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.


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 $\gamma$.


1 Answer 1


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 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 $\sup\limits_{K\in(0,\infty)}\limsup\limits_{t\to\infty}\dfrac{L(Kt)}{L(t)}<\infty$.

  • $\begingroup$ Thanks, that's a really clean example, $\endgroup$ Nov 22, 2021 at 0:52

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