TheoremStatement: 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.

Linear estimability is equivalent to row-space membership

Statement

In a fixed-design first-moment regression model with $X \in R^{n \times p}$ , a contrast $c^{T} β$ is linearly estimable if and only if $c \in row (X) = im (X^{T}),$ equivalently if and only if there exists $a \in R^{n}$ with $X^{T} a = c$ . Whenever this holds, $a^{T} Y$ is unbiased for $c^{T} β$ for every $β$ . Here $row (X)$ is defined as the image of the transpose map.

Facts & Assumptions

Given: A fixed-design first-moment regression model with parameter $β \in R^{p}$ , a contrast vector $c \in R^{p}$ , and a fixed vector $a \in R^{n}$ when one is supplied.

[F1]

The response has finite coordinate means and $E_{β} Y = X β$ for every $β$ . Fixed-design linear regression model.

[F2]

The contrast $c^{T} β$ is linearly estimable exactly when a fixed measurable statistic $a^{T} Y$ is unbiased for it at every $β$ . Linearly estimable contrasts in fixed-design regression.

[F3]

Expectation is linear on finite linear combinations of integrable random variables. Linearity, monotonicity, and the modulus bound for expectation.

[F4]

The standard coordinate vectors $e_{0}, \dots, e_{p - 1}$ form a basis of $R^{p}$ . The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ .

Proof

technique · direct

1.1F1F3algebra

For any fixed $a$ , each coordinate of $Y$ is integrable by [F1], so the finite linear combination $a^{T} Y$ is integrable. Linearity [F3] gives $E_{β} [a^{T} Y] = a^{T} E_{β} Y = a^{T} X β = (X^{T} a)^{T} β .$

2.1F2step 1.1

If $X^{T} a = c$ , step 1.1 gives $E_{β} [a^{T} Y] = c^{T} β$ for every $β$ . By [F2] the contrast is linearly estimable. The equation $X^{T} a = c$ says exactly that $c \in im (X^{T}) = row (X)$ .

2.2F2F4step 1.1choose

Conversely, suppose $c^{T} β$ is linearly estimable. By [F2] there is a fixed $a$ for which the expectation in step 1.1 equals $c^{T} β$ for every $β$ . Thus $(X^{T} a - c)^{T} β = 0$ for every $β$ ; taking $β = e_{j}$ for each standard coordinate vector in [F4] shows coordinatewise that $X^{T} a = c$ . Hence $c$ belongs to the row space.

3.1step 2.1step 2.2step 1.1∎

Steps 2.1 and 2.2 prove both implications, and step 1.1 proves the stated unbiasedness for every representative $a$ with $X^{T} a = c$ .

Depends on

Used by

Unequal error variances can make OLS fail to be BLUE Counterexample
Collinear predictors can leave a contrast estimable Example
Exact t and partial F inference in full-rank Gaussian regression Theorem
Gauss–Markov theorem for estimable contrasts Theorem

Dependency tree · two levels

41 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)