CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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A boundary truth can give an equal-tailed credible interval zero coverage

Statement refuted

Every equal-tailed posterior credible interval has frequentist coverage at least its nominal level, even when the truth is on the boundary of the parameter space.

Setup

For $0 < α < 1$ , observe $X_{1}, \dots, X_{n}$ IID Bernoulli $(θ)$ with parameter space $Θ = [0, 1]$ and prior density $π (θ) = 1$ on $[0, 1]$ . For every data vector, let $C_{n}$ be the equal-tailed posterior interval between the $α / 2$ and $1 - α / 2$ posterior quantiles. The truth $θ_{0} = 0$ lies on the boundary. Under this truth all observations are zero with probability one. The posterior density is then $(n + 1) (1 - θ)^{n}$ , with CDF $F_{n} (t) = 1 - (1 - t)^{n + 1}$ on $[0, 1]$ . Its lower endpoint is $q_{α / 2} = 1 - (1 - α / 2)^{1 / (n + 1)} > 0$ , so the interval excludes $θ_{0}$ for every $n$ .

Facts & Assumptions

Given: The Bernoulli model, uniform prior, and equal-tailed posterior interval in the Counterexample.

[F1]

Posterior credible set: a graph-measurable set rule is level- $1 - α$ credible when its posterior probability is at least $1 - α$ for each data value.

[F2]

Bayes formula for dominated kernels: a dominated prior and sampling model give posterior density proportional to prior density times likelihood when the marginal likelihood is positive and finite.

[L1]

At a fixed parameter $θ$ , the frequentist coverage of a set rule $C$ is $P_{θ} {θ \in C (Y)}$ .

[A1]

In the usual topology on $[0, 1]$ , $θ_{0} = 0$ is a boundary point and is not an interior point.

Refutation

1.1F1F2givenalgebra

For data with $k$ successes, the likelihood is $θ^{k} (1 - θ)^{n - k}$ . It is bounded by $1$ and positive for $0 < θ < 1$ , so its integral against the uniform prior is finite and strictly positive. [F2] therefore gives the posterior density proportional to that likelihood. The sample space is finite for fixed $n$ , so the interval rule's membership graph is a finite union of measurable slices. The posterior density is positive on $(0, 1)$ , making its CDF continuous and strictly increasing there; the posterior probability between its two quantiles is $(1 - α / 2) - α / 2 = 1 - α$ . Thus the rule is credible by [F1].

1.2givenalgebra

At $θ_{0} = 0$ , the data are all zero with probability one, so the likelihood is $(1 - θ)^{n}$ . Its normalizing integral is $\int_{0}^{1} (1 - θ)^{n} d θ = 1 / (n + 1)$ , giving posterior density $(n + 1) (1 - θ)^{n}$ and CDF $F_{n} (t) = 1 - (1 - t)^{n + 1}$ . Solving $F_{n} (q_{u}) = u$ yields $q_{u} = 1 - (1 - u)^{1 / (n + 1)}$ . In particular $q_{α / 2} > 0$ , so $0 \notin C_{n}$ for every $n$ .

2.1A1L1step 1.2algebra∎

Because the truth is zero and all data are zero almost surely, [L1] gives frequentist coverage $P_{0} {0 \in C_{n} (X_{1}, \dots, X_{n})} = 0$ . This contradicts the asserted lower bound $1 - α > 0$ and exhibits failure at the boundary identified in [A1].

Depends on

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