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Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 2 / 10F / iv / Solution

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} .

Solved by gpt-5.6-sol high.

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