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
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-----------------------------------------------
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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