$Y_{1...n}\sim \operatorname{Bin}(1,p)$, iid, and I need to find an unbiased estimator for $\theta=\operatorname{var}(y_i)$.

I did some calculations and I think that the answer is $p(1-p)-\frac{p(1-p)}{n}$

  • Is this correct?
  • If not, how can I find an unbiased estimator?
  • 1
    $\begingroup$ If $p=\frac{\sum{y}}{n}$, it's not unbiased, as you can check by working out its expectation. Why don't you show your calculations & perhaps someone will point out the error. $\endgroup$ Commented Jun 15, 2013 at 13:26

2 Answers 2


This answer cannot be correct. An estimator cannot depend on the values of the parameters: since they are unknown it would mean that you cannot compute the estimate.

An unbiased estimator of the variance for every distribution (with finite second moment) is

$$ S^2 = \frac{1}{n-1}\sum_{i=1}^n (y_i - \bar{y})^2.$$

By expanding the square and using the definition of the average $\bar{y}$, you can see that

$$ S^2 = \frac{1}{n} \sum_{i=1}^n y_i^2 - \frac{2}{n(n-1)}\sum_{i\neq j}y_iy_j,$$

so if the variables are IID,

$$E(S^2) = \frac{1}{n} nE(y_j^2) - \frac{2}{n(n-1)} \frac{n(n-1)}{2} E(y_j)^2. $$

As you see we do not need the hypothesis that the variables have a binomial distribution (except implicitly in the fact that the variance exists) in order to derive this estimator.

  • 1
    $\begingroup$ I think the OP is distinguishing between (small) $p$ the statistic $\frac{\sum{y}}{n}$ & (big) $P$ the binomial parameter, though perhaps not. $\endgroup$ Commented Jun 15, 2013 at 17:40
  • $\begingroup$ Ah, I did not think about this. Is it something standard? $\endgroup$
    – gui11aume
    Commented Jun 15, 2013 at 18:53
  • $\begingroup$ No, just my guess. More standard would be to use Greek letters for parameters, capital Latin letters for random variables & small Latin letters for their observed outcomes. In any case, $p$ was introduced without definition. $\endgroup$ Commented Jun 15, 2013 at 19:00

@gui11aume is right of course. An outline of a derivation specific to a $\operatorname{Bin}(1,\pi)$ distribution follows:

  1. Find the variance in terms of $\pi$ to reparameterize the probability mass function: $$\theta=\operatorname{Var}{Y_i}=\pi(1-\pi)$$
  2. Find the maximum-likelihood estimator of $\theta$: $$\hat\theta=\frac{\sum{y_i}}{n}\left(1-\frac{\sum{y_i}}{n}\right)$$
  3. Calculate its expectation: $$\newcommand{\E}{\operatorname{E}}\E\hat\theta=\theta\cdot\frac{n-1}{n}.$$ Note thankfully that the bias term is a constant.
  4. Write the unbiased estimator: $$\tilde\theta=\frac{\hat\theta}{\frac{n-1}{n}}=\frac{\sum{y_i}}{n}\left(1-\frac{\sum{y_i}}{n}\right)\cdot\frac{n}{n-1}=p(1-p)\cdot\frac{n}{n-1}$$ where $p$ is the statistic $\frac{\sum{y_i}}{n}$

Because $\sum{y}$ is sufficient & complete, $\tilde\theta$ is not just any unbiased estimator of the population variance, but the unique minimum-variance unbiased estimator.


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