For every $0<\lambda<1/\alpha$, the <Chernoff bound> and the sub-exponential moment-generating bound give
$$
\mathbb P(X\geq t)
\leq\exp\left(-\lambda t+\frac{\lambda^2\nu}{2}\right).
$$
If $0<t\leq\nu/\alpha$, choose $\lambda=t/\nu$; at the endpoint, take a limit from below. This yields
$$
\mathbb P(X\geq t)\leq e^{-t^2/(2\nu)}.
$$
If $t>\nu/\alpha$, let $\lambda\uparrow1/\alpha$. Since $\nu/(2\alpha^2)<t/(2\alpha)$,
$$
-\frac t\alpha+\frac\nu{2\alpha^2}
\leq-\frac t{2\alpha},
$$
and hence $\mathbb P(X\geq t)\leq e^{-t/(2\alpha)}$.

Solution

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

 Solution

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