $$ f(x) = - \left( \frac{1}{x} - \frac{1}{x^3} + \frac{3}{x^5}\right) \phi(x) .$$

Now show (lemma): $\frac{df}{dx} = \left(1 + \frac{15}{x^6}\right)\phi(x)$.

(To prove this, use that $\frac{d\phi}{dx} = - x \phi(x)$, the product rule, and some cancellation.)

Now suppose $x > 0$:

\begin{align*} 1 - \Phi(x) &= \int_{t=x}^{\infty} \phi(x) dx \\ &\leq \int_{t=x}^{\infty} \left(1 + \frac{15}{x^6}\right) \phi(x) dx \\ &= \lim_{t\to\infty} f(t) - f(x) \\ &= - f(x) \\ &= \left(\frac{1}{x} - \frac{1}{x^3} + \frac{3}{x^5} \right) \phi(x) . \end{align*}

Using the next term in the series gives $f(x) = -\left(\frac{1}{x} - \frac{1}{x^3} + \frac{3}{x^5} - \frac{15}{x^7}\right)\phi(x)$ and $\frac{df}{dx} = \left(1 - \frac{105}{x^8}\right)\phi(x)$. Notice because of the alternating positive/negative terms, $\frac{df}{dx}$ is now $\phi$ times something less than one, so plugging it into the same proof gives a lower bound on $1 - \Phi(x)$. I have a blog post on the general form of this (sorry to not point you to a more formal reference).

Answer 2

This is a special case of Proposition 1.3.

The case $x<0$ obtains from this by symmetry: $\Phi(-z)=1-\Phi(z)$ and $\phi(-z)=\phi(z)$ for all $z$.