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Fixed-design regression, estimability, and Gauss–Markov

1 · Prerequisites

2 · Summary

Objective

This page develops least squares for a fixed matrix X∈Rn×p and response vector Y∈Rn. “Fixed design” means the results condition on X; it does not specify how a scientific study selected that design or prove that the regression mean is adequate for an application.

Scope

For an arbitrary design rank, the least-squares coefficient set is nonempty and is a translate of ker⁡X, while the fitted vector and residual are unique. A contrast cTβ is linearly estimable exactly when c∈row⁡(X), equivalently when it has a linear unbiased representative aTY under the first-moment model EβY=Xβ.

Under full column rank, OLS has formula (XTX)−1XTY. If additionally Cov⁡β(ϵ)=σ2In with σ2>0, OLS is unbiased with covariance σ2(XTX)−1 and is BLUE for every estimable contrast. These conclusions use only first and second moments; they do not require Gaussian errors.

The page records the model-side fields of Scientific statistical model contracts at an abstract level; it is not a complete scientific model contract for an application because no target population or scientific units are specified. One analysis record is Y∈Rn; row index i is a response slot; X and β are fixed; and Y is observed under Pβ. No scientific, sampling, or assignment unit, time origin, or target population is specified; no enrollment process is modeled, so its status is not applicable here; no latent observation mechanism or randomized design is modeled. The law is on Borel Rn for the standard Euclidean metric topology, without an additional support assumption. Expectations and covariances are under Pβ with X fixed. The mean equation is a model premise that is empirically assessable in an application; no data or validation is supplied here, and no scientific adequacy is claimed.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Fixed-design linear regression model

Definition

Let n,p≥1 be integers and let X∈Rn×p be a fixed matrix. A fixed-design first-moment linear regression model is a parametric statistical experiment on (Rn,B(Rn)) with parameter set Rp, laws (Pβ)β∈Rp, and observed vector Y(y)=y, such that every coordinate of Y is integrable under Pβ and EβY=Xβ(β∈Rp). Here B(Rn) is the Borel sigma-algebra, and expectation is taken under the particular law Pβ. The residual vector ϵβ:=Y−Xβ therefore has coordinatewise mean zero. The phrase fixed design means that X is held constant in this experiment; it carries no probability law or assignment mechanism for X.

This definition states only the first-moment model. It imposes no covariance, independence, or Gaussian assumption. Such assumptions must be stated where used.

Remarks

Residual mean. For each fixed β and coordinate j, put cj=(Xβ)j. Both Yj and the constant cj are integrable under the probability law Pβ, and the constant has expectation cj. By Linearity, monotonicity, and the modulus bound for expectation, Eβ[ϵβ,j]=Eβ[Yj−cj]=Eβ[Yj]−cj=(Xβ)j−cj=0. Thus the residual-mean conclusion in the Definition uses expectation linearity coordinatewise under each law; it adds no covariance or distributional assumption.

This is an abstract mathematical regression model, not a complete scientific model contract for an application. The Definition of Scientific statistical model contracts requires a field record for a specified analysis, including its target population; this pair has no specified application or target population. The following notes record the model-side fields that can be stated at this abstract level.

The scientific application, including its target population and substantive adequacy evidence, remains unspecified.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passOpen item page →

Linear statistics on Euclidean data are Borel-measurable

Statement

Let n≥1 and a∈Rn. The linear map ℓa:Rn→R,ℓa(y)=aTy, is measurable from the Euclidean Borel space Rn to the Borel space R.

Facts & Assumptions

Given: An integer n≥1, a vector a∈Rn, and the standard Euclidean metrics and their Borel sigma-algebras on Rn and R.

[F1]

Every linear map L:Rm→Rk, with m,k≥1, is bounded: for some K≥0, ∥Lh∥2≤K∥h∥2 for every h. Linearity means preservation of linear combinations by A linear map L:Rm→Rn in Euclidean coordinates, and the bound is Every Euclidean linear map has a unique matrix and satisfies ∥Lh∥2≤K∥h∥2 for some K≥0.

[F2]

On Rm for m≥1, the norm from the Euclidean inner product agrees with d2. The inner product is bilinear, and Rm consists of functions on {0,…,m−1} with pointwise vector operations. The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn, Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it.

[F3]

For every m≥1, d2 is a metric on Rm by Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it. For any metric space, its metric-open sets form a topology: the empty set and the whole space are open by 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, and arbitrary unions and finite intersections are open by Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed. The balls B(x,r) used below are as defined by Open ball, closed ball and sphere in a metric space. Thus the metric topology gives a topological space by Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison, and its Borel sigma-algebra is generated by those open sets by The Borel sigma-algebra of a topological space.

