\documentclass{article}

\usepackage{deps/setup}
\usepackage{deps/macros}
\usepackage{showkeys}

\title{Probability Notes}
\subtitle{Based on KUL Course Notes for ``Kansrekenen I'' (2018) by Tim Verdonck}
\author{omicron}

\begin{document}
\maketitle

\section{Probability Spaces}

\begin{definition}[Sigma Algebra]\label{def:sigma-algebra}
    A collection $\mathcal A$ of subsets of $\Omega$ is called a
    \emph{sigma-algebra} (or $\sigma$-algebra) on the universe $\Omega$ if
    $\mathcal A$ satisfies the following axioms:
    \begin{enumerate}
        \item $\Omega \in \mathcal A$,
        \item $A \in \mathcal A \implies A^\complement \in \mathcal A$,
        \item $\forall n\in\bb N : A_n \in \mathcal A \implies
            \paren{\bigcup_{n\in \bb N} A_n} \in \mathcal A$.
    \end{enumerate}
    We call the pair $(\Omega, \mathcal A)$ a \emph{measurable space} and the
    elements of $\mathcal A$ \emph{events}.
\end{definition} 

\begin{observation}\label{obs:has-empty}
    Let $(\Omega, \mathcal A)$ be a measurable space, then $\emptyset \in
    \mathcal A$.
\end{observation}
\begin{proof}
    By \cref{def:sigma-algebra} we have $\Omega \in \mathcal A$ and
    $\Omega^\complement = \emptyset \in \mathcal A$.
\end{proof}

\begin{definition}[Probability Measure]\label{def:probability-measure}
    A function $\probop : \mathcal A \to \bb R$ is called a \emph{probability
    measure} if it satisfies the following axioms:
    \begin{enumerate}
        \item $\prob{\Omega} = 1$.
        \item $\forall A \in \mathcal A : \prob A \geq 0$.
        \item For a family of pairwise disjoint sets $A_1, A_2, \ldots \in
            \mathcal A$,
            \[
                \prob{\bigcup_{n \in \bb N} A_n} = \sum_{n \in \bb N}
                \prob{A_n}.
            \]
            We call this axiom the axiom of \emph{countable additivity} or
            \emph{$\sigma$-additivity}.
    \end{enumerate}
    The triple $(\Omega, \mathcal A, \probop)$ is called a \emph{probability
    space}, comprised of the universe $\Omega$, a $\sigma$-algebra $\mathcal A$
    and a probability measure $\probop$.
\end{definition}
\begin{observation}\label{obs:prob-empty}
    Let $(\Omega, \mathcal A, \probop)$ be a probability space, then
    $\prob{\emptyset} = 0$.
\end{observation}
\begin{proof}
    Note that $\emptyset = \bigcup_{n\in\bb N} \emptyset$ and that the
    right-hand side is a union of disjoint sets. By applying the
    sigma-additivity axiom we get \[ \prob{\emptyset} = \prob{\bigcup_{n\in\bb
    N} \emptyset} = \sum_{n\in\bb N} \prob{\emptyset}. \] Since $P$ takes real
    values we know the series must converge. This can only happen if
    $P(\emptyset) = 0$.
\end{proof}

\begin{definition}[Monotonous sequence of sets]\label{def:monotonous-sets}
    A sequence of sets $\paren{A_n}_{n\in\bb N_0}$ is said to be
    \emph{increasing} if $A_n \subseteq A_{n+1}$ for
    every $n \in \bb N_0$. Similarly, a sequence is called \emph{decreasing}
     if $A_n \supseteq A_{n+1}$ for every $n \in \bb
    N_0$. A sequence is called \emph{monotonous} if it is either increasing or
    decreasing. For such sequences we define
    \[
        \lim_{n\to\infty} A_n = \begin{cases}
            \bigcup_{n=1}^{\infty} A_n & \text{if $A_n$ is increasing}, \\
            \bigcap_{n=1}^{\infty} A_n & \text{if $A_n$ is decreasing}. \\
        \end{cases}
    \]
\end{definition}

