\lecture{21}{2023-06-29}{}
% TODO: replace bf

This is the last lecture relevant for the exam.
(Apart from lecture 22 which will be a repetion).

\begin{goal}
    We want to see an application of the
    optional stopping theorem \ref{optionalstopping}.
\end{goal}

\begin{notation}
    Let $E$ be a complete, separable metric space (e.g.~$E = \R$).
    Suppose that for all $x \in  E$ we have a probability measure
    $\bfP(x, \dif y)$ on $E$.
    % i.e. $\mu(A) \coloneqq \int_A \bP(x, \dif y)$ is a probability measure.
    Such a probability measure is a called
    a \vocab{transition probability measure}.
\end{notation}
\begin{examle}
    $E =\R$,
    \[\bfP(x, \dif y) = \frac{1}{\sqrt{2 \pi} } e^{- \frac{(x-y)^2}{2}} \dif y\]
    is a transition probability measure.
\end{examle}
\begin{example}[Simple random walk as a transition probability measure]
    $E = \Z$, $\bfP(x, \dif y)$
    assigns mass $\frac{1}{2}$ to $y = x+1$ and $y = x -1$.
\end{example}

\begin{definition}
    For every bounded, measurable function $f : E \to  \R$,
    $x \in E$
    define
    \[
        (\bfP f)(x) \coloneqq  \int_E f(y) \bfP(x, \dif y).
    \]
    This $\bfP$ is called a \vocab{transition operator}.
\end{definition}
\begin{fact}
    If $f \ge 0$, then $(\bfP f)(\cdot ) \ge 0$.

    If $f \equiv 1$, we have $(\bfP f) \equiv 1$.
\end{fact}

\begin{notation}
    Let $\bfI$ denote the \vocab{identity operator},
    i.e.
    \[
        (\bfI f)(x) = f(x)
    \]
    for all $f$.
    Then for a transition operator $\bfP$ we write
    \[
    \bfL \coloneqq  \bfI - \bfP.
    \]
\end{notation}

\begin{goal}
Take $E = \R$.
Suppose that $A^c \subseteq \R$ is a bounded domain.
Given a bounded function $f$ on $\R$,
we want a function $u$ which is bounded,
such that
$Lu = 0$ on $A^c$ and $u = f$ on $A$.
\end{goal}

We will show that $u(x) = \bE_x[f(X_{T_A})]$
is the unique solution to this problem.

\begin{definition}
    Let $(\Omega, \cF, \{\cF_n\}_n, \bP_x)$
    be a filtered probability space, where for every $x \in \R$,
    $\bP_x$ is a probability measure.
    Let $\bE_x$ denote expectation with respect to $\bfP(x, \cdot )$.
    Then $(X_n)_{n \ge 0}$ is a \vocab{Markov chain} starting at $x \in \R$
    with \vocab[Markov chain!Transition probability]{transition probability}
    $\bfP(x, \cdot )$ if
    \begin{enumerate}[(i)]
        \item $\bP_x[X_0 = x] = 1$,
        \item for all bounded, measurable $f: \R \to  \R$,
            \[\bE_x[f(X_{n+1}) | \cF_n] \overset{\text{a.s.}}{=}%
                \bE_{x}[f(X_{n+1}) | X_n] = %
            \int f(y) \bfP(X_n, \dif y).\]
    \end{enumerate}
    (Recall $\cF_n = \sigma(X_1,\ldots, X_n)$.)
\end{definition}
\begin{example}
    Suppose $B \in \cB(\R)$ and $f = \One_B$.
    Then the first equality of (ii) simplifies to
    \[
    \bP_x[X_{n+1} \in  B | \cF_n] = \bP_x[X_{n+1} \in B | \sigma(X_n)].
    \]
\end{example}

\begin{definition}[Conditional probability]
    \[
    \bP[A | \cG] \coloneqq \bE[\One_A | \cG].
    \]
\end{definition}

\begin{example}
    Let $\xi_i$ be i.i.d.~with$\bP[\xi_i = 1] = \bP[\xi_i = -1] = \frac{1}{2}$
    and define $X_n \coloneqq \sum_{i=1}^{n} \xi_i$.

    Intuitively, conditioned on $X_n$, $X_{n+1}$ should
    be independent of $\sigma(X_1,\ldots, X_{n-1})$.

    For a set $B$, we have
    \[
    \bP_0[X_{n+1} \in B| \sigma(X_1,\ldots, X_n)]
    = \bE[\One_{X_n + \xi_{n+1} \in B} | \sigma(X_1,\ldots, X_n)]
    = \bE[\One_{X_n + \xi_{n+1} \in B} | \sigma(X_n)].
    \]

    \begin{claim}
        $\bE[\One_{X_{n+1} \in B} | \sigma(X_1,\ldots, X_n)] = \bE[\One_{X_{n+1} \in B} | \sigma(X_n)]$.
    \end{claim}
    \begin{subproof}
    The rest of the lecture was very chaotic...
    \end{subproof}
\end{example}


%TODO







{   \huge\color{red}
    New information after this point is not relevant for the exam.
}
Stopping times and optional stopping are very relevant for the exam,
the Markov property is not.