124 lines
4.1 KiB
TeX
124 lines
4.1 KiB
TeX
\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.
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(Apart from lecture 22 which will be a repetion).
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\begin{goal}
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We want to see an application of the
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optional stopping theorem \ref{optionalstopping}.
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\end{goal}
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\begin{notation}
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Let $E$ be a complete, separable metric space (e.g.~$E = \R$).
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Suppose that for all $x \in E$ we have a probability measure
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$\mathbf{P}(x, \dif y)$ on $E$.
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% i.e. $\mu(A) \coloneqq \int_A \bP(x, \dif y)$ is a probability measure.
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Such a probability measure is a called
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a \vocab{transition probability measure}.
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\end{notation}
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\begin{example}
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$E =\R$,
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\[\mathbf{P}(x, \dif y) = \frac{1}{\sqrt{2 \pi} } e^{- \frac{(x-y)^2}{2}} \dif y\]
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is a transition probability measure.
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\end{example}
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\begin{example}[Simple random walk as a transition probability measure]
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$E = \Z$, $\mathbf{P}(x, \dif y)$
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assigns mass $\frac{1}{2}$ to $y = x+1$ and $y = x -1$.
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\end{example}
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\begin{definition}
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For every bounded, measurable function $f : E \to \R$,
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$x \in E$
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define
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\[
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(\mathbf{P} f)(x) \coloneqq \int_E f(y) \mathbf{P}(x, \dif y).
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\]
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This $\mathbf{P}$ is called a \vocab{transition operator}.
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\end{definition}
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\begin{fact}
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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}
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\begin{notation}
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Let $\mathbf{I}$ denote the \vocab{identity operator},
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i.e.
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\[
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(\mathbf{I} f)(x) = f(x)
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\]
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for all $f$.
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Then for a transition operator $\mathbf{P}$ we write
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\[
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\mathbf{L} \coloneqq \mathbf{I} - \mathbf{P}.
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\]
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\end{notation}
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\begin{goal}
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Take $E = \R$.
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Suppose that $A^c \subseteq \R$ is a bounded domain.
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Given a bounded function $f$ on $\R$,
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we want a function $u$ which is bounded,
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such that
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$\mathbf{L}u = 0$ on $A^c$ and $u = f$ on $A$.
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\end{goal}
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We will show that $u(x) = \bE_x[f(X_{T_A})]$
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is the unique solution to this problem.
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\begin{definition}
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Let $(\Omega, \cF, \{\cF_n\}_n, \bP_x)$
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be a filtered probability space, where for every $x \in \R$,
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$\bP_x$ is a probability measure.
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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 $x \in \R$
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with \vocab[Markov chain!Transition probability]{transition probability}
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$\mathbf{P}(x, \cdot )$ if
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\begin{enumerate}[(i)]
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\item $\bP_x[X_0 = x] = 1$,
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\item for all bounded, measurable $f: \R \to \R$,
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\[\bE_x[f(X_{n+1}) | \cF_n] \overset{\text{a.s.}}{=}%
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\bE_{x}[f(X_{n+1}) | X_n] = %
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\int f(y) \mathbf{P}(X_n, \dif y).\]
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\end{enumerate}
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(Recall $\cF_n = \sigma(X_1,\ldots, X_n)$.)
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\end{definition}
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\begin{example}
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Suppose $B \in \cB(\R)$ and $f = \One_B$.
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Then the first equality of (ii) simplifies to
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\[
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\bP_x[X_{n+1} \in B | \cF_n] = \bP_x[X_{n+1} \in B | \sigma(X_n)].
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\]
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\end{example}
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\begin{example}
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Let $\xi_i$ be i.i.d.~with$\bP[\xi_i = 1] = \bP[\xi_i = -1] = \frac{1}{2}$
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and define $X_n \coloneqq \sum_{i=1}^{n} \xi_i$.
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Intuitively, conditioned on $X_n$, $X_{n+1}$ should
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be independent of $\sigma(X_1,\ldots, X_{n-1})$.
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\begin{claim*}
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For a set $B$, we have
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\[
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\bE[\One_{X_{n+1} \in B} | \sigma(X_1,\ldots, X_n)] %
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= \bE[\One_{X_{n+1} \in B} | \sigma(X_n)].\]
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\end{claim*}
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\begin{subproof}
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\todo{TODO}
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% 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 $\sigma( \sigma(\One_{X_{n+1} \in B}), X_n)$.
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% Hence the claim follows from \autoref{ceroleofindependence}.
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\end{subproof}
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\end{example}
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{ \large\color{red}
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New information after this point is not relevant for the exam.
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}
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Stopping times and optional stopping are very relevant for the exam,
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the Markov property is not.
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No notes will be allowed in the exam.
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Theorems from the lecture as well as
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homework exercises might be part of the exam.
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