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\begin{example}
    Derangements (Permutation with no fixed
    points)
    \[ \Omega = \{\text{permutations of } \{1,
    \dots, n\}\} \]
    \[ D = \{\sigma \in \Omega : \sigma(i) \neq i
    \,\,\forall\,\, i = 1, \dots, n\} \]
    \ul{Question}: Is $\PP(D) =
    \frac{|D|}{|\Omega|}$ large or small (when $n
    \to \infty$)?
    \[ \forall\,\, i \in \{1, \dots, n\} : A_i =
    \{\sigma \in \Omega : \sigma(i) = i\} .\]
    \begin{itemize}
        \item $D = A_1^c \cap \cdots \cap A_n^c =
            \left( \bigcup_{i = 1}^n A_i
            \right)^c$.
        \item $\PP (A_{i_1} \cap \cdots \cap
            A_{i_k}) = \frac{(n - k)!}{n!}$
    \end{itemize}
    Now Inclusion-Exclusion Principle implies:
    \begin{align*}
        \PP \left( \bigcup_{i = 1}^n A_i \right)
        &= \sum_{k = 1}^n (-1)^{k + 1} \sum_{i_1 <
        \cdots i_k} \PP (A_{i _1} \cap \cdots \cap
        A_{i_k}) \\
        &= \sum_{k = 1}^n (-1)^{k + 1} {n \choose
        k} \frac{(n - k)!}{n!}
    \end{align*}
    So
    \begin{align*}
        \PP(D)
        &= 1 - \PP \left( \bigcup_{i = 1}^n A_i
        \right) \\
        &= 1 - \sum_{k = 1}^n \frac{(-1)^{k +
        1}}{k!} \\
        &= \sum_{k = 1}^n \frac{(-1)^k}{k!}
    \end{align*}
    And as $n \to \infty$,
    \[ \PP(D) \to \sum_{k = 0}^\infty
    \frac{(-1)^k}{k!} = e^{-1} \approx 0.37 \]
\end{example}

\subsubsection*{Comments}

What if instead we have
\[ \Omega' = \{f : \{1, \dots, n\} \to \{1, \dots,
n\}\} .\]
\[ D = \{f \in \Omega' : f(i) \neq i
\,\,\forall\,\, i = 1, \dots, n\} .\]
Then
\[ \PP(D) = \frac{(n - 1)^n}{n^n} = \left( 1 -
\frac{1}{n} \right)^n \]
which also approaches $e^{-1}$ as $n \to \infty$.
\begin{itemize}
    \item Would be nice to write as a product of
        probabilities, i.e. $\left( \frac{n -
        1}{n} \right)^n$, and we will be allowed
        to do this soon.
    \item $f(i)$ is a random quantity associated
        to $\Omega$. (Will be allowed to study
        $f(i)$ as a \emph{random variable}.)
    \item Are allowed to toss a fair coin $n$
        times.
        \[ \Omega = \{H, T\}^n \]
\end{itemize}

\subsubsection*{Independence}

$(\Omega, \mathcal{F}, \PP)$ as before.

\begin{definition*}
    \begin{itemize}
        \item Events $A, B \in \mathcal{F}$ are
            \emph{independent} if
            \[ \PP(A \cap B) = \PP(A)\PP(B) \]
            (denoted $A \ci B$).
        \item A \emph{countable} collection of
            events $(A_n)$ is \emph{independent}
            if $\forall$ distinct $i_1, \dots,
            i_k$ we have:
            \[ \PP(A_{i_1} \cap \cdots \cap
            A_{i_k}) = \prod_{j = 1}^k
            \PP(A_{i_j}) .\]
    \end{itemize}
\end{definition*}

\begin{note*}
    ``Pairwise independence'' does not imply
    independence.
\end{note*}

