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Let $ X_{1}, \cdots, X_{n} $ be a sequence of dependent random variables having the same finite mean $ \mu=\mathrm{E}\left(X_{1}\right), $ the same finite variance $ \sigma^{2}=\operatorname{Var}\left(X_{1}\right), $ and $ \left|\operatorname{Cov}\left(X_{i}, X_{j}\right)\right| \leq \frac{1}{(i-j)^{4}} $ for $ i \neq j . $ Show that $ \bar{X} $ converges in probability to $ \mu . $

I am stuck on the upper bound of the sum of the covariances: $\sum_{i=1}^{n} \sum_{j \neq i,j=1}^{n}|{\rm Cov}(X_i,X_j)|\le2 \cdot \sum_{k=1}^{n-1}(\frac{n-k}{k^4}) $

How to show that $n^2$ is a proper upper bound of the sum?

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1 Answer 1

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Since $\left\lvert\operatorname{Cov}(X_i,X_j)\right\rvert\leq (i-j)^{-4}\leq 1$, summing over all $n(n-1)$ ordered pairs $(i,j)$ with $i\neq j$ gives the easy upper bound $n(n-1)<n^2$.

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  • $\begingroup$ But to make the probability tend to $0$, I need $n^2$ to be not asymptotically tight (the summation should not have a $n^2$ term). $\endgroup$
    – Dennis
    Feb 17, 2021 at 15:15
  • $\begingroup$ In which case, $\sum_{k=1}^{n-1} (n-k)/k^4\leq (n-1)\zeta(4)=O(n)$. $\endgroup$ Feb 17, 2021 at 15:20

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