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\noindent
Interpretation: ``Finite'' eg 100 out of a large
population, ``Infinite'' $\to$ affects positive
proportion of population.

\begin{proof}[(baby proof)]
    (subcritical) $\mu < 1$
    \[ \PP(Z_n \ge 1) \le \frac{\EE[Z_n]}{1} =
    \mu^n \to 0 \]
    (Markov's Inequality) \\
    (supercritical):
    \begin{note*}
        $\EE[Z_n] \to \infty$
        does \emph{not} imply $\PP(Z_n = 0)
        \not\approx 1$.
    \end{note*}
    Reminder: $G$ the PGF of $X$, $G_n$ the PGF of
    $Z_n$. We care about $\{Z_n = 0\}$, $q_n =
    \PP(Z_n = 0)$. Also $q_n = G_n(0)$.
    \begin{claim*}
        $q$ the extinction probability, then $G(q)
        = q$.
    \end{claim*}
    \begin{proof}[1]
        $G$ continuous. Note $q_{n + 1} = G(q_n)$
        and $q_{n + 1} \to q$, and $G(q_n) \to
        G(q)$ so $q = G(q)$.
    \end{proof}
    \myskip
    \begin{proof}[2]
        LTP (revision of random sums)
        \[ \text{Total finite} \iff \text{ALl
        subtrees of 1st generation are finite} \]
        \begin{align*}
            q
            &= \PP(\text{finite}) \\
            &= \sum_{k \ge 0} \PP(\text{all
            finite} \mid Z_1 = l) \PP(Z_1 = k) \\
            &= \sum_{k \ge 0}
            [\PP(\text{finite})]^k \PP(Z_1 = k) \\
            &= \sum_{k \ge 0} q^k \PP(Z_1 = k) \\
            &= G(q)
        \end{align*}
    \end{proof}

    \noindent
    Facts about $G$:
    \begin{itemize}
        \item $G(0) = \PP(X = 0) \ge 0$
        \item $G(1) = 1$
        \item $G'(1) = \EE[X] = \mu$
        \item $G$ is \emph{smooth}, all
            derivatives $\ge 0$ on $[0, 1)$.
    \end{itemize}
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/c71795f4a2f111ec.png}
    \end{center}
    \begin{remark*}
        \begin{itemize}
            \item Exactly one solution on $[0, 1)$
            \item By IVT / Rolle on $G(z) - z$.
        \end{itemize}
    \end{remark*}
    \begin{theorem*}
        Assume $\PP(X = 1) \neq 1$. Then $q$ is
        the \emph{minimal} solution to $z = G(z)$
        in $[0, 1]$.
    \end{theorem*}
    \begin{corollary*}
        $q = 1 \iff \mu \le 1$.
    \end{corollary*}
    \begin{proof}
        Let $t$ be the minimal solution. Reminder:
        $G$ is increasing,
        \[ t \ge 0 \]
        \[ \implies G(t) \ge G(0) \]
        \[ \implies G(G(t)) \ge G(G(0)) \]
        \[ \implies G_n(t) \ge G_n(0) \]
        \[ \implies t \ge q_n \]
        \[ \implies t \ge q \]
        Note $q$ \emph{is} a solution, so we must
        have $q = t$.
    \end{proof}
\end{proof}

\subsubsection*{Continuous Probability}

Focus now: Case where $\mathrm{Im}(X)$ is an
\emph{interval} in $\RR$. \\
Why?
\begin{itemize}
    \item Natural for measuring, for example
        physical quantity, for example proportions
    \item ``Limits'' of discrete random variable
    \item Calculus tools for nice calculations
\end{itemize}
Redefinition:

\begin{definition*}
    A random variable $X$ on $(\omega,
    \mathcal{F}, \PP)$ is a function $X : \Omega
    \to \RR$ such that $\{X \le x\} \in
    \mathcal{F}$.
\end{definition*}

\noindent
Check: consistent with previous definition when
$\Omega$ countable (or $\mathrm{Im}(X)$ is
countable). \myskip
Drawback: Can't take $\mathcal{F} =
\mathcal{P}(\RR)$.

\begin{definition*}
    The \emph{cumulative distribution function}
    (CDF) of RV $X$ is $F_X : \RR \to [0, 1]$
    \[ F_X(x) = \PP(X \le x) \]
\end{definition*}

\subsubsection*{Examples}

$X$ a dive on $\{1, \dots, 6\}$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/f172563aa2f211ec.png}
\end{center}
Angle of ludo spinner:
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/0eb63b9ea2f311ec.png}
\end{center}

\subsubsection*{Properties of CDF}

\begin{itemize}
    \item $F_X$ increasing, i.e. $x \le y \implies
        F_X(x) \le F_X(y)$. Why? $F_X(x) = \PP(X
        \le x) \le \PP(X \le y) = F_X(y)$.
    \item $\PP(X > x) = 1 - F_X(x)$
    \item $\PP(a < x \le b) = F_X(b) - F_X(a)$.
        Why? $\PP(a < X \le b) = \PP(X \le b) -
        \PP(X \le a)$.
    \item $F_X$ is right-continuous and left
        limits exist, i.e.
        \[ \lim_{y \downarrow x} F_X(y) = F_X(x) \]
        and
        \[ \lim_{y \uparrow x} F_X(y) = F_X(x^-) =
        \PP(X < x) \]
    \item $\lim_{x \to \infty} F_X(x) = 1$,
        $\lim_{x \to -\infty} F_X(x) = 0$.
\end{itemize}
