Re: Normal distribtion

• To: mathgroup at smc.vnet.net
• Subject: [mg49188] Re: Normal distribtion
• From: "Kevin J. McCann" <kjm at KevinMcCann.com>
• Date: Wed, 7 Jul 2004 01:42:21 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

```A Histogram plot of this does not look very Gaussian to me. Min of zero, max

Kevin

"Roger L. Bagula" <rlbtftn at netscape.net> wrote in message
news:ccdlms\$sd5\$1 at smc.vnet.net...
> I found a better faster way to get a Gaussian/ white noise:
> In Mathematica notebook style:
>
> x[a_]=(1+Sqrt[1-a^2))/a
> Noise=Table[Exp[-x[Sin[2*Pi*Random[]]]^2/2/Sqrt[2*Pi],{n,1,500}]
> ListPlot[noise,PlotRange--> All,PlotJoined->True]
>
> It is a projective line ( circle to line random taken as the basic for a
> normal distribution's amplitude.) based algorithm.
> Since it doesn't use either the polar method with has a choice loop or
> the older 12 random method, the random operation has only to be used
> once for each noise value.
> It is faster by several factors and can be translated to almost any
> language.
> George Marsaglia wrote:
> > "philou" <philou2000 at msn.com> wrote in message
> >
> >>Hi,
> >>I heard that it was possible to get a realisation of a normal
> >>distribution from two realisations of an uniform distribution. Can
> >>someone explain me how to do that ? What transformations should I do ?
> >
> >
> > Your hazy reference may have been based on my polar method for
> > generating a pair of independent standard normal variates X and Y:
> >
> > If U and V are independent uniform in (-1,1), conditioned by
> >
> >                S = U^2+V^2 < 1
> >
> > then S is uniform in (0,1) and independent of the point
> > (U/sqrt(S),V/sqrt(S)), which is uniform on the unit circumference.
> >
> > Thus if R=sqrt(-2*ln(S)/S) then
> >       X=R*U
> >       Y=R*V
> > are a pair independent standard normal variates, obtained by
> > projecting that uniform point on the unit circumference
> > a random distance with a root-chi-square-2 distribution,
> > exploiting the uniformity of S and its independence of
> > the random point on the circumference.
> >
> > Of course one has to discard uniform (-1,1) pairs U,V
> > for which S=U^2+V^2>1, so each normal variate is produced
> > at an average cost of 4/pi=1.27 uniform variates.
> >
> > This method is sometimes improperly attributed to Box and Muller,
> > who pointed out that pairs of normal variates could be generated as
> > rho*cos(theta), rho*sin(theta) with rho root-chisquare-2 ,
sqrt(-2*ln(U)),
> >  and theta uniform in (0,2pi), a result we owe to Laplace, who showed us
> > how to find  the infinite integral of exp(-x^2) by getting its square
> > as the integral of exp(-x^2-y^2), then transforming to polar
coordinates.
> >
> > For a method faster than my polar method, requiring about 1.01 uniform
> > variates per normal variate, try the ziggurat method of
> > Marsaglia and Tsang, in volume 5, Journal of Statistical Software:
> >   http://www.jstatsoft.org/index.php?vol=5
> >
> > George Marsaglia
> >
> >
>

```

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