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Adaptive empirical-Bayes credible-set calibration and failure — Examples

1 · Prerequisites

2 · Summary

A polynomial-tail truth satisfies the polished-tail block-energy condition and has conservative frequentist coverage after sufficiently large inflation. The sparse-gap truth lies in a polynomial hyperrectangle but has coverage tending to zero along a subsequence, when its regularity cap exceeds the smoothness index and its scale and inflation satisfy the stated bounds.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedOpen item page →

A power-law truth lies in the polished-tail class

Example

Fix $β > 0$ , $M > 0$ , and $ρ \geq 2$ . Set $θ_{0, 1} = 0$ and $θ_{0, i} = \sqrt{M} i^{- 1 / 2 - β}$ for $i \geq 2$ . Let $N_{0} = 2$ and $L_{0} = (2 β + 1) / (1 - ρ^{- 2 β})$ . Then $θ_{0} \in Θ_{β} (M) \cap Θ_{p t} (L_{0}, N_{0}, ρ)$ . Consequently, under the hypotheses of Conservative coverage of adaptive empirical-Bayes credible balls on polished tails, an admissible sufficiently large inflation gives conservative frequentist coverage tending to one for this truth.

Facts & Assumptions

Given: $β > 0$ , $M > 0$ , $ρ \geq 2$ , and the sequence in the Example.

[F1]

Membership in $Θ_{β} (M)$ means $\sup_{i} i^{1 + 2 β} θ_{i}^{2} \leq M$ . Hyperrectangle classes in a Gaussian sequence model

[F2]

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

[F3]

For a truth in this polished-tail class with $θ_{0, 1} = 0$ , the source theorem gives an inflation $L$ with frequentist coverage tending to one. Conservative coverage of adaptive empirical-Bayes credible balls on polished tails

Verification

1.1F1givenalgebra

The sequence is square summable because $\sum_{i = 2}^{\infty} i^{- 1 - 2 β} \leq 1 + \int_{1}^{\infty} x^{- 1 - 2 β} d x < \infty$ . For every $i \geq 2$ , $i^{1 + 2 β} θ_{0, i}^{2} = M$ , and the first coordinate is zero. Thus [F1] holds.

1.2F2givenalgebra

For $N \geq 2$ , monotonicity of $x^{- 1 - 2 β}$ gives $\sum_{i = N}^{\infty} i^{- 1 - 2 β} \leq N^{- 1 - 2 β} + \int_{N}^{\infty} x^{- 1 - 2 β} d x \leq (1 + \frac{1}{2 β}) N^{- 2 β}$ and $\sum_{i = N}^{⌊ ρ N ⌋} i^{- 1 - 2 β} \geq \int_{N}^{⌊ ρ N ⌋ + 1} x^{- 1 - 2 β} d x \geq \frac{1 - ρ^{- 2 β}}{2 β} N^{- 2 β} .$ Multiplying by $M$ shows that the tail-to-block ratio is at most $(2 β + 1) / (1 - ρ^{- 2 β}) = L_{0}$ , so [F2] holds.

2.1F3step 1.2∎

The truth has $θ_{0, 1} = 0$ and satisfies [F2], so the theorem [F3] supplies a sufficiently large inflation for which its frequentist coverage tends to one.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedOpen item page →

A sparse-gap truth lies in a hyperrectangle but is undercovered

Example

Assume AC. Fix $β > 0$ , $p \geq 0$ , $C \geq 1$ , $γ \in (0, 1)$ , and a finite $A > β$ . Run the Gaussian sequence procedure in the cited counterexample with cap $A_{n} = A$ and take $0 < M \leq (1984 C^{8})^{- 1}$ . Put $s = 1 + 2 β + 2 p$ , $ρ_{j} = 2^{j}$ , $n_{1} = 2$ , and recursively let $n_{j + 1} = ⌈ (2 ρ_{j + 1}^{2})^{s} n_{j} ⌉$ . Put $N_{j} = n_{j}^{1 / s}$ . Then $1 < N_{1} < ρ_{1}$ and $n_{j + 1} \geq (2 ρ_{j + 1}^{2})^{s} n_{j}$ . Define $θ_{0, i} \geq 0$ by $θ_{0, i}^{2} = {\begin{cases} 0, & N_{j} / ρ_{j} < i \leq N_{j} for some j, \\ 0, & 2 N_{j} < i \leq ρ_{j} N_{j} for some j, \\ M i^{- 1 - 2 β}, & otherwise . \end{cases}$ The verification below proves $θ_{0, 1} = 0$ and $θ_{0} \in Θ_{β} (M)$ . For every deterministic nonnegative inflation sequence for which there exist $K_{L} \in (0, \infty)$ and $j_{0} \in N$ such that, for every $j \geq j_{0}$ , $L_{n_{j}} \leq K_{L} \sqrt{M} ρ_{j}^{(1 + 2 p) / (8 + 8 β + 8 p)}$ coverage of this fixed truth tends to zero along $n_{j}$ , by Sparse-gap truth causes empirical-Bayes credible-set undercoverage.

Facts & Assumptions

Given: The parameters and sparse-gap sequence in the Example.

[F1]

$Θ_{β} (M)$ consists of square-summable sequences satisfying $\sup_{i} i^{1 + 2 β} θ_{i}^{2} \leq M$ . Hyperrectangle classes in a Gaussian sequence model

[F2]

For the stated cap and small- $M$ regime, the sparse-gap counterexample proves vanishing coverage along $n_{j}$ for the specified inflation sequences. Sparse-gap truth causes empirical-Bayes credible-set undercoverage

Verification

1.1F1givenalgebra

The growth condition implies $N_{j + 1} \geq 2 ρ_{j + 1}^{2} N_{j}$ , so $N_{j + 1} / ρ_{j + 1} \geq 2 ρ_{j + 1} N_{j} \geq 2 ρ_{j} N_{j}$ . Each next first gap therefore begins beyond the preceding second gap; the two gaps with the same index are disjoint because the first ends at $N_{j}$ and the second begins strictly after $2 N_{j}$ . Every coordinate is zero or has squared value $M i^{- 1 - 2 β}$ . Hence $\sup_{i} i^{1 + 2 β} θ_{0, i}^{2} \leq M$ , and $\sum_{i} θ_{0, i}^{2} \leq M (1 + \int_{1}^{\infty} x^{- 1 - 2 β} d x) < \infty$ . Thus [F1] holds.

1.2givenalgebra

Since $N_{1} = 2^{1 / s}$ with $s > 1$ , $N_{1} / ρ_{1} < 1 < N_{1}$ . Therefore coordinate $1$ lies in the first zero gap and $θ_{0, 1} = 0$ .

2.1F2step 1.1step 1.2∎

The construction and its cap, credibility level and small- $M$ assumptions match [F2]. Thus the claimed vanishing coverage follows from [F2], while step 1.1 gives hyperrectangle membership.

Sources

Botond Szabó, Aad van der Vaart and Harry van Zanten, Frequentist coverage of adaptive nonparametric Bayesian credible sets, Annals of Statistics 43(4) (2015)