# Stationary processes that do not satisfy Gordin's central limit theorem

We are doing an assignment for our Advanced Econometrics course for which we are trying to illustrate Gordin's Central Limit Theorem by simulation. We used an AR(1) process to show that if the conditions of the theorem are satisfied, its implications hold too. Now we are trying to find counter-examples, and we were wondering if you know of any stationary and ergodic process that does not satisfy Gordin's Conditions? Here is a statement of the theorem:

If a stationary (1) and ergodic (2) process, $$y_t$$, satisfies Gordin's condition(3-5), then, (a) $$y_t$$ has zero mean, (b) absolutely summable autocovariances $$\gamma_j$$ and (c) $$\sqrt{n}\bar{y} \rightarrow_d N\left(0, \sum_{j = - \infty}^{\infty}\gamma_{j}\right)$$.

More explicitly, if a random process $$y_t$$ satisfies the following conditions:

(1) The joint distribution of $$y_t, y_{t+1}, ... y_{t+k}$$ does not depend on t (stationarity)

(2) For any positive integer k and any bounded functions $$f$$ and $$g$$ from $$R^{k=1} \rightarrow R$$, $$\lim_{n \rightarrow \infty} |E[f(y_{t}, ...,y_{t+k})g(y_{t+n}, ...,y_{t+k + n})]| = |E[f(y_{t}, ...,y_{t+k})]| |E[g(y_{t}, ...,y_{t+k})]|$$ (ergodicity)

(3) $$E(y_t^2) =$$ is finite.

(4) $$E(y_t|y_{t-j}, y_{t-j-1}, . . .) \rightarrow_{m.s.} 0$$ as $$j \rightarrow \infty$$

(5) $$\sum_{j = 0}^{\infty} \left[E\left(r_{tj}^2\right)\right]^{1/2}$$ where $$r_{ij} = E\left(y_t|y_{t-j}, y_{t-j-1}, ...\right) - E\left(y_t|y_{t-j-1}, y_{t-j-2}, ...\right)$$.

Then the following properties are true:

(a) $$E(y_t) = 0$$

(b) $$\sum_{j = - \infty}^{\infty} | \gamma_j |$$ is finite

(c) $$\sqrt{n}\bar{y} \rightarrow_d N\left(0, \sum_{j = - \infty}^{\infty}\gamma_{j}\right)$$

• In the end we opted for a process that follows a Cauchy distribution, as - as far as we know - it is stationary and ergodic, but has an infinite raw second moment, so (3) doesn't hold and all three implications break down. Any thoughts or additional ideas? – hrrrrrr5602 Dec 15 '18 at 14:32
• More specifically: (a) does not hold as the mean of a Cauchy distribution is undefined. By implication, (b) does not hold either as the variance is undefined as well. Again, (c) does not hold as the sample mean of a Cauchy distribution is itself Cauchy distributed for all $n$ and does not have finite variance in the limit. – hrrrrrr5602 Dec 15 '18 at 14:33
• I am currently also working on a simulation, as Matblab newbie, that illustrates Gordin's Theorem and was interested in how you arrived at checking conditions (2), (4) and (5) of the AR process. – Hendrik De Winter Nov 28 at 14:35