 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/iii/paper-39/3/iv/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

Continue at time zero. At time one stop on the lower branch and continue on the upper branch:

τ_{0} = {1, 2, ξ_{1} = 7, ξ_{1} = 11.

This is a stopping time since the branch decision is measurable at time one. Its expected reward is

E ξ_{τ_{0}} = \frac{3}{5} 7 + \frac{2}{5} (\frac{2}{3} 13 + \frac{1}{3} 10) = 9.

It therefore attains the upper bound in part (iii).

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