ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The exact Wishart trace and covariance likelihood ratio are different statistics

Example

Take $p = 1$ , $n = 4$ , $Σ_{0} = 1$ , and the observed sample $(\sqrt{3 / 2}, - \sqrt{3 / 2}, 0, 0) .$ The exact covariance trace statistic and the covariance likelihood-ratio statistic have different values on this observation. The trace statistic has an exact $χ_{3}^{2}$ law under the null, whereas the likelihood-ratio result in The Gaussian covariance likelihood ratio has its fixed-dimensional Wilks limit gives only an asymptotic $χ_{1}^{2}$ calibration. This example does not assert a finite-sample law for the likelihood ratio.

Facts & Assumptions

Given: The displayed one-dimensional sample, $n = 4$ , and null variance $Σ_{0} = 1$ .

[F1]

For observations $x_{i}$ , $\overset{ˉ}{x} = n^{- 1} \sum_{i} x_{i}$ , $R = \sum_{i} (x_{i} - \overset{ˉ}{x})^{2}$ , $S = R / (n - 1)$ , and $M = R / n$ . The iid multivariate-normal sample and covariance conventions

[F2]

The null likelihood-ratio statistic is $n {tr (Σ_{0}^{- 1} M) - \log \det (Σ_{0}^{- 1 / 2} M Σ_{0}^{- 1 / 2}) - p}$ ; its chi-square result is asymptotic with $p$ fixed. The Gaussian covariance likelihood ratio has its fixed-dimensional Wilks limit

[F3]

Under the null, $tr (Σ_{0}^{- 1} (n - 1) S) \sim χ_{p (n - 1)}^{2}$ , including when the scatter is singular. The exact covariance trace statistic has p(n−1) chi-square degrees of freedom

Verification

1.1givenF1algebra

The sample has mean $0$ , so its centered sum of squares is $R = 3 / 2 + 3 / 2 = 3$ . Thus $S = R / 3 = 1$ and $M = R / 4 = 3 / 4$ by [F1].

2.1F2F3step 1.1algebra

Since $p = 1$ and $Σ_{0} = 1$ , the trace statistic is $(n - 1) S = 3$ , consistent with its $χ_{1 \cdot 3}^{2} = χ_{3}^{2}$ null law [F3]. Substituting $M = 3 / 4$ in [F2] gives the observed likelihood-ratio value $4 (\frac{3}{4} - \log \frac{3}{4} - 1) = 4 (\log \frac{4}{3} - \frac{1}{4}) \approx 0.1507.$ This is not the trace value $3$ .

3.1F2step 2.1∎

The calculations show two different finite-sample statistics and their distinct calibrations; the fact that one observed value is $0.1507$ makes no claim about the likelihood-ratio statistic's null distribution at $n = 4$ . The asymptotic qualification in [F2] remains necessary.

Depends on

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Sources

Arak Mathai, Serge Provost, and Hans J. Haubold, Multivariate Statistical Analysis in the Real and Complex Domains, Chapter 5 §5.5, PDF pp. 47–66 (standard reference, not scraped)
Soumendu Mukherjee, Multivariate Analysis 2021, Lecture 7, PDF pp. 1–4 (standard reference, not scraped)
Peter D. Hoff, Multivariate Statistical Analysis, Duke STA 832 lecture notes, relevant sections in Chapters 4 and 7 (standard reference, not scraped)