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According to Kolmogorov 0-1 law, if $ \left(X_{i}\right)_{i=1}^{\infty} $ are independent, then the tail sigma algebra is trivial. I want to construct such variables which are "almost independent": k wise independent, such that the tail sigma algebra is non-trivial.

We can look at $ X_{i}\sim U\left( \left\{ 0,1\right\} \right) $ where we set $ X_{a\left(k+1\right)}=\bigoplus_{i=1}^{k}X_{a\left(k+1\right)-i} $ in order to construct k-wise independent variables, but I can't find an event in their tail sigma algebra which is not trivial. Maybe this is not a good (enough) example of k-wise independent variables, but I can't think of a better one.

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2 Answers 2

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Let $(Y_1,\dots, Y_k)$ be a random $k$-tuple uniformly distributed in $(\mathbb R/\mathbb Z)^k$.

Let $X_i = \sum_{j=1}^k i^j Y_j \in \mathbb R/\mathbb Z$.

Then the $X_i$ are $k$-wise independent (and uniformly distributed on $\mathbb R/\mathbb Z$), since the linear map sending $(Y_1,\dots, Y_k)$ to $(X_{i_1},\dots, X_{i_k})$ is given by a $k\times k$ integer matrix with nonzero (Vandermonde) determinant, and such maps preserve the uniform measure on the torus $(\mathbb R/\mathbb Z)^k$.

(One way to check this is by Weyl's criterion, that $X_{i_1},\dots, X_{i_k}$ are uniformly distributed if and only if the expectation of $e^{ 2 \pi i \sum_{j=1}^k X_{i_j} n_j}$ is $0$ for all tuples $n_1,\dots, n_k$ of integers other than $0,\dots, 0$. This follows from uniform distribution of the $Y_1,\dots, Y_k$ because $\sum_{j=1}^k X_{i_j} n_j = \sum_{\ell=1}^k (\sum_{j=1}^k n_j i_j^\ell) Y_\ell $ and $(\sum_{j=1}^k n_j i_j^\ell)$ are integers not all $0$ by the nonvanishing of the Vandermonde. Another way is to note that each possible value of $X_1,\dots, X_k$ has a number of preimages equal to the absolute value of the determinant which cancels the inverse factor of the determinant of the derivative, which is the inverse of the Vandermonde determinant, from the change-of-variables formula in integration.)

But we can write $ k! Y_k = \sum_{j=0}^{k} (-1)^j \binom{k}{j} X_{i-j}$ for any $i$ so $k! Y_k$ is a nonconstant function in the tail sigma-algebra.

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    $\begingroup$ This is a very nice construction. I have only taken the liberty to make a couple of edits in your answer. Can you detail why "such maps preserve the uniform measure on the torus"? $\endgroup$ Commented Aug 31, 2022 at 20:14
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    $\begingroup$ @PetriKattilakoski you are confusing quantifiers. conditioned on the outcome of $X_1,\dots$, $Y_k$ is determined by the tail sigma algebra. However, any event determined by the tail sigma algebra is constant when you condition on $X_1,\dots$. However, as Will points out, the event that $Y_k \in [0,1/2]$ is determined by the sigma algebra but occurs with probability $1/2$ (implying non-triviality). $\endgroup$ Commented Aug 31, 2022 at 20:20
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    $\begingroup$ @IosifPinelis Sure, added two proofs, one more elementary but sketchier. $\endgroup$
    – Will Sawin
    Commented Aug 31, 2022 at 20:21
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    $\begingroup$ Thank you for your response. Do you have a reference concerning "a number of preimages equal to the absolute value of the determinant"? This is probably something simple, and yet, it would be good to see related matters in such a reference. $\endgroup$ Commented Aug 31, 2022 at 20:29
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    $\begingroup$ @PetriKattilakoski (1) One just plugs in and simplifies. (2) I'm using a "modulo" symbol, not a "set minus" symbol. $\endgroup$
    – Will Sawin
    Commented Aug 31, 2022 at 22:00
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There is no such counterexample. Indeed, by Theorem 1, any strong mixing sequence obeys the $0$-$1$ law. Also, any $m$-dependent sequence is obviously strong mixing, for any integer $m\ge0$, and hence must obey the $0$-$1$ law.

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  • $\begingroup$ do you mean $k$-independent and any natural $k>1$? $\endgroup$ Commented Aug 31, 2022 at 19:26
  • $\begingroup$ I am not quite sure I understand the concept of strong mixing, so take this comment with a grain of salt. While my example is indeed a strong mixing, I can think of a sequence of random variables such that ANY (K+1) of them are dependent, for example using linear equations with k variables, am I wrong? $\endgroup$ Commented Aug 31, 2022 at 19:32
  • $\begingroup$ @ZachHunter : I do not know what $k$-independent would mean. I have added a link to the definition of $m$-dependence. $\endgroup$ Commented Aug 31, 2022 at 19:32
  • $\begingroup$ Naturally speaking, k-wise independence means that given a variable $X_i$, knowing the values of any (other) k-1 variables, does not give more information about $X_i$. Formally, it means that (for discrete variables) $$ \prod_{i=1}^{k}\mathbb{P}\left(X_{n_{i}}=a_{i}\right)=\mathbb{P}\left(\bigwedge_{i=1}^{k}X_{n_{i}}=a_{i}\right) $$ $\endgroup$ Commented Aug 31, 2022 at 19:39
  • $\begingroup$ @PetriKattilakoski : The definition of strong mixing is given in the paper linked in my answer. As for your example, I cannot discern any definition of a sequence of random variables there. If you meant $Y_n:=\bigoplus_{i=1}^k X_{n-i}$, where the $X_i$'s are independent $U(\{0,1\})$, then the sequence $(Y_n)$ is $m$-dependent for $m=k-1$ and therefore must obey the 0-1 law. $\endgroup$ Commented Aug 31, 2022 at 19:45

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