X_1 through X_n are iid Poisson(lambda) and you want to estimate theta = P(X = 0) = e^{-lambda}. Starting from the unbiased estimator T = 1{X_1 = 0}, condition on the sufficient statistic S = sum X_i, give the Rao-Blackwellised estimator, and evaluate it at n = 10 with S = 20.

X_1 through X_n are iid Poisson(lambda) and you want to estimate theta = P(X = 0) = e^{-lambda}. Starting from the unbiased estimator T = 1{X_1 = 0}, condition on the sufficient statistic S = sum X_i, give the Rao-Blackwellised estimator, and evaluate it at n = 10 with S = 20.

Approach: Establish sufficiency by factorisation, then compute the conditional distribution of one count given the total, which is binomial with no unknown parameter left in it.

0.1216. The joint mass function is lambda^S * e^{-n*lambda} / prod(x_i!), which factorises into a function of S and lambda times a function of the data alone, so S is a sufficient statistic. Given S = s, the counts are multinomial across n equally likely cells, so X_1 is Binomial(s, 1/n) and P(X_1 = 0 | S = s) = ((n - 1)/n)^s, free of lambda as sufficiency guarantees. The Rao-Blackwell theorem says this conditional expectation is unbiased for e^{-lambda} with variance no larger than that of T, and at n = 10 and S = 20 it gives 0.9^20 = 0.1216, against an indicator that could only report 0 or 1. Since S is complete as well as sufficient, the Lehmann-Scheffe theorem makes ((n - 1)/n)^S the unique minimum variance unbiased estimator. The maximum likelihood value e^{-xbar} = e^{-2} = 0.1353. It differs from the unbiased estimator and is biased upward.

Follow-up: What is the variance of the Rao-Blackwellised estimator, and how does it compare with the Cramer-Rao bound for estimating e^{-lambda}?

Key concepts: Rao-Blackwell theorem, sufficient statistic, completeness, minimum variance unbiased.