CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)

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Sparse-gap truths defeat uniform coverage over a hyperrectangle

Statement refuted

Assume AC. Fix $p \geq 0$ , $C \geq 1$ , $β > 0$ , finite $A > β$ , $0 < M \leq (1984 C^{8})^{- 1}$ , and $0 < γ < 1$ . For $n \geq 3$ , observe $X_{i} = κ_{i} θ_{i} + n^{- 1 / 2} Z_{i}$ in the Gaussian sequence experiment, where $θ \in ℓ^{2} (N_{+}, R)$ is fixed, $C^{- 2} i^{- 2 p} \leq κ_{i}^{2} \leq C^{2} i^{- 2 p}$ , and the $Z_{i}$ are independent $N (0, 1)$ . Define $Θ_{β} (M) = {θ \in ℓ^{2} : \sup_{i \geq 1} i^{1 + 2 β} θ_{i}^{2} \leq M}$ . For $0 < α \leq A$ , use the product prior $Π_{α} = ⨂_{i \geq 1} N (0, i^{- 1 - 2 α})$ and its product Gaussian posterior with coordinate mean ${\hat{θ}}_{n, α, i} (x) = n κ_{i} x_{i} / (i^{1 + 2 α} + n κ_{i}^{2})$ and variance $(i^{1 + 2 α} + n κ_{i}^{2})^{- 1}$ . Select ${\hat{α}}_{n}$ as a Borel global maximizer over $[0, A]$ of $ℓ_{n} (α; x) = - \frac{1}{2} \sum_{i \geq 1} [\log (1 + \frac{n κ_{i}^{2}}{i^{1 + 2 α}}) - \frac{n^{2} κ_{i}^{2} x_{i}^{2}}{i^{1 + 2 α} + n κ_{i}^{2}}],$ Assume the score series is finite on a Borel full-measure data domain for the truths considered and is upper semicontinuous in $α$ there. For ${\hat{α}}_{n} > 0$ , let $r_{n, γ} ({\hat{α}}_{n})$ be the $1 - γ$ posterior quantile of $∥ θ - {\hat{θ}}_{n, {\hat{α}}_{n}} (X) ∥_{2}$ , and let ${\hat{C}}_{n} (L)$ be the ball centered at ${\hat{θ}}_{n, {\hat{α}}_{n}} (X)$ with radius $L r_{n, γ} ({\hat{α}}_{n})$ . Set ${\hat{C}}_{n} (L) = ℓ^{2}$ if ${\hat{α}}_{n} = 0$ . There exists a finite fixed $L \geq 1$ for which ${lim inf}_{n \to \infty} \inf_{θ \in Θ_{β} (M)} P_{θ}^{(n)} {θ \in {\hat{C}}_{n} (L)} \geq 1 - γ$

Statement references

The mildly ill-posed sequence model, Gaussian prior family, selector domain, posterior ball, and zero-selector convention.

The hyperrectangle is defined by the coordinatewise bound $\sup_{i} i^{1 + 2 β} θ_{i}^{2} \leq M$ .

An explicit sparse-gap truth gives coverage tending to zero along a subsequence under a stated inflation bound.

Facts & Assumptions

Given: AC and the model, prior, selector, $p, C, β, A, M, γ$ in the Statement refuted.

[A1]

AC supplies the countable Gaussian product construction used by the empirical-Bayes procedure. The Axiom of Choice

[F1]

$Θ_{β} (M) = {θ \in ℓ^{2} : \sup_{i \geq 1} i^{1 + 2 β} θ_{i}^{2} \leq M}$ . Hyperrectangle classes in a Gaussian sequence model

[F2]

$θ \in Θ_{p t} (L_{0}, N_{0}, ρ)$ only if $\sum_{i = N}^{\infty} θ_{i}^{2} \leq L_{0} \sum_{i = N}^{⌊ ρ N ⌋} θ_{i}^{2}$ for every integer $N \geq N_{0}$ . Polished-tail class for a Gaussian sequence

[F3]

For the sparse-gap construction below, every deterministic nonnegative inflation sequence satisfying $L_{n_{j}} \leq K_{L} \sqrt{M} ρ_{j}^{(1 + 2 p) / (8 + 8 β + 8 p)}$ eventually, for some finite $K_{L} > 0$ , has $P_{θ_{0}}^{(n_{j})} {θ_{0} \in {\hat{C}}_{n_{j}} (L_{n_{j}})} \to 0$ . Sparse-gap truth causes empirical-Bayes credible-set undercoverage

[F4]

For each fixed $A, L_{0}, N_{0}, ρ$ , there is a finite inflation constant $L$ for which coverage tends to one uniformly over $Θ_{p t} (L_{0}, N_{0}, ρ)$ with $θ_{1} = 0$ . Conservative coverage of adaptive empirical-Bayes credible balls on polished tails

[F5]

For ${\hat{α}}_{n} > 0$ , the procedure uses the posterior-mean-centered $ℓ^{2}$ ball with radius $L$ times the $1 - γ$ posterior quantile; its zero-selector branch is the whole space. Empirical-Bayes credible-ball procedure in the inverse Gaussian sequence model

[F6]

