[R] Sum(Random Numbers)=100
Moshe Olshansky
m_olshansky at yahoo.com
Tue Jul 8 08:39:05 CEST 2008
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}}
>
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