 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-2/12f/b/solution

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

Write

p_{n} = P (A_{n})

. Independence gives, first for finite partial counts and then by monotone convergence,

E [2^{- N}] = \prod_{n \geq 1} E [2^{- 1_{A_{n}}}] = \prod_{n \geq 1} (1 - \frac{p _{n}}{2}) .

Using

1 - x \leq e^{- x}

factor by factor,

E [2^{- N}] \leq exp (- \frac{1}{2} n \geq 1 \sum p_{n}) = e^{- E [N] /2} .

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