# Unbiased estimator of the standard deviation of the proportion

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.

• An estimate of the (large sample) standard error of $\hat p$ is $\sqrt\frac{\hat{p}(1-\hat{p})}{n}$, because standard error of $\hat p$ is just the s.d. of $\hat p$. – StubbornAtom Mar 16 at 15:57

## 1 Answer

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.

• Thanks! Then, what would be the unbiased estimator? Moreover, is there any reason to use the standard error after replacing $p$ in the s.d. with $\hat{p}$? – user3509199 Mar 16 at 15:59