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Sparse-gap truth causes empirical-Bayes credible-set undercoverage

Statement refuted

The adaptive empirical-Bayes credible balls have nonvanishing frequentist coverage uniformly over every truth in a hyperrectangle $Θ_{β} (M)$ .

Explicit sparse-gap construction.

Use Empirical-Bayes credible-ball procedure in the inverse Gaussian sequence model, assume AC, and fix $β > 0$ , $p \geq 0$ , $C \geq 1$ , $γ \in (0, 1)$ and a finite $A > β$ . Run the procedure with the fixed cap $A_{n} = A$ and $C^{- 2} i^{- 2 p} \leq κ_{i}^{2} \leq C^{2} i^{- 2 p}$ . Choose $0 < M \leq (1984 C^{8})^{- 1}$ . Put $s = 1 + 2 β + 2 p$ , $ρ_{j} = 2^{j}$ , $n_{1} = 2$ , and $n_{j + 1} = ⌈ (2 ρ_{j + 1}^{2})^{s} n_{j} ⌉, N_{j} = n_{j}^{1 / s} .$ Define the fixed nonnegative truth by $θ_{0, i}^{2} = {\begin{cases} 0, & N_{j} / ρ_{j} < i \leq N_{j} for some j, \\ 0, & 2 N_{j} < i \leq ρ_{j} N_{j} for some j, \\ M i^{- 1 - 2 β}, & otherwise . \end{cases}$ Then $θ_{0, 1} = 0$ 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 $j \geq j_{0}$ , $L_{n_{j}} \leq K_{L} \sqrt{M} ρ_{j}^{(1 + 2 p) / (8 + 8 β + 8 p)}$ one has $P_{θ_{0}}^{(n_{j})} {θ_{0} \in {\hat{C}}_{n_{j}} (L_{n_{j}})} ⟶ 0.$ The finitely many indices with $n_{j} < 3$ are irrelevant. In particular, for every $β > 0$ this gives a counterexample for any fixed cap $A > β$ , including constant inflation factors.

Facts & Assumptions

Given: AC and the model, parameters, cap, truth and inflation bound above.

[F1]

The coordinate posterior has mean ${\hat{θ}}_{n, α, i} = n κ_{i} X_{i} / (i^{1 + 2 α} + n κ_{i}^{2})$ and variance $v_{i} (α) = (i^{1 + 2 α} + n κ_{i}^{2})^{- 1}$ . Posterior and marginal-likelihood formulas for the Gaussian sequence prior

[F2]

For every fixed $ℓ^{2}$ truth with first coordinate zero, the one-sided marginal-likelihood bracket gives $P_{θ_{0}}^{(n)} {{\hat{α}}_{n} \geq {\underline{α}}_{n}} \to 1$ , without a polished-tail assumption. Here, with $K_{α} = n^{1 / (1 + 2 α + 2 p)}$ and $τ = (16 C^{8})^{- 1}$ , $h_{n} (α) = \frac{1}{K_{α} \log K_{α}} \sum_{i \geq 1} \frac{n^{2} i^{1 + 2 α} (\log i) θ_{0, i}^{2}}{(i^{1 + 2 α + 2 p} + n)^{2}}, {\underline{α}}_{n} = \inf {α \in [0, A] : h_{n} (α) \geq τ},$ with $\inf \emptyset = A$ . Marginal-likelihood maximizer bracket on polished tails

[F3]

$Θ_{β} (M)$ is the set of $ℓ^{2}$ sequences satisfying $\sup_{i} i^{1 + 2 β} θ_{i}^{2} \leq M$ . Hyperrectangle classes in a Gaussian sequence model

[F4]

A standard normal coordinate satisfies $E Z_{i}^{2} = 1$ . Nonnegative coordinate sums may be integrated termwise by applying Tonelli to the probability space and $N$ with counting measure. Location-scale normal laws have their stated mean and variance Tonelli's theorem for nonnegative measurable functions on a sigma-finite product

