3
$\begingroup$

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 $\mathbb{E}(T' | S')$?
$\endgroup$

1 Answer 1

1
$\begingroup$

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.

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

Your Answer

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

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