You observe n = 100 independent Poisson counts with sample mean 4.0. Derive the maximum likelihood estimator of lambda, compute the Fisher information carried by one observation, and give the Cramer-Rao lower bound on the variance of any unbiased estimator of lambda at that value.

You observe n = 100 independent Poisson counts with sample mean 4.0. Derive the maximum likelihood estimator of lambda, compute the Fisher information carried by one observation, and give the Cramer-Rao lower bound on the variance of any unbiased estimator of lambda at that value.

Approach: Differentiate the Poisson log-likelihood to get the estimator, then take minus the expected second derivative to get the information per observation.

0.04. The log-likelihood is sum(x_i)*log(lambda) - n*lambda up to a constant, so the derivative sum(x_i)/lambda - n vanishes at lambda_hat = xbar = 4.0. The Fisher information for one observation is minus the expected second derivative, E[X]/lambda^2 = 1/lambda = 0.25 at lambda = 4. The Cramer-Rao bound is 1/(n*I(lambda)) = lambda/n = 4/100 = 0.04. The sample mean has Var(xbar) = lambda/n = 0.04 exactly, so it attains the bound and is an efficient unbiased estimator, which is expected because the Poisson is a one-parameter exponential family with xbar as its natural sufficient statistic.

Follow-up: What is the Cramer-Rao bound for estimating P(X = 0) = e^{-lambda}, and does the plug-in estimator attain it?

Key concepts: maximum likelihood, Fisher information, Cramer-Rao bound, efficiency.