Solution
Unbiased Estimation of a Zero Probability
Show the problem again
Four independent observations X₁, ..., X₄ come from a Poisson distribution with unknown mean λ. The quantity of interest is e^(−λ) = P(X₁ = 0). The crude estimator 1{X₁ = 0} is unbiased. Rao-Blackwellise it by conditioning on the sufficient statistic T = X₁ + ... + X₄. If the observed total is T = 5, what is the value of the resulting estimator? Give a reduced fraction.
Worked solution
Given T = t, the vector (X₁, ..., Xₙ) is multinomial with t trials and equal cell probabilities 1/n, so X₁ ~ Binomial(t, 1/n) and P(X₁ = 0 | T = t) = (1 − 1/n)ᵗ. The estimator is therefore ((n − 1)/n)ᵀ. This is unbiased, since T ~ Poisson(nλ) gives E[sᵀ] = e^(nλ(s − 1)) and with s = (n − 1)/n this is e^(−λ). By Lehmann-Scheffé it is the unique minimum-variance unbiased estimator. With n = 4 and t = 5: (3/4)⁵ = 243/1024. The answer is 243/1024.
Source: Standard Rao-Blackwell / UMVUE example for the Poisson family; statement written for AxiomIQ.