For a Poisson(λ) sample, the sample mean X̄ is the UMVUE of λ. A statistician proposes a ridge-shrinkage estimator that is slightly biased toward zero but has substantially lower mean squared error than X̄ in simulation. Should the statistician prefer the UMVUE?
AYes — the UMVUE is optimal by definition and cannot be outperformed by any estimator
BYes — unbiasedness is a non-negotiable requirement for valid statistical inference
CNo — the UMVUE is optimal only among unbiased estimators; a biased estimator with lower MSE may be preferable in practice
DNo — the ridge estimator must also be a UMVUE if it is computed from a sufficient statistic
UMVUE minimizes variance within the class of *unbiased* estimators — it is not globally optimal across all estimators. Mean squared error = bias² + variance. A biased estimator can achieve lower total MSE by trading a small increase in bias for a larger reduction in variance — which is exactly what ridge, Bayes, and shrinkage estimators do. The UMVUE is a theoretically important benchmark, but the unbiasedness constraint is a choice, not a law of nature.
Question 2 Multiple Choice
A statistician argues: 'I have a sufficient statistic T and found an unbiased function h(T) of it. By Rao-Blackwell, h(T) must be the UMVUE.' What is the critical flaw in this argument?
ANothing — any unbiased function of a sufficient statistic is the UMVUE by Rao-Blackwell
BRao-Blackwell only applies to maximum likelihood estimators, not arbitrary sufficient statistics
CRao-Blackwell shows conditioning on T cannot increase variance, so the best unbiased estimator is a function of T — but uniqueness (UMVUE status) also requires T to be complete
DSufficient statistics only exist for exponential families, so the argument fails in general
Rao-Blackwell establishes that among unbiased estimators, the best must be a function of T — but if T is not complete, there may be *multiple* unbiased functions of T, and the argument doesn't identify which minimizes variance uniformly. Completeness (the Lehmann-Scheffé condition) rules out redundancy: the only function of T with zero expectation for all θ is zero itself. This guarantees *uniqueness* — there is at most one unbiased function of T, so if you find one, it's the UMVUE.
Question 3 True / False
A UMVUE minimizes variance uniformly over all values of θ — meaning it beats every other unbiased estimator at every possible parameter value, not just on average.
TTrue
FFalse
Answer: True
'Uniformly' is the key word. UMVUE doesn't find the unbiased estimator with lowest *average* variance — it is the estimator whose variance is ≤ every other unbiased estimator's variance for *every* value of θ. This is a much stronger condition than minimizing expected variance under a prior. Some problems have no UMVUE because no single unbiased estimator uniformly dominates all others.
Question 4 True / False
A UMVUE usually achieves the Cramér-Rao lower bound.
TTrue
FFalse
Answer: False
The Cramér-Rao bound is a lower bound on variance for unbiased estimators, but it is not always tight (achievable). A UMVUE minimizes variance among unbiased estimators, but that minimum may still be strictly greater than 1/I(θ) if no unbiased estimator achieves the bound. Achieving the CR bound is sufficient for UMVUE status, but not necessary — a UMVUE is the best available within unbiased estimation regardless of whether it reaches the theoretical floor.
Question 5 Short Answer
What role does *completeness* of a sufficient statistic play in establishing a UMVUE, and why is sufficiency alone insufficient?
Think about your answer, then reveal below.
Model answer: A sufficient statistic T captures all information in the data about θ. Rao-Blackwell shows the best unbiased estimator must be a function of T — but without completeness, there may be multiple unbiased functions of T, and we cannot identify which minimizes variance uniformly. Completeness rules out redundancy: it guarantees the only function of T with zero expectation for all θ is the zero function. This means there is at most one unbiased function of T — so if you find one, it is uniquely the UMVUE (Lehmann-Scheffé theorem).
Completeness is what converts 'among the best candidates' into 'the unique best.' Without it, you may have reduced variance by conditioning on T (Rao-Blackwell), but you can't claim you've found the minimum-variance unbiased estimator. The exponential family structure guarantees completeness of the natural sufficient statistic, which is why UMVUE results are cleanest for exponential families.