Definition. Let $\{X_t, t>0\}$ be a sequence of real random variables on $(\Omega,\mathcal A,P)$ and denote by $\mathcal F_i^k=\sigma(X_t, i\leq t\leq k)$ the sigma-algebra generated. For any positive $n$, set $$\alpha(n)=\sup\left\{\lvert P(A\cap B)-P(A)P(B)\rvert: B\in\mathcal{F}_{1}^{k},A\in\mathcal{F}_{k+n}^\infty, k\in\mathbb N\right\}.$$ The sequence is said to be strongly mixing if $\alpha(n)\to 0$ as $n\to\infty$.
Consider the sets $A=\{\omega:a\leq X_i(\omega)\leq b\}$ and $B=\{\omega:c\leq X_j(\omega)\leq d\}$, where $a,b,c,d\in\mathbb R$. Suppose the sequence $\{X_t\}$ is strongly mixing.
- Does $A\in\sigma(X_i)$ and $B\in\sigma(X_j)$?
If the answer is yes, then I know that $$\alpha(\lvert i-j \rvert)\geq \lvert P(A\cap B)-P(A)P(B)\rvert \quad (Eq.1).$$ as the mixing coefficient is the supremum of all distances between the probabilities. The above result leads me to the second question:
- Suppose that $(Eq.1)$ holds. Can I conclude that $P(A\cap B)\leq P(A)P(B)+\alpha(\lvert i-j \rvert)$?
It is more about real analysis. I don't know if $P(A\cap B)$ is bigger or smaller than $P(A)P(B)$, but as $P(A)P(B)$ is inside a closed ball of center $P(A\cap B)$ and radius $\alpha(\lvert i-j \rvert)$, certainly $P(A)P(B)+\alpha(\lvert i-j \rvert)$ exceeds $P(A\cap B)$.
Can you help me to clarify these two questions? I'm trying to find strategies to lower bound a random variable, and this idea came to my mind. The result of question 2 seems straighforward but is unusual as far as |I know. So I'm feeling insecure to apply it.
As pointed out by TheBridge, $A$ belongs to $\sigma(X_i)$ because $A= X_i^{-1}([a,b])$ and by the same argument, $B\in \sigma(X_j)$. For your second question, $$ \alpha(\lvert i-j \rvert)\geqslant \lvert P(A\cap B)-P(A)P(B)\rvert \geqslant P(A\cap B)-P(A)P(B) $$ and the wanted inequality follows immediately.