Re: RandomArray from user defined distribution?
- To: mathgroup at smc.vnet.net
- Subject: [mg73334] Re: [mg73319] RandomArray from user defined distribution?
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Tue, 13 Feb 2007 06:52:09 -0500 (EST)
- Reply-to: hanlonr at cox.net
RandomArray works with PoissonDistribution
Needs["Statistics`"];
PDF[PoissonDistribution[a*t],n]
(a*t)^n/(E^(a*t)*n!)
RandomArray[PoissonDistribution[1],{10}]
{0,0,1,1,0,0,2,0,0,0}
If you mean for your distribution to be continuous in t then
p[a_,t_]:=a^2 *t* Exp[-a*t];
Integrate[p[a,t],{t,0,Infinity},Assumptions->{a>0}]
1
Mean is
Integrate[t*p[a,t],{t,0,Infinity},Assumptions->{a>0}]
2/a
Standard deviation is
Simplify[Sqrt[Integrate[t^2*p[a,t],{t,0,Infinity},
Assumptions->{a>0}]-(2/a)^2],a>0]
Sqrt[2]/a
CDF is
c[a_,t_]=Integrate[p[a,x],{x,0,t}]
1 - (a*t + 1)/E^(a*t)
Off[Reduce::ratnz];
myRandom[a_?Positive]:=
Last[Reduce[{c[a,t]==Random[],t>0},t]];
myRandomArray[a_?Positive,n_Integer]:=Table[myRandom[a],{n}];
myRandomArray[2,5]
{0.816351,1.0222,0.477733,0.425778,0.24972}
Mean[myRandomArray[2,100]]
1.06227
Bob Hanlon
---- rob <robIV at piovere.com> wrote:
> I'd like to use the RandomArray to produce some data from
> what I think is a Poisson distribution in t
> P[t] = a^2 t Exp[-a*t] where a is mean, sigma.
>
> I see one can use RandomArray to produce sample data from a
> lot of continuous distributions but the Poisson isn't among
> them (it's only available in the discrete form).
>
> I've made a bunch of crippled attempts to force my P[t] to
> put out examples but have failed. Any suggestions? Thanks.
>