 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-1/30k/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

The variable

M_{n}

F_{n}

-measurable. Positivity, independence, and the mean-one assumption give

E [M_{n}] = \prod_{i = 1}^{n} E [Y_{i}] = 1,

so it is integrable. Since

Y_{n + 1}

is independent of

F_{n}

E [M_{n + 1} ∣ F_{n}] = E [M_{n} Y_{n + 1} ∣ F_{n}] = M_{n} E [Y_{n + 1}] = M_{n} .

Thus $(M_{n})$ is the product martingale from independent mean-one factors.

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