From: Moshe Olshansky <m_olshansky_at_yahoo.com>

Date: Mon, 07 Jul 2008 23:39:05 -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 - 06:43:30 GMT

Date: Mon, 07 Jul 2008 23:39:05 -0700 (PDT)

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.

*> From: Shubha Vishwanath Karanth <shubhak_at_ambaresearch.com>
**> Subject: [R] Sum(Random Numbers)=100
*

> To: r-help@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}}
**>
**> ______________________________________________
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**> PLEASE do read the posting guide
**> http://www.R-project.org/posting-guide.html
**> and provide commented, minimal, self-contained,
**> reproducible code.
*

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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 - 06:43:30 GMT

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