1
$\begingroup$

Let $x$ be an observation from $X\sim Bin(n,p)$. I want to estimate $p$ and use ML estimator, $\widehat{p}=\frac{x}{n}$. I also want to estimate the variance of the estimator $\widehat{p}$. It equals: $Var(\widehat{p})=Var(\frac{X}{n})=\frac{Var(X)}{n^2}=\frac{p(1-p)}{n}$. As we don't know $p$, it is often replaced by $\widehat{p}$ in the formula. We obtain: $$Var(\widehat{p})=\frac{\widehat{p}(1-\widehat{p})}{n}.$$ In my case, as $x=0$, $\widehat{p}=0$ and thus $Var(\widehat{p})=0$. We thus see that by approximating $p$ with $\widehat{p}$, we obtain a null variance. I will explain now why it's wrongly estimated. The reason why I get $x=0$ is because $p$ is low (but not null) and because the sample size $n$ is low. But when estimating $Var(\widehat{p})$, we approximate $p$ with $\widehat{p}=0$, which gives us a null variance.

How can we correct for that?

$\endgroup$
  • 1
    $\begingroup$ If you are confident that "$p$ should be low" (but not zero), a Bayesian approach suggests itself in which you assign a prior probability distribution to $p$ which reflects this knowledge. The resulting posterior variance will then be different from zero. $\endgroup$ – Christoph Hanck Jan 23 '18 at 13:24
  • $\begingroup$ What kind of prior would you use? $\endgroup$ – Anthony Jan 23 '18 at 14:07
  • $\begingroup$ Probably a beta prior. $\endgroup$ – Christoph Hanck Jan 23 '18 at 14:08
  • $\begingroup$ See: stats.stackexchange.com/questions/82720/… $\endgroup$ – kjetil b halvorsen Dec 29 '18 at 17:55
0
$\begingroup$

Why do you need an estimator of the variance $np(1-p)$? Recommendations would probably depend on context. An easier problem is to construct a confidence interval, see Confidence interval around binomial estimate of 0 or 1. A simple ad-hoc solution to get a variance estimator when $X=0$ is to take the upper limit of that confidence interval, and base the variance estimate on that.

To find an unbiased variance estimate which is non-zero when $X=0$ is probably hopeless (unless it is sometimes negative!) Proving that would be an interesting exercise.

If these adhockeries are not good enough, then you could go for the Bayesian ideas mentioned in comments.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.