For the <Poisson process>,
$$
G(\lambda)=\sum_{n=0}^\infty e^{\lambda n}
\frac{(\mu t)^n}{n!}e^{-\mu t}
=\boxed{\exp[\mu t(e^\lambda-1)]}.
$$
Differentiating at zero gives
$$
\boxed{\mathbb E N_t=\mu t},
\qquad
\boxed{\operatorname{Var}(N_t)=\mu t}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 353 / 4 / a / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-353/4/a/i/solution