Unbiased estimator of the standard deviation of the proportion

Question

For the sample proportion $\hat{p}$, standard error is known as $\sqrt\frac{\hat{p}(1-\hat{p})}{n}$.

I wonder whether this is the unbiased estimator for the standard deviation of the sample proportion, which is $\sqrt\frac{p(1-p)}{n}$ where $p$ is the true proportion.

Answer 1

Suppose that $X\sim \text{Bin}(n,p)$, so that $\hat{p}=X/n$. It follows that $$E\hat{p}=np/n=p; \quad {E\hat{p}^2}=n^{-2} EX^2=\frac{np(1-p)+n^2p^2}{n^2}=\frac{p(1-p)}{n}+p^2.$$ So $$ E\frac{\hat{p}(1-\hat{p})}{n}=\frac{1}{n}E\hat{p}-\frac{1}{n}E\hat{p}^2=\frac{p(1-p)}{n}-\frac{p(1-p)}{n^2} $$ By Jensen's inequality (square root function is concave), it follows that $$ E\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\leq\sqrt{E\frac{\hat{p}(1-\hat{p})}{n}}=\sqrt{\frac{p(1-p)}{n}-\frac{p(1-p)}{n^2}}<\sqrt{\frac{p(1-p)}{n}} $$ as long as $p\neq 0,1$.

It follows that the estimator is not unbiased.