Path: utzoo!utgpu!water!watmath!clyde!rutgers!mailrus!ames!pasteur!ucbvax!agate!saturn!chromo!joseph
From: joseph@chromo.ucsc.edu (Joseph Reger)
Newsgroups: comp.lang.c
Subject: Re: random number generator
Summary: How to get approximately normal deviates
Keywords: Random numbers, normal distribution, central limit theorem
Message-ID: <1989@saturn.ucsc.edu>
Date: 17 Feb 88 17:40:43 GMT
References: <11809@brl-adm.ARPA> <225800004@uxe.cso.uiuc.edu>
Sender: usenet@saturn.ucsc.edu
Reply-To: joseph@chromo.ucsc.edu (Joseph Reger)
Organization: Physics Department, University of California, Santa Cruz
Lines: 25

In article <225800004@uxe.cso.uiuc.edu> mcdonald@uxe.cso.uiuc.edu writes:
>
>>Simplistic, but this works:
>>Generate about 12 random numbers (assuming 0.0 <= i <1.0)
>
>       ADD THEM TOGETHER
>
>>and divide by 12.  This is sufficiently close to gaussian distribution
>>for most purposes.  The more numbers you generate before you do the
>>division, the closer you will get to purely gaussian distribution.


Devide by 12??

This method uses the central limit theorem and gives nearly
gaussian (normal) random numbers if the number of uniform
random numbers to be summed up is sufficiently large:
   Take N uniform random deviates on (0,1): u(1), u(2),...,u(N).
Then g = sqrt(12/N) * ( (u(1)+u(2)+...+u(N)) - N/2 )
yields a (nearly) normal variable with mean 0 and variance 1.
N=12 seems to be sufficiently large for most purposes and
avoids computing the sqrt() factor.
(So do not devide by 12, subtract 6!). 
g' = M + D * g gives a normal random number with mean M and
standard deviation D.