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Linear feature identifiability and Gaussian MAP tests — Examples and Counterexamples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

An adjoint-range feature has an exact one-probe estimator

Example

Assume Countable Choice. Let $X = H = R^{2}$ with the Euclidean inner product, let $K (x_{1}, x_{2}) = (x_{1}, x_{2} / 2)$ , fix $ϵ > 0$ , and let $Z$ be a jointly Gaussian centered family on $H$ , linear into $L^{2}$ with covariance $E [Z (g) Z (k)] = ⟨ g, k ⟩_{H}$ . Define $Y_{x} (g) = ⟨ K x, g ⟩_{H} + ϵ Z (g)$ and consider the feature $ϕ (x_{1}, x_{2}) = x_{1}$ . The map $K$ is compact, $K^{*} (1, 0) = (1, 0)$ , and the observation $Y_{x} (1, 0)$ is an unbiased exact estimator of $ϕ (x)$ with variance $ϵ^{2}$ . Thus the feature vector lies in the adjoint range and is exactly estimable by one probe.

Example references

Exact linear feature estimability through the adjoint range

Compact linear inverse model between Hilbert spaces

Finite-rank operators are compact

Facts & Assumptions

Given: The finite-dimensional inverse model and feature in the Example, with $ϵ > 0$ .

[F1]

A feature $x \mapsto ⟨ x, h ⟩$ is exactly estimable by one probe precisely when $h \in R (K^{*})$ . Exact linear feature estimability through the adjoint range

[F2]

In a compact inverse model, the parameter and data spaces are Hilbert spaces and the forward map is compact and linear. Compact linear inverse model between Hilbert spaces

[F3]

Every bounded finite-rank operator is compact. Bounded finite rank operators are compact

Proof

1.1givenF2F3algebra

The matrix of $K$ in the standard basis is $diag (1, 1 / 2)$ , so its range is two-dimensional and it is bounded. It is compact by [F3], hence it satisfies the compact inverse-model hypothesis [F2]. The matrix is self-adjoint, so $K^{*} = K$ and $K^{*} (1, 0) = (1, 0)$ , the Riesz vector of $ϕ$ .

2.1givenF1step 1.1algebra∎

The isonormal model gives $E_{x} Y_{x} (1, 0) = ⟨ K x, (1, 0) ⟩ = x_{1}$ and ${Var}_{x} Y_{x} (1, 0) = ϵ^{2} ∥ (1, 0) ∥^{2} = ϵ^{2}$ . Since [F1] applies to the feature vector $(1, 0) = K^{*} (1, 0)$ , this probe is exact and has the asserted finite variance.

ExampleConstruction: AI-generatedVerification: AI-generatedverified 2026-10-22 (gpt-6.1-sol)Open item page →

A compact diagonal map has a closure-only feature

Example

Assume Countable Choice. Set $I = {j \in N : j \geq 1}$ . On the real Hilbert space $X = H = ℓ^{2} (I, R)$ with standard basis $(e_{j})_{j \in I}$ , define $K x = (x_{j} / j)_{j \in I}$ and the feature vector $h = (1 / j)_{j \in I}$ . Fix $ϵ > 0$ , let $Z$ be a jointly Gaussian centered family indexed by $H$ , linear into $L^{2}$ with covariance $E [Z (g) Z (k)] = ⟨ g, k ⟩_{H}$ , and define $Y_{x} (g) = ⟨ K x, g ⟩_{H} + ϵ Z (g)$ . The map $K$ is compact and self-adjoint. Its adjoint range contains every finitely supported sequence, so $\overline{R (K^{*})} = ℓ^{2} (I, R)$ . The vector $h$ belongs to $ℓ^{2} (I, R)$ but not to $R (K^{*})$ , since any preimage would have all coordinates equal to $1$ . Thus this feature lies in the closure of the adjoint range but is not exactly estimable by one probe.

Example references

Exact linear feature estimability through the adjoint range

Compact linear inverse model between Hilbert spaces

Convergence and divergence of real p-series

Giné and Nickl, Mathematical Foundations of Infinite-Dimensional Statistical Models, §1.2.2

Facts & Assumptions

Given: The real square-summable sequence Hilbert space, its standard basis, the diagonal map and feature in the Example.

