ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)

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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.

Depends on

Used by

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Sources

Kretschmann and Werner, Maximum a posteriori testing in statistical inverse problems, §1.1 (standard reference, not scraped)
Giné and Nickl, Mathematical Foundations of Infinite-Dimensional Statistical Models, §1.2.2 (standard reference, not scraped)