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?


1 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.

  • $\begingroup$ Thanks for the answer! That makes sense. Since you mentioned Lehmann-Scheffé and Rao-Blackwell, is there any chance you know the answer to my question here stats.stackexchange.com/q/519714/163242 ? Sorry to ask, but I think everyone else is stumped on it, and it seems like you might know. $\endgroup$ Apr 15, 2021 at 4:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.