how to generate random variable with constraint?
in [Matlab]
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From: Matt J on 20 Jul 2010 11:20 "jay " <ssjzdl(a)gmail.com> wrote in message <i22vhv$frn$1(a)fred.mathworks.com>... > I need to generate 5 random variables between [0,1], let's say a, b, c, d, e. The constraint is that a<b<c<d<e. How to make this happen? Please advise. thanks ========== A=cumsum(rand(1,5))/5;
From: Walter Roberson on 20 Jul 2010 11:23 Matt J wrote: > "jay " <ssjzdl(a)gmail.com> wrote in message > <i22vhv$frn$1(a)fred.mathworks.com>... >> I need to generate 5 random variables between [0,1], let's say a, b, >> c, d, e. The constraint is that a<b<c<d<e. How to make this happen? >> Please advise. thanks > ========== > > A=cumsum(rand(1,5))/5; I'm relatively sure that Roger showed in an earlier thread that that approach produces biased results.
From: Matt J on 20 Jul 2010 11:35 Walter Roberson <roberson(a)hushmail.com> wrote in message <Jbj1o.16614$Bh2.16201(a)newsfe04.iad>... > > > > A=cumsum(rand(1,5))/5; > > I'm relatively sure that Roger showed in an earlier thread that that > approach produces biased results. ================ Did the OP require that there not be bias?
From: Matt J on 20 Jul 2010 11:58 Walter Roberson <roberson(a)hushmail.com> wrote in message <Jbj1o.16614$Bh2.16201(a)newsfe04.iad>... > > > > A=cumsum(rand(1,5))/5; > > I'm relatively sure that Roger showed in an earlier thread that that > approach produces biased results. ======= Another possibility. Still biased? A=cumsum(rand(1,6)); A=A/sum(A); A(end)=[];
From: Walter Roberson on 20 Jul 2010 12:00
Matt J wrote: > Walter Roberson <roberson(a)hushmail.com> wrote in message > <Jbj1o.16614$Bh2.16201(a)newsfe04.iad>... > >> > > A=cumsum(rand(1,5))/5; >> >> I'm relatively sure that Roger showed in an earlier thread that that >> approach produces biased results. > ================ > > Did the OP require that there not be bias? You are correct, the OP placed no such restriction in the question, including not requiring that the random numbers be drawn from a uniform random distribution. In terms of what the OP demonstrably asked for, the following would also be valid: ((0:4) + rand(1,5)) ./ 5 There was, however, some ambiguity in the OP's phrasing. |