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Setting and definitions

Let $X = \{X(t), t \in T \}$, $T \subset \mathbb{Z}$, be an infinite-variance associated stochastic process, i.e. $$ \text{Cov}(f(X(I)), g(X(J))) \geq 0 $$ for all finite disjoint subsets $I, J \subset T$ and bounded, coordinate-wise increasing Borel functions $f: \mathbb{R}^{\vert I \vert} \rightarrow \mathbb{R}$, $g: \mathbb{R}^{\vert J \vert} \rightarrow \mathbb{R}$.

A stochastic process $Y$ is called (BL, $\theta$)-dependent if if there exists a non-increasing sequence $\theta = (\theta_r)_{r \in \mathbb{Z}}$ with $\theta_r \rightarrow 0$ as $r \rightarrow \infty$ and $$ \Big\vert \text{Cov} \Big( f\big(Y(I)\big), g\big(Y(J)\big) \Big) \Big\vert \leq \text{Lip}(f)\text{Lip}(g)(\vert I \vert \wedge \vert J \vert) \theta_r $$

for any bounded Lipschitz-continuous functions $f: \mathbb{R}^{\vert I \vert} \rightarrow \mathbb{R}$, $g: \mathbb{R}^{\vert J \vert} \rightarrow \mathbb{R}$ and finite disjoint subsets $I, J \subset T$ such that $\text{dist}(I, J) := \min\{ \vert i - j \vert : i \in I, j \in J \} = r$.

Question

Is $X$ (BL, $\theta$)-dependent? The finite-variance case can be proven as seen below. But the proof relies on covariances of the process and I don't know how to generalize the main inequality that was used in the proof.

Proof for finite-variance case

If $X$ had a finite-variance, then Theorem 5.3. in Bulinski & Shashkin (2007) states that $$ \Big\vert \text{Cov} \Big( f\big(X(I)\big), g\big(X(J)\big) \Big) \Big\vert \leq \sum_{i \in I} \sum_{j \in J} \text{Lip}_i(f)\text{Lip}_j(g) \text{Cov}(X(i), X(j)) $$ for all any bounded Lipschitz-continuous functions $f: \mathbb{R}^{\vert I \vert} \rightarrow \mathbb{R}$, $g: \mathbb{R}^{\vert J \vert} \rightarrow \mathbb{R}$ and finite disjoint subsets $I, J \subset T$. Hence, $X$ is (BL, $\theta$)-dependent with $$ \theta_r := \sup_{i \in I} \sum_{j \in \mathbb{Z} : \vert i - j \vert \geq r} \vert \text{Cov}(X(i), X(j)) \vert $$ under the assumption that these quantities exist and tend to zero.

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The answer is no. E.g., Let $X(t)=Z$ for all $t$, where $Z$ is any random variable with infinite variance. Then the process $(X(t)\colon t\in\mathbb Z)$ is positively associated. On the other hand, for $f_n(x):=\min(n,\max(-n,x))$ and each natural $j$ we have $$Cov(f_n(X_0),f_n(X_r))\to Var\,Z=\infty$$ as $n\to\infty$, whereas $$\text{Lip}(f_n)\text{Lip}(f_n)(|\{0\}|\wedge|\{r\}|)=1.$$

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    $\begingroup$ That's a very elegant counterexample. I should have thought of that and made my question more precise. What if $X$ is non-degenerate, e.g. alpha-stable? Is there any way to bound the covariance then? $\endgroup$
    – AlbertRapp
    Commented Oct 31, 2022 at 18:50
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    $\begingroup$ @AlbertRapp : You can approximate a degenerate process $X$ by non-degenerate ones however well, in terms of this setting. So, if we had your desired property for all non-degenerate processes, we would also have it for the degenerate ones, which is impossible, as shown in the answer. So, the non-degeneracy condition will not help. $\endgroup$ Commented Oct 31, 2022 at 19:00
  • $\begingroup$ I'm still a bit hung up about this. Wouldn't your solution imply that no infinite-variance process can be (BL, $\theta$)-dependent? This does not sound right. Where is my error? $\endgroup$
    – AlbertRapp
    Commented Nov 1, 2022 at 18:54
  • $\begingroup$ @AlbertRapp : I don't see such an implication. My answer only deals with "degenerate" processes -- constant in $t$. As for my previous comment, it only says that, if the desired property were true for all non-degenerate processes (with infinite variance), then it would hold for all processes with infinite variance -- which latter is false, by my answer. So, my comment only implies that the desired property is false for some non-degenerate process. $\endgroup$ Commented Nov 1, 2022 at 20:05
  • $\begingroup$ Oh I see. I understand your comment now. I'll create a follow-up question about associated alpha-stable moving averages. This is really the main process I'm interested in but I thought maybe the it could be answered in more generality. $\endgroup$
    – AlbertRapp
    Commented Nov 2, 2022 at 7:25

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