Turn unbiasedness into uniqueness

Try

Could two different unbiased functions of the same sufficient statistic estimate p in a binomial experiment? For one Bernoulli observation, let a(0) and a(1) be the differences between their reported values. Unbiasedness would require a(0)(1−p)+a(1)p=0 for every p in the open unit interval. A linear polynomial vanishing throughout an interval has both coefficients zero, forcing both differences to vanish. This small argument illustrates completeness: no nonzero integrable function of the statistic can have expectation zero at every parameter.

Predict

Predict why “at every parameter” is essential. At p=1/2, the nonzero function a(S)=S−1/2 has expectation zero for a single Bernoulli observation. It does not have zero expectation for all p. Checking a finite list of parameter values or seeing nearly zero simulated averages cannot establish completeness. Completeness is a statement about a family of distributions, not one realized dataset or one selected probability law. It is also distinct from sufficiency, which concerns the conditional distribution of the original data.

Experiment

For S∼Bin(2,p), write the zero-expectation condition and divide by (1−p)². Set t=p/(1−p), which ranges over all positive real numbers.

Value s	Binomial probability	Term after division
0	(1−p)²	a(0)
1	2p(1−p)	2a(1)t
2	p²	a(2)t²

The resulting polynomial is a(0)+2a(1)t+a(2)t². If it is zero for all positive t, each coefficient is zero. Extend the argument to general n, identifying the nonzero binomial coefficients that multiply the unknown a(s). This is an actual proof for the finite binomial family, not an appeal to a broad theorem name.

Understand

A family of laws of T is complete if every measurable a with Eθ∣a(T)∣ finite and Eθa(T)=0 for every θ must satisfy a(T)=0 almost surely under every family member. Suppose T is complete and sufficient, and δ(T) is a finite-variance unbiased estimator of g(θ). Take any other finite-variance unbiased estimator U. Rao–Blackwell conditioning produces V(T), still unbiased and with no larger variance than U. Completeness applied to V−δ makes these functions equal almost surely. Therefore δ has variance no larger than any competitor. This is the Lehmann–Scheffé argument for a uniformly minimum-variance unbiased estimator, abbreviated UMVU.

The theorem does not remove its comparison-class restriction. It concerns unbiased estimators with the needed moments, and “uniformly” means for every parameter in the specified model. It does not claim minimum mean squared error against arbitrary biased estimates. In finite examples, uniqueness is literal on supported values; in continuous models, equality is an almost-sure statement with the corresponding version qualifications.

Apply

Use S(S−1)/(n(n−1)) to estimate p² when n≥2. Expand S(S−1) as the sum of XᵢXⱼ over ordered distinct pairs. Each term has expectation p² by independence, and there are n(n−1) terms. The estimator is an unbiased function of the complete sufficient binomial total, so it is UMVU. For n=4 and observed S=2 its value is 1/6, whereas the squared sample proportion is 1/4. Explain why the disagreement is expected rather than a contradiction.

At p=0 or p=1, some values of S are unsupported. The interior-family polynomial proof already forces the function to vanish at every possible integer value, so including endpoints does not damage completeness in this case. Do not generalize that convenient feature to arbitrary families without checking support.

Continue

When applying the theorem, document unbiasedness, sufficiency, completeness and finite variance separately. Recognizing an exponential-family expression alone is not a substitute for verifying the required parameter-space conditions of any completeness theorem you invoke. Continue to Information and the Cramér–Rao bound for a different optimality tool based on local likelihood derivatives. Its assumptions will fail in some models where sufficiency arguments still work.