# Weak concentration bounds for averages of independent random variables in Orlicz spaces

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 $$\gamma$$.

In general, the answer is no. Moreover, the answer is no even if $$$$\phi(t)=t\ln(1+t). \tag{1}$$$$
Indeed, suppose that $$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)$$ is slowly varying as $$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$$.