2
$\begingroup$

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?


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

$\endgroup$
2
  • 1
    $\begingroup$ Is this a homework question? If not, maybe you should tell us where it comes from. $\endgroup$ Jan 24, 2016 at 3:04
  • $\begingroup$ @AnthonyQuas It is based on my other question... $\endgroup$
    – BCLC
    Jan 24, 2016 at 5:57

1 Answer 1

1
$\begingroup$

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 $P(A_1 \cap A_2) = 1/20 \ne P(A_1)P(A_2)$.

$\endgroup$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.