Is $\sqrt{\frac{p(1-p)}{n}}$ a biased but consistent estimator for proportion? Let $(B1,\ B2,\ B3,….,\ Bn)$ be a vector of i.i.d Bernoulli random variables with parameter $p$. Let $P$ be an estimator of $p$. Note that $P=\frac{B1+B2+…+Bn}{n}$ is an unbiased estimator of $p$ by the linearity of expectation. My question is we know the variance of the estimator $P$ is $\sqrt{\frac{p(1-p)}{n}}$, but what would be an unbiased estimator of the standard deviation of P, namely an estimator of $\sqrt{\frac{p(1-p)}{n}}$? I try to use $\sqrt{\frac{P(1-P)}{n}}$ as an estimator of the standard deviation, but I felt like $\sqrt{\frac{P(1-P)}{n}}$ is a biased but consistent estimator of the variance of the estimator $P$, right? I did a a simple numerical example using $p=0.5$ and $n=2$ and it seems like the estimator $\sqrt{\frac{P(1-P)}{n}}$ is extremly downward biased when $n$ is small.
 A: Note that $\text{Var}[P] = p(1 - p) / n$, whereas $\sqrt{p(1 - p)/n}$ is the standard deviation. I'm going to assume you're interested in the variance.
Since
$$E[P(1 - P) / n] = E[P]/n - E[P^2]/n = p/n - \text{Var}[P]/n - p^2/n = p(1 - p)/n - \text{Var}[P]/n,$$
the estimator $P(1 - P)/n$ is (downwards) biased. The bias converges to 0 as $n \to \infty$, though (so, as you said, it is consistent).
Since you want an unbiased estimator of $\text{Var}[P]$, you can use the following estimator of $p(1 - p)$:
$$S^2 = \frac{1}{n - 1} \sum_{i = 1}^n (B_i - P)^2.$$
So $S^2 / n$ is an unbiased estimator of $p(1 - p)/n$, i.e. the variance of $P$.
A: To show the estimator $\sqrt{\frac{P(1-P)}{n}}$ is biased, we need to show 
$$
\mathbb{E}\left[\sqrt{\frac{P(1-P)}{n}}\right]\neq \sqrt{\frac{p(1-p)}{n}}
$$
Note that
$$
\mathbb{E}\left[X^2\right]=var(X)+\mathbb{E}\left[X\right ]^2.
$$
Therefore, we have that
\begin{align}
\mathbb{E}\left[\sqrt{\frac{P(1-P)}{n}}\right]^2&=\mathbb{E}\left[\sqrt{\frac{P(1-P)}{n}}^2\right]-var\left(\sqrt{\frac{P(1-P)}{n}}\right)\\
&=\mathbb{E}\left[\frac{P(1-P)}{n}\right]-var\left(\sqrt{\frac{P(1-P)}{n}}\right)\\
&=\mathbb{E}\left[\frac{P}{n}-\frac{P^2}{n}\right]-var\left(\sqrt{\frac{P(1-P)}{n}}\right)\\
&=\mathbb{E}\left[\frac{P}{n}\right]-\mathbb{E}\left[\frac{P^2}{n}\right]-var\left(\sqrt{\frac{P(1-P)}{n}}\right)\\
&=\frac{p}{n}-\frac{\mathbb{E}\left[P^2\right]}{n}-var\left(\sqrt{\frac{P(1-P)}{n}}\right)\\
&=\frac{p}{n}- \frac{var\left(P\right) + \mathbb{E}\left[P\right]^2}{n}-var\left(\sqrt{\frac{P(1-P)}{n}}\right)\\
&=\frac{p}{n}- \frac{var\left(P\right) + p^2}{n}-var\left(\sqrt{\frac{P(1-P)}{n}}\right)\\
&=\frac{p(1-p)}{n}-\frac{var(P)}{n}-var\left(\sqrt{\frac{P(1-P)}{n}}\right)
\end{align}
Taking the square root of the last expression, it is clear that the inequality holds. Hence the estimator is biased. It is also clear that as $n\rightarrow \infty$, so equality holds as $n$ approaches infinity, so the estimator of the standard deviation of $P$ is indeed consistent. This completes the proof.
