Re: Estimating the mean of the cumulative hypergeometric?

On Jan 13, 3:43 pm, petertwocakes <petertwoca...@xxxxxxxxxxxxxx>
wrote:
This is my routine for calculating the cumulative _Binomial_
distribution, with relacement, given n and p:

double q = 1-p,  c = pow(q,n); // i.e. when k=0
double cp = c; // cumulative probability
for(double k=1; k<=n; k++)
{
    c = ((n-(k-1))/k)*c*(p/q);
    cp += c;

}

It's fast, and (for me) clear to understand, and as it's iterative,
the cost of calculating the cumulative probability is little more than
for calculating a specific probability.
So... I REALLY thought I would be able to modify this to work with no-
replacement, after all the only difference is that p&q are changing as
a result of N and m being reduced.
(m = number successes in N; k = number successes in sample(n))
Surely, in the loop above, if I could recalculate p&q by how N and m
are changing, then this would give the same result as the
hypergeometric??

I don't see that this is an efficient way to approach it. p and q are
changing, sure, but they're changing in a way that depends on the
history of the previous selections, and you don't want to be tracking
all those possibilities individually.

Your loop code for the hypergeometric should not be that different
from the binomial. You keep a running count of the product of the
binomial coefficients involving k: this is what I was alluding to
earlier. Each iteration involves multiplying by two ratios, something
like c = c*(m - k)/(k + 1)*(n - k)/(N - m - n + k + 1), depending on
exactly how you implement it. Of course, you can pre-calculate N - m -
n + 1. You do also have the overhead of initialising this multiplier
to (N - m)C(n) and calculating the denominator Ncn.
.

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