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Consider a sequence of identically distributed real-valued random variables $(X_i)_{i\in\mathbb{N}}$, with $\mathbb{E}\left[X_i\right]=0$ and $\mathbb{E}\left[X_i^2\right]=1$.

Suppose that there exists some constant $c\in(0,1)$, such that $$\mathbb{E}\left[X_iX_j\right]<c^{|i-j|}$$ for any $i,j\in\mathbb{N}$ (with $i\not= j$).

Is it true that $$n^{-1/2}\sum_{i=1}^nX_i\overset{d}{\rightarrow}\mathcal{N}(0,1)$$ as $n\rightarrow\infty$?

There are many central limit theorems for dependent random variables but I have been unable to find any quite like this. I am not attached to the specific choice of covariance bound, but would love to find a result of this form that depends on nothing more than first and second moments. The motivation for the question in part comes from this nice thread on the law of large numbers:

Weak Law of Large Numbers for Dependent Random Variables with Bounded Covariance

Davide Giraudo
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user301395
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2 Answers2

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Even uncorrelatedness is not sufficient for a central limit theorem.

  1. In Some pairwise independent sequences for which the central limit theorem fails by Janson Svante, a stationary sequence of pairwise independent random variables with finite variance is constructed in such a way that the central limit theorem fails (the limit can be degenerated or non-normal).

  2. In A strictly stationary, "causal,'' 5-tuplewise independent counterexample to the central limit theorem by Richard C. Bradley, a strictly stationary sequence $\left(X_k\right)_{k\in\mathbb Z}$ satisfying the following properties is constructed

    • $X_k$ takes the values $-1$ and $1$ with respective probabilities $1/2$;
    • if $I\subset \mathbb Z$ has five elements, then $\left(X_i\right)_{i\in I}$ is independent;
    • $X_k$ is a functional of independent identically distributed random variables;
    • for each infinite set $J$ of $\mathbb N$, there exists an infinite subset $I$ of $J$ and a non-normal non-degenerated distribution probability measure $\mu$ such that the sequence $\left(n^{-1/2}\sum_{k=1}^nX_k\right)_{n\in I}$ converges to $\mu$ in distribution.
Davide Giraudo
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  • That is incredible. Taking all covariances to be zero is nowhere near strong enough. I guess we will need to add some condition like stationarity then. – user301395 Apr 12 '18 at 07:47
  • In bradley's paper, the constructed sequence is strictly stationary. What lakes is a global dependence structure of the sequence. – Davide Giraudo Apr 12 '18 at 11:37
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This can be examined directly. Writing out the autocovariances, the expression for the variance of the sum can be compactly written as

$$\text{Var}\left (n^{-1/2}\sum_{i=1}^{n}X_i\right)= 1 + \frac 2n\Big[(n-1)c + (n-2)c^2 + (n-3)c^3 + ...+ (n-(n-2))c^{n-2} + (n-(n-1))c^{n-1}\Big]$$

$$=1+2\Big[c+c^2+c^3+...+c^{n-1}\Big] - \frac 2n\Big[c+2c^2+3c^3+...+(n-1)c^{n-1}\Big]$$

$$=1+2\frac {c-c^n}{1-c}-\frac 2n\sum_{k=1}^{n-1}kc^k$$

$$=1+2\frac {c-c^n}{1-c}- \frac {2c}{(1-c)^2}\cdot\frac {\big[1-nc^{n-1}+(n-1)c^n\big]}{n}$$

So at the limit,

$$\lim_{n\to \infty}\text{Var}\left (n^{-1/2}\sum_{i=1}^{n}X_i\right) = 1+\frac {2c}{1-c} $$

This will be also the variance of the limiting distribution under the usual regularity conditions. As should be expected, the closer is $c$ to unity, the higher the limiting variance. For example, for $c=0.9$ the variance will be equal to $19$.

Alecos Papadopoulos
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