[F1]

The real square-summable sequences form a Hilbert space with inner product $⟨ x, y ⟩ = \sum_{j} x_{j} y_{j}$ and norm squared $\sum_{j} x_{j}^{2}$ . This holds for $I$ as well as $N$ : the bijective linear reindexing $(x_{j})_{j \in I} \mapsto (x_{n + 1})_{n \in N}$ preserves inner products and norms, hence transfers completeness from $ℓ^{2} (N, R)$ . The real square-summable sequence space is complete, Hilbert space

[F2]

For a square-summable sequence indexed by $I$ , its coordinate truncations converge in norm because $\sum_{j > N} x_{j}^{2} \to 0$ . Thus the standard coordinate vectors are orthonormal with dense span, hence form a complete orthonormal basis to which Fourier expansion applies. Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ , Fourier expansion in a Hilbert space

[F3]

The series $\sum_{j \geq 1} j^{- 2}$ converges. The p-series for a real exponent p converges exactly when p is greater than one

[F4]

Every bounded finite-rank operator is compact, and under Countable Choice an operator-norm limit of compact operators into a Banach space is compact. Bounded finite rank operators are compact, Norm limit of compact operators is compact, The Axiom of Countable Choice ( ${A C}_{ω}$ )

[F5]

The adjoint identity is $⟨ K x, y ⟩ = ⟨ x, K^{*} y ⟩$ . The Hilbert-space adjoint of a bounded operator

[F6]

A feature vector is exactly estimable by one probe iff it belongs to $R (K^{*})$ . Exact linear feature estimability through the adjoint range

[F7]

The compact inverse model has Hilbert parameter and data spaces and a compact forward map. Compact linear inverse model between Hilbert spaces

[A1]

Countable Choice is inherited through the compactness of norm limits in [F4], the Hilbert adjoint in [F5], Fourier expansion in [F2], and the exact-estimability interface in [F6]. The Axiom of Countable Choice ( ${A C}_{ω}$ )

Proof

1.1givenF1F2F4F5F7A1algebraconstruct

The map is linear and $∥ K x ∥_{2}^{2} = \sum_{j} x_{j}^{2} / j^{2} \leq ∥ x ∥_{2}^{2}$ , so it is bounded. For $N \geq 1$ , let $K_{N} x$ retain only the first $N$ output coordinates. Its range is spanned by $e_{1}, \dots, e_{N}$ , and $∥ K_{N} x ∥_{2} \leq ∥ x ∥_{2}$ makes it bounded, so [F4] makes $K_{N}$ compact. Also $∥ (K - K_{N}) x ∥_{2}^{2} = \sum_{j > N} x_{j}^{2} / j^{2} \leq (N + 1)^{- 2} ∥ x ∥_{2}^{2}$ , with equality in the operator norm on $e_{N + 1}$ . Thus $∥ K - K_{N} ∥ = 1 / (N + 1) \to 0$ , and [F4] makes $K$ compact. The diagonal real entries show $⟨ K x, y ⟩ = \sum_{j} x_{j} y_{j} / j = ⟨ x, K y ⟩$ ; by [F5], $K^{*} = K$ . For every finitely supported $z$ , the sequence $x_{j} = j z_{j}$ is finitely supported and $K x = z$ , so $c_{00} \subseteq R (K^{*})$ . Truncations are dense in $ℓ^{2}$ by [F2], proving $\overline{R (K^{*})} = ℓ^{2}$ .

2.1givenF1F3F6step 1.1algebra∎

By [F3], $h = (1 / j)$ is square-summable. If $K^{*} x = h$ , self-adjointness and the coordinate equations give $x_{j} / j = 1 / j$ , hence $x_{j} = 1$ for every $j$ . But the first $N$ squared coordinates of such an $x$ sum to $N$ for every $N$ , so no finite bound can contain all partial sums; this contradicts $x \in ℓ^{2}$ . Therefore $h \notin R (K^{*})$ . Since $h \in \overline{R (K^{*})}$ by step 1.1, [F6] shows that the feature is not exactly estimable by a single probe despite lying in the closure of the adjoint range.

