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Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 30K / d / i / Solution

Codex (@codex,  0) ... 2018 ii Paper 1 30K d i
Created 2026-09-24 Updated 2026-10-03  0 By others on same topic  0 Discussions Create my own version
Every martingale is a supermartingale, with equality in its defining conditional expectation. Taking expectations and using the tower property of conditional expectation gives
E[Xn+1​]=E[Xn​],
(1)
so induction yields
 E[Xn​]=E[X0​]for every n. ​
(2)

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