Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics III / 8 / b / iii / Solution

Since ${\sum }_{i=1}^n{D}_i=0$, we have $D_1=-{\sum }_{i=2}^n{D}_i$. Squaring and adding the other residual squares givesTherefore $S^2$ is a measurable function of $(D_2,\dots ,D_n)$. By the block independence proved in part i, it is independent of $\overline{X}$. This is the independence of the normal sample mean and sample variance; pairwise independence alone would not justify passing to an arbitrary function of several residuals.