core27 min studyLesson 253 in Mathematics

The Gauss–Markov theorem and efficiency

Prove the covariance comparison behind BLUE and identify exactly which competing estimators it covers.

On this page

What you’ll understand

  • State mean, rank, and covariance assumptions separately.
  • Derive the nonnegative covariance difference for a linear unbiased competitor.
  • Explain why BLUE does not imply universal predictive optimality.

Least squares uses all observed responses in a particular linear combination. Why prefer that combination to another unbiased one? The Gauss–Markov theorem supplies a precise answer under a spherical error covariance model: no other linear unbiased estimator of the same coefficient vector has smaller variance in any contrast direction. The careful wording matters. The theorem compares a specified class of estimators under specified assumptions; it does not declare least squares the best possible procedure for every loss, data-generating process, or prediction problem.

Separate the assumptions before using the acronym

Take Y=Xβ+εY=X\beta+\varepsilon with a fixed design of full column rank. Assume E(ε)=0E(\varepsilon)=0 and Cov⁡(ε)=σ2I\operatorname{Cov}(\varepsilon)=\sigma^2 I for a finite positive variance. The covariance condition combines common marginal variance, called homoskedasticity, with zero covariance between different errors. Independence is sufficient for the latter but stronger than necessary. Gaussianity is not an assumption of this version of the theorem. With random predictors these statements should be made conditional on the realized design.

Write the least-squares estimator as AYAY, where A=(XTX)−1XTA=(X^{\mathsf T}X)^{-1}X^{\mathsf T}. Because AX=IAX=I, its expectation is the target coefficient vector for every β. Its covariance is the common error variance times the inverse Gram matrix. These are different conclusions: the conditional mean assumption establishes unbiasedness, while the covariance assumption supplies this particular variance formula. Correlated or unequal-variance errors need not destroy unbiasedness, but they invalidate the spherical formula and its optimality comparison.

AX=I,E(AY)=β,Cov⁡(AY)=σ2(XTX)−1AX=I,\qquad E(AY)=\beta,\qquad \operatorname{Cov}(AY)=\sigma^2(X^{\mathsf T}X)^{-1}
The covariance equality assumes full rank and spherical errors; Gaussianity and mutual independence are unnecessary for the BLUE comparison.↔ Scroll wide equations horizontally.

The competitor must add a perpendicular component

Any competing estimator linear in the response has form BYBY, with B fixed once the design is fixed. Unbiasedness for every β forces BX=IBX=I. Let D=B−AD=B-A. Then DX=0DX=0: the competitor differs only by a linear combination of response directions that contain no modeled mean. The covariance cross terms vanish because A is built from the columns of X and D annihilates those columns. What remains is an added positive-semidefinite matrix.

For a coefficient contrast c, multiply that covariance difference on both sides by c. The result is σ2∥DTc∥2\sigma^2\|D^{\mathsf T}c\|^2, which cannot be negative. This proves the ordering simultaneously for every contrast rather than merely comparing the diagonal entries. Equality for a particular contrast means the competitor agrees with least squares on that contrast. Equality of the entire covariance matrix, with positive σ², forces D to vanish and therefore the competing linear estimator to be identical.

B=A+D,DX=0⟹Cov⁡(BY)−Cov⁡(AY)=σ2DDT⪰0B=A+D,\quad DX=0\quad\Longrightarrow\quad \operatorname{Cov}(BY)-\operatorname{Cov}(AY)=\sigma^2DD^{\mathsf T}\succeq0
The difference is positive semidefinite, so every linear contrast has no smaller variance under a competing linear unbiased estimator.↔ Scroll wide equations horizontally.

Best has a deliberately limited meaning

The acronym BLUE means best linear unbiased estimator: linear in observed responses, unbiased for all allowed coefficient vectors, and best by variance. It does not compare biased procedures such as ridge regression. A biased estimator can reduce mean squared error by accepting a controlled displacement of its expectation. Nor does the theorem compare arbitrary nonlinear estimators under this minimal moment model. Stronger distributional assumptions can support additional results, but those require their own arguments and loss functions.

The theorem also says nothing about whether a scientific mean model is correct, whether a coefficient represents a causal intervention, or whether deployment predictors resemble training predictors. In a rank-deficient design, the corresponding statement applies to estimable linear functions rather than to arbitrary coefficient coordinates. If the error covariance is a known positive-definite matrix instead of a multiple of identity, whitening leads to the generalized least-squares extension. The model determines what efficiency means; an acronym cannot substitute for checking that model.

