92 lines
3.7 KiB
TeX
92 lines
3.7 KiB
TeX
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% TODO \begin{goal}
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% TODO We want to drop our assumptions on finite mean or variance
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% TODO and say something about the behaviour of $ \sum_{n \ge 1} X_n$
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% TODO when the $X_n$ are independent.
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% TODO \end{goal}
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\begin{theorem}[Theorem 3, Kolmogorov's three-series theorem] % Theorem 3
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\label{thm3}
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Let $X_n$ be a family of independent random variables.
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\begin{enumerate}[(a)]
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\item Suppose for some $C \ge 0$, the following three series
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of numbers converge:
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\begin{itemize}
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\item $\sum_{n \ge 1} \bP(|X_n| > C)$,
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\item $\sum_{n \ge 1} \underbrace{\int_{|X_n| \le C} X_n d\bP}_{\text{\vocab{truncated mean}}}$,
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\item $\sum_{n \ge 1} \underbrace{\int_{|X_n| \le C} X_n^2 d\bP - \left( \int_{|X_n| \le C} X_n d\bP \right)^2}_{\text{\vocab{truncated variance} }}$.
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\end{itemize}
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Then $\sum_{n \ge 1} X_n$ converges almost surely.
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\item Suppose $\sum_{n \ge 1} X_n$ converges almost surely.
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Then all three series above converge for every $C > 0$.
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\end{enumerate}
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\end{theorem}
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For the proof we'll need a slight generalization of \autoref{thm2}:
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\begin{theorem}[Theorem 4] % Theorem 4
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\label{thm4}
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Let $\{X_n\}_n$ be independent and \vocab{uniformly bounded}
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(i.e. $\exists M < \infty : \sup_n \sup_\omega |X_n(\omega)| \le M$).
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Then $\sum_{n \ge 1} X_n$ converges almost surely
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$\iff$ $\sum_{n \ge 1} \bE(X_n)$ and $\sum_{n \ge 1} \Var(X_n)$
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converge.
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\end{theorem}
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\begin{refproof}{thm3}
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Assume, that we have already proved \autoref{thm4}.
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We prove part (a) first.
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Put $Y_n = X_n \cdot \One_{\{|X_n| \le C\}}$.
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Since the $X_n$ are independent, the $Y_n$ are independent as well.
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Furthermore, the $Y_n$ are uniformly bounded.
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By our assumption, the series
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$\sum_{n \ge 1} \int_{|X_n| \le C} X_n d\bP = \sum_{n \ge 1} \bE[Y_n]$
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and $\sum_{n \ge 1} \int_{|X_n| \le C} X_n^2 d\bP - \left( \int_{|X_n| \le C} X_n d\bP \right)^2 = \sum_{n \ge 1} \Var(Y_n)$
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converges.
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By \autoref{thm4} it follows that $\sum_{n \ge 1} Y_n < \infty$
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almost surely.
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Let $A_n \coloneqq \{\omega : |X_n(\omega)| > C\}$.
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Since the first series $\sum_{n \ge 1} \bP(A_n) < \infty$,
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by Borel-Cantelli, $\bP[\text{infinitely many $A_n$ occcur}] = 0$.
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For the proof of (b), suppose $\sum_{n\ge 1} X_n(\omega) < \infty$
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for almost every $\omega$.
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Fix an arbitrary $C > 0$.
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Define
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\[
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Y_n(\omega) \coloneqq \begin{cases}
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X_n(\omega) & \text{if} |X_n(\omega)| \le C,\\
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C &\text{if } |X_n(\omega)| > C.
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\end{cases}
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\]
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Then the $Y_n$ are independent and $\sum_{n \ge 1} Y_n(\omega) < \infty$
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almost surely and the $Y_n$ are uniformly bounded.
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By \autoref{thm4} $\sum_{n \ge 1} \bE[Y_n]$ and $\sum_{n \ge 1} \Var(Y_n)$
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converge.
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Define
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\[
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Z_n(\omega) \coloneqq \begin{cases}
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X_n(\omega) &\text{if } |X_n| \le C,\\
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-C &\text{if } |X_n| > C.
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\end{cases}
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\]
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Then the $Z_n$ are independent, uniformly bounded and $\sum_{n \ge 1} Z_n(\omega) < \infty$
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almost surely.
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By \autoref{thm4} we have
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$\sums_{n \ge 1} \bE(Z_n) < \infty$
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and $\sums_{n \ge 1} \Var(Z_n) < \infty$.
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We have
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\[
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\bE(Y_n) &=& \int_{|X_n| \le C} X_n d \bP + C \bP(|X_n| \ge C)\\
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\bE(Z_n) &=& \int_{|X_n| \le C} X_n d \bP - C \bP(|X_n| \ge C)\\
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\]
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Since $\bE(Y_n) + \bE(Z_n) = 2 \int_{|X_n| \le C} X_n d\bP$
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the second series converges,
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and since
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$\bE(Y_n) - \bE(Z_n)$ converges, the first series converges.
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For the third series, we look at
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$\sum_{n \ge 1} \Var(Y_n)$ and
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$\sum_{n \ge 1} \Var(Z_n)$ to conclude that this series converges
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as well.
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\end{refproof}
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