 Solution

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

 Top articles  Latest articles New article in topic

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

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

Every martingale is a supermartingale, with equality in its defining conditional expectation. Taking expectations and using the tower property of conditional expectation gives

E [X_{n + 1}] = E [X_{n}],

so induction yields

E [X_{n}] = E [X_{0}] for every n .

 Read the full article

 New to topics? Read the docs here!