# Chaitanya's Random Pages

## October 24, 2013

### A lower bound for the tail probability of a normal distribution

Filed under: mathematics — ckrao @ 11:17 am

If the random variable $X$ has a normal distribution with mean 0 and variance 1, we know that its tail probability $\text{Pr}(X >x)$ is given by the integral

$\displaystyle \text{Pr}(X >x) = \frac{1}{\sqrt{2\pi}} \int_x^{\infty} \exp (-t^2/2)\ \text{d}t$

(that is, the area under a standard normal bell-shaped curve from $x$ up to $\infty$).

This integral does not have a closed form in terms of elementary functions (unless $x = 0$). However we can find a good lower bound as

$\displaystyle \frac{\sqrt{4 + x^2} - x}{2} \cdot \frac{1}{\sqrt{2\pi}} \exp (-x^2/2) \leq \text{Pr}(X >x) \quad x > 0. \quad \quad (1)$

We show the proof of this result based on [1]. Interestingly it involves Jensen’s inequality: recall that this states that if $f$ is a convex function and ${\mathbb E}$ denotes expectation with respect to some probability distribution, then $f({\mathbb E} Y) \leq {\mathbb E} f(Y)$ for any random variable $Y$. Applied to the convex function $f(u) = 1/u (u > 0)$ this becomes

$\displaystyle \frac{1}{{\mathbb E} Y} \leq {\mathbb E} \frac{1}{Y}. \quad \quad (2)$

For fixed $x > 0$ we now let $Y$ have the distribution

$\displaystyle f_Y(t) = \begin{cases}\frac{te^{-t^2/2}}{\int_x^{\infty} te^{-t^2/2}\ d\text{t} } & \text{if } t \geq x,\\ 0 & \text{if } t < x. \end{cases} \quad \quad (3)$

Applying this in (2) gives

$\displaystyle \frac{\int_x^{\infty} te^{-t^2/2}\ d\text{t}}{\int_x^{\infty} t^2 e^{-t^2/2} \ d\text{t}} \leq \frac{\int_x^{\infty} e^{-t^2/2} \ d\text{t}}{\int_x^{\infty} te^{-t^2/2}\ d\text{t} } \quad \quad (4)$

Evaluating these terms,

$\displaystyle \int_x^{\infty} te^{-t^2/2}\ d\text{t} = \left[-e^{-t^2/2} \right]_x^{\infty} = e^{-x^2/2}$

and

\begin{aligned} \int_x^{\infty} t^2 e^{-t^2/2}\ d\text{t} &= \left[-t e^{-t^2/2} \right]_x^{\infty} - \int_x^{\infty} e^{-t^2/2}\ d \text{t}\\ &= xe^{-x^2/2} + \int_x^{\infty} e^{-t^2/2}\ d \text{t},\end{aligned}

so (4) becomes

$\displaystyle \left(e^{-x^2/2}\right)^2 \leq \left(xe^{-x^2/2} + \int_x^{\infty} e^{-t^2/2}\ d \text{t}\right) \left(\int_x^{\infty} e^{-t^2/2} \ d\text{t}\right).$

This is a quadratic inequality in $a(x):= \left(\int_x^{\infty} e^{-t^2/2} \ d\text{t}\right)$ being of the form

$\displaystyle c(x) \leq (b(x) + a(x))a(x),$

where $c(x) = \left(e^{-x^2/2}\right)^2$ and $b(x) = xe^{-x^2/2}$.

This is equivalent to $c(x) + b^2(x)/4 \leq (a(x) + b(x)/2)^2$, or

$\displaystyle \sqrt{c(x) + b^2(x)/4} - b(x)/2 \leq a(x)$

since the quantities involved are non-negative. In other words we have

$\displaystyle e^{-x^2/2} \frac{\sqrt{4 + x^2} - x}{2} \leq \int_x^{\infty} e^{-t^2/2} \ d\text{t},$

which is equivalent to (1), as desired.

Another lower and upper bounds for the tail probability of a normal distribution are

$\displaystyle \frac{x}{x^2+1} \frac{1}{\sqrt{2\pi}} \exp(-x^2/2) \leq \text{Pr}(X >x) \leq \frac{1}{x}\frac{1}{\sqrt{2\pi}} \exp(-x^2/2),$

a proof of which can be seen, e.g. in [2].
The bound we have looked at can be combined with the following upper bound to arrive at the following tighter bounds.

$\displaystyle \frac{1}{x+ \sqrt{4 + x^2} } \leq \sqrt{\frac{\pi}{2}}\exp (x^2/2) \text{Pr}(X >x) \leq \frac{1}{\sqrt{x+ 8/\pi + x^2}}$

See [3] and [4] for a derivation of the upper bound as well as similar bounds.

#### References

[1] Z. W. Birnbaum, An Inequality for Mill’s Ratio, Ann. Math. Statist. Volume 13, Number 2 (1942), 245-246.

[2] J. D. Cook, Upper and lower bounds for the normal distribution function, 2009.

[4] L. Duembgen, Bounding Standard Gaussian Tail Probabilities, University of Bern Technical Report 76, 2010