# Inequality for the tail of normal distribution function

Let $Ф(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{x} e^{-t^2/2} \, dt$ be the cumulative distribution function of the standard normal distribution.

Numerical calculations suggest the following inequality is true: $$1 - Ф(x) > \frac{2}{\pi} \log( e^{x^2/2} -1 ) e^{-x^2}$$ for $x > 0$. The question is how to prove it?

There are some known lower bounds on $1 - Ф(x)$ like $$1 - Ф(x) > \frac{1}{\sqrt{2\pi}} \frac{x}{x^2+1} e^{-x^2/2}$$ and $$1 - Ф(x) > \frac{1}{2\sqrt{2\pi}} (\sqrt{x^2+4}-x) e^{-x^2/2}.$$ In both cases the difference of left and right sides is monotonically decreasing, but in my inequality the difference isn't monotonic.

• As Robert noted due to a different exponential your inequality is worse than known for large x. It makes sense for x near zero. But for such values it is more natural to seek for improvements in terms of powers of x, not exponentials. But on this field it seems impossible to compete with known excellent inequalities via continuous fractions and Pade approximants. – Sergei Sep 10 '14 at 6:55
• Yes, I know that my inequality is worse than the second lower bound everywhere (though there is an improvement of this inequality that is better than the second bound on some interval). But it came up in my study, so I need to prove it. – user47855 Sep 10 '14 at 11:30

For sufficiently large $x$, the inequality is true, as $$1 - \Phi(x) \sim \dfrac{e^{-x^2/2}}{\sqrt{2\pi} x}$$
while your expression (let's call it $R(x)$) is $O(x^2 e^{-x^2})$. In principle one can get explicit bounds corresponding to these asymptotics and thereby determine a finite value $B$ such that $1 - \Phi(x) > R(x)$ for all $x \ge B$. You then want to look at the interval $[0,B]$. A finite number of computations using interval arithmetic should be able to handle this. Thus if $1 - \Phi(x_0) - R(x_0) \ge a$ and you have an estimate $-\Phi'(x) - R'(x) \le b$ in an interval $[a,c]$, then $1 - \Phi(x) - R(x) \ge a + b (x-x_0)$ in that interval.