Compare two unbiased estimates of a common mean

Four uncorrelated observations share mean μ and variance σ². Compare their arithmetic mean with T=(Y1+Y2)/2T=(Y_1+Y_2)/2. Explain the efficiency difference without assuming Gaussian data.

  1. Both sets of weights sum to one, so both estimators have expectation μ for every μ. The four-observation mean assigns weight 1/41/4 to each response.

  2. Uncorrelated errors make the variance the common variance times the sum of squared weights. The full mean has variance 4σ2/16=σ2/44\sigma^2/16=\sigma^2/4, while T has variance 2σ2/4=σ2/22\sigma^2/4=\sigma^2/2.

  3. The competitor's extra weights relative to the mean are (1/4,1/4,−1/4,−1/4)(1/4,1/4,-1/4,-1/4). They sum to zero and have squared norm 1/41/4, giving variance excess σ2/4\sigma^2/4.

The result

Ignoring two observations doubles the variance while preserving unbiasedness. The conclusion uses equal variances and zero pairwise covariances, not a Gaussian distribution.

Explore the model

Interactive lab

Interactive lab · sampling

Inspect the equal-weight mean in a concrete model

Pause and Reset; set Success probability p to 0.5 and Independent samples n to 20. Predict mean 0.5 and variance 0.0125 for the sample mean, then inspect the exact distribution. Start another fresh Reset with p=0.5 and n=80 and predict variance 0.003125. Playback highlights exact outcomes, not repeated random experiments. This iid Bernoulli example illustrates the arithmetic mean's variance; it does not compare all linear estimators.

Inspect the equal-weight mean in a concrete modelExact binomial masses for the mean of n independent Bernoulli trials, compared with normal mass between adjacent half-count boundaries; normal tails outside the finite support remain outside the bars. Playback highlights outcomes, not random samples. The Berry–Esseen bound applies to standardized CDFs, not individual bars or continuity-corrected errors. At p = 0 or 1, normal comparison and that bound are disabled. Normal-CDF evaluation uses a numerical approximation with absolute error below 7.5×10⁻⁸ (at most 1.5×10⁻⁷ per interval, before rounding). Current readings: Mean sample proportion: 0.5; Sample-mean variance: 0.0125; Uniform CDF error bound: ≤ 0.1062. All parameters are adjustable with the labeled controls below.Exact distribution of the mean of 20 independent Bernoulli variables with success probability 0.5; playback highlights outcome 0.00.250.50.75100.0580.1150.1730.23Sample mean · k / nProbability massNormal bin masses are an approximationk = 0 · exact probability = 9.54e-7
Exact binomial massNormal interval massHighlighted outcome / mean
Mean sample proportion
0.5
Sample-mean variance
0.0125
Uniform CDF error bound
≤ 0.1062
t = 0 s

Exact binomial masses for the mean of n independent Bernoulli trials, compared with normal mass between adjacent half-count boundaries; normal tails outside the finite support remain outside the bars. Playback highlights outcomes, not random samples. The Berry–Esseen bound applies to standardized CDFs, not individual bars or continuity-corrected errors. At p = 0 or 1, normal comparison and that bound are disabled. Normal-CDF evaluation uses a numerical approximation with absolute error below 7.5×10⁻⁸ (at most 1.5×10⁻⁷ per interval, before rounding). Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forEfficiency compares estimator variances under a fixed model, rather than judging one realized estimate by closeness.

Check your understanding

Which additional assumption is unnecessary for the standard full-rank Gauss–Markov theorem?

Try it yourself

Three uncorrelated observations have the same mean μ and variance σ². Compare T=Y1/2+Y2/4+Y3/4T=Y_1/2+Y_2/4+Y_3/4 with their arithmetic mean. Calculate the variance excess exactly.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Square the weights, sum them, and subtract the corresponding sum for three equal weights.

Compare with the solution

Both estimators are unbiased because their weights sum to one. The variance of T is σ2(1/4+1/16+1/16)=3σ2/8\sigma^2(1/4+1/16+1/16)=3\sigma^2/8. The equal-weight mean has variance σ2/3\sigma^2/3. The excess is (3/8−1/3)σ2=σ2/24(3/8-1/3)\sigma^2=\sigma^2/24, agreeing with the nonnegative covariance comparison.

Take it with you

  • BLUE requires a specified mean and covariance model, but no Gaussian assumption.
  • A linear unbiased competitor adds a nonnegative covariance term.
  • Biased estimation, causality, and deployment prediction require separate comparisons.

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