% vim: tw=50
% 19/02/2022 11AM

\subsubsection*{Unbounded Random Walk: ``Gambler's Ruin''}

\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/985fcaca943111ec.png}
\end{center}
\begin{align*}
    \PP_x(\text{hit 0})
    &= \lim_{a \to \infty}(\text{hit 0 before
    $a$}) \\
    &= \begin{cases}
        1 - \left( \frac{q}{p} \right)^x & p > q
        \\
        1 & p < q \\
        1 & p = q = \half
    \end{cases}
\end{align*}

\[ p = \half : \EE_x[\text{time to hit 0}] \ge
\EE_x[\text{time to hit 0 or $a$}] = x(a - x) \]
which $\to \infty$ as $a \to \infty$.
\myskip
\ul{Key conclusion}: $T_x$ (time to hit 0 from
$x$) is for $p = \half$:
\begin{itemize}
    \item finite with probability $= 1$
    \item infinite expectation
\end{itemize}

\begin{note*}[non-examinable]
    Alternative derivation of $\EE[T_1] = \infty$.
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/1e23555a943211ec.png}
    \end{center}
    ``Random Walk $2 \mapsto 1$'' = ``Random Walk
    $1 \mapsto 0$'' + 1
    \[ \EE[T_1] = \half \times 1 + \half(1 +
    \ub{\EE[T_2]}_{2\EE[T_1]}) \]
    \[ \EE[T_1] = 1 + \EE[T_1] \]
    so $\EE[T_1] = \infty$.
\end{note*}

\subsubsection*{Generating Functions}

\ul{Setting}: $X$ is a random variable taking
values in $\{0, 1, 2, \dots\}$.

\begin{definition*}
    The \emph{Probability Generating Function} of
    $X$ is
    \[ G_X(z) = \EE[z^X] = \sum_{k \ge 0} z^k
    \PP(X = k) .\]
    Analytic comment: $G_X : (-1, 1) \stackrel{k
    \ge 0}{\to} \RR$.
\end{definition*}

\noindent
\ul{Idea}: ``To \emph{encode} the distribution of
$X$ as a function with nice analytic properties''.

\setcounter{customexample}{0}
\begin{example}
    $X \sim \mathrm{Bern}(p)$
    \[ G_X(z) =z^0 \PP(X = 0) + z^1 \PP(X = 1) =
    (1 - p) + pz \]
\end{example}

\begin{example*}
    $X \sim \mathrm{Bin}(n, p)$ we will save for
    later.
\end{example*}

\begin{example}
    $X \sim \mathrm{Poisson}(\lambda)$
    \begin{align*}
        G_X(z)
        &= \sum_{k \ge 0} z^k e^{-\lambda}
        \frac{\lambda^k}{k!} \\
        &= e^{-\lambda} \sum_{k \ge 0}
        \frac{(\lambda z)^k}{k!} \\
        &= e^{-\lambda} e^{\lambda z} \\
        &= e^{\lambda(z - 1)}
    \end{align*}
\end{example}

\subsubsection*{Recovering PMF (mass function) from PGF}

\begin{note*}
    $G_X(0) = 0^0 \PP(X = 0) = \PP(X = 0)$.
\end{note*}

\ul{Idea}: Differentiate $n$ times.
\begin{align*}
    \dfrac[n]{}{z} G_X(z)
    &= \sum_{k \ge 0} \dfrac[n]{}{z} (z^k) \PP(X =
    k) \\
    &= \sum_{k \ge 0} k(k - 1) \cdots (k - n + 1)
    z^{k - n} \PP(X = k) \\
    &= \sum_{k \ge n} k(k - 1) \cdots (k - n + 1)
    z^{k - n} \PP(X = k) \\
    &= \sum_{l \ge 0} (l + 1)(l + 2) \cdots (l +
    n) z^l \PP(X = l + n)
\end{align*}
\ul{Evaluate at 0}:
\[ \dfrac[n]{}{z} G_X(0) = n! \PP(X = n) .\]
\[ \PP(X = n) = \frac{1}{n!} G_X^{(n)} (0) \]
\ul{Key fact}: PGF \emph{determines} PMF /
distribution exactly.

\subsubsection*{Recovering other probabilistic quantities}

\begin{note*}
    $G_X(1) = \sum_{k \ge 0} \PP(X = k) = 1$.
\end{note*}

\noindent
\ul{Technical comment}: $G_X(1)$ means $\lim_{z
\to 1} G_X(z)$ if the domain is $(-1, 1)$ (the
limit is from below).

\begin{itemize}
    \item What about $G_X'(1)$?
        \[ G_X'(z) = \sum_{k \ge 1} kz^{k - 1}
        \PP(X = k) \]
        \[ G_X'(1) = \sum_{k \ge 1} k \PP(X = k) =
        \EE[X] \]
    \item What about $G_X^{(n)}(1)$?
        \begin{align*}
            G_X^{(n)}(1)
            &= \sum_{k \ge n} k(k - 1) \cdots
            (k - n + 1) \PP(X = k) \\
            &= \EE[X(x - 1) \cdots (X - n + 1)]
        \end{align*}
    \item Other expectations:
        \begin{align*}
            \EE[X^2]
            &= \EE[X(X - 1)] + \EE[X] \\
            &= G_X''(1) + G_X'(1)
        \end{align*}
        \begin{align*}
            \mathrm{Var}(X) = G_X''(1) + G_X'(1) -
            [G_X'(1)]^2
        \end{align*}
        Idea: Find in general $\EE[P(X)]$ using
        $\EE[\text{falling factorials of $X$}$.
\end{itemize}

\begin{note*}[Linear Algebra Aside]
    The falling factorials
    \[ 1, X, X(X - 1), X(X - 1)(X - 2) \]
    form a \emph{basis} for $\RR[X]$ (the set of
    polynomials with real coefficients).
\end{note*}

\subsubsection*{PGFs for sums of Independent Random Variables}

$X_1, \dots, X_n$ independent random variables. \\
$G_{X_1}, \dots, G_{X_n}$ are the PGFs. \\
Let $X = X_1 + \cdots + X_n$. \\
\ul{Question}: What's the PGF of $X$? (Is it
nice)?
\begin{align*}
    G_X(z)
    &= \EE[Z^X] \\
    &= \EE[z^{X_1 + \cdots + X_n}] \\
    &= \EE[z^{X_1} z^{X_2} \cdots z^{X_n}] \\
    &= \EE[z^{X_1}] \cdots \EE[z^{X_n}] \\
    &= G_{X_1}(z) \cdots G_{X_n}(z)
\end{align*}
Special case: $X_i = X_1 \to G_X(z) =
(G_{X_1}(z))^n$.

\begin{note*}
    \[ \EE[f(X) g(Y)] = \EE[f(X)] \EE[g(Y)] \]
    for independent random variables $X, Y$.
\end{note*}
