Unbiased estimator

I just derive the method
E(Y/n(1-Y/n))
=E(Y/n)-E(Y^2/n^2)
=(1/n)E(Y)-(1/n^2)E(Y^2)
=p-(1/n^2)E(Y^2)
=p-(1/n^2)(n^2p^2+npq)
=p-p^2-(pq/n)
=p(p-1)-(pq/n)
=pq-(pq/n)
=pq(n-1)/n
So if we want to have an unbiased estimator of pq
Then we can use (n/n-1)(Y/n)(1-Y/n)
Now if we want to have the unbiased estimator of V(Y) in term of Y and n.
Then we can use
(n^2/n-1)(Y/n)(1-Y/n)

I don't know whether you can use the rule of Var(Y)= E(Y^2)- E(Y)^2

Var(Y)
= E(Y^2)- E(Y)^2
= Y^2/n - Y^2/n^2
= Y^2 (n-1)/(n^2)

2007-01-25 23:57:31 補充：
This is not an estimator. Your question had already said Y is the distribution itself, *not the estimation*, so the variance of Y is calculated, not estimated.Estimation of variance based on empirical data is entirely different matter.