Is the limsup or liminf of n-wise independent events independent?

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.

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

Do we eventually have mutual independence if we let n increase?

$\lim_n A_{m,n}$ does not necessarily exist, but we can define:

$$CI_m := \liminf_n A_{m,n}$$

$$CS_m := \limsup_n A_{m,n}$$

1. Are $CI_1, CI_2, ...$ independent?

How about some subsequence $CI_{m^{*}}, CI_{m^{*}+1}, ...$ or $CI_{f(1)}, CI_{f(2)}$ for $f(m): \mathbb N \to \mathbb N$?

2. Are $CS_1, CS_2, ...$ independent?

How about some subsequence $CS_{m^{*}}, CS_{m^{*}+1}, ...$ or $CS_{f(1)}, CS_{f(2)}$ for $f(m): \mathbb N \to \mathbb N$?

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Column sums:

$\forall m \in \mathbb N, P(A_{m,1}) + P(A_{m,2}) + ... \le \infty$.

If we can find an $m^{*} \in \mathbb N$ s.t. $A_{m^{*},1}, A_{m^{*},2}, ...$ are at least pairwise independent, then by the Borel 0-1 Law, we have that $P(\limsup_n A_{m,n}) = 0$ or $1$ for $m \ge m^{*}$.

Hence, $CS_{m^{*}}, CS_{m^{*} + 1}, CS_{m^{*} + 2}, ...$ are independent.

If $m^{*} = 1$, then $CS_1, CS_2, ...$ are independent.

Is that right? How can we prove or disprove the existence of such a $m^{*}$?

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Row sums:

Not sure if this helps, but meanwhile it seems that $\forall n \ge 2, P(A_{1,n}) + P(A_{2,n}) + ... \le \infty$ and the $A_{1,n}, A_{2,n}, ...$'s are at least pairwise independent.

If so, then by the Borel 0-1 Law, we have that $P(\limsup_m A_{m,n}) = 0$ or $1$ for $n \ge 2$.

Hence, we can define

$$RI_n := \liminf_m A_{m,n}$$

$$RS_n := \limsup_m A_{m,n}$$

$RS_2, RS_3, RS_4, ...$ are independent.

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Possibly related: Stochastic matrix

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Answer 1

Here is a counterexample. Since this question is more restrictive than the sequel, it is a counterexample to that, too.

Let $\{1,2,3,4\}$ have equal probability.

Let $1 \in A_{1,n},A_{2,n}$.

Let $2 \in A_{1,2k},A_{2,2k+1}$.

Let $3 \in A_{1,2k+1},A_{2,2k}$.

Then $A_{1,n}$ is independent of $A_{2,n}.$ Each has probability $1/2$ and the intersection has probability $1/4$.

In the tail $\sigma$-algebra, $CI_1 =CI_2 = \{1\}$, and $CS_1=CS_2 = \{1,2,3\}$, and these are not independent.