[F4]

A map of topological spaces is continuous when, for each point and each open target neighborhood of its image, it maps some open source neighborhood of the point into that target neighborhood. Continuity of a map of topological spaces at a point and globally.

[F5]

A continuous map between topological spaces has Borel preimages of Borel sets. A continuous map has Borel preimages of Borel sets.

[F6]

A map between measurable spaces is measurable when the preimage of every measurable target set is measurable in the source. A measurable function between measurable spaces.

Proof

technique · direct
1.1F2F3constructalgebra

Define J:R→R1 by J(t)(0)=t. By [F2], J is a bijection with inverse x↦x0, and preserves linear combinations. The one-coordinate metric formula gives d2(J(s),J(t))=(s−t)2=∣s−t∣. Thus the usual scalar distance dR(s,t):=∣s−t∣ inherits the metric axioms from d2 on R1. By [F3], both this scalar metric and the source metric d2 give the required topologies and Borel sigma-algebras.

2.1F1F2step 1.1choosealgebra

Put L:=J∘ℓa:Rn→R1. Since aTy=⟨a,y⟩, bilinearity in [F2] and step 1.1 show that L preserves linear combinations. By [F1], choose K≥0 with ∥Lh∥2≤K∥h∥2 for every h. Hence ∣ℓa(y)−ℓa(z)∣=∥L(y−z)∥2≤Kd2(y,z) for all y,z∈Rn.

3.1F3F4step 1.1step 2.1choosealgebra

Let y∈Rn and let V be open in the scalar target topology with ℓa(y)∈V. By [F3], choose ε>0 with BdR(ℓa(y),ε)⊆V. Put δ:=ε/(K+1)>0, where K is from step 2.1. If z∈Bd2(y,δ), then step 2.1 gives ∣ℓa(z)−ℓa(y)∣≤Kd2(z,y)≤(K+1)d2(z,y)<ε. Thus ℓa[Bd2(y,δ)]⊆BdR(ℓa(y),ε)⊆V. The source ball is open by [F3], so [F4] shows that ℓa is continuous between the topological spaces established in step 1.1.

4.1step 3.1F5F6∎

By [F5], every Borel subset of the target has a Borel preimage in the source; [F6] therefore makes ℓa measurable.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Least-squares solution set

Definition

For integers n,p≥1, a matrix X∈Rn×p, and a response vector y∈Rn, define the least-squares solution set BLS(y):=arg min⁡b∈Rp∥y−Xb∥22. This is a set-valued definition; it does not select one coefficient vector. The next theorem proves that the set is nonempty and identifies the fitted vector and residual as unique, including when X is rank-deficient.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Linearly estimable contrasts in fixed-design regression

Definition

In a fixed-design first-moment regression model with parameter β∈Rp, a linear contrast is the estimand ψc(β):=cTβ,c∈Rp. The contrast is linearly estimable if there is a fixed vector a∈Rn such that the statistic aTY is unbiased for ψc at every parameter value: Eβ[aTY]=cTβfor every β∈Rp. The vector a may depend on the fixed design and on c, but not on the observed response or on the unknown parameter. The map y↦aTy is a measurable statistic by Linear statistics on Euclidean data are Borel-measurable and Statistics and estimators.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Least-squares fitted vectors and residuals are unique

Statement

Let n,p≥1, X∈Rn×p, y∈Rn, and C=col⁡(X)⊆Rn. For the least-squares solution set BLS(y) of Least-squares solution set, the following hold:

  1. BLS(y) is nonempty, and b is in it exactly when XT(Xb−y)=0.
  2. If b0∈BLS(y), then BLS(y)=b0+ker⁡X.
  3. Every least-squares solution has the same fitted vector PCy and residual y−PCy.
  4. The direction space ker⁡X of the affine solution set has dimension p−rank⁡(X).

The theorem makes no statistical moment or distributional assumption.

Facts & Assumptions

Given: Integers n,p≥1, a real matrix X∈Rn×p, and y∈Rn. Put TX:Rp→Rn, TXb:=Xb, and C=col⁡(X)=im⁡TX.

[F1]

The least-squares solution set is BLS(y)=arg min⁡b∈Rp∥y−Xb∥22; it is set-valued until uniqueness is proved. Least-squares solution set.

