I'm studying point estimation and I have found this question that seems pretty tricky to me.

If $T$ is a minimal sufficient statistic for $\theta$ with $E(T) = \tau(\theta)$, can you say that $T$ is also the UMVUE for $\tau(\theta)$?

Rao-Blackwell theorem states that an unbiased estimator $T$ for $\tau(\theta)$ can be improved using a sufficient statistic $U$ for $\theta$, i.e. $T^*=E[T|U]$ has a variance lower than the one of $T$.

Lehmann-Scheffé theorem states that $T$ must be a function of a complete sufficient statistic in order to be the unique UMVUE for $\tau(\theta)$.

But what about the fact that $T$ is minimal sufficient? Does this provide some results about $T$?

  • $\begingroup$ Hold on a second, are you looking for MVUE or the UMVUE? The 'U' in UMVUE stands for Unique, so saying you are looking for the unique UMVUE is a little confusing. $\endgroup$
    – JohnK
    Commented Jan 29, 2016 at 13:03
  • $\begingroup$ @JohnK Oh sorry, in our notation the U stays for Unbiased. The uniqueness derives from Lehmann-Scheffé theorem. $\endgroup$
    – PhDing
    Commented Jan 29, 2016 at 14:24
  • $\begingroup$ Here is what I think, the MVUE definitely has to be a sufficient statistic, otherwise you can always get a better estimator by applying the Rao-Blackwell step. The same applies to a minimial sufficient statistic as by definition it is a function of all other sufficient statistics. $\endgroup$
    – JohnK
    Commented Jan 29, 2016 at 14:28
  • $\begingroup$ Thus, we can state that being MVUE implies being a minimal sufficient statistic? $\endgroup$
    – PhDing
    Commented Jan 30, 2016 at 14:52
  • 3
    $\begingroup$ @JohnK The 'U' as I know stands for 'uniform' and not 'unique'. UMVUE is always unique whenever it exists. You know this of course, but this wasn't conveyed properly I feel. $\endgroup$ Commented Jun 27, 2018 at 11:59

1 Answer 1


Since a minimal sufficient statistic is not a complete statistic in general (cf. Is a minimal sufficient statistic also a complete statistic), the answer to your question is negative.

For a concrete example, consider $X\sim U(\theta,\theta+1)$ where $\theta$ is the parameter of interest. Here $X$ is minimal sufficient for $\theta$ but $X$ is not the UMVUE of its expectation. For details see this post.


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