lecture 21
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@ -16,7 +16,7 @@ First, let us recall some basic definitions:
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\item $\bP$ is a \vocab{probability measure}, i.e.~$\bP$ is a function $\bP: \cF \to [0,1]$
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\item $\bP$ is a \vocab{probability measure}, i.e.~$\bP$ is a function $\bP: \cF \to [0,1]$
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such that
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such that
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\begin{itemize}
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\begin{itemize}
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\item $\bP(\emptyset) = 1$, $\bP(\Omega) = 1$,
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\item $\bP(\emptyset) = 0$, $\bP(\Omega) = 1$,
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\item $\bP\left( \bigsqcup_{n \in \N} A_n \right) = \sum_{n \in \N} \bP(A_n)$
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\item $\bP\left( \bigsqcup_{n \in \N} A_n \right) = \sum_{n \in \N} \bP(A_n)$
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for mutually disjoint $A_n \in \cF$.
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for mutually disjoint $A_n \in \cF$.
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\end{itemize}
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\end{itemize}
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@ -2,7 +2,7 @@
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By the tower property (\autoref{cetower})
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By the tower property (\autoref{cetower})
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it is clear that $(\bE[X | \cF_n])_n$
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it is clear that $(\bE[X | \cF_n])_n$
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is a martingale.
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is a martingale.
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First step:
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First step:
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Assume that $X$ is bounded.
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Assume that $X$ is bounded.
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Then, by \autoref{cejensen}, $|X_n| \le \bE[|X| | \cF_n]$,
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Then, by \autoref{cejensen}, $|X_n| \le \bE[|X| | \cF_n]$,
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@ -84,7 +84,7 @@ we need the following theorem, which we won't prove here:
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L^p &\longrightarrow & (L^q)^\ast \\
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L^p &\longrightarrow & (L^q)^\ast \\
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f &\longmapsto & (g \mapsto \int g f \dif d\bP)
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f &\longmapsto & (g \mapsto \int g f \dif d\bP)
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\end{IEEEeqnarray*}
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\end{IEEEeqnarray*}
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We also have $(L^1)^\ast \cong L^\infty$,
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We also have $(L^1)^\ast \cong L^\infty$,
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however $ (L^\infty)^\ast \not\cong L^1$.
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however $ (L^\infty)^\ast \not\cong L^1$.
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\end{fact}
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\end{fact}
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@ -95,7 +95,7 @@ we need the following theorem, which we won't prove here:
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$(X_{n_k})_k$ such that for all $Y \in L^q$ ($\frac{1}{p} + \frac{1}{q} = 1$ )
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$(X_{n_k})_k$ such that for all $Y \in L^q$ ($\frac{1}{p} + \frac{1}{q} = 1$ )
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\[
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\[
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\int X_{n_k} Y \dif \bP \to \int XY \dif \bP
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\int X_{n_k} Y \dif \bP \to \int XY \dif \bP
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\]
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\]
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(Note that this argument does not work for $p = 1$,
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(Note that this argument does not work for $p = 1$,
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because $(L^\infty)^\ast \not\cong L^1$).
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because $(L^\infty)^\ast \not\cong L^1$).
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@ -116,14 +116,14 @@ we need the following theorem, which we won't prove here:
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\subsection{Stopping times}
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\subsection{Stopping times}
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\begin{definition}[Stopping time]
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\begin{definition}[Stopping time]
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A random variable $T: \Omega \to \N \cup \{\infty\}$ on a filtered probability space $(\Omega, \cF, \{\cF_n\}_n, \bP)$ is called a \vocab{stopping time},
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A random variable $T: \Omega \to \N_0 \cup \{\infty\}$ on a filtered probability space $(\Omega, \cF, \{\cF_n\}_n, \bP)$ is called a \vocab{stopping time},
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if
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if
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\[
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\[
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\{T \le n\} \in \cF_n
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\{T \le n\} \in \cF_n
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\]
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\]
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for all $n \in \N$.
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for all $n \in \N$.
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Equivalently, $\{T = n\} \in \cF_n$ for all $n \in \N$.
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Equivalently, $\{T = n\} \in \cF_n$ for all $n \in \N$.
