For $k\geq0$, take $f_k(x)=\mathbf1_{\{x\geq k+1\}}$ and write $q_k=\mathbb P(X\geq k+1)$. Since $f_k^2=f_k$,
$$
\operatorname{Ent}(f_k(X)^2)=-q_k\log q_k.
$$
Its discrete derivative is nonzero only at $k$, so
$$
\mathbb E|Df_k(X)|^2=\mathbb P(X=k)=p_k.
$$
For a <Poisson distribution>, $p_{k+1}/p_k=\lambda/(k+1)$ and $q_k\sim p_{k+1}$ as $k\to\infty$. Hence
$$
\frac{-q_k\log q_k}{p_k}
\sim\frac{\lambda}{k+1}\log\frac1{p_{k+1}}
\sim\lambda\log k\longrightarrow\infty,
$$
where the final estimate follows from <Stirling formula>. No finite constant $C$ can therefore make the proposed inequality hold for every $f$.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-208/3/b/solution