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Let $H=X(X^{T}X{)}^{-1}{X}^T$ be the orthogonal projection onto the design column space. Linear transformations of the Gaussian errors giveThe residual distribution is a singular multivariate normal supported on the orthogonal complement of that column space. Their cross-covariance is ${\sigma }^2(X^{T}X{)}^{-1}{X}^T(I-H)=0$. Joint normality even makes the two vectors independent.

Order the observations as $11,12,21,22$. The two normal linear models have design matricesRemoving each row mean leaves the information for the common slope equal to the sum of within-row squared fertilizer deviations. These sums are $1$ and $4$, respectively. ThereforeThe second design estimates the slope four times as precisely in variance terms.

The observed data in the second design are fitted exactly by $\hat{\theta }=(100,300,100{)}^T$. With $a=(2,2,0{)}^T$, the no-fertilizer total forecast isThe future four independent errors contribute variance $4{\sigma }^2$. Parameter-estimation uncertainty contributes ${\sigma }^2{a}^T(X_{II}^T{X}_{II}{)}^{-1}a=13{\sigma }^2$, so the prediction-error variance is $17{\sigma }^2$, not merely the variance of the estimated mean. Normally, with $s^2=\Vert \widehat{ϵ}{\Vert }^2/(4-3)$, the prediction interval in a normal linear model isHere the printed data have zero residual and hence $s=0$. The formal plug-in interval collapses to $[800,800]$, exhibiting zero-residual degeneracy of regression prediction. Under the stated model with unknown positive variance, this exact fit is a probability-zero sample and the Studentized statistic is undefined at it; the collapsed expression is not evidence that future yield is certain. No positive-width numerical 95% interval can be obtained from these exact data by the usual residual-variance method without extra variance information or an explicit observation-rounding model.