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Topic: Re: Question regarding claimed approximation to standard normal CDF
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 Dan Luecking Posts: 26 Registered: 11/12/08
Re: Question regarding claimed approximation to standard normal CDF
Posted: Nov 10, 2011 6:48 AM
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Resent-From: <bergv@illinois.edu>
From: Dan Luecking <LookInSig@uark.edu>
Date: November 9, 2011 12:29:50 PM MST
To: "sci-math-research@moderators.isc.org"
<sci-math-research@moderators.isc.org>
Subject: Re: Question regarding claimed approximation to standard
normal CDF

On Wed, 2 Nov 2011 16:58:37 +0000, jdm <james.d.mclaughlin@gmail.com>
wrote:

> I recently encountered the following claim in a cryptography paper
> regarding the CDF of the standard normal distribution:
>
> Phi(-b) \approx e^{-b^{2}/2}/sqrt(2*pi) when b is large.
>
> This looks almost exactly the same as the exact formula, the
> difference being that the integration db is omitted.
>
> I have been unable to find any other source for this approximation,
> and when I emailed the author I was referred to another paper which
> used the approximation without providing evidence of its validity.
>
> Could someone please supply me with a source for this approximation,
> such as a textbook in which it appears, or, if it is in fact not
> valid, tell me so?

It depends on what \approx means. In fact
b \Phi(-b) < e^{-b^{2}/2} / \sqrt{2*pi}
for all positive b.

This useful upper bound on Phi(-b) can be obtained as follows.
For x \le -b < 0  we get  b \le -x and so

b \Phi(-b) = \int_{-\infty}^{-b}  b e^{x^2/2} dx / \sqrt{2\pi}
< \int_{-\infty}^{-b} -x e^{x^2/2} dx / \sqrt{2\pi}
=  e^{b^2/2} / \sqrt{2\pi}

Dan
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