I'm trying to understand the proof of this theorem.

  1. An unbiased estimator $T$, that is a function of a complete statistic $S$, is unique, i.e. there can't be other unbiased estimators that are functions of that statistic.

  2. Take some arbitrary unbiased estimator $T'$, use the Rao-Blackwell theorem on it with $S$, i.e. the expected value of it conditioning on a sufficient statistic, but take the statistic also complete, $T = \mathbb{E}(T' | S)$

  3. Since it's unique there can only be one of those, and it's variance is better than the estimator you started with, so it's the best (UMVUE).

I have some holes in this proof:

  • Does Rao-Blackwell guarantee that $T = \mathbb{E}(T' | S)$ will be ONLY a function of $S$ ? Or do we simply not care if it's also a function of the data, since $S$ is sufficient? Or is the completeness-implying-uniqueness doesn't require it to be ONLY a function of $S$?
  • All the proof show is that there won't be another unbiased and complete statistic, but how do we know that there isn't a non-complete unbiased estimator that will be better? I'm missing the point that shows that using the Rao-Blackwell on the complete statistic is not arbitrary. Say I have a complete statistic $S$ and a non-complete statistic $S'$, both being sufficient (as there are infinite sufficient statistics) - will $\mathbb{E}(T' | S)$ always be better than $\mathbb{E}(T' | S')$?

1 Answer 1


I'm not 100% sure about this, but maybe:

Let's define $T$ as the complete and sufficient statistic, and $g(T)$ as the unbiased estimator - $\mathbb{E}(g(T)) = \theta$.

Let $T'$ be the non-complete and sufficient statistic, and $h(T')$ the unbiased estimator, $\mathbb{E}(h(T')) = \theta$.

From Rao-Blackwell let's take the non-complete statistic and improve it by conditioning on the complete and sufficient statistic: $\mathbb{E}(h(T')|T) := f(T)$. This is a function of $T$, which is at-least as good as $h(T').$

Now, $\mathbb{E}(g(T)-f(T)) = 0$, but since this is a function of $T$ and $T$ is complete, it implies that $g(T) = f(T)$, which means that $g(T)$ is at-least as good as $h(T')$.

Which proves that an unbiased estimator using a complete statistic is at least as good as one using a non-complete statistic.

  • 1
    $\begingroup$ It also proves it is unique when the statistic is both complete and sufficient. $\endgroup$
    – Xi'an
    Commented Mar 13, 2020 at 11:17

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