 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-1/30k/b/solution

 Top articles  Latest articles New article in topic

Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 30K / b / Solution by

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

Because

X

is a martingale,

E [ξ_{n} ∣ F_{n - 1}] = 0.

p_{n} = P (ξ_{n} = 1 ∣ F_{n - 1}),

and using

ξ_{n} \in {- 1, 1}

gives

0 = p_{n} - (1 - p_{n}) = 2 p_{n} - 1.

Thus

P (ξ_{n} = 1 ∣ F_{n - 1}) = P (ξ_{n} = - 1 ∣ F_{n - 1}) = \frac{1}{2} .

e_{1}, \dots, e_{m} \in {- 1, 1}

, repeated use of the tower property of conditional expectation gives

P (ξ_{1} = e_{1}, \dots, ξ_{m} = e_{m}) = E [1_{{ξ_{1} = e_{1}, \dots, ξ_{m - 1} = e_{m - 1}}} P (ξ_{m} = e_{m} ∣ F_{m - 1})] = \frac{1}{2} P (ξ_{1} = e_{1}, \dots, ξ_{m - 1} = e_{m - 1}) = 2^{- m} .

Hence the increments are IID symmetric signs, proving the symmetric signs forced by the martingale property result. Thus $X_{n} - X_{0}$ is a simple symmetric random walk.

 Read the full article

 New to topics? Read the docs here!