Suppose that $(X\_n)\_{n\geq 1}$ is a sequence of (non-negative) random variables on a probability space ($\Omega, \mathcal{A}, P)$ such that $X\_n = o\_P(n^{-\beta})$ for some $\beta \in (0,1)$.


Does it hold that $\frac{1}{n}\sum\_{i=1}^n X\_i = o\_P(n^{-\beta})$ ?


A paper I've come across claims it is based on the following reasoning. From the almost sure representation theorem, there exists a sequence of random variables $Y\_n$ such that $Y\_n \overset{d}{=} X\_n$ and $n^\beta Y\_n \xrightarrow{a.s.} 0$. Then, for almost all $\omega \in \Omega$, $n^\beta \frac{1}{n}\sum\_{i=1}^n Y\_i(\omega) \rightarrow 0$. Therefore $\frac{1}{n}\sum\_{i=1}^n Y\_i \xrightarrow{a.s.} 0$. 


The last step of their proof is what I think might be wrong. They claim that $n^\beta \frac{1}{n} \sum\_{i=1}^n Y\_i \overset{d}{=} n^\beta \frac{1}{n} \sum\_{i=1}^n X\_i$, which would imply the claim. Although $X\_n \overset{d}{=} Y\_n$ for all $n$, the dependence structure between the $X\_n$'s and $Y\_n$'s might be different, which would a priori lead to different laws for $\sum\_{i=1}^n X\_i$ and $\sum\_{i=1}^n Y\_i$.


Would this proof have an easy fix ? Or would there be an alternative proof ? Or is the result wrong, and if so, what would be a counterexample ?



Convergence in probability of Cesaro means
This claim is false. E.g., let $\beta=1$ and $X\_n=\frac1{n\ln(n+1)}$, nonrandom. Then $X\_n=o(n^{-\beta})$, whereas $\frac1n\,\sum\_{i=1}^n X\_i\sim\frac1n\,\ln\ln n \ne o\_P(n^{-\beta})$. 


Or, take any $\beta>1$ and any $X\_1>0$. Then $\frac1n\,\sum\_{i=1}^n X\_i\ge\frac1n\,X\_1\ne o\_P(n^{-\beta})$. So, $\frac1n\,\sum\_{i=1}^n X\_i\ne o\_P(n^{-\beta})$. 




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**Added in response to the edit of the original question:**


In the case $\beta\in(0,1)$, the claim is false, too, and in fact it is false for any real $\beta$ -- but this takes a bit more effort to show. E.g., let $U$ be a random variable uniformly distributed on $[1,2)$. For each natural $n$, let $X\_n:=j!$ if $2^j\le n<2^{j+1}$ and $\frac n{2^j}\le U<\frac{n+1}{2^j}$ for some $j\in\{0,1,\dots\}$, and $X\_n:=0$ otherwise. Then for any real $t>0$ we have 
$$P(X\_n>tn^{-\beta})\le P(X\_n>0)=1/2^{j\_n}<2/n\to0,
$$
where $j\_n:=\lfloor\log\_2 n\rfloor$. So, $X\_n=o\_P(n^{-\beta})$. 


However, for any natural $j$ and $n=2^{j+1}$, 
$$\frac1n\,\sum\_{i=1}^n X\_i\ge2^{-j-1}\,\sum\_{i=2^j}^{2^{j+1}-1} X\_i
=2^{-j-1}j!>n^a
$$
for any real $a$ if $j$ is large enough. So, $\frac1n\,\sum\_{i=1}^n X\_i\ne o\_P(n^{-\beta})$ for any real $\beta$. 


