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

{c}

A centered <random variable> $X$ with $\mathbb E[X^2]=\nu$ satisfies the Bernstein moment condition with parameter $b$ when
$$
|\mathbb E[X^k]|\leq\frac{k!}{2}\nu b^{k-2}
$$
for every integer $k\geq2$. Summing its moment-generating series shows that it is sub-exponential with parameters $(2\nu,2b)$.

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 Incoming links: Bernstein moment condition