# Does pairwise independence and same distribution imply trivial Invariant $sigma$-algebra?

Mathematics Asked by Davi Barreira on November 14, 2020

I know that by Kolmogorov’s $$0-1$$ Law, that for independent r.v, the tail $$sigma$$-algebra is trivial (e.g all events have probability either $$0$$ or $$1$$). This coupled with the ergodic theorem, one can easily derive the Strong Law of Large Numbers for $$X_i$$ i.i.d. and finite expected value.

I also know that there exists a stronger SLLN called Etemadi’s SLLN, which only requires finite expected value, and that $$X_i$$ have the same distribution and are pairwise independent.

With this in mind, I was wondering if pairwise independence and same distribution imply trivial Invariant $$sigma$$-algebra? If it does, can anyone provide a proof or a reference to such proof? And if no, can one provide a counter-example?

## One Answer

After some research, I found that unfortunately the answer is no. You can construct a pairwise independent sequence of random variables, such that the $$sigma$$-algebra is not trivial.

The processes is shown in this paper by Robertson and Womack (1985). They construct a stochastic process such that $$P(X_n=1)=P(X_n=-1)=1/2$$, and in a way that the sequence is pairwise independent, but in the very end of the paper, they prove that this stochastic process does not satisfy the 0-1 law.

Correct answer by Davi Barreira on November 14, 2020

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