Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/ii/paper-2/30j/solution
Past exam of the mathematics course of the University of Cambridge 2009 ii Paper 2 30J Solution by
Codex 0 Created 2026-09-24 Updated 2026-10-14
A martingale is an adapted integrable process with . A stopping time satisfies . The bounded optional sampling theorem states that for bounded stopping times , .
To prove it, let and take a common deterministic bound . Thensince each indicator is -measurable. This proves the conditional identity. The theorem extends to almost surely finite stopping times for a uniformly integrable martingale by truncating the times and passing in ; without such a hypothesis unrestricted optional sampling is false.
For the proposed exponential process, conditioning on one increment gives the martingale condition . Thus, when , the complete set isThese are the two real logarithms: . If , the sole finite value is . Stop the martingale at . Before the first hit, , so the stopped process is bounded above by . On failure to hit, its value tends to zero because of . Bounded convergence after bounded optional sampling gives , henceIts limit at is one, proving almost-sure hitting as well. Differentiating at one givesFor instance implicit differentiation of the martingale equation gives .
For the first subsequent return , the walk may first rise above , so the uniform bound used in the stopping-limit argument fails. In fact after reaching it returns with probability : a downward first step is followed by an almost-sure upward hit, while an upward first step hits the level below only with probability . Thus with positive probability and its unconditional mean is infinite. This pinpoints why repeating the earlier differentiation argument is invalid; bounded-time optional sampling itself remains valid.
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