Does central limit theorem hold for general weakly dependent variables? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T13:43:44Z http://mathoverflow.net/feeds/question/20019 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20019/does-central-limit-theorem-hold-for-general-weakly-dependent-variables Does central limit theorem hold for general weakly dependent variables? John Jiang 2010-04-01T00:31:41Z 2010-04-01T02:56:43Z <p>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? </p> <p>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 $c$?</p> http://mathoverflow.net/questions/20019/does-central-limit-theorem-hold-for-general-weakly-dependent-variables/20030#20030 Answer by Yuri Bakhtin for Does central limit theorem hold for general weakly dependent variables? Yuri Bakhtin 2010-04-01T02:38:40Z 2010-04-01T02:46:06Z <p>Not necessarily. One has to impose more restrictive mixing and moment conditions. A classical book is:</p> <p>Ibragimov I.A., Linnik Yu.V. <em>Independent and stationary sequences of random variables</em></p> <p>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?</p> <p>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. </p> <p>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.</p> http://mathoverflow.net/questions/20019/does-central-limit-theorem-hold-for-general-weakly-dependent-variables/20031#20031 Answer by Tom LaGatta for Does central limit theorem hold for general weakly dependent variables? Tom LaGatta 2010-04-01T02:56:43Z 2010-04-01T02:56:43Z <p>Your double subscripts are extraneous. Let's consider a simpler situation, where we have a single family of random variables <code>$\{X_i\}$</code>.</p> <p>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 <code>$\{X_i\}$</code>. If you're interested, you can push it farther to prove a central limit theorem. Suppose that $X_i$ has mean $\mu$ for each $i$.</p> <p>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 <code>$$S_{RN} = \tfrac{1}{R} \left( S_{RN}^{(1)} + \dots + S_{RN}^{(R)} \right).$$</code>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, $S_{RN} \to \mu$. </p> <p>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.</p>