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

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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.

Depends on

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Sources

Kretschmann and Werner, Maximum a posteriori testing in statistical inverse problems, Theorem 2.6 (standard reference, not scraped)