Independent <Poisson process>[Poisson processes] have stationary independent increments, so their weighted sum does too and is stochastically continuous. Moreover,
$$
\mathbb E e^{iuX_t}
=\prod_{k=1}^n
\exp\{t\lambda_k(e^{iua_k}-1)\}
=e^{t\psi(u)},
$$
where
$$
\psi(u)=\sum_{k=1}^n\lambda_k(e^{iua_k}-1).
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 201 / 6 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-201/6/c/solution