 Solution

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

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

The process is adapted to its natural filtration. At each fixed time it has finite support and is therefore integrable. Conditional on

X_{n - 1} = 0

E [X_{n} ∣ X_{n - 1} = 0] = \frac{1}{2 n} - \frac{1}{2 n} = 0 = X_{n - 1} .

Conditional on

X_{n - 1} \neq = 0

E [X_{n} ∣ X_{n - 1}] = \frac{1}{n} (n X_{n - 1}) + (1 - \frac{1}{n}) 0 = X_{n - 1} .

E [X_{n} ∣ F_{n - 1}] = E [X_{n} ∣ X_{n - 1}] = X_{n - 1},

(X_{n})

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