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I'm learning Kolmogorov's zero-one law in probability theory:

Let $(Ω,{\mathcal F},P)$ be a probability space and let $F_n$ be a sequence of mutually independent $\sigma$-algebras contained in $\mathcal{F}$. Let $$G_n=\sigma\bigg(\bigcup_{k=n}^\infty F_k\bigg)$$ be the smallest $\sigma$-algebra containing $F_n, F_{n+1}, \dots$. Then Kolmogorov's zero-one law asserts that for any event $$ F\in \bigcap_{n=1}^\infty G_n$$ one has either $P(F) = 0$ or $1$.

I've no idea how $G_n$ and $\bigcap_{n=1}^{\infty} G_n$ would look like. What's the point of such construction? Could any one come up with some concrete examples of how this theorem works?

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  • $\begingroup$ Small question, how do we know that $\sigma(\cup_{k=n}^\infty F_k)$ is "the smallest $\sigma$-algebra containing $F_n,F_{n+1},...$"? For example, $\sigma(A\cup B)$ is not the smallest $\sigma$ algebra containing $A,B$, is it? $\endgroup$ – BlueBuck Oct 4 '15 at 16:23
  • $\begingroup$ @BlueBuck: en.wikipedia.org/wiki/… $\endgroup$ – Jack Oct 5 '15 at 0:09
  • $\begingroup$ Sorry but I don't see how that article answers my question. I understand what the $\sigma$-algebra generated by a family (such as $F_n,F_{n+1},...$) means, I just don't understand why the $\sigma$-algebra generated by the union of sets in that family gives the $\sigma$-algebra generated by the family. $\endgroup$ – BlueBuck Oct 5 '15 at 18:21
  • $\begingroup$ @BlueBuck I think: 1 The smallest $\sigma$-algebra containing $A,B$ is $\sigma(A,B) := \sigma(\sigma(A) \cup \sigma(B))$, if $A$ and $B$ are events. 2 $\sigma(A,B) = \sigma(A \cup B)$ if $A$ and $B$ are $\sigma$-algebras $\endgroup$ – BCLC Jan 24 '16 at 13:18
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The usual name for the sigma algebra you have there is the "tail" $\sigma-$algebra. It is all the stuff that does not depend on finitely many of the $\sigma-$ algebras.

I usually think about the Kolmogorov 0-1 law like this: if you have a sequence of independent random variables and an event that is invariant if you ignore finitely many of the variables, then the probability of that event is either 0 or 1.

A typical example of how you can actually use this is to show that the convergence in the classical central limit theorem cannot be almost sure.

Let $X_i$ be a sequence of $iid$ random variables with $EX_i = 0$ and $EX_i^2 = 1$ and let $$S_n = \sum_{i=1}^n X_i$$

It is easy to see that for any fixed $n$ and $M>0$ the event $$\{ \limsup_k \frac{S_k}{\sqrt{k}} >M\}$$ lies in $$\sigma(\cup_{k=n}^\infty F_k)$$ since it only depends on the "tail" of our sequence (this is clear if you write out the definition of $\limsup$). Therefore by independence, the probability of this event is either 0 or 1. Using the central limit theorem, it is not hard to see that this is probability is positive for any fixed $M$ and therefore it is $1$. We then obtain that $$P(\limsup \frac{S_k}{\sqrt{k}} = \infty) = 1$$

Symmetry gives that $$P(\liminf \frac{S_k}{\sqrt{k}} = - \infty) = 1$$ so we cannot have almost sure convergence (or convergence in probability).