\begin{example*}
    $\Omega = \{(H, H), (H, T), (T, H), (T, T)\}$,
    $\PP(\{\omega\}) = \frac{1}{4}
    \,\,\forall\,\, \omega \in \Omega$. Now define
    \[ A = \text{first coin in $H$} = \{(H, H),
    (H, T)\} \]
    \[ B = \text{second coin $H$} = \{(H, H), (T,
    H)\} \]
    \[ C = \text{same outcome} = \{(H, H), (T,
    T)\} .\]
    Then we have that
    \[ \PP(A) = \PP(B) = \PP(C) = \half \qquad A
    \cap B = A \cap C = B \cap C = \{(H, H)\} \]
    \[ \implies \PP(A \cap B) = \PP(A \cap C) =
    \PP(B \cap C) = \frac{1}{4} \]
    so pairwise independent, however
    \[ \PP(A \cap B \cap C) = \frac{1}{4} \neq
    \PP(A)\PP(B)\PP(C) \]
    so the events are not independent.
\end{example*}

\subsubsection*{Example(s) of Independence}

\begin{itemize}
    \item Define
        \[ \Omega' = \{f : \{1, \dots, n\} \to
        \{1, \dots, n\}\} .\]
        \[ A_i := \{f \in \Omega' : f(i) = i\} .\]
        \[ \PP(A_i) = \frac{n^{n - 1}}{n^n} =
        \frac{1}{n} \]
        \[ \PP(A_{i_1} \cap \cdots \cap A_{i_k}) =
        \frac{n^{n - k}}{n^n} = \frac{1}{n^k} =
        \prod_{j = 1}^k \PP(A_{i_j}) \]
        Here: $(A_i)$ independent events.
    \item Define
        \[ \Omega = \{\sigma : \text{permutation
        of} \{1, \dots, n\}\} \]
        \[ A_i = \{\sigma \in \Omega : \sigma(i) =
        i\} \]
        For $i \neq j$,
        \[ \PP(A_i \cap A_j) = \frac{(n - 2)!}{n!}
        = \frac{1}{n(n - 1)} \neq \PP(A_i)\PP(A_j) \]
        So here, $(A_i)$ are not independent.
\end{itemize}

\subsubsection*{Properties}

\begin{claim}
    If $A$ is independent of $B$, then $A$ is also
    independent of $B^c$.
\end{claim}

\begin{proof}
    \begin{align*}
        \PP(A \cap B^c)
        &+ \PP(A) - \PP(A \cap B) \\
        &= \PP(A) - \PP(A)\PP(B) \\
        &= \PP(A)[1 - \PP(B)] \\
        &= \PP(A)\PP(B^c)
    \end{align*}
\end{proof}

\begin{claim}
    $A$ is independent of $B = \Omega$ and of $C =
    \phi$.
\end{claim}

\begin{proof}
    \[ \PP(A \cap \Omega) = \PP(A) =
    \PP(A)\PP(\Omega) .\]
    And by claim 1, this implies that $A \ci
    \emptyset$.
\end{proof}

\myskip
As an exercise, one can further prove that if
$\PP(B) = 0 \text{ or } 1$, then $A$ is
independent of $B$.

\subsubsection*{Conditional Probability}

$(\Omega, \mathcal{F}, \PP)$ as before. \myskip
Consider $B \in \mathcal{F}$ with $\PP(B) > 0$, $A
\in \mathcal{F}$.

\begin{definition*}
    The \emph{conditional probability of $A$ given
    $B$} is
    \[ P(A \mid B) := \frac{\PP(A \cap B)}{\PP(B)} \]
    ``The probability of $A$ if we know $B$
    happened''. (for example revealing info in
    succession).
\end{definition*}

\begin{example*}
    If $A$, $B$ independent,
    \[ \PP(A \mid B) = \frac{\PP(A \cap
    B)}{\PP(B)} = \frac{\PP(A)\PP(B)}{\PP(B)} =
    \PP(A) .\]
    ``Knowing whether $B$ happened doesn't affect
    the probability of $A$.''
\end{example*}