lecture 21

inputs/lecture_1.tex

@@ -16,7 +16,7 @@ First, let us recall some basic definitions:
         \item $\bP$ is a \vocab{probability measure}, i.e.~$\bP$ is a function $\bP: \cF \to [0,1]$
             such that
             \begin{itemize}
-                \item $\bP(\emptyset) = 1$, $\bP(\Omega)  = 1$,
+                \item $\bP(\emptyset) = 0$, $\bP(\Omega)  = 1$,
                 \item $\bP\left( \bigsqcup_{n \in \N} A_n \right) = \sum_{n \in \N} \bP(A_n)$
                     for mutually disjoint $A_n \in \cF$.
             \end{itemize}

inputs/lecture_20.tex

@@ -116,7 +116,7 @@ we need the following theorem, which we won't prove here:
 \subsection{Stopping times}
 
 \begin{definition}[Stopping time]
-    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},
+    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},
     if
     \[
     \{T \le  n\} \in  \cF_n

inputs/lecture_21.tex

@@ -0,0 +1,136 @@
+\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.

jrpie-math.sty

@@ -40,6 +40,10 @@
 \RequirePackage{mkessler-faktor}
 \RequirePackage{mkessler-mathsymb}
 \RequirePackage[extended]{mkessler-mathalias}
+% \makeatletter
+% \expandafter\MakeAliasesForwith\expandafter\mathbf\expandafter{\expandafter bf\expandafter}\expandafter{\mkessler@mathalias@all}
+% \makeatother
+
 \RequirePackage{mkessler-refproof}
 
 % mkessler-mathfont has already been imported

probability_theory.tex

@@ -44,6 +44,7 @@
 \input{inputs/lecture_18.tex}
 \input{inputs/lecture_19.tex}
 \input{inputs/lecture_20.tex}
+\input{inputs/lecture_21.tex}
 
 \cleardoublepage