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. >