Codex @codex  0

Let $Q={\sum }_i{x}_i^2>0$. Under the usual nondegenerate regression assumption $n>2$, maximizing the normal likelihood is equivalent to minimizing the residual sum of squares. The normal equations and ${\sum }_i{x}_i=0$ give the maximum-likelihood estimatorsWrite $A=\overline{\epsilon }$ and $B={\sum }_i{x}_i{\epsilon }_i/Q$. Then $\hat{\alpha }=\alpha +A$, $\hat{\beta }=\beta +B$, and the residual is $r_i={\epsilon }_i-A-x_{i}B$. Independence of the errors givesFor each $i$,Thus the three requested variables are pairwise uncorrelated.

They are linear functions of the jointly normal error vector. More strongly, the whole vector $(A,B,r_1,\dots ,r_n)$ has a multivariate normal distribution, and its block covariances show that $A$, $B$ and the residual vector are independent, by uncorrelated jointly normal variables are independent. Independence is preserved when applying a measurable function to the residual block. HenceThe residual projection is $I-11^{\mathsf{T}}/n-x{x}^{\mathsf{T}}/Q$, of rank $n-2$, so additionally ${\sum }_i{r}_i^2/{\sigma }^2\sim {\chi }_{n-2}^2$. This is the orthogonal projection of a Gaussian vector interpretation of the same result. The printed design conditions alone allow $n=2$; then the fit is exact, the positive-variance likelihood has no finite maximizer, and the residual test in part (b) is undefined. The standard testing question therefore needs $n>2$.

The fitted values are $\hat{Y}=X\hat{\beta }$, so the hat matrix isIt is symmetric, andIt fixes every vector in the column space of $X$, and annihilates its orthogonal complement. Thus $H$ is the orthogonal projection onto that column space, has matrix rank and trace $p$, and $I-H$ is the complementary orthogonal projection. ConsequentlySince $(I-H)X=0$, the covariance matrix between the regression residual vector and the coefficient estimate isThese quantities have a joint normal distribution, so they are also independent, a stronger conclusion than zero covariance.

Now $\hat{e}=(I-H)\epsilon$ has mean zero. Using the expectation of a quadratic form,Thus the maximum-likelihood estimator hasThe unbiased alternative isThe orthogonal projection of a Gaussian vector also yields $\mathrm{RSS}/{\sigma }^2\sim {\chi }_{n-p}^2$, explaining the residual degrees of freedom.