# Efficient estimators and CRLB

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?

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.

• 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? – PhDing Dec 29 '15 at 9:41
• @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. – JohnK Dec 29 '15 at 9:43
• 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. – PhDing Dec 29 '15 at 10:21
• @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. – JohnK Dec 29 '15 at 10:23
• 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 – PhDing Dec 29 '15 at 10:33