 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-4/29k/d/solution

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

Given the present state

(S_{t}^{1}, M_{t}) = (x, m)

, take an independent fresh copy of the stock path and denote its initial value, terminal value after

T - t

steps, and running maximum by

(y, z, w) = (S_{0}^{1}, S_{T - t}^{1}, M_{T - t})

. Scaling the fresh path to start at

x

makes its terminal value

x z / y

and its future running maximum

x w / y

. Including the maximum already attained gives

M_{T} = max {m, \frac{x w}{y}} .

g (x, m, y, z, w) = h (\frac{x z}{y}, max {m, \frac{x w}{y}}) .

V_{t} (ω) = v_{t} (S_{t}^{1} (ω), M_{t} (ω)),

with

v_{t} (x, m) = E_{Q} [g (x, m, S_{0}^{1}, S_{T - t}^{1}, M_{T - t})] .

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