# Variance in central limit theorem for dependent random variables.

I am reading some notes and struggling to show something which should be basic. I will write out the problem below. We have a measure preserving dynamical system $(X,T,\mu)$, and a measurable map $f:X\to \mathbb{R}$. We consider the sequence of identically distributed (although dependent) random variables $f,f\circ T,f\circ T^2$.... Moreoverm we suppose that $\mu$ is mixing. I.e., $\lim\limits_{n\to\infty}\int_X (\psi\circ T^n)\phi d\mu=(\int\limits_X \psi d\mu)(\int\limits_X \phi d\mu)$.

Under some conditions (which are detailed here https://vaughnclimenhaga.wordpress.com/2013/03/17/spectral-methods-3-central-limit-theorem/ (chapter 2)) we have the central limit theorem holds with variance $\sigma^2=\sum\limits_{n\in\mathbb{Z}}\int_X f\cdot (f\circ T^n)d\mu$.

If we write $S_nf=\sum\limits_{k=0}^{n-1}f\circ T^k$, the notes say that $\sigma^2$ can be written as $\sigma^2=\lim\limits_{n\to\infty}\frac{1}{n}\int (S_nf)^2d\mu$.

I am struggling to show why this is true. I feel an argument such as this should work but am evidently going wrong somewhere:

$\sigma^2=\lim\limits_{n\to\infty}\sum_{k=0}^{n-1}\int\limits_Xf\cdot f\circ T^kd\mu=\lim\limits_{n\to\infty}\int_X f\cdot S_nf d\mu=\lim\limits_{n\to\infty}\int_X\frac{S_n}{n}S_nd\mu=\lim\limits_{n\to\infty}\frac{1}{n}\int_X (S_nf)^2d\mu$

where for the third equality I fudged over and used the $T$ - invariance of $\mu$.

Could someone please tell me where(if) I am going wrong?Thanks!

If $\sum_{n\in\Bbb{Z}}|\int_Xf\cdot(f\circ T^n)d\mu|<\infty$, then \begin{align} \frac1n\int_X(S_nf)^2d\mu&=\frac1n\sum_{i,j=1}^n\int_X (f\circ T^i)\cdot (f\circ T^j) d\mu =\frac1n\sum_{i,j=1}^n \int_Xf\cdot (f\circ T^{j-i})d\mu\\ &=\sum_{k=-n+1}^{n-1}\frac{n-|k|}n\int_Xf\cdot (f\circ T^{k})d\mu\to \sum_{k\in\Bbb{Z}}\int_Xf\cdot(f\circ T^k)d\mu\qquad \text{as $n\to\infty$}. \end{align}