TITLE: What's the difference between Rao-Blackwell Theorem and Lehmann-Scheffé Theorem?
QUESTION [16 upvotes]: I know that the Rao-Blackwell theorem states that an unbiased estimator given a sufficient statistic will yield the best unbiased estimator. Is the only difference between Lehmann-Scheffé and Rao-Blackwell that in Lehmann-Scheffé, you need an unbiased estimator that is based on a complete sufficient statistic? I am also having a hard time conceptually understanding the definition of a complete statistic.

REPLY [3 votes]: As Michael Hardy said, Rao-Blackwell only guarantees to improve (or not hurt) your original unbiased estimator. I.e. you start with some unbiased estimator $T(\underline x)$, then you improve it by taking the expected value conditioned on a sufficient statistic $T'(\underline x) := \mathbb{E}[T(\underline x) | S(\underline x)]$. Improve meaning that it's variance will be less or equal than the original estimator variance. 
But something funny happens if you add the completeness to the statistic ($S(\underline x)$) - you get uniqueness. I.e. there will only be one unbiased estimator that will be be a function of $S$. And so if you start with some unbiased estimator, and use the Rao-Blackwell method with a sufficient & complete statistic, the result will be the only unbiased estimator for that parameter, and it will have a better variance than any of the original estimator. So it's (U)MVUE (Uniform-ly minimal variance unbiased estimator). 
Completeness is a restriction for a statistic that if $\mathbb{E}(g(\underline x)) = 0$ then $g(\underline x) = 0$ always, with probability 1 for any $x$. Take a Bernoulli distribution - and $g(\underline x) = x_1 - x_2$. $\mathbb{E}(x_1 -x_2) = 0$ but $g(x) \neq 0$ when there are different results between the first and second experiment. It's only equal zero when they are the same. So it's not a complete statistic. $h(\underline x) = \sum x_i$ however is a complete statistic, as if it's expected value is equal to zero, it means that it itself is equal to zero with probability of 1.