151 lines
6.4 KiB
TeX
151 lines
6.4 KiB
TeX
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% Lecture 8 2023-05-02
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\subsection{Kolmogorov's 0-1-law}
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Some classes of events always have probability $0$ or $1$.
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One example of such a 0-1-law is the Borel-Cantelli Lemma
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and its inverse statement.
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We now want to look at events that capture certain aspects of long term behaviour
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of sequences of random variables.
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\begin{definition}
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Let $X_n, n \in \N$ be a sequence of random variables
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on a probability space $(\Omega, \cF, \bP)$.
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Let $\cT_i \coloneqq \sigma(X_i, X_{i+1}, \ldots )$
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be the $\sigma$-algebra generated by $X_i, X_{i+1}, \ldots$.
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Then the \vocab{tail-$\sigma$-algebra} is defined as
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\[
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\cT \coloneqq \bigcap_{i \in \N} \cT_i.
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\]
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The events $A \in \cT \subseteq \cF$ are called \vocab[Tail event]{tail events}.
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\end{definition}
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\begin{remark}
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\begin{enumerate}[(i)]
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\item Since intersections of arbitrarily many $\sigma$-algebras
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is again a $\sigma$-algebra, $\cT$ is indeed a $\sigma$-algebra.
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\item We have
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\[
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\cT = \{A \in \cF ~|~ \forall i \exists B \in \cF^{\otimes \N} : A = \{\omega | (X_i(\omega), X_{i+1}(\omega), \ldots) \in B\} \}. % TODO?
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\]
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\end{enumerate}
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\end{remark}
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\begin{example}[What are tail events?]
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Let $X_n, n \in \N$ be a sequence of independent random variables on a probability
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space $(\Omega, \cF, \bP)$. Then
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\begin{enumerate}[(i)]
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\item $\left\{\omega | \sum_{n \in \N} X_n(\omega) \text{ converges} \right\}$ is a tail event,
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since for all $\omega \in \Omega$ we have
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\begin{IEEEeqnarray*}{rCl}
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&& \sum_{i=1}^\infty X_i(\omega) \text{ converges}\\
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&\iff& \sum_{i=2}^\infty X_i(\omega) \text{ converges}\\
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&\iff& \ldots \\
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&\iff& \sum_{i=k}^\infty X_i(\omega) \text{ converges}.\\
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\end{IEEEeqnarray*}
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(Since the $X_i$ are independent, the convergence
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of $\sum_{n \in \N} X_n$ is not influenced by $X_1,\ldots, X_k$
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for any $k$.)
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\item $\left\{\omega | \sum_{n \in \N} X_n(\omega) = c\right\} $
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for some $c \in \R$
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is not a tail event,
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because $\sum_{n \in \N} X_n$ depends on $X_1$.
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\item $\{\omega | \lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^{n} X_i(\omega) = c\}$
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is a tail event, since
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\[
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c = \lim_{n \to \infty} \sum_{i=1}^{n} X_i = \underbrace{\lim_{n \to \infty} \frac{1}{n} X_1}_{= 0} + \lim_{n \to \infty} \frac{1}{n} \sum_{i=2}^n X_i = \ldots = \lim_{n \to \infty} \frac{1}{n} \sum_{i=k}^n X_i.
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\]
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\end{enumerate}
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\end{example}
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So $\cT$ includes all long term behaviour of $X_n, n \in \N$,
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which does not depend on the realisation of the first $k$ random variables
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for any $k \in \N$.
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\begin{theorem}[Kolmogorov's 0-1 law]
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\label{kolmogorov01}
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Let $X_n, n \in \N$ be a sequence of independent random variables
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and let $\cT$ denote their tail-$\sigma$-algebra.
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Then $\cT$ is \vocab{$\bP$-trivial}, i.e.~$\bP[A] \in \{0,1\}$
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for all $A \in \cT$.
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\end{theorem}
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\begin{idea}
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The idea behind proving, that a $\cT$ is $\bP$-trivial is to show that
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for any $A, B \in \cF$ we have
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\[
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\bP[A \cap B] = \bP[A] \cdot \bP[B].
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\]
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Taking $A = B$, it follows that $\bP[A] = \bP[A]^2$, hence $\bP[A] \in \{0,1\}$.
