Past exam of the options I course of the University of Oxford / 2008 / O / O2 / Solution

The normal equations for linear least squares are $X^{T}X\hat{\theta }=X^{T}Y$. Under full column rank, $\hat{\theta }=(X^{T}X{)}^{-1}{X}^{T}Y$. Since $\mathrm{Cov}(Y)={\sigma }^{2}I$,Here $\bar{x}=0$ and $\bar{z}=2$, so the design matrix isIts columns are pairwise orthogonal and have squared lengths $6,28,4$. Consequently $X^{T}X=\mathrm{diag}(6,28,4)$, givingBoth estimators are linear functions of the Jointly Gaussian observation vector, so they are Jointly Gaussian. Zero covariance then implies independence. This final implication uses the normal error model, whereas the variance calculation itself needs only its second moments.