Let $\phi$ be an $N$-function, (i.e. $\phi : \mathbb{R}\_{\geq 0} \to \mathbb{R}\_{\geq 0}$ is convex and satisfies $\lim\_{t \to 0} \frac{\phi(t)}{t} = 0, \lim\_{t\to \infty} \frac{\phi(t)}{t} = \infty$).


We can define the associated Luxemburg norm on the appropriate subspace of $\mathbb{R}$-valued random variables (Orlicz space $L\_\phi$), by $\|Z\|\_{\phi} := \inf \{ \lambda : \mathbb{E} \phi(|Z|/\lambda) \leq 1\}$. The space $L\_\phi$ is just a space of all random variables for which this infimum is finite.


For example, when $\phi(t) = t^p$ for $p > 1$, this is just the $L\_p$ norm, and when $\phi(t) = e^{t^2} -1$ the corresponding Orlicz space is the space of sub-gaussian random variables.


**Question**


Is the following statement true:
Consider a sequence $Z\_1, Z\_2, \ldots Z\_n$ of independent random variables with $\mathbb{E} Z\_i=0$ and $\|Z\_i\|\_\phi \leq 1$ for all $i$. Then
$$ \|\frac{1}{n} \sum\_{i \leq n} Z\_i\|\_\phi \leq \lambda\_\phi(n)$$
for some $\lambda\_\phi$ s.t. $\lambda\_\phi(n) \to 0$ as $n\to \infty$.


**A bit more context**


I am mostly interested in the scenario in which $\phi$ grows only slightly faster than linear, say $\phi(t) = t \ln(1+t)$.


Together with Markov inequality, this would imply that for any $\varepsilon, \delta$, there is some $n\_\phi(\varepsilon, \delta)$ , s.t. $\mathbf{Pr}(|\sum\_{i\leq n} \frac{1}{n} X\_i| > \varepsilon) \leq \delta$. Law of large numbers asserts that as long as the first moment is bounded, the sample average $\sum \frac{1}{s} X\_i$ converges to $0$ - and I hope to be able to quantify the speed of convergence under just slightly stronger assumption.


Note that this is true in $L\_p$ spaces: for any $1< p \leq 2$, we have $\|\frac{1}{n}\sum\_i Z\_i\|\_p \lesssim n^{1/p - 1}$, so the interesting case is $\phi$ growing slower than $t^{1+\gamma}$ for any $\gamma$.



Weak concentration bounds for averages of independent random variables in Orlicz spaces
In general, the answer is no. Moreover, the answer is no even if
\begin{equation}
 \phi(t)=t\ln(1+t). \tag{1}
\end{equation}


Indeed, suppose that $P(Z\_i=0)=1-2p$ and $P(Z\_i=b)=p=P(Z\_i=-b)$ for all $i$, where
\begin{equation\*}
 p:=\frac1{2\phi(b)},
\end{equation\*}
$\phi$ is as given by (1),
and $b$ is a large enough positive real number so that $p\in(0,1/2)$.


Then for all $i$ we have $EZ\_i=0$ and $E\phi(|Z\_i|)=1$, so that $\|Z\_i\|\_\phi\le1$. On the other hand, for all real $c>0$ and all natural $n\ge2$
\begin{equation\*}
\begin{aligned}
 &E\phi\Big(\Big|\frac1n\sum\_{i=1}^n Z\_i\Big|/c\Big) \\ 
 &\ge\sum\_{i=1}^n \phi\Big(\frac b{cn}\Big)P(|Z\_i|=b,\ Z\_j=0\ \forall j\ne i) \\ 
 &=n \phi\Big(\frac b{cn}\Big)2p(1-2p)^{n-1} \\ 
 &=\frac{2pb}c\,\ln\Big(1+\frac b{cn}\Big)(1-2p)^{n-1}\to\frac1{2c}>1 
\end{aligned}
\end{equation\*}
as $n\to\infty$, if $b=n^2$ and $c\in(0,1/2)$. So, for all large enough $n$ we have $E\phi\big(\big|\frac1n\sum\_{i=1}^n Z\_i\big|/c\big)>1$ and hence $\|\frac1n\sum\_{i=1}^n Z\_i\|\_\phi\ge c$ and hence
\begin{equation\*}
 \Big\|\frac1n\sum\_{i=1}^n Z\_i\Big\|\_\phi\not\to0 
\end{equation\*}
as $n\to\infty$.




---


More generally, the answer will remain no if $\phi(t)=t \ell(t)$, where $\ell$ is any function such that $\ell(t)$ is [slowly varying](https://en.wikipedia.org/wiki/Slowly_varying_function) as $t\to\infty$. Yet more generally, the answer will remain no if $\phi(t)=t L(t)$, where $L$ is any function such that $\sup\limits\_{K\in(0,\infty)}\limsup\limits\_{t\to\infty}\dfrac{L(Kt)}{L(t)}<\infty$.


