Suppose that one can construct an unbiased estimator $X$ of the quantity $E$, is there a way of getting an unbiased and positive estimator of $E^2$? Indeed, if $X_1$ and $X_2$ are two independent copies of the estimator $X$, the quantity $X_1 X_2$ is an unbiased estimator of $E^2$, but is not positive in general.

Motivation: when interested in the marginal distribution $\pi(\theta) = \int_{X} \pi(\theta, x) \, dx$ of an intractable distribution $\pi(\theta,x)$, one can use an MCMC algorithm if one can compute $\pi(\theta)$. This is generally not the case. Nevertheless, it is now well-known that this can also be done if one can only construct an unbiased and positive estimator of $\pi(\theta)$, hence the question.

  • 1
    $\begingroup$ Yeah, that is always a problem with unbiased estimators, they are not invariant and you have to find 'ad hoc' estimators. $\endgroup$ – user10525 Apr 26 '12 at 23:19
  • 1
    $\begingroup$ In the fully nonparametric case, the uniformly minimum variance unbiased estimator is $\frac{2}{n(n-1)} \sum_{i < j} X_i X_j$. It's not guaranteed to be positive, though. $\endgroup$ – cardinal Apr 27 '12 at 1:34

Your Answer

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

Browse other questions tagged or ask your own question.