How can one show that $\bar{X}$ is the best unbiased estimator for $\lambda$ without using the Cramèr-Rao lower bound?

Question

Assume we have the random sample $X_1, \dots, X_n$ with mean $\mu$ and variance $\sigma^2 < \infty$. We have that $E[S^2] = \sigma^2$, where $S^2 = \sum_{i = 1}^n \dfrac{(X_i - \bar{X})^2}{n - 1}$ and $\bar{X} = \sum_{i = 1}^n \dfrac{X_i}{n}$. Now assume the $X_i$ are Poisson random variables with parameter $\lambda$. I was trying to find the best unbiased estimator $\bar{X}$ for $\lambda$ using the Cramèr-Rao lower bound. But, apparently, this can also be done without using the Cramèr-Rao lower bound. How can one show that $\bar{X}$ is the best unbiased estimator for $\lambda$ without using the Cramèr-Rao lower bound?

Answer 1

The Lehmann-Scheffé Theorem says that any unbiased estimator that is a function of a complete sufficient statistic is the minimum variance unbiased estimator.

Here $\sum_i X_i$ is a complete sufficient statistic for $\lambda$ and $\bar X$ is a function of it, and is unbiased for $\lambda$, so it is best unbiased.

You might then ask if this really avoids using Cramèr-Rao bound under the hood somewhere. It does. The proof relies on the Rao-Blackwell theorem, which says the conditional expectation of any estimator given a sufficient statistic has smaller mean squared error than the estimator you started with. The Cramèr-Rao bound works with the covariance of the estimator and the score function, and uses the Cauchy-Schwarz inequality.