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\end{idea}
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\begin{refproof}{kolmogorov01}
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Let $\cF_n \coloneqq \sigma(X_1,\ldots,X_n)$
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and remember that $\cT_{n} = \sigma(X_{n}, X_{n+1},\ldots)$.
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The proof rests on two claims:
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\begin{claim}
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For all $n \ge 1$, $A \in \cF_n$ and $B \in \cT_{n+1}$
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we have $\bP[A \cap B] = \bP[A]\bP[B]$.
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\end{claim}
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\begin{subproof}
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This follows from the independence of the $X_i$.
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It is
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\[
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\sigma\left( X_1,\ldots,X_n \right) = \sigma\left(\underbrace{\{X_{1}^{-1}(B_1) \cap \ldots \cap X_n^{-1}(B_n)\} | B_1,\ldots,B_n \in \cB(\R)\}}_{\text{\reflectbox{$\coloneqq$}}\cA} \right).
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\]
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$\cA$ is a semi-algebra, since
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\begin{enumerate}[(i)]
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\item $\emptyset, \Omega \in \cA$,
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\item $A, B \in \cA \implies A \cap B \in \cA$,
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\item for $A \in \cA$, $A^c = \bigsqcup_{i=1}^n A_i$
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for disjoint sets $A_1,\ldots,A_n \in \cA$.
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\end{enumerate}
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Hence it suffices to show the claim for sets $A \in \cA$.
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Similarly
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\[
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\sigma(\cT_{n+1}) = \sigma \left( \underbrace{ \{X_{n+1}^{-1}(M_1) \cap \ldots \cap X_{n+k}^{-1}(M_k) | k \in \N, M_1,\ldots, M_k \in \cB(\R)\}}_{\text{\reflectbox{$\coloneqq$}} \cB} \right).
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\]
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Again, $\cB$ is closed under intersection.
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So let $A \in \cA$ and $B \in \cB$.
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Then
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\[
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\bP[A \cap B] = \bP[A] \cdot \bP[B)
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\]
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by the independence of $\{X_1,\ldots,X_{n+k}\}$,
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and since $A$ only depends on $\{X_1,\ldots,X_n\}$
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and $B$ only on $\{X_{n+1},\ldots, X_{n+k}\}$.
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\end{subproof}
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\begin{claim}
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$\bigcup_{n \in \N} \cF_n$ is an algebra
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and
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\[
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\sigma\left( \bigcup_{n \in \N} \cF_n \right) = \sigma(X_1,X_2,\ldots) = \cT_1.
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\]
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\end{claim}
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\begin{subproof}
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``$\supseteq$ '' If $A_n \in \sigma(X_n)$, then $A_n \in \cF_n$.
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Hence $A_n \in \bigcup_{n \in \N} \cF_n$.
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Since $\sigma(X_1,X_2,\ldots)$ is generated by $\{A_n \in \sigma(X_n) : n \in \N\}$,
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this also means $\sigma(X_1,X_2,\ldots) \subseteq\sigma\left( \bigcup_{n \in \N} \cF_n \right)$.
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``$\subseteq$ '' Since $\cF_n = \sigma(X_1,\ldots,X_n)$,
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obviously $\cF_n \subseteq \sigma(X_1,\ldots,X_n)$
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for all $n$.
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It follows that $\bigcup_{n \in \N} \cF_n \subseteq \sigma(X_1,X_2,\ldots)$.
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Hence $\sigma\left( \bigcup_{n \in \N} \cF_n \right) \subseteq\sigma(X_1,X_2,\ldots)$.
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\end{subproof}
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Now let $T \in \cT$.
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Then $T \in \cT_{n+1}$ for any $n$.
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Hence $\bP[A \cap T] = \bP[A] \bP[T]$
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for all $A \in \cF_n$ by the first claim.
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It follows that the same folds for all $A \in \bigcup_{n \in \N} \cF_n$,
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hence for all $A \in \sigma\left( \bigcup_{n \in \N} \cF_n \right)$,
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and by the second claim for all $A \in \sigma(X_1,X_2,\ldots) = \cT_1$.
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But since $T \in \cT$, in particular $T \in \cT_1$,
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so by choosing $A = T$, we get
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\[
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\bP[T] = \bP[T \cap T] = \bP[T]^2
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\]
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hence $\bP[T] \in \{0,1\}$.
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\end{refproof}
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