Counterexample

1.1F3givenalgebra

The recursion gives $N_{j + 1} \geq 2 ρ_{j + 1}^{2} N_{j}$ . The preceding last gap endpoint satisfies $ρ_{j - 1} N_{j - 1} \leq N_{j} / (2 ρ_{j}) < N_{j}$ , while the next first gap begins at $N_{j + 1} / ρ_{j + 1} \geq 2 ρ_{j + 1} N_{j} > 2 N_{j}$ . Thus no other zero gap intersects $N_{j} < i \leq 2 N_{j}$ , where $θ_{0, i}^{2} = M i^{- 1 - 2 β}$ . Every squared coordinate is zero or $M i^{- 1 - 2 β}$ , so $\sum_{i} θ_{0, i}^{2} \leq M (1 + (2 β)^{- 1}) < \infty$ and [F3] holds. Finally $1 < N_{1} = 2^{1 / s} < 2 = ρ_{1}$ , so the first gap contains coordinate $1$ .

1.2givenalgebra

Put $δ_{j} = s \log ρ_{j} / (4 \log n_{j})$ , $a_{j} = β + δ_{j}$ , $t_{j} = s + 2 δ_{j}$ and $d = 1 + 2 p$ . Iterating the growth bound gives $\log n_{j} \geq \log 2 + s (\log 2) (j^{2} + 2 j - 3)$ for $j \geq 2$ , hence $δ_{j} \to 0$ . Consequently, eventually $0 < δ_{j} \leq 1 / 4$ , $a_{j} < A$ , $N_{j} / ρ_{j} \geq 2$ and $N_{j} \geq 2$ . Restrict to these indices and put $T = s + 1 / 2$ , so $s < t_{j} \leq T$ and $n_{j}^{δ_{j}} = ρ_{j}^{s / 4}$ .

1.3algebra

For $x \geq 2$ and $q \geq 2$ , split the integers $i > x$ into intervals $(2^{l} x, 2^{l + 1} x]$ . Each contains at most $3 \cdot 2^{l} x$ integers, and $\log i \leq (l + 2) \log x$ . Hence $\sum_{i > x} i^{- q} \log i \leq 3 x^{1 - q} \log x \sum_{l \geq 0} (l + 2) 2^{- l (q - 1)} \leq 18 x^{1 - q} \log x$ , because $\sum_{l \geq 0} (l + 2) 2^{- l} = 6$ .

2.1F2step 1.1step 1.3algebra

Set $n = n_{j}$ and first take $0 \leq α \leq β$ . Write $t = 1 + 2 α + 2 p$ , $K = n^{1 / t}$ and $q = 2 + 2 α + 4 p + 2 β \geq 2$ . In [F2], use the denominator bounds $(i^{t} + n)^{2} \geq n^{2}$ for $i \leq K$ and $(i^{t} + n)^{2} \geq i^{2 t}$ for $i > K$ , together with $θ_{0, i}^{2} \leq M i^{- 1 - 2 β}$ . The first part of $h_{n} (α)$ is at most $M$ , since $i^{2 α - 2 β} \leq 1$ and $\sum_{i \leq K} \log i \leq K \log K$ . By step 1.3 the tail is at most $18 M n^{2} K^{- q} = 18 M K^{2 α - 2 β} \leq 18 M$ . Thus $h_{n} (α) \leq 19 M$ throughout $[0, β]$ .

2.2F2step 1.1step 1.2algebra

Now take $β < α \leq a_{j}$ , and write $δ = α - β$ , $t = s + 2 δ$ , $K = n^{1 / t}$ , $N = N_{j}$ and $ρ = ρ_{j}$ . The zero gaps split $h_{n}$ into three nonnegative parts, over $i \leq N / ρ$ , $N < i \leq 2 N$ and $i > ρ N$ . Bounding the denominator by $n^{2}$ , the first part satisfies $H_{1} \leq M (\log N / \log K) (N / ρ)^{1 + 2 δ} / K$ . Here $\log N / \log K = t / s \leq T / s$ and $N^{1 + 2 δ} / K = n^{2 δ (t + 1) / (s t)} \leq ρ^{(s + 1) / (2 s)}$ , using $δ \leq δ_{j}$ and $(t + 1) / t \leq (s + 1) / s$ . Therefore $H_{1} \leq M (T / s) ρ^{- (s - 1) / (2 s)}$ .

