From: Moshe Olshansky <m_olshansky_at_yahoo.com>

Date: Tue, 08 Jul 2008 00:15:04 -0700 (PDT)

R-help_at_r-project.org mailing list

https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Tue 08 Jul 2008 - 07:18:33 GMT

Date: Tue, 08 Jul 2008 00:15:04 -0700 (PDT)

For arbitrary lambda_i it can take years until the sum of 50 such random variables is 100!
But if one makes lambda_i = 2, then the probability that the sum of 50 of them equals 100 is about 1/sqrt(2*pi*100), so on average that sequence of 50 numbers must be generated about sqrt(2*pi*100)) ~ 25 times, which is very reasonable.

- On Tue, 8/7/08, Peng Jiang <jp021_at_sjtu.edu.cn> wrote:

*> From: Peng Jiang <jp021_at_sjtu.edu.cn>
**> Subject: Re: [R] Sum(Random Numbers)=100
**> To: m_olshansky_at_yahoo.com
**> Cc: r-help_at_stat.math.ethz.ch, "Shubha Vishwanath Karanth" <shubhak_at_ambaresearch.com>
**> Received: Tuesday, 8 July, 2008, 4:56 PM
**> Hi,
*

> I am afraid there is no other way except using brute

*> force, that
**> is , loop until their sum reaches your expectation.
**> it is easy to figure out this probability by letting
**> their sum to be
**> a new random variable Z and Z = X_1 + \ldots + X_n
**> where X_i ~ Poisson({\lambda}_i) . By calculating
**> their moment
**> generate function we can find the pmf of Z which is
**> a new Poisson random variable with the parameter
**> \sum_{i}{{\lambda}_i}.
**>
**> and Moshe Olshansky's method is also correct except
**> it is based on
**> the conditioning.
**> On 2008-7-8, at 下午1:58, Shubha Vishwanath Karanth
**> wrote:
**> On 2008-7-8, at 下午2:39, Moshe Olshansky wrote:
**>
**> > If they are really random you can not expect their sum
**> to be 100.
**> > However, it is not difficult to get that given that
**> the sum of n
**> > independent Poisson random variables equals N, any
**> individual one
**> > has the conditional binomial distribution with size =
**> N and p = 1/n,
**> > i.e.
**> > P(Xi=k/Sn=N) = (N over k)*(1/n)^k*((n-1)/n)^(N-k).
**> > So you can generate X1 binomial with size = 100 and p
**> = 1/50; if X1
**> > = k1 then the sum of the rest 49 must equal 100 - k1,
**> so now you
**> > generate X2 binomial with size = 100-k1 and p = 1/49;
**> if X2 = k2
**> > then generate X3 binomial with size = 100 -(k1+k2) and
**> p = 1/48, etc.
**> >
**> > Why do you need this?
**> >
**> >
**> > --- On Tue, 8/7/08, Shubha Vishwanath Karanth
**> <shubhak_at_ambaresearch.com
**> > > wrote:
**> >
**> >> From: Shubha Vishwanath Karanth
**> <shubhak_at_ambaresearch.com>
**> >> Subject: [R] Sum(Random Numbers)=100
**> >> To: r-help_at_stat.math.ethz.ch
**> >> Received: Tuesday, 8 July, 2008, 3:58 PM
**> >> Hi R,
**> >>
**> >>
**> >>
**> >> I need to generate 50 random numbers (preferably
**> poisson),
**> >> such that
**> >> their sum is equal to 100. How do I do this?
**> >>
**> >>
**> >>
**> >>
**> >>
**> >> Thank you,
**> >>
**> >> Shubha
**> >>
**> >>
**> >>
**> >> This e-mail may contain confidential and/or
**> privileged
**> >> i...{{dropped:13}}
**> >>
**> >> ______________________________________________
**> >> R-help_at_r-project.org mailing list
**> >> https://stat.ethz.ch/mailman/listinfo/r-help
**> >> PLEASE do read the posting guide
**> >> http://www.R-project.org/posting-guide.html
**> >> and provide commented, minimal, self-contained,
**> >> reproducible code.
**> >
**> > ______________________________________________
**> > R-help_at_r-project.org mailing list
**> > https://stat.ethz.ch/mailman/listinfo/r-help
**> > PLEASE do read the posting guide
**> http://www.R-project.org/posting-guide.html
**> > and provide commented, minimal, self-contained,
**> reproducible code.
**>
**> -----------------------------------------------
**> Peng Jiang 江鹏 ,Ph.D. Candidate
**> Antai College of Economics & Management
**> 安泰经济管理学院
**> Department of Mathematics
**> 数学系
**> Shanghai Jiaotong University (Minhang Campus)
**> 800 Dongchuan Road
**> 200240 Shanghai
**> P. R. China
*

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