Hi Peter, H Berwin,
thanks a lot for your clarifications, it makes more sense now. But having
our input and thinking a little bit more about the problem, I realized that
I am simply interested in the pdf p(y) that y *number* of entities (which
ones is irrelevant) in N are are *not* drawn after the sampling process has
been completed. Even simpler (I guess), in a first step, I would only need
the mean number of expected non-drawn entities in N (pMean).
The range of my values:
N is in the range of 1 --- 100 000
x is in the range of 10 --- 40 000 000
n is in the range of 20
I guess in cases where n*x is substantially smaller then N, I could simply
use a binominal distribution for n*x samples to approximate it -- right?
For cases where n*x is substantially bigger then N, I can safely (especially
in the context of my simulation) assume that all entities in N are drawn at
least once.
But what about the range in between?
Thanks again,
Cheers,
Rainer
On Sat, Sep 25, 2010 at 5:19 PM, Peter Dalgaard <pda...@gmail.com> wrote:
> On 09/25/2010 04:24 PM, Rainer M Krug wrote:
> > Hi
> >
> > This is OT, but I need it for my simulation in R.
> >
> > I have a special case for sampling with replacement: instead of sampling
> > once and replacing it immediately, I sample n times, and then replace all
> n
> > items.
> >
> >
> > So:
> >
> > N entities
> > x samples with replacement
> > each sample consists of n sub-samples WITHOUT replacement, which are all
> > replaced before the next sample is drawn
> >
> > My question is: which distribution can I use to describe how often each
> > entity of the N has been sampled?
> >
> > Thanks for your help,
> >
> > Rainer
> >
>
> How did you know I was in the middle of preparing lectures on the
> variance of the hypergeometric distribution and such? ;-)
>
> If you look at a single item, the answer is of course that you have a
> binomial with size=x and prob=n/N. The problem is that these binomials
> are correlated between items.
>
> If you can make do with a 2nd order approximation, then the covariances
> between the indicators for two items being selected is easily found from
> the symmetry and the fact that if you sum all N indicators you get the
> constant n. I.e. the variance is p(1-p) and the covariance is
> -p(1-p)/(N-1). For sums over repeated samples, just multiply everything
> by the number x of samples.
>
> If you intend to just count the frequency of a particular feature in
> each of your n-samples, i.e., you have x replications of a
> hypergeometric experiment, then you can do somewhat better by computing
> the explicit convolution of x hypergeometrics (convolve(x, rev(y),
> type="o") and Reduce() are your friends). I'm not sure this is actually
> worth the trouble, but it should be doable for decent-sized N and x.
>
>
>
> --
> Peter Dalgaard
> Center for Statistics, Copenhagen Business School
> Phone: (+45)38153501
> Email: pd....@cbs.dk Priv: pda...@gmail.com
>
--
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Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology,
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