Do we have independence if we let the indices of the events increase?

Question

Let $(\Omega, \mathscr F, \mathbb P)$ be a probability space.

Consider events indexed by $m, n \in \mathbb N$:

$ \ \ \ \ \ \ \ \ \ \ \ A_{1,n}, A_{2,n}, A_{3,n} ...$ are n-wise independent.

$A_{m,1}: A_{1,1}, A_{2,1}, A_{3,1}, ...$ are 1-wise independent.

$A_{m,2}: A_{1,2}, A_{2,2}, A_{3,2}, ...$ are 2-wise independent.

$A_{m,3}: A_{1,3}, A_{2,3}, A_{3,3}, ...$ are 3-wise independent.

$\vdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ddots$

Now consider $A_1, A_2, ...$ s.t.

$$A_i \in \tau_{A_{i,n}} := \bigcap_{n=1}^{\infty} \sigma(A_{i,n}, A_{i,n+1}, ...)$$

Are $A_1, A_2, ...$ independent?

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Well one sufficient condition:

If $A_{1,n}, A_{1,n+1}, ...$ are independent, $A_{2,n}, A_{2,n+1}, ...$ are independent, ..., then, each $A_i$ has a probability of 0 or 1 by Kolmogorov 0-1 Law

Answer 1

Here is a counterexample. Let the $0-1$ random variables $X_n$ follow a Pólya urn model: Choose $U$ uniformly on $[0,1]$ and then let the $X_i$ be independent Bernoulli trials with success rate $U$. $P(X_n=1) = 1/2$.

Flip a fair coin $Y$. If $Y=\textrm{heads}$ let $A_{1,n}=A_{2,n} = \{X_n = 1\}$. If $Y=\textrm{tails}$ then let $A_{1,n}=A_{2,n}^c = \{X_n = 1\}$. $A_{1,n}$ and $A_{2,n}$ are independent.

The limiting frequencies of $A_{1,n}$ and of $A_{2,n}$ are measurable in the tails, but they are not independent variables since they are either equal or complementary. In particular, let $A_i$ be the event that the limiting frequency of the $A_{i,n}$ events is in $[0,1/10]$ for $i=1,2$. Then $P(A_1)=P(A_2)=1/10$ and $P(A_1 \cap A_2) = 1/20 \ne P(A_1)P(A_2)$.