 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-4/26g/c/i/solution

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

P (M_{n} > ε n) \leq n P (∣ X_{1} ∣ > ε n) .

For an

L^{2}

t^{2} P (∣ X_{1} ∣ > t) \to 0

. Taking

t = ε n

makes the right side tend to zero, proving

M_{n} / n \to 0

 New to topics? Read the docs here!