For $q\geq2$, $q!\geq2\,3^{q-2}$. Hence, for $0\leq\lambda<3$,
$$
e^\lambda-1-\lambda
=\sum_{q=2}^\infty\frac{\lambda^q}{q!}
\leq\frac{\lambda^2}{2}\sum_{r=0}^\infty\left(\frac\lambda3\right)^r
=\frac{\lambda^2}{2(1-\lambda/3)}.
$$
Substitution into the defining <moment-generating function> bound shows that $X$ is <Sub-Gamma random variable in the right tail>[sub-Gamma in the right tail] with variance parameter $\sigma^2$ and scale parameter $1/3$.

Solution

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

 Solution

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