Past exam of the mathematics course of the University of Cambridge 2018 ii Paper 3 30K b Solution Created 2026-09-24 Updated 2026-10-03
Assume . Conditional on , the random part of the next Kalman filter estimate is times the innovation. Its variance isTherefore the Bellman equation givesCompleting the square givesStarting from yieldsand hence
Past exam of the mathematics course of the University of Cambridge 2022 ii Paper 1 30K a Solution Created 2026-09-24 Updated 2026-10-03
Let be the minimum expected remaining cost after time , given . The terminal condition is , and the Bellman equation is
Past exam of the mathematics course of the University of Cambridge 2023 ii Paper 1 30K c Solution Created 2026-09-24 Updated 2026-09-25
Let the value function at time be the greatest conditional expected terminal utility attainable from wealth . Independence of the innovations gives the Bellman equationfor . A policy obtained from the maximizing choices is admissible because each time- holding is a predictable process, hence is known before is observed.
We prove the claimed properties by backward induction. They hold at because is increasing and concave. Suppose they hold at time . For fixed , the mapis increasing because , and taking a supremum preserves this inequality.
For concavity, let and be optimal at wealths and . For , use the admissible portfolioThe next wealth is the same convex combination of the next wealths generated from and . ThereforeThis completes the induction and proves monotonicity and concavity of a portfolio value function.
If discovery in box earns reward and stopping earns zero, the Bellman equation isIf , every posterior state has some with , so stopping is never optimal.