Fix $0\leq\lambda<c^{-1}$. Since $X$ is centered, the elementary inequality $\log u\leq u-1$ gives
$$
\log\mathbb E e^{\lambda X}
\leq\mathbb E(e^{\lambda X}-1-\lambda X).
$$
On ${X\leq0}$, use $e^u-1-u\leq u^2/2$ for $u\leq0$; on ${X>0}$, expand the <exponential function> into its <power series>. The hypotheses therefore give
$$
\begin{aligned}
\mathbb E(e^{\lambda X}-1-\lambda X)
&\leq\frac{\lambda^2}{2}\mathbb E X^2
 +\sum_{q=3}^\infty\frac{\lambda^q}{q!}\mathbb E X_+^q\\
&\leq\frac{\sigma^2\lambda^2}{2}
 +\frac{\sigma^2}{2}\sum_{q=3}^\infty\lambda^qc^{q-2}
=\frac{\sigma^2\lambda^2}{2(1-c\lambda)}.
\end{aligned}
$$
This is precisely the <Sub-Gamma random variable in the right tail> bound with variance parameter $\sigma^2$ and scale parameter $c$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 210 / 1 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-210/1/c/solution