\renewcommand{\coddate}{{\small \today}}
\renewcommand{\sheweek}{1. Events and their Probabilities}

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\section{Events and their Probabilities}
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\subsection{Sample Space}
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\begin{enumerate}
\item \textbf{Experiment}: a procedure, that we usually think of as
being repeatable any number of times,  which has a well-defined set of
possible outcomes.

\item \textbf{Sample outcome}: a
potential eventuality of the experiment. The notation $\omega$ is
used for an outcome.

\item \textbf{Sample space}:
the set of all possible outcomes. The notation $\Omega$ is used
for the sample space of an experiment. An outcome $\omega$ is a
member of the sample space $\Omega$, that is, $\omega \in \Omega$.

\item \textbf{A simple experiment}: A fair six-sided die is thrown
twice. The outcomes are pairs of numbers between $1$ and $6$. For
example, $(3,5)$ denotes a $3$ on the first throw and $5$ on the
second. The sample space is given by $\Omega = \{ (i,j) \ : \ i =
1,\ldots,6, \ j=1,\ldots,6 \}$. In this example the sample space is
finite so can be written out in full:
 \[\left\{\begin{array}[c]{cccccc}
(1,1), & (1,2), & (1,3), & (1,4), & (1,5), & (1,6), \\
 (2,1), & (2,2), & (2,3),& (2,4), & (2,5), & (2,6), \\
  (3,1), & (3,2), & (3,3), & (3,4), & (3,5), &(3,6), \\
   (4,1), & (4,2), & (4,3), & (4,4), & (4,5), & (4,6), \\
    (5,1), &(5,2), & (5,3), & (5,4), & (5,5), & (5,6), \\
     (6,1), & (6,2), & (6,3), & (6,4),& (6,5), & (6,6)
\end{array}\right\}.\]
\end{enumerate}

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\subsection{Events} \label{events:sb}
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For any experiment, the events form a collection of subsets of
$\Omega$ which we denote ${\cal F}$. The collection ${\cal F}$ has
the following properties; \vS
\begin{enumerate}
\item[i] $\emptyset \in {\cal F}$, \vS
\item[ii] if $A \in {\cal F}$ then $A^c \in {\cal F}$, \vS
\item[iii] if $A_1,A_2, \ldots, \in {\cal F}$ then
$\bigcup_{i=1}^{\infty} A_i \in {\cal F}$. \vS
\end{enumerate}
[Aside: Any collection of subsets with these properties is known as a
sigma field or a sigma algebra. It is also, less helpfully, sometimes
called a Borel field,] We associate the pair $(\Omega, {\cal F})$ with
our experiment; the sample space consisting of all of the possible
outcomes, and a collection of events which are just subsets of the
outcomes.

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\subsection{Probability Space}
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\begin{enumerate}
\item \textbf{Probability measure}: $P$ on $(\Omega,
{\cal F})$ is a function $P:{\cal F} \longrightarrow [0,1]$
satisfying;
\begin{enumerate}
\item[i] $P(A) \geq 0$,
\item[ii] $P(\Omega)=1$,
\item[iii] if $A_1,A_2,\ldots,$ is a collection of mutually
exclusive members of ${\cal F}$ then $P(\bigcup_{i=1}^{\infty}
A_i) = \sum_{i=1}^{\infty} P(A_i)$.
\end{enumerate}
We can associate a probability space $(\Omega, {\cal F}, P)$ with
any experiment.

\newpage
\item \textbf{Properties of probability measures}
\begin{enumerate}
\item $P(A^c) = 1 - P(A)$.
\item If $A \subseteq B$ then $P(B) = P(A) + P(B \backslash A) \geq
P(A)$.
\item $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
\item More generally if $A_1,\ldots,A_n$ are events then
$$P(\bigcup_{i=1}^n A_i) = \sum_i P(A_i) - \sum_{i<j} P(A_i \cap
A_j) + \sum_{i<j<k} P(A_i \cap A_j \cap A_k) + \ldots + (-1)^{n+1}
P(A_1 \cap \ldots \cap A_n).$$
\item Boole inequality: $P(\bigcup_{i=1}^n A_i) \leq \sum_{i=1}^n
P(A_i)$.
\end{enumerate}

