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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.

Depends on

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