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\end{definition}
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\end{definition}
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\begin{example}
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\begin{example}
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@ -131,21 +131,21 @@ we need the following theorem, which we won't prove here:
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\end{example}
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\end{example}
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\begin{example}[Hitting times]
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\begin{example}[Hitting times]
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For an adapted process $(X_n)_n$
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For an adapted process $(X_n)_n$
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with values in $\R$ and $A \in \cB(\R)$, the \vocab{hitting time}
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with values in $\R$ and $A \in \cB(\R)$, the \vocab{hitting time}
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\[
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\[
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T \coloneqq \inf \{n \in \N : X_n \in A\}
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T \coloneqq \inf \{n \in \N : X_n \in A\}
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\]
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\]
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is a stopping time,
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is a stopping time,
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as
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as
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\[
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\[
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\{T \le n \} = \bigcup_{k=1}^n \{X_k \in A\} \in \cF_n.
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\{T \le n \} = \bigcup_{k=1}^n \{X_k \in A\} \in \cF_n.
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\]
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\]
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However, the last exit time
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However, the last exit time
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\[
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\[
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T \coloneqq \sup \{n \in \N : X_n \in A\}
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T \coloneqq \sup \{n \in \N : X_n \in A\}
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\]
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\]
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is not a stopping time.
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is not a stopping time.
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\end{example}
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\end{example}
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@ -158,7 +158,7 @@ we need the following theorem, which we won't prove here:
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Then
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Then
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\[
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\[
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T \coloneqq \inf \{n \in \N : S_n \ge A \lor S_n \le B\}
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T \coloneqq \inf \{n \in \N : S_n \ge A \lor S_n \le B\}
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\]
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\]
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is a stopping time.
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is a stopping time.
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\end{example}
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\end{example}
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@ -173,11 +173,11 @@ we need the following theorem, which we won't prove here:
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are stopping times.
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are stopping times.
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Note that $T_1 - T_2$ is not a stopping time.
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Note that $T_1 - T_2$ is not a stopping time.
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\end{example}
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\end{example}
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\begin{remark}
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\begin{remark}
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There are two ways to interpret the interaction between a stopping time $T$
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There are two ways to interpret the interaction between a stopping time $T$
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and a stochastic process $(X_n)_n$.
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and a stochastic process $(X_n)_n$.
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\begin{itemize}
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\begin{itemize}
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\item The behaviour of $ X_n$ until $T$,
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\item The behaviour of $ X_n$ until $T$,
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@ -193,22 +193,22 @@ we need the following theorem, which we won't prove here:
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If we look at a process
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If we look at a process
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\[
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\[
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S_n = \sum_{i=1}^{n} X_i
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S_n = \sum_{i=1}^{n} X_i
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\]
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\]
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for some $(X_n)_n$, then
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for some $(X_n)_n$, then
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\[
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\[
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S^T = (\sum_{i=1}^{T \wedge n} X_i)_n
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S^T = (\sum_{i=1}^{T \wedge n} X_i)_n
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\]
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\]
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and
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and
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\[
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\[
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S_T = \sum_{i=1}^{T} X_i.
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S_T = \sum_{i=1}^{T} X_i.
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\]
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\]
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\end{example}
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\end{example}
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\begin{theorem}
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\begin{theorem}
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If $(X_n)_n$ is a supermartingale and $T$ is a stopping time,
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If $(X_n)_n$ is a supermartingale and $T$ is a stopping time,
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then $X^T$ is also a supermartingale,
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then $X^T$ is also a supermartingale,
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and we have $\bE[X_{T \wedge n}] \le \bE[X_0]$ for all $n$.
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and we have $\bE[X_{T \wedge n}] \le \bE[X_0]$ for all $n$.
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If $(X_n)_n$ is a martingale, then so is $X^T$
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If $(X_n)_n$ is a martingale, then so is $X^T$
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and $\bE[X_{T \wedge n}] \le \bE[X_0]$.
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and $\bE[X_{T \wedge n}] \le \bE[X_0]$.
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\end{theorem}
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\end{theorem}
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\begin{proof}
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\begin{proof}
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@ -222,7 +222,7 @@ we need the following theorem, which we won't prove here:
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It is also clear that $X^T_n$ is integrable since
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It is also clear that $X^T_n$ is integrable since
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\[
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\[
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\bE[|X^T_n|] \le \sum_{k=1}^{n} \bE[|X_k|] < \infty.
