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 ?