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. 



Hoeffding to bound Orlicz norm
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.


