Because $\mathbb EX=0$, the <moment-generating function> series and the <Bernstein moment condition> imply, for $|\lambda|b<1$,
$$
\begin{aligned}
\mathbb Ee^{\lambda X}
&=1+\sum_{k=2}^{\infty}\frac{\lambda^k\mathbb E[X^k]}{k!}\\
&\leq1+\frac{\nu\lambda^2}{2}
\sum_{k=2}^{\infty}(|\lambda|b)^{k-2}\\
&=1+\frac{\nu\lambda^2}{2(1-|\lambda|b)}\\
&\leq\exp\left(\frac{\nu\lambda^2}{2(1-|\lambda|b)}\right).
\end{aligned}
$$
For $|\lambda|<1/(2b)$, the denominator satisfies $1-|\lambda|b>1/2$, so
$$
\mathbb Ee^{\lambda X}\leq e^{\nu\lambda^2}
=e^{\lambda^2(2\nu)/2}.
$$
Thus $X$ is sub-exponential with parameters $(2\nu,2b)$.

Solution

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

 Solution

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