An estimator is efficient if it reaches the Cramér-Rao Lower Bound and since it is efficient, it is also the UMVU estimator of the parametric function $\tau(\theta)$. But Cramér-Rao inequality and the related lower bound hold if and only if two assumptions are satisfied: 1) the support of $X's$ does not depend on $\theta$ and 2) the first derivative wrt to $\theta$ and the intgral wrt to $\mathbf{x}$ are interchangeable.

If we are in a case in which, instead, the support of the $X's$ depends on $\theta$, e.g. if $f(x;\theta)\sim U(0,\theta)$, can we state that an efficient estimator does not exist since the Cramér-Rao inequality does not hold? Or there are some other ways to find an efficient estimator?


1 Answer 1


The regularity conditions for the CLRB indeed do not hold for the $U(0,\theta)$ since as you said the support depends on the unknown parameter and hence it is not common. For a full list of the required conditions you can check:

What are the regularity conditions for Likelihood Ratio test

You can still get a good estimator using the maximum likelihood estimator and the sufficient statistic though. It is known that the mle is a function of the sufficient statistic and in this case they coincide. Hence if we can find an unbiased function of it by the Rao-Blackwell theorem we have an MVUE. Additionally it can be shown that this family is complete and so by the Lehmann-Scheffe theorem this estimator would be the Unique MVUE.

I leave it to you to find that estimator.

  • $\begingroup$ But the estimator I find cannot be efficient, right? So, is it true that if the regularity conditions of CRLB are not satisfied, an efficient estimator (i.e. that reaches CRLB) does not exist? $\endgroup$
    – PhDing
    Dec 29, 2015 at 9:41
  • $\begingroup$ @Alessandro In cases of violation of the regularity conditions, you can even find estimators that have lower variance that what the CRLB would predict. It's quite a nuissance. $\endgroup$
    – JohnK
    Dec 29, 2015 at 9:43
  • $\begingroup$ Ok, I think I've understood. My doubt arose from the solution of an exercise that asked for the $\tau(\theta)$ that could be estimated using an efficient estimator (i.e. that reaches CRLB). I was proceeding decomposing the score function but then I read that since the support of X contained $\theta$ no parametric function could be estimated efficiently. $\endgroup$
    – PhDing
    Dec 29, 2015 at 10:21
  • $\begingroup$ @Alessandro The CRLB requires the regularity conditions so it's a bit restrictive. What you can take from this is that an efficient estimator might exist in the class of unbiased estimators even if they do not hold. $\endgroup$
    – JohnK
    Dec 29, 2015 at 10:23
  • 1
    $\begingroup$ Ok, it's clear! Indeed, my book refers to CRLB as a way to easily find the UMVUE once the CRLB can be computed and once an unbiased statistic for $\tau(\theta)$ is easily located. It should not be seen as a unique way to find efficient estimator but just one of the ways. Thank you $\endgroup$
    – PhDing
    Dec 29, 2015 at 10:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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