Sufficiency in Lehmann Scheffe We are wondering what sufficiency in the Lehmann Scheffe Theorem is needed for. 
Our reasoning was:


*

*If an unbiased estimator is uncorrelated with all unbiased estimators of 0, it is UMVUE

*If the estimator is from a complete family, it is uncorrelated with all unbiased estimators of 0

*Therefore an unbiased estimator from a complete family is UMVUE

*UMVUE is unique


So what exactly is sufficiency needed for? Why is completeness not enough?
Would be nice if anyone could point out the flaws in our reasoning and explain!
 A: If your Uniformly Minimum Variance Unbiased Estimator $U$ of $\theta$ were not a function of a sufficient statistic $S$ alone (a.s.), then, by the Rao-Blackwell Theorem, the random variable $V=\textrm{E}_\theta[U\mid S]$ would be an unbiased estimator of $\theta$ that has variance uniformly smaller than $U$, which contradicts the fact that $U$ is an Uniformly Minimum Variance Unbiased Estimator of $\theta$. A Lehmann-Scheffé style result without the sufficiency assumption would leave that window open.
A: The problem is (2), as others have noted. Let $X_1, ..., X_n$ be iid normal with mean $\theta$ and variance $1$. The statistic $T(X_1, ..., X_n) = X_1$ is complete, because the normal family is complete. But it is not uncorrelated with all unbiased estimators of $0$; take $\hat 0 = X_1 - X_2$. In constructing unbiased estimators of $0$ we are allowed to use the entire data, whereas completeness is only a property of the marginal distribution of $T(X_1, ..., X_n)$.
The logic in Casella and Berger is this: if $T = T(X_1, ..., X_n)$ is sufficient then it suffices to only consider the distribution of $T$ when looking for unbiased estimators, by Rao-Blackwell. This is what sufficiency is giving you - it allows you to ignore everything except $T$. So it suffices to show, if $g(T)$ is unbiased for $\theta$, that $g(T)$ is uncorrelated with every unbiased estimate of $0$, $\hat 0(T)$ [note: because of sufficiency, we have reduced the problem of showing uncorrelatedness with every estimator of $0$ to only have to show it for estimators that depend only on $T$]. But if $T$ is complete then there are no unbiased estimators $\hat 0(T)$ other than $0$ which $g(T)$ is uncorrelated with, so we are done. Of course, now that we've established that $g(T)$ is the UMVUE, it follows a posteriori that $g(T)$ is uncorrelated with all unbiased estimators of $0$, $\hat 0(X_1, ..., X_n)$, that depend on the entire sample.
A: Even though this is an old question I thought it would be helpful to have a more detailed answer.
Step 2 only follows for sure if your complete estimator $T$ is also sufficient. Suppose $U$ is any random variable with $E(U)=0$. Then
$$
E(E(U|T)) = E(U) = 0 \Rightarrow E(U|T) = 0
$$
Note that this result depends on sufficiency, because $E(U|T)$ needs to be the same function for all $\theta$, so generally it can't depend on $\theta$. It follows that
$$
Cov(T,U) = E(TU) -E(T)E(U) = E(TU) = E(TE(U|T)) = E(T*0) = 0.    
$$
Take the example by guy above. There we have $E(U|T) = E(X_1-X_2|X_1) = X_1-\theta$, a function that depends on $\theta$, and thus we don't necessarily get zero covariance with every mean zero random variable.
