probability-theory/inputs/lecture_21.tex

\lecture{21}{2023-06-29}{}
\subsection{An Application of the Optional Stopping Theorem}

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
    $\mathbf{P}(x, \dif y)$ 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{example}
    $E =\R$,
    \[\mathbf{P}(x, \dif y) = \frac{1}{\sqrt{2 \pi} } e^{- \frac{(x-y)^2}{2}} \dif y\]
    is a transition probability measure.
\end{example}
\begin{example}[Simple random walk as a transition probability measure]
    $E = \Z$, $\mathbf{P}(x, \dif y)$
    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
    \[
        (\mathbf{P} f)(x) \coloneqq  \int_E f(y) \mathbf{P}(x, \dif y).
    \]
    This $\mathbf{P}$ is called a \vocab{transition operator}.
\end{definition}
\begin{fact}
    If $f \ge 0$, then $(\mathbf{P} f)(\cdot ) \ge 0$.

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

\begin{notation}
    Let $\mathbf{I}$ denote the \vocab{identity operator},
    i.e.
    \[
        (\mathbf{I} f)(x) = f(x)
    \]
    for all $f$.
    Then for a transition operator $\mathbf{P}$ we write
    \[
    \mathbf{L} \coloneqq  \mathbf{I} - \mathbf{P}.
    \]
\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
$\mathbf{L}u = 0$ 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 $\mathbf{P}(x, \cdot )$.
    Then $(X_n)_{n \ge 0}$ is a \vocab{Markov chain} starting at $x \in \R$
    with \vocab[Markov chain!Transition probability]{transition probability}
    $\mathbf{P}(x, \cdot )$ if
    \begin{enumerate}[(i)]
        \item $\bP_x[X_0 = x] = 1$,
        \item for all bounded, measurable $f: \R \to  \R$,
            \[\bE_x[f(X_{n+1}) | \cF_n] \overset{\text{a.s.}}{=}%
                \bE_{x}[f(X_{n+1}) | X_n] = %
            \int f(y) \mathbf{P}(X_n, \dif y).\]
    \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{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 $\sigma(X_1,\ldots, X_{n-1})$.

    \begin{claim*}
        For a set $B$, we have
        \[
        \bE[\One_{X_{n+1} \in B} | \sigma(X_1,\ldots, X_n)] %
        = \bE[\One_{X_{n+1} \in B} | \sigma(X_n)].\]
    \end{claim*}
    \begin{subproof}
        \todo{TODO}
        % We have $\sigma(\One_{X_{n+1} \in B}) \subseteq \sigma(X_{n}, \xi_{n+1})$.
        % $\sigma(X_1,\ldots,X_{n-1})$
        % is independent of $\sigma( \sigma(\One_{X_{n+1} \in B}), X_n)$.
        % Hence the claim follows from \autoref{ceroleofindependence}.
    \end{subproof}
\end{example}

{   \large\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.
No notes will be allowed in the exam.
Theorems from the lecture as well as
homework exercises might be part of the exam.