Say $X\_1, \cdots, X\_n$ are i.i.d random variables with mean zero, let $S\_n = \sum\_{i=1}^n X\_i$, we know by SLLN $$\frac{S\_n}{n}\rightarrow 0\text{ a.s}$$


We could further know that the sequence of random variables $\{\frac{S\_n}{n}\}$ are uniformly integrable. Hence u.i + converge in prob(weaker than a.s) implies that $$\mathbb{E}\left(\left|\frac{S\_n}{n}\right|\right)\rightarrow 0$$


For classical large deviation theory, such as Cramer's rule(though for r.vs that has moment generating function existed), and roughly speaking, it tells us the speed of the probability convergence, $$P\left(\frac{S\_n}{n}\in A\right) \sim e^{-nI(A)}$$(Please forgive me not rigorous here).


So for the L1 convergence as I mentioned above, could we have a convergence speed statement? Such as $$\mathbb{E}\left(\left|\frac{S\_n}{n}\right|\right)= O(\frac{1}{n})$$(I just pick one decreasing order as $O(\frac{1}{n})$, it definitely could be something else)


I would like to find the result of the above type. Could anyone give me a hint on what literature or paper should I look at?



Sample average L1 convergence speed
$\newcommand{\ep}{\epsilon}$Somehow, I have only now recalled about Latala's inequalities for moments of the sums of positive independent random variables (r.v.'s), which, in particular, allow one to easily obtain the order of magnitude of $E|S\_n|$.


Indeed, by the [Marcinkiewicz--Zygmund inequalities](https://en.wikipedia.org/wiki/Marcinkiewicz%E2%80%93Zygmund_inequality),
\begin{equation\*}
 E|S\_n|\asymp E\Big(\sum\_1^n X\_i^2\Big)^{1/2},
\end{equation\*}
where $a\asymp b$ means that $a/C\le b\le Ca$ for some universal real constant $C>0$.


By Latala's [Theorem 1](https://projecteuclid.org/euclid.aop/1024404522) (with $X\_i^2$ in place of $X\_i$ in that theorem),
\begin{equation\*}
 E\Big(\sum\_1^n X\_i^2\Big)^{1/2}\asymp b\_n^{1/2}, 
\end{equation\*}
where
\begin{equation\*}
 b\_n:=\inf\{t>0\colon E\sqrt{1+X^2/t}\le e^{1/(2n)}\} \tag{-1}
\end{equation\*}
and $X:=X\_1$.


Thus,
\begin{equation\*}
 E|S\_n|\asymp b\_n^{1/2}, \tag{0}
\end{equation\*}
which determines $E|S\_n|$ up to a universal real constant factor.


In particular, this implies
\begin{equation\*}
 E\Big|\frac{S\_n}n\Big|\to0\tag{1}
\end{equation\*}
whenever $E|X|<\infty$.
Indeed, for real $t>0$, let
\begin{equation\*}
 Y\_t:=(\sqrt{1+X^2/t}-1)\sqrt t. 
\end{equation\*}
Then $0\le Y\_t\le|X|$ and $Y\_t\to0$ (since $0\le Y\_t\le X^2/(2\sqrt t))$. So, by the condition $E|X|<\infty$ and dominated convergence,
\begin{equation\*}
 EY\_t\to0
\end{equation\*}
as $t\to\infty$. So, for each real $\ep>0$ and all large enough $n$ (depending on $\ep$), letting $t=(2\ep n)^2$, we have $EY\_t\le\ep$, that is,
\begin{equation\*}
 E\sqrt{1+X^2/t}\le1+\ep/\sqrt t=1+1/(2n)<e^{1/(2n)}. 
\end{equation\*}
In view of (-1), this yields
\begin{equation\*}
 b\_n\le(2\ep n)^2.
\end{equation\*}
So, (0) implies
\begin{equation\*}
 \limsup\_n E\Big|\frac{S\_n}n\Big|\le2C\ep
\end{equation\*}
for some real $C>0$ and every $\ep>0$. Thus, (1) follows.




---


Using (0), one can show that actually the bound $c\_n$ on $E\big|\frac{S\_n}n\big|$ in formula (2) in my other answer here is optimal up to a universal real constant factor, that is, one has
\begin{equation\*}
 E|S\_n|\asymp nc\_n, \tag{2}
\end{equation\*}
provided that $P(X=0)<1$.


Indeed, by Jensen's inequality, $E|S\_n|=E|S\_n^X|\le E|S\_n^Z|=E|S\_n^X-S\_n^Y|\le2E|S\_n|$, where $S\_n^X,S\_n^Y,S\_n^Z$ are as in my other answer on this page. So, (2) can be rewritten as
\begin{equation\*}
 E|S\_n^Z|\asymp nc\_n. \tag{2Z}
\end{equation\*}
Also, by (0),
\begin{equation\*}
 E|S\_n^Z|\asymp\sqrt{b\_n^Z}, \tag{0Z}
\end{equation\*}
where $b\_n^Z$ is obtained from $b\_n$ by replacing $X$ by $Z:=Z\_1$ in the definition (-1) of $b\_n$.


Next, for real $u$
\begin{equation\*}
 \sqrt{1+u^2}-1\asymp\min(u^2,|u|)=u^2\,1(|u|\le1)+|u|\,1(|u|>1). 
\end{equation\*}
On the other hand (cf. (-1)),
\begin{equation\*}
 E\sqrt{1+Z^2/b\_n^Z}-1=e^{1/(2n)}-1\asymp1/n,
\end{equation\*}
so that
\begin{equation\*}
 E\frac{Z^2}{b\_n^Z}\,1(|Z|\le b\_n^Z)+E\frac{|Z|}{\sqrt{b\_n^Z}}\,1(|Z|>b\_n^Z)\asymp\frac1n,
\end{equation\*}
whence
\begin{equation\*}
 EZ^2\,1(|Z|\le b\_n^Z)\ll\frac{{b\_n^Z}}n,\quad 
 E|Z|\,1(|Z|>b\_n^Z)\ll\frac{\sqrt{b\_n^Z}}n,
\end{equation\*}
where $a\ll b$ means $a\le Cb$ for some universal positive real constant $C$.
So, by formula (2) in my other answer on this page,
\begin{align\*}
 nc\_n&\le K(b\_n^Z)\sqrt n+nL(b\_n^Z) \\ 
 &=\sqrt{EZ^2\,1(|Z|\le b\_n^Z)}\sqrt n+n E|Z|\,1(|Z|>b\_n^Z) \\ 
 &\ll\sqrt{\frac{{b\_n^Z}}n}\sqrt n+n \frac{\sqrt{b\_n^Z}}n
 =2\sqrt{b\_n^Z}. 
\end{align\*}
Therefore and in view of formulas (0Z) above and (2) in my other answer on this page,
\begin{equation\*}
 nc\_n\ll\sqrt{b\_n^Z}\asymp E|S\_n^Z|\le2E|S\_n|\ll nc\_n. 
\end{equation\*}
Thus, formula (2) in this answer has been proved.


