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probability-theory/inputs/lecture_21.tex

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\lecture{21}{2023-06-29}{}
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\subsection{An Application of the Optional Stopping Theorem}
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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 $E$.
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% 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}
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$E =\R$,
\[\mathbf{P}(x, \dif y) = \frac{1}{\sqrt{2 \pi} } e^{- \frac{(x-y)^2}{2}} \dif y\]
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is a transition probability measure.
\end{example}
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\begin{example}[Simple random walk as a transition probability measure]
$E = \Z$, $\mathbf{P}(x, \dif y)$
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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
\[
(\mathbf{P} f)(x) \coloneqq \int_E f(y) \mathbf{P}(x, \dif y).
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\]
This $\mathbf{P}$ is called a \vocab{transition operator}.
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\end{definition}
\begin{fact}
If $f \ge 0$, then $(\mathbf{P} f)(\cdot ) \ge 0$.
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If $f \equiv 1$, we have $(\mathbf{P} f) \equiv 1$.
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\end{fact}
\begin{notation}
Let $\mathbf{I}$ denote the \vocab{identity operator},
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i.e.
\[
(\mathbf{I} f)(x) = f(x)
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\]
for all $f$.
Then for a transition operator $\mathbf{P}$ we write
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\[
\mathbf{L} \coloneqq \mathbf{I} - \mathbf{P}.
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\]
\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
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$\mathbf{L}u = 0$ on $A^c$ and $u = f$ on $A$.
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\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 $\mathbf{P}(x, \cdot )$.
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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
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\begin{enumerate}[(i)]
\item $\bP_x[X_0 = x] = 1$,
\item for all bounded, measurable $f: \R \to \R$,
\[\bE_x[f(X_{n+1}) | \cF_n] \overset{\text{a.s.}}{=}%
\bE_{x}[f(X_{n+1}) | X_n] = %
\int f(y) \mathbf{P}(X_n, \dif y).\]
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\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 $\sigma(X_1,\ldots, X_{n-1})$.
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\begin{claim*}
For a set $B$, we have
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\[
\bE[\One_{X_{n+1} \in B} | \sigma(X_1,\ldots, X_n)] %
= \bE[\One_{X_{n+1} \in B} | \sigma(X_n)].\]
\end{claim*}
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\begin{subproof}
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\todo{TODO}
% We have $\sigma(\One_{X_{n+1} \in B}) \subseteq \sigma(X_{n}, \xi_{n+1})$.
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% $\sigma(X_1,\ldots,X_{n-1})$
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% is independent of $\sigma( \sigma(\One_{X_{n+1} \in B}), X_n)$.
% Hence the claim follows from \autoref{ceroleofindependence}.
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\end{subproof}
\end{example}
{ \large\color{red}
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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.
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No notes will be allowed in the exam.
Theorems from the lecture as well as
homework exercises might be part of the exam.