 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-3/29k/a/solution

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

T \leq N

, write

M_{T} = M_{0} + \sum_{n = 1}^{N} 1_{{T \geq n}} (M_{n} - M_{n - 1}) .

The indicator is measurable at time

n - 1

. Taking conditional expectations term by term and using the supermartingale inequality gives

E M_{T} \leq E M_{0}

. The same calculation gives equality for the martingale

M

. Since their initial values agree,

E M_{T} \geq E M_{T}

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