Should the unbiased estimator of the variance of the sample proportion have (n-1) in the denominator? I know that the variance of the sample mean is:
$$\frac{\sigma^2}{n}$$
And that the unbiased estimator for that expression is:
$$\frac{\sigma^2}{n-1}$$
The variance of the sample proportion is:
$$\frac{p(1-p)}{n}$$
Does an unbiased estimator of variance of the sample proportion also need to have (n-1) in the denominator, like the unbiased estimator of the variance of the sample mean?
I feel the resoning for the denominator should be applicable for both situations, but I haven't seen it mentioned anywhere...
 A: If $\hat{p}$ is the sample proportion, then $n\hat{p} \sim \text{Binomial}(n, p)$ so
$$E[\hat{p}(1-\hat{p})] = E[\hat{p}] - E[\hat{p}^2] = E[\hat{p}] - \text{Var}(\hat{p}) - E[\hat{p}]^2 =  p(1-p) (1 - \frac{1}{n})$$
so it seems that the unbiased estimator of $\frac{p(1-p)}{n}$ would be
$$\frac{\hat{p}(1-\hat{p})}{n-1}.$$
I might have made a miscalculation somewhere though.
A: If you have a random variable $X$ with a binomial distribution with parameters $n$ and $p$, then $\mathbb E\left[\frac{X}{n} \right]=p$, so the sample proportion $Y=\frac{X}{n}$ is an unbiased estimator of $p$.
Meanwhile $\mathrm{Var}\left(Y \right) = \mathrm{Var}\left(\frac{X}{n} \right) = \frac{p(1-p)}{n}$ is the variance of the sample proportion,
implying $\mathbb E[Y^2] = \frac{p(1-p)}{n}+\frac{p^2}{n^2}$,
so $\mathbb E\left[Y(1-Y) \right] = p - \frac{p(1-p)}{n}-\frac{p^2}{n^2} = p(1-p)\frac{n-1}{n} $ and thus   $\mathbb E\left[\frac{Y(1-Y)}{n-1} \right] = \frac{p(1-p)}{n}$,
showing $\frac{Y(1-Y)}{n-1} $ is an unbiased estimator of the variance of the sample proportion $\frac{p(1-p)}{n}$
