# Convergence in probability of Cesaro means

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 ?

• Does your notation $X_n = o_P(n^{-\beta})$ mean $n^\beta X_n \to 0$ in probability? – 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 $$\beta$$.

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