# Convergence in probability from Cesaro mean

If $X_n \ge 0$ be sequence of non-negative random variables (not necessarily independent) such that: $$\displaystyle \frac{1}{n}\sum\limits_{k=1}^{n} X_k \overset{P}{\longrightarrow} 1$$

Then does it follow that $$\displaystyle \frac{1}{n}\max\limits_{1 \le k \le n} X_k \overset{P}{\longrightarrow} 0 \qquad \text{ or } \qquad \frac{1}{n}\min\limits_{1 \le k \le n} X_k \overset{P}{\longrightarrow} 0 \quad ?$$

It's an exercise in Amir Dembo's lecture notes (Exercise 2.3.27 here). If the first result for 'max' can be verified the remaining exercise, asks us to show $\displaystyle \frac{1}{n^r}\sum\limits_{k=1}^{n} X_k^r \overset{P}{\longrightarrow} 0$ for any fixed $r>1$ (which follows by passing to a a.s. convergent subsequence).

The result for independent sequence of random variables has been mentioned here to be affirmative (Dugue 1957).

• Just a remark : the two statements in your "or" are equivalent (the question is homogeneous in the vector $(X_1,...,X_n)$ and you can just multiply by $-1$ to go from one statement to the other). – Patrick Da Silva Nov 7 '17 at 16:12
• @PatrickDaSilva: No the given condition is not symmetric under that mapping. $1$ is positive. – user21820 Nov 7 '17 at 16:13
• But the second conclusion is almost certainly trivially true, and should be removed. – user21820 Nov 7 '17 at 16:14
• I'm sorry, I didn't see the non-negative assumption. My bad! (But do you have a good reason to assume it?) – Patrick Da Silva Nov 7 '17 at 16:15
• @user21820 : I agree, since the minimum is smaller than the average and the average is bounded in probability, dividing it by $n$ surely makes it go to $0$. – Patrick Da Silva Nov 7 '17 at 16:16

Let $M_n=\frac 1 n \sum_{k=1}^nX_k.$ Let $\epsilon>0$ be given. Let $N,R\in\mathbb N$ and $0<\lambda<1$ be parameters to be chosen later depending only on the distribution of $(X_k)_{k\geq 1}$ and on $\epsilon.$ Consider the event that $M_{\lfloor \lambda^rn\rfloor}\in[1-\epsilon,1+\epsilon]$ for all integers $0\leq r\leq R.$ The probability of this event tends to $1$ as $n\to\infty$ so it occurs with probability at least $1-\epsilon$ for $n\geq N.$ We will show that for approriate parameters this event implies $X_i\leq 2n\epsilon$ for all $1\leq i\leq n.$
Since $M_{\lfloor \lambda^Rn\rfloor}\leq 1+\epsilon$ we know that $X_i\leq \lfloor \lambda^Rn\rfloor (1+\epsilon)$ for $i\leq \lfloor \lambda^Rn\rfloor.$ We will want to ensure this upper bound is at most $2\epsilon n.$
Since $M_{\lfloor \lambda^{r+1} n\rfloor}\geq 1-\epsilon$ and $M_{\lfloor \lambda^r n\rfloor}\leq 1+\epsilon$ we know that $X_i\leq \lfloor \lambda^r n\rfloor(1+\epsilon)-\lfloor\lambda^{r+1} n\rfloor(1-\epsilon)$ for $\lfloor\lambda^{r+1} n\rfloor<i\leq \lfloor\lambda^r n\rfloor.$ We will want to ensure this upper bound is also at most $2\epsilon n.$
Apart from a small error from the floors, these upper bounds can both be ensured by choosing $\lambda\geq 1-\epsilon$ and $R\geq \log \epsilon^{-1}/\log \lambda^{-1}.$