Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 208 / 3 / b / Solution

Put $Z_i=f_i(X^{(i)})$ and ${\Delta }_i=Z-Z_i$. The weakly self-bounding function assumptions give ${\Delta }_i\ge 0$ and ${\sum }_i{\Delta }_i^2\le Z$. For $0\le \lambda <2$, the bound $\varphi (-x)\le x^2/2$ and the modified logarithmic Sobolev inequality implyLet $H(\lambda )=\log \mathbb{E}{e}^{\lambda Z}$. Dividing by $\mathbb{E}{e}^{\lambda Z}$ turns this intoand thereforeSince $H(\lambda )\sim \lambda \mathbb{E}Z$ as $\lambda \downarrow 0$, integration givesSubtracting $\lambda \mathbb{E}Z$ from both sides yieldsas required. This is a Herbst argument with a variance proxy controlled by $Z$ itself.