183 lines
7.6 KiB
TeX
183 lines
7.6 KiB
TeX
\documentclass{article}
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\usepackage{deps/setup}
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\usepackage{deps/macros}
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\usepackage{showkeys}
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\title{Probability Notes}
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\subtitle{Based on KUL Course Notes for ``Kansrekenen I'' (2018) by Tim Verdonck}
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\author{omicron}
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\begin{document}
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\maketitle
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\section{Probability Spaces}
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\begin{definition}[Sigma Algebra]\label{def:sigma-algebra}
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A collection $\mathcal A$ of subsets of $\Omega$ is called a
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\emph{sigma-algebra} (or $\sigma$-algebra) on the universe $\Omega$ if
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$\mathcal A$ satisfies the following axioms:
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\begin{enumerate}
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\item $\Omega \in \mathcal A$,
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\item $A \in \mathcal A \implies A^\complement \in \mathcal A$,
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\item $\forall n\in\bb N : A_n \in \mathcal A \implies
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\paren{\bigcup_{n\in \bb N} A_n} \in \mathcal A$.
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\end{enumerate}
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We call the pair $(\Omega, \mathcal A)$ a \emph{measurable space} and the
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elements of $\mathcal A$ \emph{events}.
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\end{definition}
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\begin{observation}\label{obs:has-empty}
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Let $(\Omega, \mathcal A)$ be a measurable space, then $\emptyset \in
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\mathcal A$.
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\end{observation}
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\begin{proof}
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By \cref{def:sigma-algebra} we have $\Omega \in \mathcal A$ and
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$\Omega^\complement = \emptyset \in \mathcal A$.
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\end{proof}
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\begin{definition}[Probability Measure]\label{def:probability-measure}
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A function $\probop : \mathcal A \to \bb R$ is called a \emph{probability
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measure} if it satisfies the following axioms:
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\begin{enumerate}
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\item $\prob{\Omega} = 1$.
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\item $\forall A \in \mathcal A : \prob A \geq 0$.
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\item For a family of pairwise disjoint sets $A_1, A_2, \ldots \in
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\mathcal A$,
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\[
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\prob{\bigcup_{n \in \bb N} A_n} = \sum_{n \in \bb N}
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\prob{A_n}.
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\]
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We call this axiom the axiom of \emph{countable additivity} or
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\emph{$\sigma$-additivity}.
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\end{enumerate}
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The triple $(\Omega, \mathcal A, \probop)$ is called a \emph{probability
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space}, comprised of the universe $\Omega$, a $\sigma$-algebra $\mathcal A$
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and a probability measure $\probop$.
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\end{definition}
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\begin{observation}\label{obs:prob-empty}
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Let $(\Omega, \mathcal A, \probop)$ be a probability space, then
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$\prob{\emptyset} = 0$.
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\end{observation}
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\begin{proof}
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Note that $\emptyset = \bigcup_{n\in\bb N} \emptyset$ and that the
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right-hand side is a union of disjoint sets. By applying the
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sigma-additivity axiom we get \[ \prob{\emptyset} = \prob{\bigcup_{n\in\bb
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N} \emptyset} = \sum_{n\in\bb N} \prob{\emptyset}. \] Since $P$ takes real
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values we know the series must converge. This can only happen if
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$P(\emptyset) = 0$.
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\end{proof}
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\begin{definition}[Monotonous sequence of sets]\label{def:monotonous-sets}
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A sequence of sets $\paren{A_n}_{n\in\bb N_0}$ is said to be
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\emph{increasing} if $A_n \subseteq A_{n+1}$ for
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every $n \in \bb N_0$. Similarly, a sequence is called \emph{decreasing}
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if $A_n \supseteq A_{n+1}$ for every $n \in \bb
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N_0$. A sequence is called \emph{monotonous} if it is either increasing or
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decreasing. For such sequences we define
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\[
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\lim_{n\to\infty} A_n = \begin{cases}
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\bigcup_{n=1}^{\infty} A_n & \text{if $A_n$ is increasing}, \\
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\bigcap_{n=1}^{\infty} A_n & \text{if $A_n$ is decreasing}. \\
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\end{cases}
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\]
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\end{definition}
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\begin{theorem}
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Let $(\Omega, \mathcal A, \probop)$ be a probability space.
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\begin{enumerate}
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\item \emph{Finite additivity} for a pairwise disjoint family of sets
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$\set{A_n \in \mathcal A \smid n \in \set{1, \dots, N}}$
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\[
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\prob{\bigcup_{n=1}^{N} A_n} = \sum_{n=1}^{N} \prob{A_n}.
