Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 202 1 c Solution Created 2026-10-03 Updated 2026-10-05
A simple predictable process has the form , where are deterministic and is bounded and -measurable. A possible -measurable value at zero is irrelevant to this Itô integral. DefineEach coefficient is known before the corresponding increment. The conditional expectation of each subsequent increment is zero, so the result is a continuous martingale starting at zero. It is an L2-bounded continuous martingale because it is a finite sum of bounded coefficients times stopped increments of an L2-bounded continuous martingale. This definition is independent of the chosen subdivision: splitting an interval simply splits its increment into a telescoping sum. General admissible predictable processes are then integrated by completion using the Itô isometry; unbounded step coefficients are allowed when the weighted condition in the quadratic-variation measure in the next part holds.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 202 1 d Solution Created 2026-10-03 Updated 2026-10-05
Fix a L2-bounded continuous martingale . Its quadratic-variation measure on the predictable sigma-algebra isIt is finite, since . The source Hilbert space is : predictable processes with finite , identified when equal -almost everywhere. Its norm is .
The target consists of L2-bounded continuous martingales starting at zero, identified up to indistinguishability of stochastic processes, with norm . It is a Hilbert space: terminal values belong to the closed linear subspace of whose conditional expectation given is zero and whose associated martingales have continuous versions. Closure of that continuous-version subspace follows from the Doob L2 maximal inequality and an almost surely uniformly convergent subsequence.
The Itô isometry says that integration extends uniquely from simple predictable processes to a linear isometry , withFor every finite , the same identity holds with the integral restricted to and the left side . No claim of surjectivity onto all of is needed: that would require an additional Martingale representation theorem.