[F2]

For a finite-dimensional linear map T and target y, minimizers of ∥Tb−y∥ exist, are characterized by T∗(Tb−y)=0, and form b0+ker⁡T. For a linear map T:V→W between finite-dimensional inner-product spaces, x minimises ∥Tx−b∥ if and only if T∗(Tx−b)=0, equivalently T∗Tx=T∗b; minimisers exist and any two differ by an element of ker⁡T.

[F3]

The Euclidean inner product on Rm is the dot product ⟨x,y⟩=∑k<mxkyk. The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn.

[F4]

In standard real coordinate bases, the adjoint matrix of a linear map is its transpose. In orthonormal bases, the matrix of the adjoint is the conjugate transpose of the matrix.

[F5]

The orthogonal projection PCy is the C-component of y and y−PCy∈C⊥. The orthogonal projection PWv is the W-component in V=W⊕W⊥.

[F6]

PCy is the unique point of C nearest to y. The orthogonal projection is the unique nearest point in the subspace.

[F7]

For a linear map with finite-dimensional domain, dim⁡V=dim⁡ker⁡T+dim⁡im⁡T. Rank-nullity: dim⁡FV=nullity⁡T+rank⁡T.

[F8]

Squaring is strictly increasing on nonnegative reals: s<t if and only if s2<t2 for s,t≥0. Squaring is monotone on the nonnegatives.

[F10]

The matrix rank of X equals the rank of TX, which is dim⁡im⁡TX. The rank of a matrix equals the rank of the linear map x↦Ax.

Proof

technique · direct
1.1F1F2F3F4F8algebra

Regard X as the linear map TX:Rp→Rn, TXb=Xb, with the Euclidean inner products of [F3]. Every residual norm is nonnegative, and [F8] says squaring is strictly increasing on the nonnegative reals. Therefore the minimizers of ∥y−Xb∥22 are exactly the minimizers of ∥Xb−y∥2. Applying [F2] and identifying TX∗=XT by [F4] gives a minimizer, the normal equation XT(Xb−y)=0, and the full solution set b0+ker⁡X.

1.2F2F5F6choose

The image of TX is C. Every minimizer maps to a point of C nearest to y, and every point of C has a preimage under TX. By [F5] and [F6] that nearest point is exactly PCy and is unique. Thus every minimizing b has fitted vector Xb=PCy, and its residual is the unique vector y−PCy.

2.1F7F9F10step 1.1algebra∎

By step 1.1 the solution set is the translate b0+ker⁡X, whose direction space is ker⁡X. By rank-nullity [F7] and dim⁡Rp=p from [F9], dim⁡ker⁡X=p−dim⁡im⁡TX. Fact [F10] identifies dim⁡im⁡TX=rank⁡(X), giving the claimed direction-space dimension.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Linear estimability is equivalent to row-space membership

Statement

In a fixed-design first-moment regression model with X∈Rn×p, a contrast cTβ is linearly estimable if and only if c∈row⁡(X)=im⁡(XT), equivalently if and only if there exists a∈Rn with XTa=c. Whenever this holds, aTY is unbiased for cTβ 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∈Rp, a contrast vector c∈Rp, and a fixed vector a∈Rn when one is supplied.

[F1]

The response has finite coordinate means and EβY=Xβ for every β. Fixed-design linear regression model.

[F2]

The contrast cTβ is linearly estimable exactly when a fixed measurable statistic aTY 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.

Proof

technique · direct
1.1F1F3algebra

For any fixed a, each coordinate of Y is integrable by [F1], so the finite linear combination aTY is integrable. Linearity [F3] gives Eβ[aTY]=aTEβY=aTXβ=(XTa)Tβ.

2.1F2step 1.1

If XTa=c, step 1.1 gives Eβ[aTY]=cTβ for every β. By [F2] the contrast is linearly estimable. The equation XTa=c says exactly that c∈im⁡(XT)=row⁡(X).

2.2F2F4step 1.1choose

Conversely, suppose cTβ is linearly estimable. By [F2] there is a fixed a for which the expectation in step 1.1 equals cTβ for every β. Thus (XTa−c)Tβ=0 for every β; taking β=ej for each standard coordinate vector in [F4] shows coordinatewise that XTa=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 XTa=c.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Full-rank OLS is unbiased with its classical covariance

Statement

Let X∈Rn×p have full column rank (its columns are linearly independent) in a fixed-design first-moment regression model. Suppose for some σ2>0 the error vector ϵβ:=Y−Xβ has finite second moments, Eβϵβ=0, and Cov⁡β(ϵβ)=σ2In for every β, where the vector covariance matrix has entries Cov⁡β(ϵβ,i,ϵβ,j). Then the unique least-squares estimator is β^=(XTX)−1XTY, it is unbiased for β, and its covariance matrix is Cov⁡β(β^)=σ2(XTX)−1. No Gaussian assumption is made.

