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.

Collinear predictors can leave a contrast estimable

Example

Take the two-observation design and response $X = (\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array}), y = (\begin{array}{c} 1 \\ 3 \end{array}) .$ Both columns of $X$ are identical. The least-squares fit is unique, but the coefficient pair is not. The contrast $β_{1} + β_{2}$ is linearly estimable; neither $β_{1}$ nor $β_{2}$ is.

Facts & Assumptions

Given: A fixed-design first-moment regression with $X$ and observed vector $y$ as in the Example.

[F1]

Every least-squares solution has fitted vector $P_{col (X)} y$ and residual $y - P_{col (X)} y$ ; its coefficient solution set is one translate of $\ker X$ . Least-squares fitted vectors and residuals are unique.

[F2]

A contrast $c^{T} β$ is linearly estimable exactly when $c \in row (X) = im (X^{T})$ . Linear estimability is equivalent to row-space membership.

Verification

technique · direct

Given: The matrix $X$ and response $y$ displayed above.

1.1givenF1algebra

For $b = (b_{1}, b_{2})^{T}$ , put $s = b_{1} + b_{2}$ . Then $X b = (s, s)^{T}$ , and the squared residual is $(1 - s)^{2} + (3 - s)^{2} = 2 (s - 2)^{2} + 2.$ It is minimized exactly when $s = 2$ . Thus the full coefficient solution set is $(1, 1)^{T} + t (1, - 1)^{T}$ for $t \in R$ , its fitted vector is $(2, 2)^{T}$ , and its residual is $(- 1, 1)^{T}$ , in agreement with [F1].

2.1F2algebra∎

The row space of $X$ is $span {(1, 1)^{T}}$ . Hence $c = (1, 1)^{T}$ is in the row space and is estimable; indeed $X^{T} (1 / 2, 1 / 2)^{T} = (1, 1)^{T}$ , so $(Y_{1} + Y_{2}) / 2$ estimates $β_{1} + β_{2}$ unbiasedly. The vector $(1, 0)^{T}$ is not in that row space, since every vector $X^{T} a$ has equal coordinates. Therefore neither single coefficient is linearly estimable by [F2].

Remarks

The finite check verifies this stated witness: for the two-column design $X = (\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array})$ and $v = (1, - 1)^{T}$ , $X v = 0$ , the Gram matrix $X^{T} X / 2$ differs from $I_{2}$ , and $v_{1} \neq 0$ . This check illustrates the rank-deficient direction; the algebraic proof above supplies the general interpretation.

Depends on

Used by

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Dependency tree · two levels

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Sources

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