 Solution

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

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

Given

0 < z < 1

, take

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

Choose

A, B

so that

A w^{- a} + B w^{a} = 1, A w^{b} + B w^{- b} = 1.

Solving,

A = \frac{w ^{a} - w ^{- b}}{w ^{a + b} - w ^{- (a + b)}}, B = \frac{w ^{b} - w ^{- a}}{w ^{a + b} - w ^{- (a + b)}} .

The stopped discounted martingale is bounded, and

T

is finite, so part (b) gives

A + B = E [z^{T} (A w^{X_{T}} + B w^{- X_{T}})] = E (z^{T}) .

E (z^{T}) = \frac{w ^{a} + w ^{b} - w ^{- a} - w ^{- b}}{w ^{a + b} - w ^{- (a + b)}}, w = \frac{1 + 1 - z ^{2}}{z} .

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