Past exam of the mathematics course of the University of Cambridge 2001 iii Paper 24 2 Solution Created 2026-10-03 Updated 2026-10-14
For a simple predictable process, define its stochastic integral byCommon refinement of the deterministic partitions makes this definition independent of the representation. Each summand is a martingale increment multiplied by information already available at its left endpoint, so the resulting process is a continuous square-integrable martingale starting at zero. Its terminal value is the same sum with replaced by .
Write . If , then is -measurable. By the conditional expectation identity for increments of a martingale,The conditional bracket isometry for stopped martingale increments givesFor completeness, this identity follows by applying the optional stopping theorem to and and expanding the square. The latter process is a true uniformly integrable martingale here: the Doob L2 maximal inequality bounds , stopped square identities and Fatou's lemma give , and its absolute supremum is bounded by . Consequently all the stopped identities pass to the limit.
Multiplying the conditional identity by , taking expectations, and using the vanished cross terms proves the Itô isometry:
To extend the stochastic integral, use the finite quadratic-variation measure on the predictable sigma-algebra. The density of simple predictable processes for finite measures gives approximations to every . The Itô isometry makes the corresponding terminal integrals Cauchy in , and the Doob L2 maximal inequality makes the whole processes Cauchy in expected squared supremum. Their limit defines a continuous martingale, and the isometry persists. For a locally bounded predictable process, choose increasing stopping times tending to infinity on which it is bounded, also restrict the time horizon, and perform this construction after stopping. The isometry shows that the stopped definitions agree on overlaps, so they patch into a uniquely defined continuous local martingale .
Past exam of the mathematics course of the University of Cambridge 2009 iii Paper 29 1 c Solution Created 2026-10-03 Updated 2026-10-14
First justify optional sampling for the square-minus-bracket process. The Doob L2 maximal inequality gives . Stop the local martingale using a localizing sequence and bounded bracket levels. The stopped expectation identity, followed by Fatou, bounds by the expected squared path supremum. Hence has an integrable absolute supremum, and localization with conditional dominated convergence makes it a uniformly integrable martingale. The optional stopping theorem stated in part (b) is therefore applicable to both and .
For ordered stopping times , expand the squared increment and condition on . The mean increment is zero, soThis is the conditional bracket isometry for stopped martingale increments. Apply it with , and multiply by the bounded -measurable variable . The orthogonality proved in part (b) gives the Itô isometry
To identify the whole bracket process, let and . For deterministic and , apply the proved isometry to the simple integrand . It yieldsThus . Since is a martingale, the cross term has conditional mean zero. Consequently is a martingale. The process is continuous, adapted, nondecreasing and starts at zero, so uniqueness in the definition of quadratic variation provesThis deduction uses only the proved elementary isometry, not a previously assumed formula for the quadratic variation of stochastic integrals.