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?

  • $\begingroup$ Do your notes claim this is an unbiased estimator? $\endgroup$
    – whuber
    Commented May 2, 2020 at 18:13
  • $\begingroup$ Nope it does not, I was just wondering if it was. Thanks so much for the answer! Just saw it now, will read through it carefully. $\endgroup$
    – Hello
    Commented May 3, 2020 at 5:34

2 Answers 2


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).

  • $\begingroup$ Why then isn't the unbiased estimator used? $\endgroup$
    – Jon
    Commented Mar 17, 2021 at 17:19
  • 1
    $\begingroup$ @Jon Since the parameter $p$ is estimated by the sample mean (in an unbiased way), and that in turn determines the population variance, it is rare for anyone to need a separate estimate of the population variance in a Binomial setting or to be able to do anything useful with it that cannot already be carried out with the estimate of $p.$ $\endgroup$
    – whuber
    Commented Mar 17, 2021 at 17:58
  • $\begingroup$ What about for example for a confidence interval for $p$? $\endgroup$
    – Jon
    Commented Mar 17, 2021 at 18:28
  • 1
    $\begingroup$ @Jon Most of them just use the estimate of $p.$ (There are many confidence interval procedures for Binomial data: search our site for "Clopper" to find the best summaries.) Intuitively it wouldn't make much sense to employ a separate estimate of the variance that was inconsistent with the estimate of $p$--and the simple mathematical fact is that any estimate of the variance that is a (known) constant multiple of another will give the same procedure. The only thing that changes is how you describe it. $\endgroup$
    – whuber
    Commented Mar 17, 2021 at 18:34
  • 1
    $\begingroup$ @darkgbm That's a standard technique in combinatorics. Research formal power series. $\endgroup$
    – whuber
    Commented Oct 4, 2023 at 13:12

Here's an approach using the following variance formula and rule


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.



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

$E[\hat{p}(1-\hat{p})]=p - p^2 - \frac{p(1-p)}{n} = \frac{n-1}{n}p(1-p)$

  • $\begingroup$ +1 Simple and elegant. $\endgroup$
    – whuber
    Commented Nov 17, 2021 at 18:13

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