Set
$$
Z=\frac{e^{\lambda X}}{M(\lambda)},
\qquad \mathbb EZ=1.
$$
Under the probability measure with density $Z$, the <Jensen inequality> for the concave logarithm gives
$$
\frac{\operatorname{Ent}(e^{\lambda X})}{M(\lambda)}
=\mathbb E[Z\log Z]
\leq\log\mathbb E[Z^2]
=\log\frac{M(2\lambda)}{M(\lambda)^2}.
$$
The sub-Gaussian assumption with variance parameter $\nu/4$ gives $M(2\lambda)\leq e^{\nu\lambda^2/2}$. A second application of <Jensen inequality> gives $M(\lambda)\geq e^{\lambda\mathbb EX}=1$. Consequently
$$
\operatorname{Ent}(e^{\lambda X})
\leq\frac{\nu\lambda^2}{2}M(\lambda),
$$
as required.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 208 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-208/1/b/solution