Show that $\frac{1}{x}-\frac{1}{x^3}<\frac{P(X>x)}{\phi(x)}<\frac{1}{x}$ for $x>0$ where $X$~$N(0,1)$ with pdf $\phi(x)$.

Question

I tried taking exponentials and using Markov's Inequality, but this only gave me an upper bound of $\sqrt{2\pi}$. I'm not sure how to begin to approach this question - can anyone give a hint?

Answer 1

You can use these known "double inequality" for the Gaussian (Bounds and Approximations in "Q-function", Wikipedia)

Applying this double inequality to your exercise you immediately get

$\frac{x}{x^2+1}<\frac{\mathbb{P}[X>x]}{\phi(x)}<\frac{1}{x}$

To prove what you need it is enough to prove that, for $x>0$,

$\frac{1}{x}-\frac{1}{x^3}<\frac{x}{x^2+1}$