# Rao-Blackwell part of the Lehmenn-Scheffe theorem

I'm trying to understand something about the proof of this theorem. As I understand the proof is something like this:

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

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 $$h(T')$$.