Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 208 / 3 / d / Solution

Put $Y=g(X)-\mathbb{E}g(X)$ and $H(\lambda )=\log \mathbb{E}{e}^{\lambda Y}$. Apply the assumed Poisson log-Sobolev inequality to $f=e^{\lambda g}$. Since $|Dg|\le 1$ and $|e^a-1|\le \lambda e^{\lambda }$ for $|a|\le \lambda$,After division by $\mathbb{E}{e}^{\lambda g}$, the entropy bound becomesBecause $(H(\lambda )/\lambda {)}^{\prime }=(\lambda H^{\prime }-H)/{\lambda }^2$ and $H^{\prime }(0)=0$, integration from $0$ to $\lambda$ givesThe Chernoff bound and optimization over $\lambda \ge 0$ therefore yieldwhere ${\psi }^{\ast }$ is the Legendre transform of a cumulant-generating function, also called the Chernoff-Cramér transform.