At a fixed truth, frequentist coverage is the sampling probability of the membership event ${θ \in C_{n} (Y)}$ . Posterior credibility and frequentist coverage

Counterexample

Put $s = 1 + 2 β + 2 p$ , $ρ_{j} = 2^{j}$ , $n_{1} = 2$ , and $n_{j + 1} = ⌈ (2 ρ_{j + 1}^{2})^{s} n_{j} ⌉$ . Set $N_{j} = n_{j}^{1 / s}$ and define the fixed truth by $θ_{0, i}^{2} = {\begin{cases} 0, & N_{j} / ρ_{j} < i \leq N_{j} for some j, \\ 0, & 2 N_{j} < i \leq ρ_{j} N_{j} for some j, \\ M i^{- 1 - 2 β}, & otherwise. \end{cases}$ For every fixed $L \geq 1$ , this truth belongs to $Θ_{β} (M)$ , has $θ_{0, 1} = 0$ , is outside every fixed polished-tail class, and has subsequence coverage tending to zero.

Given: The hypotheses in the Statement refuted and the construction above.

1.1F1givenalgebra

Every squared coordinate is either zero or $M i^{- 1 - 2 β}$ , so $\sup_{i} i^{1 + 2 β} θ_{0, i}^{2} \leq M$ and $\sum_{i} θ_{0, i}^{2} \leq M \sum_{i} i^{- 1 - 2 β} < \infty$ . Also $N_{1} = 2^{1 / s} \in (1, 2)$ and $ρ_{1} = 2$ , hence $N_{1} / ρ_{1} < 1 \leq N_{1}$ and the first zero gap contains $i = 1$ . Thus $θ_{0} \in Θ_{β} (M)$ and $θ_{0, 1} = 0$ .

2.1F2givenstep 1.1algebra

Fix any polished-tail parameters $L_{0} \geq 1$ , integer $N_{0} \geq 1$ , and $ρ_{*} \geq 2$ . The recursion gives $N_{j} / ρ_{j} \geq 2 ρ_{j} N_{j - 1} > ρ_{j - 1} N_{j - 1}$ , so every earlier gap ends below $N_{j} / ρ_{j}$ . Also $N_{j}^{'} = ⌊ 2 N_{j} ⌋ + 1 \leq 2 N_{j} + 1$ ; because $N_{j} \to \infty$ and $ρ_{j} \to \infty$ , for all sufficiently large $j$ we have $N_{j}^{'} \geq N_{0}$ and $ρ_{*} (2 N_{j} + 1) \leq ρ_{j} N_{j}$ , whence $⌊ ρ_{*} N_{j}^{'} ⌋ \leq ρ_{j} N_{j}$ . Since $N_{j}^{'} > 2 N_{j}$ , the block $[N_{j}^{'}, ⌊ ρ_{*} N_{j}^{'} ⌋]$ lies in the current zero gap $(2 N_{j}, ρ_{j} N_{j}]$ . For all sufficiently large $j$ , $i_{j} = ⌊ ρ_{j} N_{j} ⌋ + 1 \leq ρ_{j} N_{j} + 1 < 4 ρ_{j} N_{j} \leq N_{j + 1} / ρ_{j + 1}$ , so $i_{j}$ is before the next first gap; applying the same separation at later indices puts all still later gaps farther right. Since $ρ_{j} N_{j} > 2 N_{j} + 1 \geq N_{j}^{'}$ eventually, $i_{j} > N_{j}^{'}$ ; earlier gaps end below $N_{j} / ρ_{j} < i_{j}$ . Hence $i_{j}$ lies outside every zero gap and is in the tail from $N_{j}^{'}$ , with $θ_{0, i_{j}}^{2} = M i_{j}^{- 1 - 2 β} > 0$ . The block sum is zero while that tail is positive, violating [F2]. Thus $θ_{0}$ is outside every fixed polished-tail class.

2.2A1F3F5F6givenstep 1.1algebra

For a fixed $L \geq 1$ , take $L_{n_{j}} = L$ and $K_{L} = L / \sqrt{M}$ . Since $(1 + 2 p) / (8 + 8 β + 8 p) > 0$ and $ρ_{j} \geq 1$ , we have $L_{n_{j}} = L \leq K_{L} \sqrt{M} ρ_{j}^{(1 + 2 p) / (8 + 8 β + 8 p)}$ for every $j$ . Applying [F3] gives $P_{θ_{0}}^{(n_{j})} {θ_{0} \in {\hat{C}}_{n_{j}} (L)} \to 0$ . By [F6] this is the frequentist coverage at the fixed truth; because $θ_{0} \in Θ_{β} (M)$ by step 1.1, the infimum coverage over that hyperrectangle cannot have lower limit at least $1 - γ$ .

3.1F2F4step 2.1step 2.2algebra∎

The positive coverage statement [F4] has the additional polished-tail hypothesis, which fails for $θ_{0}$ by step 2.1. Thus the sparse-gap undercoverage does not contradict the polished-tail theorem; it refutes only the broader fixed-inflation claim over the full hyperrectangle.

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Sources

Botond Szabó, A. W. van der Vaart, and J. H. van Zanten, Frequentist coverage of adaptive nonparametric Bayesian credible sets, Annals of Statistics 43(4) (2015) (standard reference, not scraped)