 Solution

ID: 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 gives

i = 1 \sum n D_{i}^{2} = (i = 2 \sum n D_{i})^{2} + i = 2 \sum n D_{i}^{2} .

Therefore

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.

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