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 ?

  • $\begingroup$ Does your notation $X_n = o_P(n^{-\beta})$ mean $ n^\beta X_n \to 0$ in probability? $\endgroup$ – Marcus M Nov 27 '19 at 18:58

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, $\frac1n\,\sum_{i=1}^n X_i\ne o_P(n^{-\beta})$.

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 $\beta$.

  • $\begingroup$ Sorry, I meant $\beta \in (0,1)$. I am editing the question accordingly. $\endgroup$ – Aurelien Nov 27 '19 at 20:00
  • $\begingroup$ @Aurelien : The claim is false for any real $\beta$. $\endgroup$ – Iosif Pinelis Nov 27 '19 at 20:51

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