# Hoeffding to bound Orlicz norm

I have been reading from Weak Convergence and Empirical Processes, and came across the following: Let $$a_1,\ldots,a_n$$ be constants and $$\epsilon_1,\ldots,\epsilon_n\sim$$Rademacher. Then

$$\mathbb{P}\left(\left|\sum_i\epsilon_i a_i\right|>x\right)\leq 2\exp\left(-\frac{x^2}{2||a||^2_2}\right)$$

Consequently, $$||\sum_i\epsilon_ia_i||_{\Psi_2}\leq\sqrt{6}||a||_2$$.

How does this follow (relation between Orlicz norm of Rademacher average and L2 norm of constants)? Thank you in advance for your time.

As I understand these notations, $$\Psi_2(x)=\exp(x^2)-1$$ and $$\|Z\|_{\Psi}=\inf \{c>0 : \mathbb{E} (\Psi(|Z| / c)) \leqslant 1\}$$. So the inequality $$\|Z\|_{\Psi}\leqslant c$$ means that $$\mathbb{E} (\Psi(|Z| / c)) \leqslant 1$$. Denote $$c=\sqrt{6}\|a\|_2$$, $$Z=|\sum_i \epsilon_i a_i|$$. We have $$\mathbb{E} (\Psi_2(Z/c))=\mathbb{E} \exp(Z^2/c^2)-1=-1+\int_0^\infty\mathbb{P} (\exp(Z^2/c^2)> t)dt=\\ \int_1^\infty\mathbb{P}( Z> c\sqrt{\log t})dt\leqslant \int_1^\infty \min(1,2\exp(-3\log t))dt<2\int_1^\infty t^{-3}dt=1,$$ as we need.
Explanations: we estimated the probability $$\mathbb{P}( Z> c\sqrt{\log t})$$ from above as $$2\exp(-3\log t)$$, this is equivalent to $$\mathbb{P}\left(Z>x\right)\leqslant 2\exp\left(-\frac{x^2}{2||a||^2_2}\right)$$ for $$x=c\sqrt{\log t}$$. Sometimes this is worse than the trivial upper estimate $$\mathbb{P}( Z> c\sqrt{\log t})\leqslant 1$$, that's why the minimum.