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The normal-model log-likelihood is $-\frac{n}{2}\log {\sigma }^2-\Vert Y-X\beta {\Vert }^2/(2{\sigma }^2)$ up to a constant. Differentiating gives the maximum-likelihood estimatorsLet $P=X(X^{T}X{)}^{-1}{X}^T$. The fitted component $P\epsilon$ and residual $(I-P)\epsilon$ are orthogonal jointly normal projections and hence independent. ThereforeThe prediction error $\tilde{y}-y^{\ast }=x^{\ast T}(\hat{\beta }-\beta )-{\epsilon }^{\ast }$ is normal with mean zero and variance ${\sigma }^2{\tau }^2$, and is independent of the residual variance estimate. Dividing by the independent estimated standard deviation yieldsThus the prediction interval in a normal linear model isThe extra $1$ includes the new observation's independent noise; omitting it would give a confidence interval for its conditional mean instead.

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.

For a new independent tumour grown for ten days in B, use $x_0=(1,10,10,0{)}^T$. Its fitted mean isWrite $h_0=x_0^T(X^{T}X{)}^{-1}{x}_0$. The prediction interval in a normal linear model iswhere $s^2=\mathrm{RSS}/26$ and $t_{26,1-\alpha /2}$ is the indicated Student's t-distribution quantile. The design information needed to evaluate $h_0$ is not supplied, so a fully numerical interval cannot be recovered from the coefficient standard errors alone: the coefficient covariances are also needed.

To prove coverage, write $Y_0=x_0^T\beta +{\epsilon }_0$, with ${\epsilon }_0\sim N(0,{\sigma }^2)$ independent of the original sample. The error $Y_0-x_0^T\hat{\beta }$ has a normal distribution with mean zero and variance ${\sigma }^2(1+h_0)$. By the joint distribution of least-squares and variance estimators, $26s^2/{\sigma }^2\sim {\chi }_{26}^2$ and is independent of $\hat{\beta }$; it is also independent of ${\epsilon }_0$. Hence the pivotal quantityConsequently $\mathrm{Pr}(-t_{26,1-\alpha /2}\le T\le t_{26,1-\alpha /2})=1-\alpha$, which is exactly the claimed prediction interval coverage. This is repeated-sampling coverage conditional on the design, assuming the model; it need not hold conditional on the particular observed training responses. The extra $1$ represents new-observation noise and must not be omitted as in a mean confidence interval.

A $(1-\alpha )$ prediction interval is a pair of statistics $L(Y),U(Y)$ such that, for every $\beta$ and ${\sigma }^2>0$, an independent future response $Y^{\ast }$ at the specified input satisfies

In the normal linear model, assume $X$ has full column rank and writeFor a new observationindependent of the training sample, the prediction error isThe two terms on the right are independent centered normal random variables, whileConsequently, with the new-point regression leveragewe have

The assumption in the question givesThis variable is independent of $\hat{\beta }$; it is also independent of the new error ${\epsilon }^{\ast }$. Therefore it is independent of the standardized prediction error, and the defining normal-over-chi-square representation of Student's t-distribution givesIf $t_{n-p,1-\alpha /2}$ is the corresponding upper quantile, symmetry yieldsSolving these inequalities for $Y^{\ast }$ proves that the prediction interval in a normal linear model isIts center and endpoints depend only on the observed data and $x^{\ast }$, and the pivotal calculation proves the required coverage uniformly in the unknown $\beta$ and ${\sigma }^2$.