An unbiased estimate for population variance Specifically, my notes claim that when calculating confidence intervals for population proportion $p$, the estimator for the population variance is given by $\frac{P_sQ_s}{n}$, where $P_s=\frac{X}{n}$ is the random variable for the proportion of successes. Thus we have $E(P_s)=p$ and Var$(P_s)=\frac{pq}{n}$. My question is, shouldn't we have $\frac{P_sQ_s}{n} \times \frac{n}{n-1}$ as the unbiased estimator instead?
Also, from my understanding, we say that $T$ is an (unbiased) estimator of the population parameter $\theta$ if we have $E(T)=\theta$. So is there any way to prove/disprove $E(\frac{P_sQ_s}{n})=\sigma^2$, where $\sigma^2$ is the population variance?  
 A: Presumably $Q_s(X) = 1 - P_s(X) = (n-X)/n.$
Writing $q=1-p$, let's work out the expectation of $n^2P_s(X)Q_s(X)$ using the definition of expectation, the formula for Binomial probabilities, and the Binomial Theorem:
$$\eqalign{
E\left[n^2P_s(X)Q_s(X)\right] &= E\left[X(n-X)\right] \\
&= \sum_x \Pr(X=x)\, x(n-x) & \text{(Definition of expectation)} \\
&= \sum_{x=0}^n \binom{n}{x}p^x q^{n-x}\, x(n-x) &\text{(Binomial distribution)} \\
&=\sum_{x=0}^n \binom{n}{x}\, pq \frac{\partial^2}{\partial p\partial q} \left(p^x\,q^{n-x}\right) \\
&=  pq \frac{\partial^2}{\partial p\partial q}\sum_{x=0}^n \binom{n}{x}\, p^x\,q^{n-x} & \text{(Linearity of differentiation)}\\
&= pq \frac{\partial^2}{\partial p\partial q}\left(p+q\right)^n &\text{(Binomial Theorem)}\\
&= pq\,n(n-1)(p+q)^{n-2}.
}$$
(When $n=1$ or $n=0$ the result is just $0.$)  Plugging in $p+q=1$ gives
$$E\left[n^2P_sQ_s\right]  = n(n-1)pq$$
for all $n,$ whence for $n\gt 1,$
$$E\left[\frac{1}{n-1}\,P_s(X)Q_s(X)\right] = \frac{pq}{n}=\operatorname{Var}\left(P_s(X)\right).$$
Therefore $P_s(X)Q_s(X)/(n-1)$ is an unbiased estimator of the variance of $X/n$ (and so obviously $P_s(X)Q_s(X)/n$ is not: it is biased).
A: Here's an approach using the following variance formula and rule
$Var(\hat{p})=\frac{p(1-p)}{n}=E[\hat{p}^2]-E[\hat{p}]^2$
where $\hat{p}$ is the sample proportion of times an indicator variable is 1 in a simple random sample of size $n$, i.e. the mean of an indicator variable, and $p$ is the corresponding population proportion for that indicator variable.
Suppose we estimate the population variance for that indicator variable, which is $p(1-p)$ (in terms of the population proportion $p$), using the estimator $\hat{p}(1-\hat{p})$ (which uses the sample statistics only). This estimator is biased and multiplying it by $n/(n-1)$ would make it unbiased.
Proof:
$E[\hat{p}(1-\hat{p})]=E[\hat{p}-\hat{p}^2]=p-E[\hat{p}^2]$
Rearranging the variance formula and rule above, we get $E[\hat{p}^2]=p^2 + \frac{p(1-p)}{n}$. Plugging that in we get
$E[\hat{p}(1-\hat{p})]=p - p^2 - \frac{p(1-p)}{n} = \frac{n-1}{n}p(1-p)$
