% vim: tw=50
% 27/10/2022 10AM

\begin{corollary*}
    All states in a communicating class are either
    all recurrent or all transient.
\end{corollary*}

\begin{theorem*}
    If $C$ is a recurrent communicating class,
    then $C$ is closed.
\end{theorem*}

\begin{proof}
    Let $x \in C$ and $x \to y$, but $y \not\in
    C$. Since $x \to y$, $\exists m \ge 0$ such
    that $p_{xy}(m) < 0$, so
    \[ \PP_x(V_x < \infty) \ge p_{xy}(m) > 0 \]
    SO this shows that $x$ is transient,
    contradiction.
\end{proof}

\begin{theorem*}
    A \emph{finite} closed class is recurrent.
\end{theorem*}

\begin{proof}
    Let $x \in C$. Since $X$ is finite, $\exists y
    \in C$ such that
    \[ \PP_x(X_n = y \text{ for infinitely many
    $n$}) > 0 \]
    by the pigeonhole principle.
    \begin{align*}
        \PP(X_n = y \text{ for infinitely many
        $n$})
        &\ge \PP_y(X_m = x, X_n = y \text{ for
        infinitely many $n \ge m$} \\
        &= \PP_y(X_m = y \text{ for infinitely
        many $n \ge m$} \mid X_m = x) \PP_y(X_m =
        x) \\
        &= \PP_x(X_n = y \text{ for infinitely
        many $n$}) p_{yx}(m) \\
        &> 0
    \end{align*}
    so $\PP_y(X_n = y \text{ for infinitely many
    $n$}) > 0$. So $y$ is recurrent.
\end{proof}

\begin{theorem*}
    Let $P$ be irreducible and recurrent. Then for
    all $x, y$
    \[ \PP_x(T_y < \infty) = 1 \]
\end{theorem*}

\begin{proof}
    \begin{align*}
        \PP_x(X_n = y \text{ infinitely many
        times})
        &= \PP_X(T_y < \infty, X_n = y \text{ for
        infinitely many $n \ge T_y$}) \\
        &= \PP_x(X_n = y \text{ infiniteley many }
        n \ge T_y \mid T_y < \infty) \\
        &~~~\cdot \PP_x(T_y <
        \infty) \\
        &= \PP_y(X_n = y \text{ infinitely many
        $n$}) \cdot \PP_x(T_y < \infty) \\
        &= \PP_x(T_y < \infty)
    \end{align*}
    Suppose $\PP_x(T_y < \infty) < 1$. Then
    $\PP_x(T_y = \infty) > 0$. Pick $p_{yx}(m) >
    0$, and define
    \[ \tilde{T}_y = \inf\{n \ge m : X_n = y\} \]
    Then
    \begin{align*}
        \PP_y(V_y < \infty)
        &\ge \PP_y(X_m = x,
        \tilde{T}_y = \infty) \\
        &= \PP_y(\tilde{T}_y = \infty \mid X_m =
        x) \PP_y(X_m = x) \\
        &= \PP_x(T_y = \infty) p_{yx}(m) \\
        &> 0
    \end{align*}
    so $y$ is transient.
\end{proof}

\newpage
\section{Simple random walks on $\ZZ^d$}

\begin{definition*}
    A simple random walk on $\ZZ^d$ is a Markov
    chain with transition matrix
    \[ p(x, x + e_i) = p(x, x - e_i) =
    \frac{1}{2d} \quad \forall x \in \ZZ^d, ~
    \forall i = 1, \dots, d \]
    where $e_i$ is the standard basis of $\RR^d$.
\end{definition*}

\begin{theorem*}[P\'olya]
    A simple random walk is recurrent when $d \le
    2$ and it is transient when $d \ge 3$.
\end{theorem*}

\begin{proof}
    \renewcommand\qedsymbol{}
    \begin{enumerate}
        \item[$d = 1$] Need to show 0 is
            recurrent, i.e. we want to show
            \[ \sum p_{00}(n) = \infty \]
            \[ p_{00}(n) = \PP_0(X_n = 0) \]
            \[ \PP_0(X_{2n} = 0) = {2n \choose n}
            \cdot \left( \half \right)^{2n} =
            \frac{(2n)!}{n!n!} \frac{1}{2^{2n}} \]
            Recall Stirling's formula: $n! \sim
            n^n e^{-n} \cdot e^{-n} \cdot
            \sqrt{2\pi n}$ so
            \[ \PP_0(X_{2n} = 0) \sim
            \frac{1}{\sqrt{\pi n}} \]
            so $\sum_n p_{00}(2n) = \infty$. So
            simple random walk on $\ZZ$ is
            recurrent. \myskip
            Now consider a random walk where we
            move right with probability $p$, and
            left with probability $q = 1 -
            p$, with $p \neq q$. Then
            \[ \PP_0(X_{2n} = 0) = {2n \choose n}
            \cdot p^n \cdot q^n \sim
            \frac{(4pq)^n}{\sqrt{\pi n}} \]
            Since $p \neq q$, $4pq < 1$, so
            \[ \sum_n \frac{(4pq)^n}{\sqrt{\pi n}}
            < \infty \]
        \item[$d = 2$] Consider projecting the
            random walk as follows:
            \begin{center}
                \includegraphics[width=0.6\linewidth]
                {images/ed2f194055db11ed.png}
            \end{center}
            Define a function $f : \ZZ^2 \to
            \RR^2$
            \[ f(x, y) = \left( \frac{x +
            y}{\sqrt{2}}, \frac{x - y}{\sqrt{2}} \right) \]
            $(X_n)$ simple random walk on $\ZZ^2$,
            $f(X_n) = (X_n^+, X_n^-)$. We claim
            that $(X_n^+)$ and $(X_n^-)$ are 2
            independent simple random walks on
            $\frac{\ZZ}{2}$. Let $(\xi_i)$ be an
            iid sequence
            \[ \PP(\xi_i = (0, 1)) = \PP(\xi_2 =
            (1,0)) = \cdots = \frac{1}{4} \]
            So let $X_n = \sum_{i = 1}^n \xi_i$
            and $\xi_i = (\xi_i^1, \xi_i^2)$.
            \[ f(X_n) = \left( \sum_{i = 1}^n
            \frac{\xi_i^1 + \xi_i^2}{\sqrt{2}},
            \sum_{i = 1}^n \frac{\xi_i^1 -
            \xi_i^2}{\sqrt{2}} \right) \]
            So we want to show that $\xi_i^1 +
            \xi_i^2$ is independent of $\xi_i^1 -
            \xi_i^2$. This can be done by checking
            lots of calculations / cases. So
            $(X_n^+)$ and $(X_n^-)$ are
            independent. Now
            \begin{align*}
                \PP_0(X_{2n} = 0)
                &= \PP_0(X_{2n}^+ = 0, X_{2n}^- =
                0) \\
                &= \PP_0(X_{2n}^+ = 0)
                \PP_0(X_{2n}^- = 0) \\
                &\sim \left( \frac{1}{\sqrt{n}}
                \right)^2 \\
                &= \frac{1}{n}
            \end{align*}
            so $\sum_n \frac{1}{n} = \infty$ so
            recurrent.
    \end{enumerate}
\end{proof}
