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](http://en.wikipedia.org/wiki/Central_limit_theorem#CLT_under_weak_dependence) 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](http://en.wikipedia.org/wiki/Mixing_%28mathematics%29#Mixing_in_stochastic_processes).


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 $\mathbb{E} Y\_0^{12} < \infty$. 



The $\sigma > 0$ condition in the Central Limit Theorem
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. 


