Put $\mu=\mathbb EZ$ and $h(u)=(1+u)\log(1+u)-u$. For $\lambda>0$, the <Chernoff bound> and the assumed cumulant-generating-function estimate give
$$
\mathbb P(Z-\mu\geq t)
\leq\inf_{\lambda>0}
\exp\{-\lambda t+\mu(e^\lambda-\lambda-1)\}.
$$
The optimizer satisfies $e^\lambda=1+t/\mu$, and hence
$$
\mathbb P(Z-\mu\geq t)
\leq e^{-\mu h(t/\mu)}
\leq\exp\!\left(-\frac{t^2}{2\mu+2t/3}\right).
$$

For the left tail, apply the same argument at a negative parameter. If $0<t<\mu$ the optimizer satisfies $e^{-\lambda}=1-t/\mu$, giving
$$
\mathbb P(Z-\mu\leq-t)
\leq\exp\{-\mu[(1-t/\mu)\log(1-t/\mu)+t/\mu]\}
\leq e^{-t^2/(2\mu)}.
$$
For $t\geq\mu$, nonnegativity of $Z$ makes the strict lower-tail event empty, with the boundary handled directly.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 208 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-208/2/a/solution