I raised some questions when read Nickl's book "Mathematical Foundation of Infinite-Dimensional Statistical Models".
The first question is how to prove the following statement mentioned in page 38:
If $\xi$ which is sub-gaussian random variable is symmetric, then $Ee^{\lambda \xi} \leq e^{\lambda^2\Vert\xi\Vert_{\psi_2}^2/2}$. where $\psi_2(x) = e^{x^2} - 1$ and $\Vert\xi\Vert_{\psi_2} = \inf\{c > 0: E_{\psi_2}(\vert\xi\vert/c) \leq 1\}$.
The second question is about the third step of proving process for following lemma:
Lemma 2.3.3 Let $\xi_i \in L^{\psi_2}, i = 1, \ldots, N, 2 \leq N \leq \infty$.Then $$ \Bigg\Vert{\max_{i \leq N}\vert \xi_i\vert}\Bigg\Vert_{\psi_2} \leq 4 \sqrt{\log{N}}\max_{i \leq N}\Vert{\xi_i}\Vert_{\psi_2} $$ and, inparticular, there exist $K_p < \infty, 1 \leq p < \infty$ such that $$\Bigg\Vert\max_{i \leq N}\vert\xi_i\vert\Bigg\Vert_{L^p}\leq K_p\sqrt{\log{N}}\max_{i \leq N}\Vert\xi_i\Vert_{\psi_2}$$.
The proving process is shown as below:
proof. We assume that $\max{\Vert\xi_i\Vert}_{\psi_2} = 1$. Then the defintion of the $\psi_2$ norm together with the exponential Chebyshev's inequality gives \begin{align*} E{\max_{i \leq N}}e^{\xi_i^2/(16\log{N})} &= \int_{0}^{\infty}Pr\Bigg\{\max_{i \leq N}e^{\xi_i^2/(16\log N)} \geq t\Bigg\}dt \\ &\leq e^{1/8} + \sum_{i = 1}^N \int_{e^{1/8}}^{\infty}Pr\Bigg\{e^{\xi_i^2/(16\log N)} \geq t\Bigg\}dt \\ &\leq e^{1/8} + 2N\int_{e^{1/8}}^{\infty} e^{-8(\log N)(\log t)}dt = e^{1/8} + 2N\int_{e^{1/8}}^{\infty} t^{-8\log N}dt \\ &= e^{1/8}\Bigg(1 + \frac{2}{8(\log N - 1)}\Bigg) < 2 \end{align*}
This can proving the first inequaility. My question is how to get the second inequaility in above proof? And the author says that we can get the second inequaility by $\Vert\xi\Vert_{L^{2k}} \leq (k!)^{1/2k}\Vert\xi\Vert_{L^{\psi_2}}$. How to conclude this relationship is my final confusion.
Thanks for your help in advanced.
