Lehmann–Scheffé theorem

Remarks

Completeness gives uniqueness: if $h_1(T)$ and $h_2(T)$ are both unbiased for $g(\theta)$, then $E_\theta[h_1(T) - h_2(T)] = 0$ for all $\theta$, so $h_1(T) = h_2(T)$ almost surely. Given any unbiased $\delta$ with finite variance, the Rao–Blackwell theorem gives the unbiased estimator $E[\delta \mid T]$, a function of $T$, with variance at most $\operatorname{Var}_\theta(\delta)$. By uniqueness this function is $h(T)$, which proves the claim.

Example: let $X_1, \dots, X_n$ be independent Poisson variables with mean $\lambda > 0$. The sum $T = X_1 + \cdots + X_n$ is complete and sufficient, and $T/n$ is unbiased for $\lambda$. So the sample mean is the unique minimum-variance unbiased estimator of $\lambda$.

Completeness cannot be dropped from the uniqueness argument. If $T$ is not complete, two different functions of $T$ can be unbiased for the same $g(\theta)$. Then the Rao–Blackwell step alone does not single out a best one.

Going deeper

For a student

This is the standard route to minimum-variance unbiased estimators. First find a complete sufficient statistic, often the natural statistic of an exponential family of full rank. Then find any function of it that is unbiased, for example by solving $E_\theta[h(T)] = g(\theta)$ or by Rao–Blackwellizing a simple unbiased estimator. The theorem then certifies optimality without comparing with other estimators one by one.