# Conditions for Central Limit Theorem for dependent sequences

Cumbersome technical assumptions (e.g., mixing properties) are used in the literature to prove Central Limit Theorems for dependent sequences. I sketched a proof that does not require any of these technical assumptions. Can you help me figure out what is wrong with this proof? The proof is at: http://www.statlect.com/central_limit_theorem_for_correlated_sequences.htm. Thanks in advance to all those who will be so generous and patient to read it.

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One thing that is not explicitly stated is that $V$ needs to be finite, so you would have to impose some sort of conditions on the covariance structure.  This is usually what those "technical conditions" are doing, making sure that the variance doesn't get "too big" – probabilityislogic May 1 '11 at 14:14
@user4422, hand-waving is good, but I did not see any proof. All the technical details usually come up, after you start to write rigorous proof. As the text stands now it is just an idea or strategy of proof, not a proof. The 2 things which stand out is the first claim that omiting some elements changes nothing, another thing is the claim that the remaining sequence is iid, it is not and assymptotic closeness to iid does not immediately allows application of CLT for iid variables. – mpiktas May 1 '11 at 14:38
OK. Right, we must ensure that V stays finite. Is there any counter-example where V goes to infinity for a stationary ergodic process? – user4422 May 1 '11 at 15:50

## 1 Answer

Additional conditions are needed. (A near-proof of this fact is that many incredibly smart individuals have been thinking deeply about these issues for over 100 years. It is highly unlikely that something like this would have escaped all of them.)

First of all, note that the formula for $V$ that you give is part of the conclusion of the associated central limit theorem. See, for example, Theorem 7.6 on pages 416–417 of R. Durrett, Probability: Theory and Examples, 3rd. ed., which based on your link, you appear to have access to.

At any rate, here is a simple counterexample to your claim.

Let $X_0$ equal $+1$ with probability $1/2$ and $-1$ with probability $1/2$. Define $X_n = (-1)^n X_0$. Then $\{X_n\}$ is a stationary ergodic process with mean 0 and variance 1, but the Central Limit Theorem fails.

The properties of stationarity and ergodicity should be pretty easy to see as we can construct this process by defining a function over the states of a two-state Markov chain with stationary probability measure $\pi(x) = 1/2$ for $x \in \{0,1\}$.

Observe that this process yields a sequence of the form $-X_0, X_0, -X_0, \ldots$ and so

1. Even without appealing to any notions about ergodicity, it is easy to see that $\newcommand{\e}{\mathbb{E}}\bar{X}_n \to \e X_0 = 0$ almost surely, and,
2. $\newcommand{\Var}{\mathbb{V}\mathrm{ar}}\Var(S_n) = 0$ if $n$ is even and $1$ if $n$ is odd.

This already is enough to conclude that there is no way that any rescaling of $S_n$ can make it converge in distribution to a normal random variable. In fact, for every function $f$ such that $f(n) \to \infty$, $S_n / f(n) \to 0$ almost surely no matter how slowly $f$ diverges.

Note also that this example should make it clear that the formula for $V$ is a conclusion of the theorem. Indeed, for the example above, $$V_n = 1 + 2 \sum_{i = 1}^n \e X_0 X_i = \left\{ \begin{array}{rl} -1, & n \text{ odd}, \\ 1, & n \text{ even}, \end{array} \right.$$ which, of course, (a) makes no sense as a variance, (b) does not have a limit, and (c) is not asymptotically equivalent to $\Var(S_n)$. (NB: I use a slightly different form for $V_n$ than you do where mine matches that given in Durrett.)

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First of all thanks for your long and careful answer, I really appreciate that. As for the example, it is very interesting, but I am struggling to understand why the process is ergodic: the notion of ergodicity I have in mind, coupled with stationarity, implies strong mixing, i.e. asymptotic independence of the terms of the sequence, which in this example seem perfectly dependent. – user4422 May 1 '11 at 22:56
Sorry, I was wrong again. Mixing is stronger. Then it might be ergodic even if it's not mixing. – user4422 May 1 '11 at 23:05
OK. Now I understood why it is ergodic. Trivially all shift invariant subsets have either probability zero or one. – user4422 May 1 '11 at 23:33
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