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The Bernoulli count is minimal sufficient

Example

Assume AC. Let X1,…,Xn be iid Bernoulli(p) with p∈(0,1), sample space {0,1}n, and T(x)=∑ixi. Then T is minimal sufficient.

Facts & Assumptions

Given: AC, n≥1, p∈(0,1), the Bernoulli product family, and counting measure as the common reference, so its likelihood densities equal the point masses.

[A1]

The Axiom of Choice: AC is the axiom branch of the discrete minimal-sufficiency theorem.

[F1]

Likelihood-ratio equivalence on common discrete support: points are equivalent exactly when their point-mass ratio is constant in the parameter; with counting measure these are also the likelihood-density ratios.

[F2]

Likelihood-ratio classes are minimal sufficient on a common discrete support: the equivalence-class statistic is minimal sufficient on a countable common positive support.

Proof

1.1F1givenalgebra

Every sample point has probability mp(x)=pt(x)(1−p)n−t(x)>0, so the common support is the whole finite sample space. Counting measure assigns mass one to each point, so the densities equal these point masses; for x,y, their ratio is mp(x)/mp(y)=[p/(1−p)]t(x)−t(y).

2.1A1F1F2step 1.1givenalgebra∎

If t(x)=t(y), this ratio is one for all p. If d=t(x)−t(y)≠0, its values at p=1/2 and p=1/3 are respectively 1 and (1/2)d, which differ. Thus equivalence classes are exactly the level sets of T. By [F2], T is minimal sufficient.

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