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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 .

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1 Answer 1

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This is a problem that can be found in the text High-Dimensional Statistics: A Non-Asymptotic Viewpoint by Martin Wainwright. An important hint is the first part of the question asking you to prove that $\phi'(z) + z\phi(z) = 0$, which comes directly from computing $\phi'(z) = \frac{-z}{\sqrt{2\pi}} e^{-z^2/2}$.

Using the above, we may note first that $1 - \Phi(z) = \mathbb{P}[Z\geq z] = \int^\infty_z \phi(t) dt$, and substituting $\phi(z) = \frac{-\phi'(z)}{z}$ and using integration by parts ($u = \frac{1}{t}$, $dv=\phi'(t)$) we get $$ \int^\infty_z \phi(t) dt = \int^\infty_z \frac{-\phi'(t)}{t} dt = \left[ \frac{-\phi(t)}{t}\right]^\infty_z - \int^\infty_z \frac{\phi(t)}{t^2} dt $$ Since $\lim_{t\to \infty} \frac{-\phi(t)}{t} = 0$, we get the top as $\frac{\phi(z)}{z} - \int_z^\infty \frac{\phi(t)}{t^2} dt$, where we may use the same substitution and apply integration by parts again: $$ \frac{\phi(z)}{z} - \int_z^\infty \frac{-\phi'(t)}{t^3} dt = \frac{\phi(z)}{z} + \left[ \frac{\phi(t)}{t^3}\right]^\infty_z - \int^\infty_z \frac{-3\phi(t)}{t^4}dt $$ $$ = \frac{\phi(z)}{z} - \frac{\phi(z)}{z^3} + \int^\infty_z \frac{3\phi(t)}{t^4}dt $$ Thus since $\int^\infty_z \frac{3\phi(t)}{t^4}dt>0$ we get $\phi(z)\left(\frac{1}{z} - \frac{1}{z^3}\right) < \mathbb{P}[Z\geq z]$. Applying the trick again to $\int^\infty_z \frac{3\phi(t)}{t^4}dt$ yields $$ \int^\infty_z \frac{-3\phi'(t)}{t^5}dt = \left[ \frac{-3\phi(t)}{t^5}\right]^\infty_z - \int^\infty_z \frac{15\phi(t)}{t^6}dt $$ $$ = \frac{3\phi(z)}{z^5} - \int^\infty_z \frac{15\phi(t)}{t^6}dt $$ and since $- \int^\infty_z \frac{15\phi(t)}{t^6}dt<0$, we get $\mathbb{P}[Z\geq z] < \phi(z)\left(\frac{1}{z} - \frac{1}{z^3} + \frac{3}{z^5}\right) $, which proves the claim.

I believe its straightforward to derive a similar bound for $\Phi(X)$ for $x<0$.

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