Facts & Assumptions

Given: Integers n,p≥1, a fixed-design first-moment regression model, full-column-rank X∈Rn×p, and the finite second-moment assumptions in the Statement.

[F1]

The fixed design has response mean EβY=Xβ. Fixed-design linear regression model.

[F2]

Least-squares minimizers satisfy XT(Xb−Y)=0; when ker⁡X={0} the minimizer is unique. Least-squares fitted vectors and residuals are unique.

[F3]

For the endomorphism G(v)=XTXv of Rp, rank-nullity gives dim⁡Rp=dim⁡ker⁡G+dim⁡im⁡G. Rank-nullity: dim⁡FV=nullity⁡T+rank⁡T.

[F4]

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

[F5]

Covariance is bilinear in finite linear combinations of square-integrable random variables. Covariance is symmetric and bilinear in finite linear combinations.

[F6]

Variance and covariance are expectations of centered products. Moments, variance, and covariance on a probability space.

[F8]

The Euclidean inner product is positive definite and ∥z∥22=zTz; in particular z≠0 implies ∥z∥22>0. The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn.

[A1]

The error coordinates are square-integrable, have mean zero, and satisfy Cov⁡β(ϵβ)=σ2In for σ2>0.

Proof

technique · direct
1.1F2F3F7F8algebra

Full column rank means Xv≠0 for every nonzero v∈Rp. By [F8], vTXTXv=∥Xv∥22>0 for such v, so Gv=0 forces v=0. Thus G has trivial kernel. The domain Rp has dimension p by [F7]; rank-nullity [F3] gives dim⁡im⁡G=p, so G is surjective and invertible. The normal equation in [F2] now gives the unique minimizer β^=(XTX)−1XTY.

2.1F1A1F4step 1.1algebra

Write A=(XTX)−1XT. By the fixed-design model [F1], ϵβ=Y−Xβ, so Y=Xβ+ϵβ. Step 1.1 gives β^=β+Aϵβ. By [F4] and [A1], Eββ^=β+AEβϵβ=β.

3.1A1F5F6step 1.1step 2.1algebra∎

By [F5] and [F6], the (r,s) entry of the covariance matrix of Aϵβ is ∑i<n∑j<nAriAsjCov⁡β(ϵβ,i,ϵβ,j)=σ2∑i<nAriAsi. Thus Cov⁡β(β^)=σ2AAT by step 2.1. Since XTX is symmetric, AAT=(XTX)−1XTX(XTX)−1=(XTX)−1. No step uses the shape of the error distribution beyond its stated moments.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Gauss–Markov theorem for estimable contrasts

Statement

Let X∈Rn×p have full column rank in a fixed-design first-moment regression model. Suppose Cov⁡β(ϵβ)=σ2In for a fixed σ2>0, where the vector covariance matrix has entries Cov⁡β(ϵβ,i,ϵβ,j). For every contrast cTβ, the OLS estimate cTβ^ 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∈Rp, and any competing linear estimator aTY.

[F1]

Full-rank OLS is β^=(XTX)−1XTY and is unbiased. Full-rank OLS is unbiased with its classical covariance.

[F2]

A contrast is linearly estimable exactly when XTa=c for some fixed a∈Rn; every such aTY 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 σ2In with σ2>0.

Proof

technique · direct
1.1F1F2algebra

Define a0:=X(XTX)−1c∈Rn. Then XTa0=c and a0TY=cT(XTX)−1XTY=cTβ^. By [F2], cTβ is estimable and a0TY is unbiased.

2.1givenF2step 1.1algebra

Let aTY be any linear unbiased estimator of the same contrast. By [F2], XTa=c. Set q:=a−a0. Then XTq=0, while a0∈col⁡(X); hence qTa0=0.

3.1givenA1F3F4step 1.1step 2.1algebra∎

From Y=Xβ+ϵβ, [A1], [F3], and [F4], Var⁡β(aTY)=σ2∥a∥22=σ2∥a0+q∥22=σ2∥a0∥22+σ2∥q∥22≥Var⁡β(a0TY). Since σ2>0, equality holds exactly when q=0. Therefore the OLS contrast estimator has minimum variance among all linear unbiased estimates.

5 · Examples, counterexamples and false statements

None yet.

Sources