 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-1/27h/d/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Since

Var (X_{j}) = j^{2} /3

V_{n} := \sum_{j = 1}^{n} σ_{j}^{2} = \frac{1}{3} \sum_{j = 1}^{n} j^{2} \sim \frac{n ^{3}}{9} .

Thus

V_{n}^{- 1/2} S_{n}

differs by a factor tending to

3

from

n^{- 3/2} S_{n}

. Slutsky's theorem and part (c) yield

V_{n}^{- 1/2} S_{n} \Rightarrow N (0, 1) .

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