# Weak law of large numbers, covariance, sequence

Let $(X_i)_{i\in\mathbb{N}}$ be a sequence of $L^2$ random variables in a probability space $(\Omega, \mathcal{A}, P)$. Define $\overline{X_n}:=\frac{1}{n}\sum_{i=1}^n X_i$. Show that (2) follows from (1):

(1) $\mathrm{Cov}(X_i,X_j)\leq c_{|i-j|}$ for a sequence $c_n$ with limit $0$.

(2) $\overline{X_n}$ converge in probability to the expected value of it (weak law of large numbers).

Without loss of generality I take $E(X_j)=0 \ \forall j$.

Be $\epsilon>0$. It exists an $N: \forall n \ge N: c_n<\epsilon$. A convergent sequence is bounded: $\exists M: c_n< M < \infty \forall n \in \mathbb{N}$.

\begin{align} \mathrm{Var}(\overline{X_n}) &= \frac{1}{n^2} *( \sum_{i=1}^n \sum_{j=1}^n Cov(X_j,X_k)) \\ &= \frac{1}{n^2} *( \sum_{j,k: |j-k|\le N} Cov(X_j,X_k) + \sum_{j,k:|j-k|>N} Cov(X_j,X_k)) \\ &\le \frac{1}{n^2} * ( \sum_{j,k: |j-k|\le N}M + \sum_{j,k:|j-k|>N} *\epsilon) \\ &\le \frac{1}{n^2} (M*n*(1+2N)+ \epsilon* n^2) \\ &= \frac{M(1+2N)}{n}+ \epsilon \\ &\rightarrow \epsilon \quad (n\rightarrow \infty) \end{align}

Because I count for the entries with $|j-k|\le N$ the number $$n+2*(n-1)+2*(n-2)+..+2*(n-N)\\=n+2nN-2*(1+2+..+N)= n+2nN-2(\frac{(N+1)N}{2}) \le n*(1+2N)$$

I get $Var(\overline{X_n}) \rightarrow 0 \quad ( n \rightarrow \infty)$.

Be $\epsilon_1, \delta_1 >0$, so $\exists N: Var(\overline{X_n}) < \epsilon_1^2 \delta_1$ for $n \ge N$.

$$P(|\overline{X_n}|>\epsilon_1) \le \frac{Var(\overline{X_n})}{\epsilon_1^2} < \frac{\delta_1}{\epsilon_1^2} *\epsilon_1^2= \delta_1$$

I think i do some mistakes because the covariance don't must be positive so also the sequence? I think only if all is positive the proof is right.

• With no hypothesis on the expectations, it is impossible to conclude that $\bar X_n$ converges (and note that $\bar X_n\to E(\bar X_n)$ is absurd unless $E(\bar X_n)$ is constant). Hence (1) does not imply (2). What is your source? – Did Sep 28 '16 at 15:49
• ((No answer, this was only to be expected, I guess.)) – Did Sep 28 '16 at 21:09
• thanks for your answer. Sry my source is not in english.. – Lauren Veganer Sep 29 '16 at 6:15
• And why does this prevent you from giving it? – Did Sep 29 '16 at 6:16
• Oh sry: mat.univie.ac.at/~gerald/teaching/wts_ue.pdf exercise 72 – Lauren Veganer Sep 29 '16 at 6:24

Your approach works. The covariances may not be positive but it is not a problem when we write $$\operatorname{Var}\left(\overline{X_n}\right)=\frac 1{n^2}\sum_{i,j=1}^n \operatorname{Cov}\left(X_i,X_j\right).$$ The sum is non-negative but the terms are not necessarily all non-negative .