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\subsubsection*{Discrete Probability Distributions}

$\Omega$ finite.

\subsubsection*{1. Bernoulli Distribution}

(``(biased) coin toss''). \\
$X \sim \mathrm{Bern}(p)$, $p \in [0, 1]$.
\[ \mathrm{Im}(x) = \{0, 1\} \]
\[ p_X(1) = \PP(X = 1) = p \]
\[ p_X(0) = \PP(X = 0) = 1 - p .\]
\ul{Key example}: $\mathbbm{1}_A \sim
\mathrm{Bern}(p)$ with $p = \PP(A)$.

\subsubsection*{2. Binomial Distribution}

$X \sim \mathrm{Bin}(n, p)$, $n \in \ZZ^+$, $p \in
[0, 1]$. \\
(`Toss coin $n$ times, count number of heads''.)
\[ \mathrm{Im}(X) = \{0, 1, \dots, n\} \]
\[ p_X(k) = \PP(X = k) = {n \choose k} p^k (1 -
p)^{n - k} \]
check:
\[ \sum_{k = 0}^n p_X(k) = (p + (1 - p))^n = 1 \]

\subsubsection*{More than one Random Variable}

\textbf{Motivation}: Doll a dice. Outcome $X \in
\{1, 2, \dots, 6\}$. Events:
\[ A = \{1 \text{ or } 2\}, \qquad B = \{1
\text{ or } 2 \text{ or } 3\}, \qquad C = \{1
\text{ or } 3 \text{ or } 5\} .\]
\[ \mathbbm{1}_A \sim \mathrm{Bern} \left(
\frac{1}{3} \right), \qquad \mathbbm{1}_B \sim
\mathrm{Bern} \left( \frac{1}{2} \right), \qquad
\mathbbm{1}_C \sim \mathrm{Bern} \left( \frac{1}{2}
\right) \]
\begin{note*}
    $\mathbbm{1}_A \le \mathbbm{1}_B$ for all
    outcomes, but $\mathbbm{1}_A \le \mathbbm{1}_C$
    for outcomes is \emph{false}.
\end{note*}

\begin{definition*}
    $X_1, \dots, X_n$ discrete random variables.
    Say $X_1, \dots, X_n$ are \emph{independent}
    if
    \[ \PP(X_1 = x_1, \dots, X_n = x_n) = \PP(X_1
    = x_1) \cdots \PP(X_n = x_n) \qquad
    \forall\,\, x_1, \dots, x_n \in \RR \]
    (suffices to check $\forall\,\, x_i \in
    \mathrm{Im}(X_i)$).
\end{definition*}

\begin{example*}
    $X_1, \dots, X_n$ independent random variables
    each with the Bernoulli$(p)$ distribution.
    Study $S_n = X_1 + \cdots + X_n$. Then
    \begin{align*}
        \PP(S_n = k)
        &= \sum_{\substack{X_1 + \cdots X_n =
        k\\X_i \in \{0, 1\}}} \PP(X_1 = x_1,
        \dots, X_n = x_n) \\
        &= \sum_{X_1 + \cdots + X_n = k} \PP(X_1 =
        x_1) \cdots \PP(X_n = x_n) \\
        &= \sum_{X_1 + \cdots + X_n = k} p^{|\{i :
        x_i = 1\}|} (1 - p)^{|\{i : x_i = 0\}|} \\
        &= \sum_{X_1 + \cdots + X_n = k} p^k (1 -
        p)^{n - k} \\
        &= {n \choose k} p^k (1 - p)^{n - k}
    \end{align*}
    so $S_n \sim \mathrm{Bin}(n, k)$.
\end{example*}

\begin{example*}[Non-example]
    $(\sigma(1), \sigma(2), \dots, \sigma(n))$
    uniform in $\sum_n$.
    \begin{claim*}
        $\sigma(1)$ and $\sigma(2)$ are \emph{not}
        independent.
    \end{claim*}
    \noindent
    Suffices to find $i_1$, $i_2$ such that
    \[ \PP(\sigma(1) = i, \sigma(2) = i_2) \neq
    \PP(\sigma(1) = i_1) \PP(\sigma(2) = i_2) \]
    for example
    \[ \PP(\sigma(1) = 1, \sigma(2) = 1) = 0 \neq
    \frac{1}{n} \times \frac{1}{n} = \PP(\sigma(1)
    = 1) \PP(\sigma(2) = 1) \]
\end{example*}

\noindent
\ul{Consequence of definition} \\
$X_1, \dots, X_n$ independent then $\forall\,\,
A_1, \dots, A_n \subset \RR$ countable, then
\[ \PP(X_1 \in A_1, \dots, X_n \in A_n) = \PP(X_1
\in A_1) \cdots \PP(X_n \in A_n) \]

\subsubsection*{$\Omega = \NN$}

``Ways of choosing a random integer'' \\

\subsubsection*{3. Geometric distribution}

(``waiting for success'') \\
$X \sim \mathrm{Geom}(p)$, $p \in (0, 1]$. \\
(``Toss a coin with $\PP(\text{heads}) = p$ until
a head appears. Count how many trials were
needed.'')
\[ \mathrm{Im}(X) = \{1, 2, \dots\} \]
\[ p_X(k) = \PP((k - 1) \text{ failures, then
success on $k$-th}) = (1 - p)^{k - 1}p \]
Check:
\[ \sum_{k \ge 1} (1 - p)^{k - 1} p = p \sum_{l
\ge 0} (1 - p)^l = \frac{p}{1 - (1 - p)} = 1 \]

\begin{note*}
    We could alternatively ``count how many
    failures before a success''.
    \[ \mathrm{Im}(Y) = \{0, 1, 2, \dots\} \]
    \[ p_Y(k) = \PP(k \text{ failures, then
    success on $(k + 1)$-th}) = (1 - p)^k p \]
    Check:
    \[ \sum_{k \ge 0} (1 - p)^k p = 1 \]
\end{note*}

\subsubsection*{4. Poisson Distribution}

$\lambda \in (0, \infty)$.
\[ X \sim \mathrm{Po}(\lambda) \]
\[ \mathrm{Im}(X) = \{0, 1, 2, \dots\} \]
\[ \PP(X = k) = e^{-\lambda} \frac{\lambda^k}{k!}
\qquad \forall\,\,k \ge 0 \]
\begin{note*}
    \[ \sum_{k \ge 0} \PP(X = k) = e^{-k} \sum_{k
    \ge 0} \frac{\lambda^k}{k!} = e^{-\lambda}
    e^\lambda = 1 \]
\end{note*}

\noindent
\ul{Motivation}: Consider $X_n \sim \mathrm{Bin}
\left( n, \frac{\lambda}{n} \right)$.
.image
\begin{itemize}
    \item Probability of an arrival in each
        interval is $p$, independently across
        intervals.
    \item Total arrivals is $X_n$.
\end{itemize}
\[ \PP(X_n = k) = {n \choose k} \left(
\frac{\lambda}{n} \right)^k \left( 1 -
\frac{\lambda}{n} \right)^{n - k} \]
Fix $k$, let $n \to \infty$:
\[ \PP(X_n = k) = \ub{\frac{n!}{n^k(n - k)!}}_{\to
1} \times \frac{\lambda^k}{k!} \times \ub{\left( 1 -
\frac{\lambda}{n} \right)^n}_{\to e^{-\lambda}}
\times \ub{\left( 1 - \frac{1}{n}
\right)^{-k}}_{\to 1} \]
so
\[ \PP(X_n = k) \to e^{-\lambda}
\frac{\lambda^k}{k!} \]