\begin{theorem}
    Let $(\Omega, \mathcal A, \probop)$ be a probability space.
    \begin{enumerate}
        \item \emph{Finite additivity} for a pairwise disjoint family of sets
            $\set{A_n \in \mathcal A \smid n \in \set{1, \dots, N}}$
            \[
                \prob{\bigcup_{n=1}^{N} A_n} = \sum_{n=1}^{N} \prob{A_n}.
            \]
        \item $\forall A \in \mathcal A : \prob{A^\complement} = 1 - \prob{A}$.
        \item For a monotonous sequence $\paren{A_n}_{n\in\bb N_0}$
            \[
                \prob{\lim_{n\to\infty} A_n} = \lim_{n\to\infty} \prob{A_n}
            \]
    \end{enumerate}
\end{theorem}
\begin{proof}
    \begin{enumerate}
    \item We start by showing finite additivity as defined above holds. Consider the
    finite family of sets defined above, we define a related infinite family of
    sets as follows
    \[
            B_n = \begin{cases}
                A_n & \text{if $n \leq N$}, \\
                \emptyset & \text{otherwise}.
            \end{cases}
    \]
    Note that since we only added empty sets this new family is also pairwise
    disjoint and that the countable union over $B_n$ is equal to the finite
    union over $A_n$. This leads to
    \begin{align*}
        \prob{\bigcup_{n=1}^{N} A_n}
            &= \prob{\bigcup_{n=1}^{\infty} B_n}. \\
        \intertext{Next we apply countable additivity by \cref{def:probability-measure} and get}
            &= \sum_{n=1}^{\infty} \prob{B_n} \\
            &= \sum_{n=1}^{N} \prob{A_n} + \sum_{n=N+1}^{\infty} \prob{\emptyset}.
        \intertext{In \cref{obs:prob-empty} we concluded $\prob{\emptyset} = 0$ so}
            &= \sum_{n=1}^{N} \prob{A_n}.
    \end{align*}
    This concludes the proof of finite additivity.

    \item Let $A \in \mathcal A$ be arbitrary. By
        \cref{def:probability-measure}, finite additivity and the definition of
            complement we get
        \[
            1 = \prob{\Omega} = \prob{A \cup A^\complement} = \prob{A} + \prob{A^\complement}.
        \]
        Subtracting $\prob{A}$ from both sides gives us
        \[
            \prob{A^\complement} = 1 - \prob{A}.
        \]
        This concludes the proof of the second point.
    \item Let $(A_n)_{n\in\bb N_0}$ be a monotonous sequence in $\mathcal A$.
        We first consider the case where $(A_n)$ is increasing. We define a
        related sequence as follows
        \[
            B_n = \begin{cases}
                A_1             & \text{if $n=1$,} \\
                A_n \setminus A_{n-1}   & \text{otherwise.}
            \end{cases}
        \]
        Note that $(B_n)$ is pairwise disjoint and that for any $n \in \bb N_0$
        \[
            \bigcup_{i=1}^n A_i = \bigcup_{i=1}^n B_i = A_n
        \]
        By using the properties of this related sequence and by applying sigma
        additivity we get
        \begin{align*}
            \prob{\lim_{n\to\infty} A_n}
                &= \prob{\bigcup_{i=1}^\infty A_i} \\
                &= \prob{\bigcup_{i=1}^\infty B_i} \\
                &= \sum_{i=1}^\infty \prob{B_i}. \\
            \intertext{Next we use the definition of a series and apply finite
            additivity (in reverse) as follows}
                &= \lim_{n\to\infty} \sum_{i=1}^n \prob{B_i} \\
                &= \lim_{n\to\infty} \prob{\bigcup_{i=1}^n B_i} \\
                &= \lim_{n\to\infty} \prob{A_n}.
        \end{align*}
        This concludes of the proof when $(A_n)$ is increasing. Next assume
        that $(A_n)$ is decreasing. Note that the complement sequence
        $(A_n^\complement)$ is an increasing sequence. Combined with De
        Morgan's laws and the earlier formula for the probability of a
        complement we get
        \begin{align*}
            \prob{\lim_{n\to\infty} A_n}
                &= \prob{\bigcap_{i=1}^\infty A_i} \\
                &= 1 - \prob{\bigcup_{i=1}^\infty A_i^\complement} \\
                &= 1 - \prob{\lim_{n\to\infty} A_n^\complement}. \\
            \intertext{since the complement sequence is an increasing sequence
            we can apply the result from before and we get}
                &= 1 - \lim_{n\to\infty} \prob{A_n^\complement} \\
                &= 1 - \lim_{n\to\infty} 1 - \prob{A_n} \\
                &= \lim_{n\to\infty} \prob{A_n}.
        \end{align*}
        This concludes the proof for the final case. \qedhere
    \end{enumerate}
\end{proof}
\end{document}