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.
Best Answer
For the equality $\|S_n\|_{L^2}=\|X_i\|_{L^2}$ you need the zero-mean condition -- that $EX_i=0$.
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*}
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.