 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-3/29l/a/solution

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

E_{Q} ξ_{n} = 1 + r .

q = Q (ξ_{n} = 1 + b)

gives

q (1 + b) + (1 - q) (1 + a) = 1 + r, q = \frac{r - a}{b - a} .

Equivalence requires

0 < q < 1

, hence

a < r < b

. At every node the risk-neutral condition determines this same conditional probability

q

, so the multipliers remain independent and identically distributed under

Q

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