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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}


\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.


[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.

[3] calculus – An inequality from the handbook of mathematical functions (by Abramowitz and Stegun) – Mathematics Stack Exchange

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


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