In the version of central limit theorem for strictly stationary but weakly dependent (for instance $\alpha$-mixing with fast decaying mixing coefficient) random variables $X_1, X_2, \cdots$, the theorem in this Wikipedia page states (see also Billingsley 1995 Theorem 27.4):

Theorem. Suppose that $X_1, X_2, \cdots$ is stationary and $\alpha$-mixing with $\alpha_n = O(n^{−5})$ and that $\mathbb{E} X_n = 0$ and $\mathbb{E} X_n^{12} < \infty$. Denote $S_n = X_1 + \cdots + X_n$, then the limit $\sigma^2 = \lim_{n\to \infty} \mathbb{E} S_n^2/n$ exists, and if $\sigma \ne 0$ then $S_n/(\sigma \sqrt{n})$ converges in distribution to the standard Gaussian distribution $\mathcal{N}(0, 1)$.

Here $\alpha_n$ is the mixing coefficients defined as e.g. in this Wikipedia page.

My question is: Is there a proof for the necessity of $\sigma > 0$? What can we conclude (if possible) when $\sigma = 0$?

In the wikipedia page, there is a little remark trying to explain the necessity of $\sigma > 0$ (note that $\sigma^2 = \mathbb{E} X_1^2 + 2 \sum_{j=1}^\infty \mathbb{E} X_1 X_{1+j} \ge 0$). Namely, if we take $Y_0, Y_1, \cdots$ to be an i.i.d. sequence, and $X_i = Y_i - Y_{i-1}$ for all $i = 1, 2,\cdots$. Then $X_1, X_2 \cdots$ is stationary and $\alpha$-mixing with $\alpha_n = 0$ for all $n \ge 2$. However, $\sigma^2 = 0$, violating the condition $\sigma^2 > 0$. I don't think this example prohibits us from concluding: $$ \frac{S_n}{\sqrt{n}} \xrightarrow[n\to \infty]{distribution} \ \sigma \mathcal{N}(0,1). $$ In the above example though the right hand side becomes the Dirac measure at zero, the left hand side is $S_n/\sqrt{n} = (Y_n - Y_0)/\sqrt{n}$, which indeed converges to zero in probability (hence in distribution), say by applying Chebyshev and assuming $\mathbb{E} Y_0^{12} < \infty$.


1 Answer 1


How is the general case different than this example?

If $\sigma=0$ then the variance of $S_n/\sqrt{n}$ goes to 0 so $S_n/\sqrt{n} \to 0$ in distribution.

  • $\begingroup$ ... and therefore $S_n / \sqrt{n} \to 0$ in probability as well, by this fact, with no further moment assumptions needed. $\endgroup$ Apr 10, 2014 at 7:04

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