Counterexample for absolute summability of autocovariances of strictly stationary strongly mixing sequence

Question

Suppose $(X_i)_{i\in\mathbb{Z}}$ is a strictly stationary, strongly (i.e. $\alpha-$)mixing sequence of real random variables. If we have $\mathbb{E}[|X_1|^{2+\epsilon}]<\infty$ for some $\epsilon>0$ and if we have for the mixing coefficients that there exists $C>0, \gamma\geq \frac{(2+\epsilon)(1+\epsilon)}{\epsilon}$ such that for large enough $k:$ $$\alpha(k) := \alpha\left(\sigma((X_i)_{i\leq 0}), \sigma((X_i)_{i\geq k})\right) \leq Ck^{-\gamma},$$ then Davydov's inequality (see for example Corollary A2 in Hall and Heyde (1980)) implies that the autocovariances are absolutely summable: $$\sum_{k=1}^\infty\left\lvert \operatorname{Cov}\left(X_0, X_k\right)\right\rvert<\infty.$$

I'm currently reading a paper where we only assume $\mathbb{E}[X_1]=0, \mathbb{E}[X_1^2] = 1,$ but higher moments might be infinite. It seems to me that this is not enough to guarantee absolutely summable autocovariances, but I find it hard to give a counterexample. In particular, I'm interested in an example of a strictly stationary and strongly mixing sequence with, $\mathbb{E}X_1^2<\infty$ and (for $k$ large enough) $\alpha(k)\leq Ck^{-\gamma}$ for some $C>0, \gamma\geq 3+2\sqrt{2}$ (the minimum of $\epsilon\mapsto\frac{(2+\epsilon)(1+\epsilon)}{\epsilon}$), but with autocovariances that are not absolutely summable. I find it hard to construct an example and the literature I found seems to either assume a higher moment exists without giving a counterexample to show its necessity, or assume some other condition to guarantee absolute summability. An example or a reference to one would be much appreciated.

Answer 1

We can construct a strictly stationary sequence $\left(X_k\right)_{k\in\mathbb Z}$ having the following properties:

1. $X_0$ has finite moments of any order.
2. $\beta(k)\leqslant Ck^{-1+\delta}$ for some $\delta\in (0,1)$.
3. The series $\sum_{k\in\mathbb Z}\lvert\operatorname{cov}\left(X_0,X_k\right)\rvert$ diverges.

Note that it is not exactly the wanted counterexample since the mixing rates are not as fast as the opening poster wants, but it is almost the best that we can get with finite moment of any order.

Let $\left(e_{k,i},k\geqslant 1,i\in\mathbb Z\right)$ be an independent family of random variables such that for all $k$ and $i$, $e_{k,i}$ takes the values $\pm 1$ with probability $1/(2n_k^2)$ and $0$ with probability $1-1/n_k^2$, where $(n_k)$ is an increasing sequence of integers such that $\sum_{k\geqslant 1}n_k^{-2}$ converges. Let $$ X_j=\sum_{k\geqslant 1}\sum_{i=1}^{n_k}e_{k,j-i}, \quad j\in\mathbb Z. $$ Item 1. can be seen by an application of Rosenthal's inequality. For a fixed $j$, let us compute $\operatorname{cov}\left(X_0,X_j\right)$ for $j>0$. By definition, since $e_{k,i}$ is centered for all $k$ and $i$, $$ \operatorname{cov}\left(X_0,X_j\right)=\sum_{k\geqslant 1}\sum_{i=1}^{n_k} \sum_{k'\geqslant 1}\sum_{i'=1}^{n_k}\mathbb E\left[e_{k,-i}e_{k',j-i'}\right]. $$ Observe that if $k\neq k'$, then $e_{k,-i}$ is independent of $e_{k',j-i'}$ hence $$ \operatorname{cov}\left(X_0,X_j\right)=\sum_{k\geqslant 1}\sum_{i,i'=1}^{n_k} \mathbb E\left[e_{k,-i}e_{k ,j-i'}\right]. $$ If $j-i'=-i$, then $ \mathbb E\left[e_{k,-i}e_{k ,j-i'}\right]=n_k^{-2}$ and if not, then $ \mathbb E\left[e_{k,-i}e_{k ,j-i'}\right]=0$ hence $$ \operatorname{cov}\left(X_0,X_j\right)=\sum_{k\geqslant 1}\frac 1{n_k^2}\sum_{i,i'=1}^{n_k} \mathbf{1}\{j-i'=-i\}. $$ Moreover, $\sum_{i =1}^{n_k} \mathbf{1}\{j-i'=-i\}=\mathbf{1}\{1\leqslant i'-j\leqslant n_k \}$. If $j>n_k$, then $\sum_{i'=1}^{n_k}\mathbf{1}\{1\leqslant i'-j\leqslant n_k \}=0$ and if $j\leqslant n_k$, this sum is $n_k-j$. We thus got $$ \operatorname{cov}\left(X_0,X_j\right)=\sum_{k\geqslant 1}\frac{n_k-j}{n_k^2} \mathbf 1\{j\leqslant n_k\}. $$ Summing this over $j$ gives a divergent series.

For the mixing rates, we can use the steps in the papers

Giraudo, Davide, and Dalibor Volný. “A Strictly Stationary β-Mixing Process Satisfying the Central Limit Theorem but Not the Weak Invariance Principle.” Stochastic Processes and their Applications 124.11 (2014): 3769–3781.

Giraudo, Davide. An improvement of the mixing rates in a counter-example to the weak invariance principle. C. R. Math. Acad. Sci. Paris 353 (2015), no. 10, 953--958

We have to choose $n_k$ such that $n_{k+1}\geqslant n_k^{1+\gamma}$ for some positive $\gamma$.

Answer 2

I have found a paper that proves the existence of a counterexample (Theorem 2 of Bradley 1983 "Information regularity and the central limit question"). The construction seems quite complicated, but it does provide exactly what I'm looking for.