LemmaStatement: Literature-sourcedProof: 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.

Quadratic likelihood optimization is squared Fisher distance

Statement

Let $d \geq 1$ , let $I$ be an invertible real symmetric matrix such that $v^{T} I v > 0$ for every nonzero $v \in R^{d}$ , let $z \in R^{d}$ , and let $C \subseteq R^{d}$ be nonempty and closed in the Euclidean metric. Define $q_{z} (h) : = z^{T} h - \frac{1}{2} h^{T} I h$ . Then there is a constant $λ > 0$ such that $h^{T} I h \geq λ ∥ h ∥_{2}^{2}$ for all $h$ , $q_{z}$ attains its maximum on $C$ , and

$\sup_{h \in C} q_{z} (h) = \frac{1}{2} z^{T} I^{- 1} z - \frac{1}{2} d_{I}^{2} (I^{- 1} z, C) .$

Facts & Assumptions

Given: $d \geq 1$ , $I$ , $z$ , and $C$ as in the Statement.

[F1]

The Euclidean inner product is $⟨ u, v ⟩ = u^{T} v$ and its norm is $∥ u ∥_{2} = \sqrt{u^{T} u}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[F2]

A Euclidean closed ball is ${\overline{B}}_{2} (c, r) = {x : ∥ x - c ∥_{2} \leq r}$ , and the unit sphere is $S^{d - 1} = {x : ∥ x ∥_{2} = 1}$ (Euclidean spheres and closed balls as subspaces of $R^{n}$ ).

[F3]

Closed means that the complement is open in the metric topology (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

[F4]

In finite-dimensional Euclidean space, a set is compact exactly when it is closed and bounded (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[F5]

A continuous real-valued function on a nonempty compact metric space attains its maximum and minimum (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

[F6]

$∣ u^{T} v ∣ \leq ∥ u ∥_{2} ∥ v ∥_{2}$ (Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation).

[F7]

For symmetric positive-definite $I$ , $d_{I} (x, C) = \inf_{c \in C} \sqrt{(x - c)^{T} I (x - c)}$ (Fisher-information metric distance to a set).

Proof

Given: The matrix, vector and set in the Statement.

1.1F1F2F4F5given

On the unit sphere $S^{d - 1}$ of [F2], the function $u \mapsto u^{T} I u$ is continuous and strictly positive. The sphere is nonempty, closed and bounded, hence compact by [F4]; [F5] gives a minimum $λ > 0$ , since a zero minimum would be attained at a unit vector and contradict positive definiteness. Rescaling any nonzero $h$ to the unit sphere gives $h^{T} I h \geq λ ∥ h ∥_{2}^{2}$ , also valid for $h = 0$ .

1.2F1F6givenalgebra

Completing the square gives $q_{z} (h) = \frac{1}{2} z^{T} I^{- 1} z - \frac{1}{2} (h - I^{- 1} z)^{T} I (h - I^{- 1} z)$ . By [F1] and [F6], $q_{z} (h) \leq ∥ z ∥_{2} ∥ h ∥_{2} - \frac{1}{2} λ ∥ h ∥_{2}^{2}$ , which tends to $- \infty$ as $∥ h ∥_{2} \to \infty$ .

2.1F2F3F4F5step 1.2givenchoose

Choose $c_{0} \in C$ . By step 1.2 there is $R > ∥ c_{0} ∥_{2}$ such that $q_{z} (h) < q_{z} (c_{0})$ whenever $∥ h ∥_{2} \geq R$ . The set $C \cap {\overline{B}}_{2} (0, R)$ is nonempty, closed and bounded, hence compact by [F2]–[F4]. The continuous quadratic $q_{z}$ attains a maximum there by [F5]; values outside are below $q_{z} (c_{0})$ , so this is a global maximum on $C$ .

3.1F7step 1.2step 2.1∎

The identity of step 1.2 says that maximizing $q_{z}$ over $C$ is equivalent to minimizing $(h - I^{- 1} z)^{T} I (h - I^{- 1} z)$ . Step 2.1 proves that this minimum is attained; by [F7] it equals $d_{I}^{2} (I^{- 1} z, C)$ . Substitution gives the stated supremum formula.

Depends on

Used by

Likelihood-ratio limit for local cones in a QMD model Theorem

Dependency tree · two levels

55 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

Peter Bartlett, Theoretical Statistics, Stat 210B Lecture 26 (standard reference, not scraped)
Steven G. Self and Kung-Yee Liang, Asymptotic Properties of Maximum Likelihood Estimators and Likelihood Ratio Tests Under Nonstandard Conditions (standard reference, not scraped)