$\cosh t ≤ \exp(t^2/2)$

hbghlyj

A8LectureNotes_MT22_24Sep2022.pdf 第3.3节 Using moment generating functions for tail bounds 用到了这个不等式:

The inequality $\cosh t ≤ \exp(t^2/2)$ can be checked directly by expanding the exponential functions and comparing coefficients in the power series).

hbghlyj

en.wikipedia.org/wiki/Hyperbolic_functions#Inequalities也有这个不等式:
The following inequality is useful in statistics: $\operatorname{cosh} (t)\leq e^{t^2/2}$ [21]
It can be proved by comparing term by term the Taylor series of the two functions.
ProjectEuclid | PDF Audibert, Jean-Yves (2009). "Fast learning rates in statistical inference through aggregation". The Annals of Statistics. p. 1627.
(10.2) $\begin{array}{l}\leq \frac{1}{2} \mathbb{E}_{g^{\prime} \sim \rho} \mathbb{E}_{g \sim \rho} e^{\alpha\left(g^{\prime}, g\right)-\alpha^{2}\left(g^{\prime}, g\right) / 2}+\frac{1}{2} \mathbb{E}_{g^{\prime} \sim \rho} \mathbb{E}_{g \sim \rho} e^{-\alpha\left(g^{\prime}, g\right)-\alpha^{2}\left(g^{\prime}, g\right) / 2} \\ \leq \mathbb{E}_{g^{\prime} \sim \rho} \mathbb{E}_{g \sim \rho} \cosh \left(\alpha\left(g, g^{\prime}\right)\right) e^{-\alpha^{2}\left(g^{\prime}, g\right) / 2} \leq 1\end{array}$
where in the last inequality we used the inequality $\cosh(t) ≤ e^{t^2/2}$ for any $t ∈\Bbb R$.
The result then follows from Theorem 3.1.