 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-2/30l/c/solution

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

For the symmetric increment

ξ_{n + 1}

E (w^{X_{n + 1}} ∣ F_{n}) = w^{X_{n}} \frac{w + w ^{- 1}}{2} .

The same multiplier holds for

w^{- X_{n}}

. Thus choose

z = \frac{2}{w + w ^{- 1}} > 0 .

Then each of

z^{n} w^{X_{n}}

and

z^{n} w^{- X_{n}}

(A w^{X_{n}} + B w^{- X_{n}}) z^{n}

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