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The Kolmogorov zero-one law says that for a sequence of independent events, any event belonging to the tail $\sigma$-field has probability either $0$ or $1$. However the converse is not true, because you can find dependent sequence of events, for which the tail events have probability zero or one. Two such interesting examples i found are:

  • Consider a probability space $(\Omega,\mathcal{A},P)$Let $E_1,E_2,\cdots$ be a sequence of events defined as $$E_1=E_2=A,~~E_3=E_4=\cdots=\phi$$ where $\phi$ denotes the empty set, and $A\in\mathcal{A}$.
  • Consider a prob space $(\Omega,\mathcal{A},P)$, and take any independent sequence of events, $\{B_n\}$. Then construct an interleaving sequence of events, $\{E_n\}$ as $$E_1=B_1,E_2=B_2,E_3=B_1\cup B_2,E_4=B_3,E_5=B_2\cup B_3,E_6=B_4,E_7=B_3\cup B_4,\cdots$$ Then $\{E_n\}$ is a sequence of dependent events and the tail $\sigma$-field generated by $\{E_n\}$ is same as that generated by $\{B_n\}$ and hence every tail event has probability either $0$ or $1$.

Seeing these examples I guess that if every event in the tail $\sigma$-field of a sequence of events has probability either $0$ or $1$, then there must exist a sequence of independent events from the same probability space which has the same tail $\sigma$-field.

Is this true, or there is some very obvious counterexample?

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    $\begingroup$ If $A$ is the whole space, the events in your first example are independent. And this means that your conjecture is true, but rather trivial. $\endgroup$ – Robert Israel Dec 13 '13 at 10:08
  • $\begingroup$ How is that? $E_1$ and $E_2$ being both $A$ can never be independent! Moreover, I cannot decide whther a conjecture is true depending on just one trivial example. Either their is a counterexample to the fact or there must be a rigorous proof. $\endgroup$ – Abishanka Saha Dec 13 '13 at 10:12
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    $\begingroup$ If $P(E_1) = 1$ then $P(E_1 \cap E_2) = P(E_2) = P(E_1) P(E_2)$ so $E_1$ and $E_2$ are independent. $\endgroup$ – Robert Israel Dec 13 '13 at 10:32
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    $\begingroup$ @RobertIsrael But how is that the same tail $\sigma-$field? $\endgroup$ – BCLC Jul 9 '16 at 2:15
  • $\begingroup$ Ah, I see... If the tail $\sigma$-field is countably generated, you can just take a sequence of events $A_i$ that generate it and then repeat each infinitely often (e.g. $A_1, A_1, A_2, A_1,A_2, A_3, \ldots$). But we don't know that the tail $\sigma$-field is countably generated: a countably generated $\sigma$-field can have a sub-$\sigma$-field that is not countably generated. $\endgroup$ – Robert Israel Jul 10 '16 at 20:40

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