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 <[EMAIL PROTECTED]> wrote:
> From: Peng Jiang <[EMAIL PROTECTED]>
> Subject: Re: [R] Sum(Random Numbers)=100
> To: [EMAIL PROTECTED]
> Cc: [EMAIL PROTECTED], "Shubha Vishwanath Karanth" <[EMAIL PROTECTED]>
> 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
> <[EMAIL PROTECTED]
> > > wrote:
> >
> >> From: Shubha Vishwanath Karanth
> <[EMAIL PROTECTED]>
> >> Subject: [R] Sum(Random Numbers)=100
> >> To: [EMAIL PROTECTED]
> >> 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@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@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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