# Showing weak law of large numbers holds

My question:

$\{X_n\}$ is a sequence of random variables. Var$(X_n)\le C\ \ \forall \ n$ and $\rho_{ij}=$Cov$(X_i,X_j)\to 0$ as $|i-j|\to \infty$ . Show WLLN holds.

In my book there are 3 theorems , and everyone of them includes some kind of expectation calculation or involvement(like finite mean).And moreover none has any association with covariance. How am I to solve?

• Some intuition: the standard WLLN says that if you average independent identically distributed variables with finite mean, then the sample means converge in probability to the mean. This turns out to be stronger than you need. Still, you need something like this. You can replace them being identically distributed with them having the same mean and an assumption to ensure that $X_n$ doesn't deviate more and more from $\mu$ as $n$ grows. You can replace them being independent with some "weak dependence" condition. There are multiple weak dependence conditions out there. – Ian Jun 17 '16 at 16:01
• There are similar variants of CLT out there as well; one of the most useful is the Lindeberg CLT. – Ian Jun 17 '16 at 16:02

## 1 Answer

One can try to show the convergence in $\mathbb L^2$. For $n$ greater than some fixed $R$,
$$\mathbb E\left[\frac 1{n^2}\left(\sum_{i=1}^nX_i\right)^2\right]=\frac 1{n^2}\sum_{i,j=1}^n\rho_{i,j}=\frac 1{n^2}\sum_{i,j=1}^n\rho_{i,j}\mathbf 1\{|i-j|\leqslant R\}+\frac 1{n^2}\sum_{i,j=1}^n\rho_{i,j}\mathbf 1\{|i-j|\gt R\}.$$ The first sum can be estimated in the following way: by Cauchy-Schwarz inequality, $\rho_{i,j}\leqslant C$ for each $i,j$ and the number of $(i,j)$ with $i,j\in\{1,\dots,n\}$ and $|i-j|\leqslant R$ is of order $nR$.

For the second sum, note that $\rho_{i,j}\mathbf 1\{|i-j|\gt R\}\leqslant \sup_{|u-v|\gt R}\rho_{u,v}$ hence the second sum does not exceed $\sup_{|u-v|\gt R}\rho_{u,v}$.

• This approach is completely new to me (as I am a beginner) and looks very interesting an d promising to me. But unfortunately, I have not been able to understand some parts of your answer. Specifically 1)the use Cauchy Schwarz inequality to derive $\rho_{ij}\le C$ ,2) order $nR$ 3)the supremum inequality.. Will you care to explain your answer in some more detail? – Qwerty Jun 18 '16 at 9:04
• 1) C-S inequality gives $\rho_{i,j}\leqslant \sqrt{\rho_{i,i}}\cdot \sqrt{\rho_{j,j}}$. 2) We have to consider the cases $i-j=0,1,2\dots,R$. There are $n$ couples $(i,j)$ such that $i,j\in\{1,\dots,n\}$ and $i=j$; $2(n-1)$ couples $(i,j)$ such that $i,j\in\{1,\dots,n\}$ and $i-j=1$ and so on. 3) For the supremum: if $|i-j|\lt R$, the inequality is trivially satisfied. Otherwise, the left hand side is $\rho_{i,j}$ and $(i,j)$ appear in the set where we take the supremum. I think now you can fill all the details. – Davide Giraudo Jun 18 '16 at 9:38