The <Chernoff bound> gives, for $0<\lambda<1/c$,
$$
\mathbb P(X\geq t)
\leq\exp\left\{-\lambda t+
\frac{\nu\lambda^2}{2(1-c\lambda)}\right\}.
$$
Writing $u=ct/\nu$, elementary differentiation shows that the exponent is minimized at
$$
\lambda_*=\frac1c\left(1-\frac1{\sqrt{1+2u}}\right).
$$
Substitution gives
$$
\mathbb P(X\geq t)\leq
\exp\left\{-\frac{\nu}{c^2}h(u)\right\},
\qquad
h(u)=1+u-\sqrt{1+2u}.
$$
Thus $h$ is the <Legendre transform> generated by the sub-Gamma cumulant bound after its natural rescaling.

Solution

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

 Solution

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