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

A non-eigenfeature gives Gaussian MAP rejection probabilities tending to one

Example

Assume the Axiom of Choice. Let $X = H = R^{2}$ , $K = diag (1, 2)$ , $C_{0} = I$ , $m_{0} = 0$ , and $ϵ = 1$ . Let $U$ be a Borel Gaussian prior with mean $0$ and covariance $I$ , independent of a jointly Gaussian centered family $Z$ on $H$ that is linear into $L^{2}$ with covariance $E [Z (g) Z (k)] = ⟨ g, k ⟩_{H}$ , and set $Y_{x} (g) = ⟨ K x, g ⟩_{H} + Z (g)$ . For $h = (1, 1)$ , the MAP feature operator is $B = K^{*} K = diag (1, 4)$ , so $h$ is not an eigenvector. The test probe is $Φ_{h} = (1 / 2, 2 / 5)$ and the threshold is zero. For every $t > 0$ , the parameter $x_{t} = t (- 1, 1)$ is null for the feature $⟨ x, h ⟩$ , and the MAP test rejection probability is $G (3 t / \sqrt{41})$ , which tends to one as $t \to \infty$ . Thus this unrestricted null class has supremum rejection probability one, although no finite $t$ attains it.

Example references

Gaussian MAP feature test

Level, power, and weighted regularization of Gaussian MAP feature tests

A non-eigenfeature can give an unrestricted Gaussian MAP test level one

Facts & Assumptions

Given: The finite-dimensional Gaussian inverse model and feature in the Example.

[F1]

The MAP probe and threshold are $Φ_{h} = K C_{0}^{1 / 2} (C_{0}^{1 / 2} K^{*} K C_{0}^{1 / 2} + ϵ^{2} I)^{- 1} C_{0}^{1 / 2} h$ and $c_{h} = ⟨ K m_{0}, Φ_{h} ⟩ - ⟨ m_{0}, h ⟩$ . Gaussian MAP feature test

[F2]

The exact rejection probability for a nonzero probe is the Gaussian tail determined by its mean and noise variance. Level, power, and weighted regularization of Gaussian MAP feature tests

[F3]

If the commuting MAP operator has a non-eigenfeature, the unrestricted null rejection supremum is one. A non-eigenfeature can give an unrestricted Gaussian MAP test level one

Proof

1.1givenF1F3algebra

Since $C_{0}^{1 / 2} = I$ and $K^{*} K = diag (1, 4)$ , one has $R = (B + I)^{- 1} = diag (1 / 2, 1 / 5)$ and $Φ_{h} = K R h = (1 / 2, 2 / 5)$ . Also $c_{h} = 0$ . The vector $(1, 1)$ is not an eigenvector because $B (1, 1) = (1, 4)$ is not a scalar multiple of it. Thus the compact finite-dimensional example satisfies the commutation and non-eigenfeature hypotheses of [F3].

2.1givenF2F3step 1.1algebra∎

For $x_{t} = t (- 1, 1)$ , $⟨ x_{t}, h ⟩ = 0$ . The test mean is $⟨ K x_{t}, Φ_{h} ⟩ = - t / 2 + 4 t / 5 = 3 t / 10$ , while $∥ Φ_{h} ∥^{2} = 1 / 4 + 4 / 25 = 41 / 100$ . Hence [F2] gives rejection probability $G ((3 t / 10) / (\sqrt{41} / 10)) = G (3 t / \sqrt{41})$ , which tends to one. By [F3] the supremum over the unrestricted null is one; the finite $t$ tail is strictly below one.

Linear feature identifiability and Gaussian MAP tests — Examples and Counterexamples

2 · Logical flowchart

3 · Definitions, theorems and proofs

4 · Examples, counterexamples and false statements

An adjoint-range feature has an exact one-probe estimator

Example

Example references

Facts & Assumptions

Proof

A compact diagonal map has a closure-only feature

Example

Example references

Facts & Assumptions

Proof

A non-eigenfeature gives Gaussian MAP rejection probabilities tending to one

Example

Example references

Facts & Assumptions

Proof

Sources