Let $(X_i)_{i\in\mathbb{N}}$ be a sequence of $L^2$ random variables with expected value $m$ for all $n$. Let $S_n=\sum_{i=1}^n X_i$ and $|\mathrm{Cov}(X_i,X_j)|\leq\epsilon_{|i-j|}$ for finite, non-negative constants $\epsilon_k$. Show that:

(1) If $\lim_{n\to\infty} \epsilon_n=0$ then $S_n/n\to m$ in $L^2$ and probability

(2) If $\sum_{k=1}^\infty \epsilon_k<\infty$, then $\mathrm{Var}(S_n/n)$ is of order $O(1/n)$ and $S_n/n\to m$ almost surely

(1) First of, I found a similar looking question here, but we don't have, that the constants are bounded by $1$, so I don't know if the Chebyshev-inequality approach works here.

(2) We have $\mathrm{Var}(S_n/n)=\dfrac{1}{n^2}Var(S_n)=\dfrac{1}{n^2}\sum_{i\ne j}\mathrm{Cov}(X_i,X_j)=\dfrac{1}{n^2}(\sum_{k=1}^n\mathrm{Cov}(X_i,X_i)+\sum_{i=1}^{n-1}\sum_{j=i}^n \mathrm{Cov}(X_i,X_j))\leq\dfrac{1}{n^2}(n\cdot\epsilon_0+\sum_{i=1}^{n-1}\sum_{j=i}^n \mathrm{Cov}(X_i,X_j))$.
This is where I'm stuck.


1 Answer 1


We can assume that $m=0$. Then take $\delta\gt 0$ write for a fixed $C$: $$\frac 1{n^2} \mathbb E[S_n^2]\leqslant \frac 1{n^2}\sum_{i,j=1}^n\varepsilon_{|i-j|} [|i-j|\leqslant C]+\frac 1{n^2}\sum_{i,j=1}^n\varepsilon_{|i-j|} [|i-j|\gt C].$$ If we choose it such that $\varepsilon_k \leqslant \delta$ if $k\geqslant C$, then we have $$\frac 1{n^2} \mathbb E[S_n^2]\leqslant \frac{2Cn}{n^2}\sup_k \varepsilon_k+\delta,$$ hence the sequence $\left(\mathbb E[S_n^2]/n^2\right)_{n\geqslant 1}$ converges to $0$.

Once we showed the good asymptotic of $\mathbb E[S_n^2]$, an application of the Borel-Cantelli's lemma yields $2^{-n}S_{2^n}\to 0$ almost surely. Indeed, define for a fixed $j$ the event $A_n:=\{|S_{2^n}|\gt 2^n/j\}$ for a fixed $\delta$. Then the series $\sum_n\mathbb P(A_n)$ is convergent, and we get $\mathbb P\left(\limsup_{n\to\infty} A_n \right)=0$. As a consequence, on a set $\Omega_j$ of probability one, we have $|S_{2^n}(\omega)| 2^n\lt1/j $ for $n\geqslant N(j,\omega)$.

In order to deduce the result for the whole sequence, note that $$\mu\left\{\max_{2^n\leqslant k\lt 2^{n+1}}\frac{|S_k|}{2^n}>\delta\right\} \leqslant 2^n\max_{2^n\leqslant k\lt 2^{n+1}}\mu\left\{\frac{|S_k|}{2^n}>\delta\right\}\\ \leqslant 2^n\max_{2^n\leqslant k\lt 2^{n+1}}\frac 1{\delta^2}\mathbb E\left[\frac{S_k^2}{2^{2n}}\right].$$

  • $\begingroup$ If I understand the first part correctly, we need $e_k\to 0$ to be able to choose such a $\delta$. I can't really follow you for the second part, could you expand on it? Edit: By that I mean the application of Borell-Cantelli. $\endgroup$
    – blst
    Jan 6, 2015 at 17:36
  • $\begingroup$ @blst I have edited. $\endgroup$ Jan 6, 2015 at 22:29
  • $\begingroup$ How do we know that $\sum_n\mathbb P(A_n)$ is convergent? $\endgroup$
    – blst
    Jan 7, 2015 at 13:13
  • $\begingroup$ Use Chebychev's inequality to bound $\mathbb P(A_n)$. $\endgroup$ Jan 7, 2015 at 13:53
  • $\begingroup$ @DavideGiraudo Regarding the part for the whole sequence: Shouldn't we prove that $\mu\left\{\max_{2^n\leqslant k\lt 2^{n+1}}{|S_k|}/{2^n}>\delta\right\}$ is less than or equal to a summable sequence (in $n$)? It seems to me that, since $\mathbb E[S_k^2]=O(k)$ and $2^n \leq k<2^{n+1}$, the last step is bounded by $B/\delta^2$ for some $B>0$, hence it is bounded by a constant (because $\delta$ is fixed). Am I missing something? $\endgroup$ Aug 10, 2015 at 10:49

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