From: Peng Jiang <jp021_at_sjtu.edu.cn>

Date: Tue, 08 Jul 2008 14:56:40 +0800

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

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 - 07:06:00 GMT

Date: Tue, 08 Jul 2008 14:56:40 +0800

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

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 - 07:06:00 GMT

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