Noise and $\psi$-mixing condition.

Question

Let $(Z_t:\,t\in\mathbb{Z})$ be noise in the following sense: $\mathbb{E}(Z_t)=0$ for all $t$ and $\mathbb{E}(Z_t Z_s) = 0$ if $t\neq s$. Does this imply the $\psi$-mixing condition $$\lim \psi(k) = 0$$ where $$ \psi(k) = \sup \left\vert 1 - \frac{\mathbb{P}(A\cap B)}{\mathbb{P}(A)\mathbb{P}(B)}\right\vert$$ where the supremum is taken for $A \in \sigma(Z_t:\, t\leq 0)$ and $B\in \sigma(Z_t: t\geq k)$ with $\mathbb{P}(A),\mathbb{P}(B) > 0$?

Answer 1

Take $(e_t)_{t\in\mathbb Z}$ a sequence of independent random variables such that for each $t$, $\mathbb P\{e_t=1\}=\mathbb P\{e_t=-1\}=1/2$ and consider a non-negative random variable $X$ independent of $(e_t)_{t\in\mathbb Z}$. Then define $Z_t:=e_t\cdot X$. In this way, the sequence $\left(Z_t\right)_{t\in\mathbb Z}$ is a noise in the sense of the opening post.

Now, for a fixed $k$, define $A:=\{Z_0\gt R\}$ and $B:=\{Z_k\gt R\}$ where $R\gt 0$ will be determinate later. Since $X$ is non-negative, we have $$A=\{e_0=1\}\cap \{X\gt R\}\mbox{ and }B=\{e_k=1\}\cap \{X\gt R\}.$$ Consequently, by independence of $(e_0,e_k,X)$, we have $$\frac{\mathbb P\left(A\cap B\right)}{\mathbb P\left(A\right) \mathbb P\left(B\right)}=\frac{\mathbb P\{X\gt R\}}{\mathbb P\{X\gt R\}^2}$$ and it suffices to chose $R$ such that $\mathbb P\{X\gt R\}\in (0,1)$ (for example, take $X$ such that $\mathbb P\{X=1\}=\mathbb P\{X=3\}=1/2$ and $R=2$).