 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-2/10f/iv/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Independence and the geometric variance formula yield

var (T_{n}) = j = 1 \sum n \frac{1 - j / n}{( j / n ) ^{2}} = j = 1 \sum n \frac{n ( n - j )}{j ^{2}} .

Since

n (n - j) \leq n^{2}

var (T_{n}) \leq n^{2} \sum_{j = 1}^{n} \frac{1}{j ^{2}} \leq C n^{2} .

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