ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-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.

Leverage is determined by the design

Example

For a simple-regression design with an intercept, take predictor values $x = (0, 1, 2, 3, 20)^{T}$ and let $X = [1 x]$ . Define the hat matrix $H = P_{col (X)}$ and the leverage of row $i$ by $h_{i i} = e_{i}^{T} H e_{i}$ . The fifth row has leverage $h_{55} = 687 / 697 \approx 0.986$ . This value depends only on $X$ , not on the response.

Facts & Assumptions

Given: The five-point predictor vector and intercept design in the Example.

[F1]

The least-squares fitted vector is the unique orthogonal projection $P_{col (X)} y$ for every response $y$ . Least-squares fitted vectors and residuals are unique.

[F2]

$P_{W} y$ is characterized by $P_{W} y \in W$ and $y - P_{W} y \in W^{⊥}$ . The orthogonal projection $P_{W} v$ is the $W$ -component in $V = W \oplus W^{⊥}$ .

Verification

technique · direct

Given: $1 = (1, 1, 1, 1, 1)^{T}$ and $x = (0, 1, 2, 3, 20)^{T}$ .

1.1givenalgebra

The predictor mean is $\overset{ˉ}{x} = 26 / 5$ . Put $z = x - \overset{ˉ}{x} 1 = (- 26, - 21, - 16, - 11, 74)^{T} / 5$ . Then $1^{T} z = 0$ and $z^{T} z = \frac{676 + 441 + 256 + 121 + 5476}{25} = \frac{1394}{5} .$ Since $x = z + \overset{ˉ}{x} 1$ , the column space is $C = span {1, z}$ , with these two generators nonzero and orthogonal.

2.1F1F2step 1.1algebra

For every $y \in R^{5}$ , the vector $p (y) = \frac{1^{T} y}{5} 1 + \frac{z^{T} y}{z^{T} z} z$ lies in $C$ , and $y - p (y)$ is orthogonal to both generators, hence to all of $C$ . By [F2], $p (y) = P_{C} y$ ; [F1] identifies this as the fitted vector. Therefore $H = \frac{11^{T}}{5} + \frac{z z^{T}}{z^{T} z}, h_{55} = \frac{1}{5} + \frac{(74 / 5)^{2}}{1394 / 5} = \frac{1}{5} + \frac{2738}{3485} = \frac{687}{697} .$

3.1step 2.1∎

Since $h_{55} = 1 - 10 / 697$ is close to one and its formula involves only the design vector, this is a high-leverage design row. The calculation makes no claim that the fifth response has a large residual or is influential; those are separate diagnostics.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 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

Statistics: Lecture Three — Linear Models (standard reference, not scraped)
Statistics for Applications, Section 15: Multiple Linear Regression (standard reference, not scraped)