2.3F1step 1.1step 1.2algebra

For fixed $α$ write $B_{i} (α) = i^{1 + 2 α} θ_{0, i} / (i^{1 + 2 α} + n κ_{i}^{2})$ and $W_{i} (α) = \sqrt{n} κ_{i} Z_{i} / (i^{1 + 2 α} + n κ_{i}^{2})$ , so the posterior-mean error is $- B (α) + W (α)$ . If $α \geq a_{j} > β$ and $N_{j} < i \leq 2 N_{j}$ , then $i^{1 + 2 α + 2 p} \geq N_{j}^{s} = n$ and $n κ_{i}^{2} / i^{1 + 2 α} \leq C^{2}$ . There are at least $N_{j} / 2$ integers in this block, each with squared truth at least $M (2 N_{j})^{- 1 - 2 β}$ . Therefore, uniformly for $α \geq a_{j}$ , $∥ B (α) ∥^{2} \geq b^{2} M N_{j}^{- 2 β}$ , where $b = 2^{- 1 - β} / (1 + C^{2}) > 0$ .

2.4F1F4step 1.2algebra

Set $R_{j} = n_{j}^{- a_{j} / t_{j}}$ and $D = C^{2} (2 + 2^{2 β + 1 / 2} / (2 β))$ . At $α = a_{j}$ , put $K = n_{j}^{1 / t_{j}} \geq 2$ . Since $v_{i} (a_{j}) \leq C^{2} i^{2 p} / (i^{t_{j}} + n_{j})$ , summing over $i \leq K$ gives at most $2 C^{2} K^{- 2 a_{j}}$ . The tail is at most $\sum_{i > K} i^{- 1 - 2 a_{j}} \leq ⌊ K ⌋^{- 2 a_{j}} / (2 a_{j}) \leq 2^{2 β + 1 / 2} K^{- 2 a_{j}} / (2 β)$ . Thus $V_{j} = \sum_{i} v_{i} (a_{j}) \leq D R_{j}^{2}$ . Each variance decreases in $α$ , so [F4] bounds the posterior expected squared centered norm by $V_{j}$ whenever $α \geq a_{j}$ . For a nonnegative random variable $Y$ , integrating $u 1_{{Y > u}} \leq Y$ gives $u P (Y > u) \leq E Y$ . Apply this with $u = V_{j} / γ$ to bound the posterior norm quantile: $\sup_{α \geq a_{j}} r_{n_{j}, γ} (α) \leq \sqrt{D / γ} R_{j}$ .

3.1step 1.2step 2.2algebra

For the middle part use $(i^{t} + n)^{2} \geq i^{2 t}$ , $\log i \leq \log (2 N)$ and at most $2 N$ indices, obtaining $H_{2} \leq 2 M (\log (2 N) / \log K) n^{2} N^{1 - 2 s - 2 δ} / K$ . Since $N \geq 2$ , the logarithm ratio is at most $2 T / s$ . Also $n^{2} N^{1 - 2 s - 2 δ} / K = n^{- 2 δ (t - 1) / (s t)} \leq 1$ . Consequently $H_{2} \leq 4 M T / s$ .

3.2F4step 2.4algebra

For every realization of the normal coordinates, $∣ W_{i} (α) ∣ \leq ∣ W_{i} (a_{j}) ∣$ whenever $α \geq a_{j}$ . Hence $\sup_{α \geq a_{j}} ∥ W (α) ∥ \leq ∥ W (a_{j}) ∥$ . By [F4], $E ∥ W (a_{j}) ∥^{2} = \sum_{i} n_{j} κ_{i}^{2} / (i^{1 + 2 a_{j}} + n_{j} κ_{i}^{2})^{2} \leq V_{j} \leq D R_{j}^{2}$ . The elementary probability bound proved in step 2.4 therefore gives $P_{θ_{0}}^{(n_{j})} {\sup_{α \geq a_{j}} ∥ W (α) ∥ > b \sqrt{M} N_{j}^{- β} / 2} \leq 4 D R_{j}^{2} / (b^{2} M N_{j}^{- 2 β})$ . These bounds also show almost sure square-summability simultaneously for every $α \geq a_{j}$ .