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  • $\begingroup$ can you explain how you got that probability is positive using Central limit theorem? I'm new to this topic. $\endgroup$ – Sahiba Arora Nov 4 '16 at 19:40
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    $\begingroup$ You wrote: "It is easy to see that for any fixed $n$ and $M>0$ the event $$\{ \limsup_k \frac{S_k}{\sqrt{k}} >M\}$$ lies in $$\sigma(\cup_{k=n}^\infty F_k)$$ since it only depends on the 'tail' of our sequence (this is clear if you write out the definition of lim sup)". As is customary when the phrase "It is easy to see that..." is used in mathematical writing, the following statement is either false or so difficult that the author was unable to actually provide a proof (likewise "it is clear that..."). I, for one, do not see why the statement should be true. Would you mind elaborating? $\endgroup$ – Evan Aad Nov 8 '16 at 11:16
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    $\begingroup$ @SahibaArora For each $\omega$, $1_{\{\limsup \frac{S_k}{\sqrt{k}} > M\}}(\omega) = \limsup_{k \to \infty} 1_{\{\frac{S_k}{\sqrt{k}} > M\}} (\omega)$. We have $ 0 \leq 1_{\{\frac{S_k}{\sqrt{k}} > M\}} \leq 1$ and the function $1$ is integrable. Apply the version of Fatou's lemma for limsup and note $P(\frac{S_k}{\sqrt{k}} > M)$ converges to a positive number. $\endgroup$ – Chris Janjigian Nov 8 '16 at 13:09
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    $\begingroup$ @EvanAad I don't check this often, so I deleted my previous comment and just wrote out the details. Sorry if you got two notifications for this. Note that for fixed $n$, $\lim_{k\to\infty} \frac{S_n}{\sqrt{k}} = 0$. You can check that if you have two sequences $a_k,b_k$ and $a_k \to 0$, then $\limsup_{k \to \infty} (a_k + b_k) = \limsup b_k$ ($\leq$ is free, then take a subsequence along which $b_{k_j} \to \limsup_k b_k$ for the reverse inequality). It follows that pointwise, $\limsup_k \sqrt{k}^{-1} \sum_{j=1}^{k} X_j = \limsup_{k\to\infty} \sqrt{k}^{-1} \sum_{j=n+1}^k X_j$. $\endgroup$ – Chris Janjigian Nov 8 '16 at 13:31
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    $\begingroup$ $F_k = \sigma(X_k)$ here. It then suffices to show that $\limsup \sqrt{k}^{-1} \sum_{j=n+1}^k X_j$ is $\sigma(\cup_{j=n}^\infty F_j)$ measurable. Note that $\sqrt{k}^{-1} \sum_{j=n+1}^k X_j$ is $\sigma(\cup_{j=n}^\infty F_j)$ measurable for each $k\geq n+1$. Then $\sup_{i \geq k} \sqrt{i}^{-1} \sum_{j=n+1}^i X_j$ is measurable for each $k\geq n+1$ and thus $\lim_{k \to \infty} \sup_{i \geq k} \sqrt{i}^{-1} \sum_{j=n+1}^i X_j = \limsup_k \sqrt{k}^{-1} \sum_{j=n+1}^k X_j$ is measurable. The last step is $\limsup f_k(\omega) = \lim_{k \to \infty} \sup_{i \geq k}f_i(\omega)$. $\endgroup$ – Chris Janjigian Nov 8 '16 at 13:58
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Even though the tail-$\sigma$-algebra is a sub-$\sigma$-algebra of $\mathcal{F}$, it can be in some sense much more complex. In particular, the tail-$\sigma$-algebra may not countably generated even in the most well-behaved cases. This implies that there cannot be any (real-valued) random variable representing the informational content of the tail-$\sigma$-algebra in general.

An atom in a $\sigma$-algebra is a nonempty measurable set with no nonempty measurable proper subset. A measurable space is atomic if every point is contained in some atom (the atoms form a cover).

Lemma: If $(\Omega,\mathcal{F})$ is countably generated, then it is atomic. If there is a $0-1$-valued measure $\mu$ on $(\Omega,\mathcal{F})$, then there exists an atom $A$ such that $\mu(A)=1$.

Proof (Sketch): Let $\mathcal{C}$ be a countable family such that $\sigma(\mathcal{C})=\mathcal{F}$. We can assume without loss of generality that $\mathcal{C}$ is closed under complements. Then for each $\omega\in\Omega$, the set $A(\omega)=\bigcap\{A\ni\omega:A\in\mathcal{C}\}$ is measurable and the atom containing $\omega$. Now if $\mu$ is $0-1$-valued, the for each $A\in\mathcal{C}$ either $\mu (A)=1$ or $\mu (A^C)=1$. The intersection of all measure one events in $\mathcal{C}$ is an atom with measure one.

Corollary: The tail-$\sigma$-algebra on $\Omega=\{0,1\}^\infty$ is not countably generated.

Proof: Endowe the space with the fair coin-flipping measure. Let $\omega=\{\omega_1,\omega_2,\omega_3,\ldots\}$ be any element in $\Omega$. Let $A_\omega^n$ be the set of all sequences in $\Omega$ that coincide with $\omega$ in all coordinates except for, maybe, the first $n$ coordinates. Clearly, $A_\omega^n$ is countable and so is $B_\omega=\bigcup_n A_\omega^n$. So $\mu (B_\omega)=0$. Also, $B_\omega$ is in the tail $\sigma$-algebra. But since $\omega$ was arbitrary, this shows that every atom would have measure zero. By the Lemma, the tail-$\sigma$-algebra is not countably generated.

The argument is taken from Borel Spaces by Rao and Rao (1981).

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    $\begingroup$ May I know how you define "atomic" here? Thanks $\endgroup$ – Theta33 Feb 8 '13 at 7:46
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    $\begingroup$ I refer to "atomic" as used in Lemma. $\endgroup$ – Theta33 Feb 8 '13 at 8:22
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    $\begingroup$ @Theta30 Oh, sorry- Iedited it now. Atomic means that every point lies in some atom. $\endgroup$ – Michael Greinecker Feb 8 '13 at 8:26
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    $\begingroup$ @EvanAad Let $\mu$ be the involved measure, which is $0-1$-valued on the tail-$\sigma$-algebra. Suppose for the sake of contradiction that the tail-$\sigma$-algebra is countably generated. By the Lemma,there exists an atom $A$ in the $\sigma$-algebra such that $\mu(A)=1$. Pick some $\omega\in A$. $\endgroup$ – Michael Greinecker Nov 8 '16 at 19:12
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    $\begingroup$ We know there is a set $B_\omega$ that contains $\omega$, is measurable in the tail-$\sigma$-algebra, and has $\mu$-measure $0$. A subset of a measure zero set has measure zero, so $B_\omega\cap A\neq A$. Therefore, $A\cap B_\omega$ is a proper subset of $A$ measurable in the tail-$\sigma$-algebra. It is also clearly nonempty, since $\omega\in A\cap B_\omega$. This contradicts $A$ being an atom. $\endgroup$ – Michael Greinecker Nov 8 '16 at 19:12

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