 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-2/27g/b/i/solution

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

Fix

ε > 0

. Since a rate-one exponential variable has

P (X_{n} > x) = e^{- x}

\sum_{n} P {X_{n} > (1 + ε) lo g n} = \sum_{n} n^{- (1 + ε)} < \infty.

The first Borel--Cantelli lemma gives the upper bound

lim sup X_{n} / lo g n \leq 1

. Conversely,

\sum_{n} P {X_{n} > (1 - ε) lo g n} = \sum_{n} n^{- (1 - ε)} = \infty.

These events are independent, so the second lemma makes them occur infinitely often. Hence the lower bound is

1 - ε

. Letting rational

ε ↓ 0

proves

n \to \infty lim sup \frac{X _{n}}{lo g n} = 1 almost surely.

 New to topics? Read the docs here!