4.1F2step 1.2step 1.3step 2.1step 2.2step 3.1algebra

For the last part set $q = 2 s + 2 δ \geq 2$ and apply step 1.3 at $x = ρ N$ . This gives $H_{3} \leq 18 M (\log (ρ N) / \log K) n^{2} (ρ N)^{1 - q} / K$ . Because $\log ρ / \log N = 4 δ_{j} \leq 1$ , the logarithm ratio is at most $2 T / s$ . Furthermore $n^{2} N^{1 - q} / K = n^{- 2 δ (t - 1) / (s t)} \leq 1$ and $ρ^{1 - q} \leq ρ^{1 - 2 s}$ . Thus $H_{3} \leq 36 M (T / s) ρ^{1 - 2 s}$ . Combining these estimates yields $\sup_{0 \leq α \leq a_{j}} h_{n} (α) \leq \max (19, 41 T / s) M \leq 62 M \leq (32 C^{8})^{- 1} < τ$ , since $s > 1$ . Hence ${\underline{α}}_{n_{j}} \geq a_{j}$ , and [F2] gives $P_{θ_{0}}^{(n_{j})} (E_{j}) \to 1$ for $E_{j} = {{\hat{α}}_{n_{j}} \geq a_{j}}$ .

4.2givenstep 1.2step 2.4step 3.2algebra

Subtracting exponents and using $n_{j}^{δ_{j}} = ρ_{j}^{s / 4}$ gives $R_{j} / N_{j}^{- β} = ρ_{j}^{- d / (4 t_{j})}$ . Put $e = d / (4 (s + 1))$ and $η = d / (8 (s + 1) (s + 1 / 2)) > 0$ . Since $t_{j} \leq T = s + 1 / 2$ , one has $d / (4 t_{j}) - e \geq η$ . The inflation assumption and step 2.4 imply $\sup_{α \geq a_{j}} L_{n_{j}} r_{n_{j}, γ} (α) / (\sqrt{M} N_{j}^{- β}) \leq K_{L} \sqrt{D / γ} ρ_{j}^{- η} \to 0$ . Step 3.2's probability bound is at most $(4 D / (b^{2} M)) ρ_{j}^{- d / (2 T)} \to 0$ .

5.1step 4.1step 2.3step 4.2∎

On $E_{j}$ the selected regularity is positive and at least $a_{j}$ . Coverage then implies $∥ B ({\hat{α}}_{n_{j}}) ∥ \leq L_{n_{j}} r_{n_{j}, γ} ({\hat{α}}_{n_{j}}) + ∥ W ({\hat{α}}_{n_{j}}) ∥$ . The left side is at least $b \sqrt{M} N_{j}^{- β}$ by step 2.3; the first term on the right is eventually strictly below half that amount by step 4.2, while the second exceeds the same half-threshold with probability tending to zero. Consequently coverage is at most $P_{θ_{0}}^{(n_{j})} (E_{j}^{c}) + o (1) \to 0$ . This fixed truth in $Θ_{β} (M)$ refutes the proposed uniform nonvanishing coverage.

Remarks

The cap condition is essential to the argument: the selected admissible range must contain $a_{j} > β$ eventually. Here $a_{j} \to β$ , so every fixed $A > β$ suffices. The printed source theorem leaves this qualification implicit. The lower likelihood bracket used here holds for all fixed square-summable truths with first coordinate zero; the polished-tail restriction needed for a matching upper bracket does not apply to this one-sided use.

Depends on

Used by

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

Dependency tree · two levels

31 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

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) (standard reference, not scraped)
B. T. Knapik, A. W. van der Vaart and J. H. van Zanten, Bayesian inverse problems with Gaussian priors, Annals of Statistics 39(5) (2011) (standard reference, not scraped)