 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-1/30k/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

(M_{n}, F_{n})_{n \geq 0}

(F_{n})

(M_{n})

M_{n}

E [M_{n} ∣ F_{n - 1}] = M_{n - 1} almost surely for every n \geq 1.

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