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\]
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\item $\forall A \in \mathcal A : \prob{A^\complement} = 1 - \prob{A}$.
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\item For a monotonous sequence $\paren{A_n}_{n\in\bb N_0}$
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\[
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\prob{\lim_{n\to\infty} A_n} = \lim_{n\to\infty} \prob{A_n}
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\]
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\end{enumerate}
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\end{theorem}
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\begin{proof}
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\begin{enumerate}
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\item We start by showing finite additivity as defined above holds. Consider the
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finite family of sets defined above, we define a related infinite family of
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sets as follows
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\[
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B_n = \begin{cases}
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A_n & \text{if $n \leq N$}, \\
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\emptyset & \text{otherwise}.
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\end{cases}
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\]
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Note that since we only added empty sets this new family is also pairwise
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disjoint and that the countable union over $B_n$ is equal to the finite
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union over $A_n$. This leads to
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\begin{align*}
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\prob{\bigcup_{n=1}^{N} A_n}
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&= \prob{\bigcup_{n=1}^{\infty} B_n}. \\
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\intertext{Next we apply countable additivity by \cref{def:probability-measure} and get}
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&= \sum_{n=1}^{\infty} \prob{B_n} \\
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&= \sum_{n=1}^{N} \prob{A_n} + \sum_{n=N+1}^{\infty} \prob{\emptyset}.
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\intertext{In \cref{obs:prob-empty} we concluded $\prob{\emptyset} = 0$ so}
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&= \sum_{n=1}^{N} \prob{A_n}.
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\end{align*}
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This concludes the proof of finite additivity.
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\item Let $A \in \mathcal A$ be arbitrary. By
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\cref{def:probability-measure}, finite additivity and the definition of
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complement we get
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\[
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1 = \prob{\Omega} = \prob{A \cup A^\complement} = \prob{A} + \prob{A^\complement}.
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\]
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Subtracting $\prob{A}$ from both sides gives us
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\[
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\prob{A^\complement} = 1 - \prob{A}.
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\]
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This concludes the proof of the second point.
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\item Let $(A_n)_{n\in\bb N_0}$ be a monotonous sequence in $\mathcal A$.
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We first consider the case where $(A_n)$ is increasing. We define a
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related sequence as follows
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\[
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B_n = \begin{cases}
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A_1 & \text{if $n=1$,} \\
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A_n \setminus A_{n-1} & \text{otherwise.}
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\end{cases}
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\]
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Note that $(B_n)$ is pairwise disjoint and that for any $n \in \bb N_0$
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\[
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\bigcup_{i=1}^n A_i = \bigcup_{i=1}^n B_i = A_n
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\]
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By using the properties of this related sequence and by applying sigma
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additivity we get
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\begin{align*}
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\prob{\lim_{n\to\infty} A_n}
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&= \prob{\bigcup_{i=1}^\infty A_i} \\
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&= \prob{\bigcup_{i=1}^\infty B_i} \\
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&= \sum_{i=1}^\infty \prob{B_i}. \\
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\intertext{Next we use the definition of a series and apply finite
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additivity (in reverse) as follows}
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&= \lim_{n\to\infty} \sum_{i=1}^n \prob{B_i} \\
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&= \lim_{n\to\infty} \prob{\bigcup_{i=1}^n B_i} \\
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&= \lim_{n\to\infty} \prob{A_n}.
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\end{align*}
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This concludes of the proof when $(A_n)$ is increasing. Next assume
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that $(A_n)$ is decreasing. Note that the complement sequence
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$(A_n^\complement)$ is an increasing sequence. Combined with De
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Morgan's laws and the earlier formula for the probability of a
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complement we get
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\begin{align*}
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\prob{\lim_{n\to\infty} A_n}
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&= \prob{\bigcap_{i=1}^\infty A_i} \\
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&= 1 - \prob{\bigcup_{i=1}^\infty A_i^\complement} \\
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&= 1 - \prob{\lim_{n\to\infty} A_n^\complement}. \\
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\intertext{since the complement sequence is an increasing sequence
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we can apply the result from before and we get}
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&= 1 - \lim_{n\to\infty} \prob{A_n^\complement} \\
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&= 1 - \lim_{n\to\infty} 1 - \prob{A_n} \\
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&= \lim_{n\to\infty} \prob{A_n}.
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\end{align*}
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This concludes the proof for the final case. \qedhere
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\end{enumerate}
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\end{proof}
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\end{document}
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