TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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Gauss–Markov theorem for estimable contrasts

Statement

Let $X \in R^{n \times p}$ have full column rank in a fixed-design first-moment regression model. Suppose ${Cov}_{β} (ϵ_{β}) = σ^{2} I_{n}$ for a fixed $σ^{2} > 0$ , where the vector covariance matrix has entries ${Cov}_{β} (ϵ_{β, i}, ϵ_{β, j})$ . For every contrast $c^{T} β$ , the OLS estimate $c^{T} \hat{β}$ is unbiased and has variance no larger than any other linear unbiased estimator of that contrast. Thus OLS is best linear unbiased (BLUE) for every estimable contrast. Here “best” means minimum variance within the class of linear unbiased estimators; Gaussianity is not assumed.

Facts & Assumptions

Given: The fixed-design model and moment hypotheses in the Statement, a contrast $c \in R^{p}$ , and any competing linear estimator $a^{T} Y$ .

[F1]

Full-rank OLS is $\hat{β} = (X^{T} X)^{- 1} X^{T} Y$ and is unbiased. Full-rank OLS is unbiased with its classical covariance.

[F2]

A contrast is linearly estimable exactly when $X^{T} a = c$ for some fixed $a \in R^{n}$ ; every such $a^{T} Y$ is unbiased. Linear estimability is equivalent to row-space membership.

[F3]

Covariance is bilinear in finite linear combinations. Covariance is symmetric and bilinear in finite linear combinations.

[F4]

Variance is the covariance of a variable with itself. Moments, variance, and covariance on a probability space.

[A1]

The error has finite second moments, mean zero, and covariance $σ^{2} I_{n}$ with $σ^{2} > 0$ .

Proof

technique · direct

1.1F1F2algebra

Define $a_{0} : = X (X^{T} X)^{- 1} c \in R^{n}$ . Then $X^{T} a_{0} = c$ and $a_{0}^{T} Y = c^{T} (X^{T} X)^{- 1} X^{T} Y = c^{T} \hat{β}$ . By [F2], $c^{T} β$ is estimable and $a_{0}^{T} Y$ is unbiased.

2.1givenF2step 1.1algebra

Let $a^{T} Y$ be any linear unbiased estimator of the same contrast. By [F2], $X^{T} a = c$ . Set $q : = a - a_{0}$ . Then $X^{T} q = 0$ , while $a_{0} \in col (X)$ ; hence $q^{T} a_{0} = 0$ .

3.1givenA1F3F4step 1.1step 2.1algebra∎

From $Y = X β + ϵ_{β}$ , [A1], [F3], and [F4], ${Var}_{β} (a^{T} Y) = σ^{2} ∥ a ∥_{2}^{2} = σ^{2} ∥ a_{0} + q ∥_{2}^{2} = σ^{2} ∥ a_{0} ∥_{2}^{2} + σ^{2} ∥ q ∥_{2}^{2} \geq {Var}_{β} (a_{0}^{T} Y) .$ Since $σ^{2} > 0$ , equality holds exactly when $q = 0$ . Therefore the OLS contrast estimator has minimum variance among all linear unbiased estimates.

Depends on

Used by

Unequal error variances can make OLS fail to be BLUE Counterexample
Known-covariance GLS is BLUE Theorem

Dependency tree · two levels

26 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)