[Date Index] [Thread Index] [Author Index]

RE: Re: help in generating a gaussian random variable

• To: mathgroup at smc.vnet.net
• Subject: [mg35530] RE: [mg35502] Re: [mg35488] help in generating a gaussian random variable
• From: "Wolf, Hartmut" <Hartmut.Wolf at t-systems.com>
• Date: Thu, 18 Jul 2002 03:06:13 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

> -----Original Message-----
> From: BobHanlon at aol.com [mailto:BobHanlon at aol.com]
> Sent: Wednesday, July 17, 2002 8:09 AM
> Subject: [mg35530] [mg35502] Re: [mg35488] help in generating a gaussian random
> variable
> 
> 
> 
> In a message dated 7/16/02 5:50:43 AM, dsalman at itee.uq.edu.au writes:
> 
> >I need to generate a Gaussian random variable y having mean 
> =0 and variance
> >=x.
> >
> >The variance x is itself a gaussian random variable having a 
> known mean
> >and
> >variance.
> >e.g. mean of x =5;
> >variance of x = 10;
> >
> >Can anyone suggest how to use this information to generate y ?
> >
> 
> Needs["Statistics`NormalDistribution`"];
> 
> data=RandomArray[
>       NormalDistribution[0, Random[
>           NormalDistribution[5, Sqrt[ 10]]]], {100}];
> 
> 
> Bob Hanlon
> Chantilly, VA  USA
> 

Bob,

sorry, but I'm wary of your proposed solution. This is because RandomArray
is an optimising function to speed up over Table Random.

A look into the package Statistics`ContinuousDistributions` reveals that the
arguments to NormalDistribution are evaluated only once. So the variance x
is not a random variable. A statistics test of

data2 = Table[
      Random[NormalDistribution[0, 
          Random[NormalDistribution[5, Sqrt[10]]]]], {100}];

should reveal this. To make it simple for me: taking samples of 10001 events
respectively, and comparing the cumulants e.g.

Block[{$DisplayFunction = Identity}, 
        ListPlot[
          Take[Transpose[{Sort[#1], Range[Length[#1]]/Length[#1]}], 
               {1, -1, 100}], 
          PlotJoined -> True, 
          PlotStyle -> #2]] & @@@ {{data, {}}, {data2, Hue[.7]}} // Show

shows a great discrepancy. Perhaps you like to draw multiple samples of data
(not neccessarily of data2) to compare.

This also shows that the desired distribution is not a normal distribution
(when x is a random variable).


Even more revealing is to compare the distributions by eye view

GraphicsArray[
    Block[{$DisplayFunction = Identity, step = 100, scale}, 
          scale = step/Length[#1]; 
          ListPlot[{Plus[##]/2, -scale/Subtract[##]} & @@@ 
              Partition[Take[Sort[#1], {1, -1, step}], 2, 1], 
            PlotJoined -> True, PlotStyle -> #2, 
            PlotRange -> {{-20, 20}, {0, .2}}]] & @@@ {{data, {}}, {data2, 
          Hue[.7]}}] // Show


--
Hartmut