Let
$$
M(\lambda)=\mathbb Ee^{\lambda X},
\qquad
\psi(\lambda)=\log M(\lambda).
$$
The <entropy functional> satisfies
$$
\frac{\operatorname{Ent}(e^{\lambda X})}{M(\lambda)}
=\lambda\psi'(\lambda)-\psi(\lambda).
$$
Hence the assumed inequality gives
$$
\left(\frac{\psi(\lambda)}{\lambda}\right)'
=\frac{\lambda\psi'(\lambda)-\psi(\lambda)}{\lambda^2}
\leq\frac\nu2.
$$
Because $\mathbb EX=0$, $\psi(\lambda)/\lambda\to0$ as $\lambda\to0$. Integrating from zero to $\lambda$ when $\lambda>0$, and from $\lambda$ to zero and then multiplying by the negative number $\lambda$ when $\lambda<0$, gives in both cases
$$
\psi(\lambda)\leq\frac{\nu\lambda^2}{2}.
$$
Thus $\mathbb Ee^{\lambda X}\leq e^{\nu\lambda^2/2}$ for every real $\lambda$, which is precisely the <sub-Gaussian random variable> bound with variance parameter $\nu$. This integration is the <Herbst argument>.

Solution

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

 Solution

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