\item \textbf{Probability from counting outcomes}
\begin{enumerate}
\item Multiplication rule for counting ordered sequences:
Operation $A_i$ can be performed in $n_i$ different ways for
$i=1,\ldots,k$. The ordered sequence (operation $A_1$, operation
$A_2$, \ldots, operation $A_k$) can be performed in $n_1 \cdot n_2
\cdot \ldots \cdot n_k$ ways. We write this product as $\prod
_{i=1}^k n_i$.
\item Permutations and combinations:
\begin{enumerate}
\item Permutations of $k$ out of $n$ distinct objects:
The number of permutations of length $k$ that can be formed from
$n$ distinct elements (not allowing repetition) is denoted $_n
P_k$, where
$$
_n P_k = \frac{n!}{(n-k)!}.
$$
\item Permutations of $n$ objects that are not all
distinct: The numbers of ways of arranging $n$ objects where $n_1$
are of type 1, $n_2$ are of type 2, $\ldots$, $n_r$ are of type
$r$ is
$$
\frac{n!}{n_1! n_2! \ldots n_r!},
$$
where $\sum_{i=1}^r n_i = n$.
\item Combinations of $k$ out of $n$ distinct objects:
The number of combinations of length $k$ that can be formed from
$n$ distinct elements is denoted $_n C_k$, where
$$
_n C_k = \frac{n!}{k! (n-k)!}.
$$
\end{enumerate}
\end{enumerate}
\end{enumerate}

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\subsection{Conditional Probability}
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Let $A$ and $B$ be events with $P(B)>0$. The conditional
probability of $A$ given $B$ is the probability that $A$ will
occur give that $B$ has occurred;
$$
P(A|B) = \frac{P(A \cap B)}{P(B)}.
$$

\begin{enumerate}
\item \textbf{Multiplication rule for intersections}: Let
$A_1,\ldots,A_n$ be a set of events defined on $\Omega$,
$$
P(\bigcap_{i=1}^n A_i) = \prod_{j=1}^n P(A_j| \bigcap_{i=0}^{j-1}
A_i),
$$
where we define $A_0 = \Omega$.

\item \textbf{Partitions}: A set of events $B_1,\ldots,B_n$ defined
on $\Omega$ is a partition if
\begin{enumerate}
\item Exhaustive: $\bigcup_{i=1}^n B_i = \Omega$,
\item Mutually exclusive: $B_i \cap B_j = \emptyset$ for $i \neq
j$,
\item Non-zero probability: $P(B_i)>0$ for $i=1,\ldots,n$.
\end{enumerate}

\item \textbf{Law of total probability}: If $B_1,\ldots,B_n$ is a
partition of $\Omega$ and $A$ is any other event defined on
$\Omega$, then
$$
P(A) = \sum_{i=1}^n P(A|B_i)P(B_i).
$$

\item \textbf{Bayes' theorem}:
If $B_1,\ldots,B_n$ is a partition of $\Omega$ and $A$ is any
other event defined on $\Omega$, then
$$
P(B_j|A) = \frac{P(A|B_j)P(B_j)}{P(A)} =
\frac{P(A|B_j)P(B_j)}{\sum_{i=1}^n P(A|B_i) P(B_i)}.
$$

\item \textbf{Independence}
\begin{enumerate}
\item
Two events $A$ and $B$ are independent ($A \perp B$) iff $P(A \cap
B) = P(A)P(B)$.
\item
Events $A_1,\ldots, A_n$ (mutually) independent if every subset is
a subset of independent events and $P(A_1 \cap A_2 \cap \ldots
\cap A_n) = P(A_1)P(A_2) \ldots P(A_n)$ (note -- this definition
is recursive).
\end{enumerate}

\item \textbf{Some properties of independent events}
\begin{enumerate}
\item \hspace{-0.2cm}\begin{tabular}[t]{l}
If $P(A) > 0$ then  $P(B|A) = P(B) \Longleftrightarrow A \perp B$. \\
If $P(B) > 0$ then  $P(A|B) = P(A) \Longleftrightarrow A \perp B$.
\end{tabular}
\item If $A \perp B$ then $A^c \perp B^c$, $A^c \perp B$ and $A
\perp B^c$.
\end{enumerate}
\end{enumerate}
