Because the <moment-generating function> is finite in a neighborhood to the right of zero and $\mathbb EX=0$, its <cumulant-generating function> satisfies
$$
\psi_X(\lambda)=\frac{\lambda^2}{2}\operatorname{Var}(X)+o(\lambda^2)
\qquad(\lambda\downarrow0).
$$
Divide the defining <Sub-Gamma random variable in the right tail> inequality by $\lambda^2/2$ and let $\lambda\downarrow0$. The right side converges to $\nu$, proving $\operatorname{Var}(X)\leq\nu$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 208 / 1 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-208/1/a/solution