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Unequal error variances can make OLS fail to be BLUE

Statement refuted

For every full-column-rank fixed design and every mean-zero error law with finite second moments, OLS is the best linear unbiased estimator of every contrast, regardless of the error covariance.

Facts & Assumptions

[F1]

OLS is BLUE for every estimable contrast when the fixed full-rank design has mean-zero errors with covariance $σ^{2} I_{n}$ , $σ^{2} > 0$ . Gauss–Markov theorem for estimable contrasts.

[F2]

Expectation is linear, and covariance is bilinear on finite linear combinations. Linearity, monotonicity, and the modulus bound for expectation, Covariance is symmetric and bilinear in finite linear combinations.

[F3]

On a finite probability space, expectation is the sum of values weighted by their point probabilities. Expectation agrees with the published finite weighted sum.

[F4]

A probability measure is a measure of total mass one. Probability measures and probability spaces.

[F5]

Variance and covariance are centered second moments. Moments, variance, and covariance on a probability space.

Counterexample

technique · direct

Given: Let $n = 2$ , $p = 1$ , $X = (1, 1)^{T}$ , and $β \in R$ . Let $Ω = {- 1, 1}^{2}$ with $F = P (Ω)$ and $P (A) = ∣ A ∣ / 4$ for $A \subseteq Ω$ ; write the coordinate signs as $U, V$ . Set $ϵ = (U, 3 V)^{T}, Y = X β + ϵ .$ Each of the four pairs $(U, V) \in {- 1, 1}^{2}$ has probability $1 / 4$ . Both errors have mean zero, but their variances are $1$ and $9$ . The single column of $X$ is nonzero, so the design has full column rank and estimates the intercept contrast $β$ . For each $β$ , define its observation law on $B (R^{2})$ by $P_{β} (B) : = \frac{1}{4} \sum_{(u, v) \in Ω} 1_{B} (β + u, β + 3 v) .$

1.1givenF2F3F4F5algebra

The weights $1 / 4$ are nonnegative and sum to one. For a disjoint sequence of subsets of $Ω$ , at most four sets are nonempty, so additivity of cardinality gives countable additivity; thus $P$ is a probability measure by [F4]. In the displayed formula, $P_{β} (R^{2}) = 1$ ; for any disjoint Borel sequence, each of the four outcomes contributes to at most one term, so $P_{β}$ is countably additive and hence a probability measure on $B (R^{2})$ . Direct weighted sums using [F3] give $E U = E V = 0$ , $E U^{2} = E V^{2} = 1$ , and $E (U V) = 0$ . Thus $E ϵ = 0$ and, by [F5], $Cov (ϵ) = diag (1, 9)$ , which is not a scalar multiple of $I_{2}$ . By linearity [F2], $E_{β} Y = X β$ .

1.2F2F5algebra

Here $X^{T} X = 2$ , so the unique OLS estimate is $\hat{β} = \frac{Y_{1} + Y_{2}}{2} = β + \frac{U + 3 V}{2} .$ It is unbiased by [F2], and bilinearity gives $Var (\hat{β}) = (1 + 9) / 4 = 5 / 2$ .

2.1givenF1F2F5step 1.1step 1.2algebra∎

The alternative estimator $\tilde{β} = \frac{9}{10} Y_{1} + \frac{1}{10} Y_{2} = β + \frac{9}{10} U + \frac{3}{10} V$ is linear and unbiased because its coefficients sum to one and both error means vanish. Its variance is $\frac{81}{100} Var (U) + \frac{9}{100} Var (V) + \frac{54}{100} Cov (U, V) = \frac{9}{10} < \frac{5}{2} .$ So OLS is not BLUE in this model. The covariance premise in [F1] fails, which is the stated boundary of the Gauss–Markov theorem.

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Sources

Statistics: Lecture Three — Linear Models (standard reference, not scraped)
Statistics for Applications, Section 15: Multiple Linear Regression (standard reference, not scraped)