Re: random positions of fixed length

[David L Lorenz <lorenz@usgs.gov>]

Jewel,

  I think you are trying to describe a hypergeometric distribution. Because of the size of the sample, in this case it is very similar to the binomial distribution, except that sampling is done without replacement.

  The probability that any one element is selected is n/N. That is also true for the binomial distribtuion, the difference between the hypergeometric and binomial distributions would appear when trying to describe the probability of observing a specific large group of elements.
Dave

-----s-news-owner@lists.biostat.wustl.edu wrote: -----

To: s-news@lists.biostat.wustl.edu
From: Jewel Bright <jwlbright@yahoo.com>
Sent by: s-news-owner@lists.biostat.wustl.edu
Date: 04/14/2007 06:19PM
Subject: [S] random positions of fixed length

Folks:

 

Sorry for a sort of dumb question. Suppose I have a sequence of random vectors A[N,n] each of length N (say, N=10000). Suppose that n<<N (say, n=100) are ones, and the rest (N-n) are zeros. Positions of "ones" in each of these vectors are random. I want to represent the vectors A[N,n]  as if they are drawn (independently, for simplicity) from some probabilistic distribution. Which distribution is it? Some people say it is binomial. I say that it is not. In binomial, the probability , p, is fixed whereas the number of "successes", n, is not fixed. In my case, n is fixed and known a priori. What is random are the positions of "ones" amongst zeros.

 

So, what is the correct way to present this kind of randomness?

It is probably a statistical textbook question. But I am not a professional statistician, more closer to mathematical biophysics, so simple statistical questions are sometimes difficult to resolve.

 

thanks

 

Jewel   

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