       Normal Product Distribution

• To: mathgroup at smc.vnet.net
• Subject: [mg49342] Normal Product Distribution
• From: DrBob <drbob at bigfoot.com>
• Date: Thu, 15 Jul 2004 07:00:05 -0400 (EDT)
• References: <7228735a.0407050100.4695fc68@posting.google.com> <QaednZQbSYcwpnTdRVn-vA@comcast.com> <ccdlms\$sd5\$1@smc.vnet.net> <200407141129.HAA26886@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```The kurtosis turns out to be 9; see below.

f[u_] := BesselK[0, Sqrt[u^2]/(sx*sy)]/(Pi*sx*sy)
Table[Integrate[u^n f[u], {u, -Infinity, Infinity},
Assumptions -> sx > 0 && sy > 0], {n, 1, 4}];
Append[%, %[]/%[]^2]
TableForm[%, TableHeadings -> {{"Mean", "Variance", "3rd
Moment", "4th Moment", "Kurtosis"}}]

{0, sx^2*sy^2, 0, 9*sx^4*sy^4, 9}
(TableForm suppressed.)

Bobby

On Wed, 14 Jul 2004 07:29:32 -0400 (EDT), Roger L. Bagula <rlbtftn at netscape.net> wrote:

> Abrupt peak due to product of Gaussian distributions:
> http://mathworld.wolfram.com/NormalProductDistribution.html
> Probably doesn't have Kurtosis near zero either.
> Roger L. Bagula wrote:
>> 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
>>>
>>>
>>
>>
>
>

--
DrBob at bigfoot.com
www.eclecticdreams.net

```

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