Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 29 6 iii Solution Created 2026-10-03 Updated 2026-10-06
First, convergence in probability at gives almost surely, since every . For any finite set of times, the corresponding vectors converge in probability: the union bound controls the probability that any coordinate differs by more than a fixed tolerance. The same holds for their increment vectors, hence also for their convergence in distribution.
For , set and . The characteristic function of a random vector factors for the independent increments of each :Pass to the limit using convergence in distribution and bounded continuous functions. The resulting factorization of the characteristic function of a random vector, with the uniqueness theorem for characteristic functions, proves independence of the . Similarly passes to the limit, proving stationary increments for .
It remains to prove stochastic continuity. For , the triangle inequality and the union bound imply, for every fixed ,The second term vanishes by stochastic continuity of . Now let and use the additional near-zero approximation hypothesis. We obtain in probability as . The stationary increments transfer this to every time: both and have the law of for when defined. Thus has all the intrinsic Lévy process properties, proving closure of Lévy processes under locally controlled convergence in probability.
If one requires the supplied process itself to be càdlàg, the hypotheses justify a càdlàg modification, rather than that stronger pathwise assertion. To see the distinction, take , let have uniform distribution on , and put . At each fixed time , almost surely, so both approximation hypotheses hold with zero error probability. Nevertheless every path has an isolated spike at and is not right-continuous there. Its identically zero modification of a stochastic process is a Lévy process with càdlàg paths. The conclusion is exact under the intrinsic definition, and exact up to modification under the convention requiring càdlàg paths.
Past exam of the mathematics course of the University of Cambridge 2019 ii Paper 1 27K b ii Solution Created 2026-09-24 Updated 2026-09-29
Independence of the coordinates gives factorization of their joint characteristic function of a random vector. Orthogonal invariance under coordinate permutations makes all coordinate marginals identical. Hence, for and ,Therefore