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I'm currently stuck on the following problem which involves proving the weak law of large numbers for a sequence of dependent but identically distributed random variables. Here's the full statement:

  • Let $(X_n)$ be a sequence of dependent identically distributed random variables with finite variance.

  • Let $\displaystyle S_n = \sum_{i=1}^n X_i $ denote the $n^\text{th}$ partial sum of the random variables $(X_n)$.

  • Assume that Cov$(X_i,X_j) \leq c^{|i-j|}$ for $i, j \leq n$ where $|c| \leq 1$.

Is it possible to show that $\displaystyle \frac{S_n}{n} \rightarrow \mathbb{E}[X_1]$ in probability? In other words, is it true that given any $\epsilon>0$,

$$ \lim_{n\rightarrow \infty} \mathbb{P}\bigg[\Big|\frac{S_n}{n} - \mathbb{E}[X_1]\Big| > \epsilon\bigg] = 0$$

EDIT: Following some comments, it turns out that I had the right approach so I've gone ahead and answered my own question below.

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I should add that I've tried using the Chebyshev inequality, but can't get the right kind of bound - so I suspect that there must be another way. –  Elements Nov 26 '12 at 23:44
no, that's right, you should be able to show $\sigma^2(\frac { S_n} n) \rightarrow 0$ –  mike Nov 26 '12 at 23:55
@mike Hmm I suspect I may not be using the right bounds... See above –  Elements Nov 27 '12 at 0:31

2 Answers 2

up vote 3 down vote accepted

Fix $\epsilon > 0$ and $n \in \mathbb{N}$, then we can use Chebyshev's inequality to see that

$$\mathbb{P}\bigg[\Big|\frac{S_n}{n} - \mathbb{E}[X_1]\Big| > \epsilon\bigg] \leq \frac{\text{Var}\Big(\frac{S_n}{n}\Big)}{\epsilon^2}$$


$$\displaystyle \text{Var}\Big(\frac{S_n}{n}\Big)= \frac{\text{Var}(S_n)}{n^2} \leq \frac{\sum_{i=1}^n\sum_{j=1}^n \text{Cov}{(X_i,X_j)}}{n^2} \leq \frac{\sum_{i=1}^n\sum_{j=1}^n c^{|i-j|}}{n^2} $$

We can then explicitly calculate the double sum $\sum_{i=1}^n\sum_{j=1}^n c^{|i-j|}$ as follows:

$$\begin{align} \sum_{i=1}^n\sum_{j=1}^n c^{|i-j|} &= \sum_{i=1}^n c^{|i-i|} + 2\sum_{i=1}^n\sum_{j=1}^{i-1} c^{|i-j|} \\ &= n + 2\sum_{i=1}^n\sum_{j=1}^{i-1} c^{|i-j|} \\ &= n + 2\sum_{i=1}^n\sum_{j=1}^{i-1} c^{i-j} \\ &= n + 2\sum_{i=1}^n c^i \frac{1 - c^{-i}}{1-c^{-1}} \\ &= n + 2\sum_{i=1}^n \frac{c^i + 1}{1-c^{-1}} \\ &= n + \frac{2c}{c-1} \sum_{i=1}^n c^{i}-1 \\ &= n + \frac{2c}{c-1} \big(\frac{1-c^{n+1}}{1-c} -n \big)\\ &= n + \frac{2c}{(c-1)^2}(c^{n+1}+1) + \frac{2c}{c-1}n\\ \ \end{align}$$


$$\lim_{n\rightarrow\infty} \mathbb{P}\bigg[\Big|\frac{S_n}{n} - \mathbb{E}[X_1]\Big| > \epsilon\bigg] = \lim_{n\rightarrow\infty} \frac{\text{Var}\Big(\frac{S_n}{n}\Big)}{\epsilon^2} \leq \lim_{n\rightarrow\infty} \frac{n + \frac{2c}{(c-1)^2}(c^{n+1}+1) + \frac{2c}{c-1}n}{n^2 \epsilon^2} = 0 $$

Seeing how our choice of $\epsilon$ was arbitrary, the statement above holds for any $\epsilon > 0 $ and shows that $\frac{S_n}{n} \rightarrow E[X_1]$ in probability, as desired.

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I just wanted to add a reference for a similar result. This appears in Some new applications of Riesz products by Gavin Brown. I have adapted the notation to suit your question.

Proposition 1: Suppose that $(X_n)$ is a sequence of random variables of bounded modulus, $E(X_n) = \mu$ for all $n$ and that $$ \sum_{N=1}^\infty \frac{1}{N} E(|Y_N|^2) < \infty $$ where $Y_N = \frac{1}{N} \sum_{n=1}^N (X_n - \mu)$. Then $Y_N \to 0$ almost surely.

Assume there exists some finite $M$ such that $|X_n -\mu| \leq M$ for all $n \in \mathbb Z^+$. Since $$E(|Y_N|^2) = \frac{1}{N^2}\sum_{i=1}^N\sum_{j=1}^N \text{Cov}(X_i,X_j) = \text{Var}\left(\frac{S_N}{N}\right)$$ Then $$ \sum_{N=1}^\infty \frac{1}{N} E(|Y_N|^2) = \sum_{N=1}^\infty \frac{1}{N^3} \text{Var}(S_N) $$ As shown in the previous response, a consequence of the weak dependence assumption is that the right hand side is finite. By the above proposition, $Y_N \to 0$ almost surely. So $$ Y_n = \frac{1}{N} \sum_{n=1}^N (X_n - \mu) = S_N - \mu \to 0 $$ so $S_N \to \mu$ almost surely.

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