Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.
|
|
|
|
Re: Question regarding claimed approximation to standard normal CDF
Posted:
Nov 10, 2011 6:48 AM
|
|
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
To reply by email, change LookInSig to luecking
|
|
|
|
