Let $X\_{i}$ be a collection of iid random variables of cardinality $n$, and let $S\_{n}=\frac{1}{\sqrt{n}}\sum\_{i=1}^{n}X\_{i}.$


Let $|| X||:=\inf\_{B}\{E[\exp(X/B)-1]\leq 1\}$. This is the so-called sub-exponential norm.


What can be said about $|| S\_{n}||$ in terms of $||X\_{i}||$?


In the $L^{2}$ norm, we have $||S\_{n}||\_{L^{2}}=||X\_{i}||\_{L^{2}}$ by choice of normalizing constant.



Independent Sums and Orlicz Norms
1. For the equality $\|S\_n\|\_{L^2}=\|X\_i\|\_{L^2}$ you need the zero-mean condition -- that $EX\_i=0$.
2. Let $X,X\_1,\dots,X\_n$ be any random variables with the same norm: $\|X\|=\|X\_1\|=\cdots=\|X\_n\|$.  

Then, by the norm inequality for the sub-exponential norm $\|\cdot\|$,
\begin{equation\*}
 \|S\_n\|\le\frac1{\sqrt n}\,\sum\_{i=1}^n \|X\_i\|=\sqrt n\,\|X\|. \tag{1}\label{1}
\end{equation\*}
3. Let us show that the trivial upper bound $\sqrt n\,\|X\|$ on $\|S\_n\|$ in \eqref{1} cannot be asymptotically improved (for $n\to\infty$), even if it is assumed that $X,X\_1,\dots,X\_n$ are iid zero-mean random variables. Toward this end, assume that  

\begin{equation\*}
 P(X=-a)=p=1-P(X=1),
\end{equation\*}
where
\begin{equation\*}
 a:=n^2,\quad p:=\frac1{a+1},\quad n\to\infty. 
\end{equation\*}
Then $EX=0$ and, for each real $B>0$,
\begin{equation\*}
 Ee^{X/B}=pe^{-a/B}+(1-p)e^{1/B}\to e^{1/B}. 
\end{equation\*}
Equating now $e^{1/B}$ with $2$, we see that  

\begin{equation\*}
 \|X\|\to\frac1{\ln2}. \tag{2}\label{2}
\end{equation\*}
Further, for each real $c>0$,
\begin{equation\*}
 Ee^{S\_n/(c\sqrt n)}=(pe^{-n^2/(cn)}+(1-p)e^{1/(cn)})^n
 =(1+(1+o(1))/(cn))^n\to e^{1/c}. 
\end{equation\*}
Equating now $e^{1/c}$ with $2$, we see that  

\begin{equation\*}
 \|S\_n\|\sim\frac{\sqrt n}{\ln2}, 
\end{equation\*}
so that, in view of \eqref{2},
\begin{equation\*}
 \|S\_n\|\sim\sqrt n\,\|X\|, 
\end{equation\*}
as claimed.