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\bE[|X^T_n|] \le \sum_{k=1}^{n} \bE[|X_k|] < \infty.
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\]
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\]
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We have
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We have
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\begin{IEEEeqnarray*}{rCl}
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\begin{IEEEeqnarray*}{rCl}
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@ -259,7 +259,7 @@ we need the following theorem, which we won't prove here:
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Then $\bP[T < \infty] = 1$, but
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Then $\bP[T < \infty] = 1$, but
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\[
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\[
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1 = \bE[S_T] \neq \bE[S_0] = 0.
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1 = \bE[S_T] \neq \bE[S_0] = 0.
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\]
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\]
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\end{example}
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\end{example}
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\begin{theorem}[Optional Stopping]
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\begin{theorem}[Optional Stopping]
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136
inputs/lecture_21.tex
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inputs/lecture_21.tex
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\lecture{21}{2023-06-29}{}
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% TODO: replace bf
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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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$\bfP(x, \dif y)$ 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{examle}
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$E =\R$,
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\[\bfP(x, \dif y) = \frac{1}{\sqrt{2 \pi} } e^{- \frac{(x-y)^2}{2}} \dif y\]
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is a transition probability measure.
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\end{examle}
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\begin{example}[Simple random walk as a transition probability measure]
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$E = \Z$, $\bfP(x, \dif y)$
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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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\[
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(\bfP f)(x) \coloneqq \int_E f(y) \bfP(x, \dif y).
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\]
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This $\bfP$ 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 $(\bfP f)(\cdot ) \ge 0$.
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If $f \equiv 1$, we have $(\bfP f) \equiv 1$.
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\end{fact}
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\begin{notation}
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Let $\bfI$ denote the \vocab{identity operator},
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i.e.
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\[
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(\bfI f)(x) = f(x)
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\]
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for all $f$.
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Then for a transition operator $\bfP$ we write
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\[
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\bfL \coloneqq \bfI - \bfP.
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\]
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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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$Lu = 0$ 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 $\bfP(x, \cdot )$.
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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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$\bfP(x, \cdot )$ 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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\[\bE_x[f(X_{n+1}) | \cF_n] \overset{\text{a.s.}}{=}%
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\bE_{x}[f(X_{n+1}) | X_n] = %
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\int f(y) \bfP(X_n, \dif y).\]
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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{definition}[Conditional probability]
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\[
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\bP[A | \cG] \coloneqq \bE[\One_A | \cG].
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\]
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\end{definition}
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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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For a set $B$, we have
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\[
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\bP_0[X_{n+1} \in B| \sigma(X_1,\ldots, X_n)]
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= \bE[\One_{X_n + \xi_{n+1} \in B} | \sigma(X_1,\ldots, X_n)]
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= \bE[\One_{X_n + \xi_{n+1} \in B} | \sigma(X_n)].
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\]
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\begin{claim}
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$\bE[\One_{X_{n+1} \in B} | \sigma(X_1,\ldots, X_n)] = \bE[\One_{X_{n+1} \in B} | \sigma(X_n)]$.
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\end{claim}
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\begin{subproof}
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The rest of the lecture was very chaotic...
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\end{subproof}
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\end{example}
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%TODO
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{ \huge\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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\RequirePackage{mkessler-faktor}
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\RequirePackage{mkessler-faktor}
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\RequirePackage{mkessler-mathsymb}
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\RequirePackage{mkessler-mathsymb}
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\RequirePackage[extended]{mkessler-mathalias}
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\RequirePackage[extended]{mkessler-mathalias}
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% \makeatletter
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% \expandafter\MakeAliasesForwith\expandafter\mathbf\expandafter{\expandafter bf\expandafter}\expandafter{\mkessler@mathalias@all}
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% \makeatother
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\RequirePackage{mkessler-refproof}
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\RequirePackage{mkessler-refproof}
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% mkessler-mathfont has already been imported
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% mkessler-mathfont has already been imported
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\input{inputs/lecture_18.tex}
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\input{inputs/lecture_18.tex}
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\input{inputs/lecture_19.tex}
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\input{inputs/lecture_19.tex}
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\input{inputs/lecture_20.tex}
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\input{inputs/lecture_20.tex}
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\input{inputs/lecture_21.tex}
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\cleardoublepage
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\cleardoublepage
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