Because the variables are <independent random variables>[independent], the moment-generating function of the sum factors. Whenever $|\lambda|<1/\max_i\alpha_i$,
$$
\mathbb E\exp\left(\lambda\sum_{i=1}^nX_i\right)
=\prod_{i=1}^n\mathbb Ee^{\lambda X_i}
\leq\exp\left(\frac{\lambda^2}{2}\sum_{i=1}^n\nu_i\right).
$$
Therefore $\sum_iX_i$ is sub-exponential with parameters
$$
\nu=\sum_{i=1}^n\nu_i,
\qquad
\alpha=\max_{1\leq i\leq n}\alpha_i.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 208 / 1 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-208/1/c/solution