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Say I have $X_{ij}$, $j \le i$ with the property that $X_{ij}$ are centered and identically distributed and $E(X_{ij} X_{ij'}) = o(\exp(-i)))$. Then does $\sum_j X_{ij}$ have Gaussian domain of attraction?

As a related question, if $X_1, X_2, X_3$ are identically distributed and centered and $E(X_i X_j) = c$, what bound can I get for $E(X_1 X_2 X_3)$ in terms of $c$?

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John, I just remembered that I asked a similar question here about six months ago. Maybe some information on this page can help you out:… – Tom LaGatta Apr 2 '10 at 23:40
up vote 5 down vote accepted

Not necessarily. One has to impose more restrictive mixing and moment conditions. A classical book is:

Ibragimov I.A., Linnik Yu.V. Independent and stationary sequences of random variables

There is a long-standing question asked by Ibragimov: is $\phi$-mixing and finiteness of second moment sufficient for CLT to hold for a stationary sequence?

Also, there are various concepts of dependence. For example, if your r.v.'s are associated (i.e. satisfy FKG inequalities) and the covariance decays as you describe, then CLT holds.

UPD. As for the second part of your question: you cannot estimate higher-order moments in terms of lower-order ones unless the joint distributions have some special structure.

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Thanks Yuri. Do you happen to also know a reference for CLT in the non-stationary setting? I indeed found a lot of papers on stationary CLT before online, but couldn't find much with the general situation, even the associated case as you described, which I am interested in. – John Jiang Apr 1 '10 at 3:47
look into the recent Bulinski & Shashkin book "Limit theorems for associated random fields and related systems". I don't have it at hand, but I think they cover nonstationary fields. The Cox-Grimmett coefficient is a useful tool, and C.Newman's inequality (and its extensions and analogues) for characteristic functions of sums of associated r.v.'s. (It estimates how much the ch.f. of the sum deviates from the product of ch.f.'s of individual associated terms). Even if your situation is not covered exactly by the book, it might happen that Newman's inequality solves your problem. – Yuri Bakhtin Apr 1 '10 at 14:36

Your double subscripts are extraneous. Let's consider a simpler situation, where we have a single family of random variables $\{X_i\}$.

As Yuri Bakhtin says above, your condition is not sufficient for a CLT to hold. Here is a simpler situation, however: suppose that $X_i$ and $X_j$ satisfy finite-range dependence. That is, there exists a positive integer $R$ such that if $|i-j| \ge R$, then $X_i$ and $X_j$ are independent. We will prove a law of large numbers for $\{X_i\}$. If you're interested, you can push it farther to prove a central limit theorem. Suppose that $X_i$ has mean $\mu$ for each $i$.

Let $S_N = \tfrac{1}{N} \sum_{i=1}^N X_i$ as usual. Without loss of generality, we may consider indices only divisible by $R$: $S_{RN} = \tfrac{1}{RN} \sum_{i=1}^{RN} X_i$. Let $$S_{RN}^{(k)} = \tfrac{1}{N} \sum_{j=0}^{N-1} X_{Rj+k}$$ for $k= 1, \dots, R$, so that $$S_{RN} = \tfrac{1}{R} \left( S_{RN}^{(1)} + \dots + S_{RN}^{(R)} \right).$$Each sum $S_{RN}^{(k)}$ is comprised of independent random variables, so the classical law of large numbers applies and $S_{RN}^{(k)} \to \mu$ both in probability and almost surely. Consequently, $S_{RN} \to \mu$.

Obviously, this argument breaks down when $R = \infty$. In that case, the problem is no longer trivial and you will have to be more cautious with your assumptions.

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To whomever downvoted this post: why don't you leave a comment explaining how it can be improved? – Tom LaGatta Apr 1 '10 at 4:01

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