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% 01/11/2022 10AM

\vspace{-1\baselineskip}
\begin{enumerate}
    \item[$d = 3$] We will prove that $\sum_n
        p_{00}(n) < \infty$, which will imply
        it is transient. Let's compute
        $p_{00}(2n)$. In order to be back at 0
        after $2n$ steps it must make $i$
        steps to the right, $i$ to the left,
        $j$ north and $j$ south, $k$ west and
        $k$ east for some $i, j, k \ge 0$ and
        $i + j + k = n$. So
        \begin{align*}
            p_{00}(2n) = \sum_{\substack{i, j,
            k \ge 0\\i + j + k = n}} {2n
            \choose i, i, j, j, k, k} \left(
            \frac{1}{6} \right)^{2n} \\
            &= {2n \choose n} \left( \half
            \right)^{2n} \sum_{\substack{i, j,
            k \ge 0\\i + j + k = n}} {n
            \choose i, j, k}^2 \left(
            \frac{1}{3} \right)^{2n} \\
            \sum_{\substack{i, j, k \ge
            0\\i +  j + k=n}} {n \choose i, j,
            k} \left( \frac{1}{3} \right)^n
            &= 1
        \end{align*}
        Let $n = 3m$. We claim that
        \[ {n \choose i, j, k} \le {n \choose
        m, m, m} \]
        To prove this, suppose the maximum
        over $i, j, k$ is attained at some $i,
        j, k$ with $i > j + 1$. Then
        \[ {n \choose i, j, k} < {n \choose i
        - 1, j + 1, k} \]
        because $i! j! > (i - 1)!(j + 1)!$. So
        for $n = 3m$,
        \[ p_{00}(2n) \le {2n \choose n}
        \left( \half \right)^{2n} \left(
        \frac{1}{3} \right)^n {n \choose m, m,
        m} = 1 \]
        Stirling's formula gives
        \[ p_{00}(2n) \le \frac{A}{n^{3/2}} \]
        for some $A > 0$. So $\sum_m
        p_{00}(6m) < \infty$. But also
        $p_{00}(6m) \ge p_{00}(6m - 2) \left(
        \frac{1}{6} \right)^2$ and $p_{00}(6m)
        \ge p_{00}(6m - 4) \left( \frac{1}{6}
        \right)^4$. So $\sum_n p_{00}(2n) <
        \infty$ so it's transient.
        \hfill\qedsymbol
\end{enumerate}

\subsubsection*{Invariant distribution}

\begin{flashcard}[probability-distribution]
\begin{definition*}
    $I$ discrete (countable / finite) set.
    $\lambda = (\lambda_i : i \in I)$ is a
    probability distribution if \cloze{$\lambda_i \ge 0$
    for all $i$ and $\sum_{i \in I} \lambda_i =
    1$.}
\end{definition*}
\end{flashcard}

\begin{example*}
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/dee919c859d011ed.png}
    \end{center}
    Then $p_{11}(n) \to \half$. So the Markov
    chain will converge to $(\half, \half)$.
\end{example*}

\noindent
Want to find a distribution $\pi$ such that if
$X_0 \sim \pi$, then $X_n \sim \pi$ for all $n$.
\begin{align*}
    \PP(X_1 = j)
    &= \sum_i \PP(X_0 = i, X_1 = y) \\
    &= \sum_i \PP(X_1 = j \mid X_0 = i) \PP(X_0 =
    1) \\
    &= \sum_i P(i, j) \pi(i)
\end{align*}
so $\pi(j) = \sum_i \pi(i) P(i, j)$ for all $j$.
$\pi = \pi P$. ($\pi$ as a row vector).

\begin{flashcard}[InvDist]
\begin{definition*}
    A probability distribution $\pi$ is called
    \emph{invariant} / \emph{stationary}
    / \emph{equilibrium} if \cloze{$\pi = \pi P$.}
\end{definition*}
\end{flashcard}

\begin{theorem*}
    Let $\pi$ be invariant and $X_0 \sim \pi$.
    Then $X_n \sim \pi$ for all $n$.
\end{theorem*}

\begin{proof}
    $n = 0$ is done.
    \begin{align*}
        \PP(X_{n + 1} = j)
        &= \sum_i \PP(X_{n + 1} = j, X_n = i) \\
        &= \sum_i \PP(X_{n + 1} = j \mid X_n = i)
        \PP(X_n = i) \\
        &= \sum_i P(i, j) \pi(i) \\
        &= \pi(j)
        &&(\pi = \pi P)
        \qedhere
    \end{align*}
\end{proof}

\begin{flashcard}[FiniteLimitIsInvDist]
\begin{theorem*}
    Let $I$ be a \cloze{\emph{finite} set}
    and $\exists i \in I$
    such that $p_{ij}(n) \to \pi(j)$ as $n \to
    \infty$. Then \cloze{$\pi = (\pi_i : i \in I)$ is an
    invariant distribution.}
\end{theorem*}
\end{flashcard}

\begin{proof}
    \begin{align*}
        \sum_{j \in I} \pi_j
        &= \sum_{j \in I} \lim_{n \to \infty}
        p_{ij}(n) \\
        &= \lim_{n \to \infty} \sum_{j \in I}
        p_{ij}(n) \\
        &= 1
    \end{align*}
    (we changed the order of sum and limit because
    the sum is finite). So $\pi$ is a distribution.
    \begin{align*}
        \pi_j
        &= \lim_{n \to \infty} p_{ij}(n) \\
        &= \lim_{n \to \infty} \sum_{k \in I}
        p_{ik}(n - 1) P(k, j) \\
        &= \sum_{k \in I} \lim_{n \to \infty}
        p_{ik}(n - 1) P(k, j) \\
        &= \sum_{k \in I} \pi_k P(k, j)
    \end{align*}
    i.e. $\pi = \pi P$.
\end{proof}

\begin{remark*}
    $I$ finite is essential: Consider a simple
    random walk on $\ZZ$. Then $p_{00}(2n) \sim
    \frac{A}{\sqrt{n}} \to 0$. Similarly
    $p_{0x}(n) \to \infty$ as $n \to \infty$.
\end{remark*}

\begin{remark*}
    $P$ is a stochastic matrix, so 1 is always an
    eigenvalue. If $P$ is irreducible, on a finite
    state space, then the Perron-Frobenius theorem
    from linear algebra ensures the existence of
    the invariant distribution.
\end{remark*}

\begin{definition*}
    $k \in I$, $T_k = \inf\{n \ge 1 : X_n = k\}$.
    First return time to $k$. $i \in I$
    \[ \nu_k(i) = \EE_k \left[ \sum_{l = 0}^{T_k
    - 1} \mathbbm{1}(X_l = i) \right] \]
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/e0dbbb4859d311ed.png}
    \end{center}
    So $\nu_k(i)$ is the expected number of visits
    to $i$ during an excursion from $k$. So
    $\nu_k$ is a measure on $I$.
\end{definition*}