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

@@ -2,7 +2,7 @@
     By the tower property (\autoref{cetower})
     it is clear that $(\bE[X | \cF_n])_n$
     is a martingale.
-    
+
     First step:
     Assume that $X$ is bounded.
     Then, by \autoref{cejensen}, $|X_n| \le  \bE[|X| | \cF_n]$,
@@ -84,7 +84,7 @@ we need the following theorem, which we won't prove here:
         L^p &\longrightarrow & (L^q)^\ast \\
         f &\longmapsto & (g \mapsto \int g f \dif d\bP)
     \end{IEEEeqnarray*}
-    
+
     We also have $(L^1)^\ast \cong L^\infty$,
     however $ (L^\infty)^\ast \not\cong L^1$.
 \end{fact}
@@ -95,7 +95,7 @@ we need the following theorem, which we won't prove here:
     $(X_{n_k})_k$ such that for all $Y \in L^q$ ($\frac{1}{p} + \frac{1}{q} = 1$ )
     \[
     \int X_{n_k} Y \dif \bP \to  \int XY \dif \bP
-    \] 
+    \]
     (Note that this argument does not work for $p = 1$,
     because $(L^\infty)^\ast \not\cong  L^1$).
 
@@ -116,14 +116,14 @@ 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
     \[
     \{T \le  n\} \in  \cF_n
-    \] 
+    \]
     for all $n \in \N$.
     Equivalently, $\{T = n\}  \in \cF_n$ for all $n \in \N$.
-    
+
 \end{definition}
 
 \begin{example}
@@ -131,21 +131,21 @@ we need the following theorem, which we won't prove here:
 \end{example}
 
 \begin{example}[Hitting times]
-    For an adapted process $(X_n)_n$ 
+    For an adapted process $(X_n)_n$
     with values in $\R$ and $A \in  \cB(\R)$, the \vocab{hitting time}
     \[
-    T \coloneqq \inf \{n \in \N : X_n \in A\} 
+    T \coloneqq \inf \{n \in \N : X_n \in A\}
     \]
     is a stopping time,
     as
     \[
     \{T \le  n \} = \bigcup_{k=1}^n \{X_k \in A\} \in \cF_n.
-    \] 
+    \]
 
     However, the last exit time
     \[
-    T \coloneqq  \sup \{n \in \N : X_n \in  A\} 
+    T \coloneqq  \sup \{n \in \N : X_n \in  A\}
-    \] 
+    \]
     is not a stopping time.
 
 \end{example}
@@ -158,7 +158,7 @@ we need the following theorem, which we won't prove here:
     Then
     \[
     T \coloneqq \inf \{n \in \N : S_n \ge A \lor S_n \le  B\}
-    \] 
+    \]
     is a stopping time.
 \end{example}
 
@@ -173,11 +173,11 @@ we need the following theorem, which we won't prove here:
     are stopping times.
 
     Note that $T_1 - T_2$ is not a stopping time.
-    
+
 \end{example}
 
 \begin{remark}
-    There are two ways to interpret the interaction between a stopping time $T$ 
+    There are two ways to interpret the interaction between a stopping time $T$
     and a stochastic process $(X_n)_n$.
     \begin{itemize}
         \item The behaviour of $ X_n$ until $T$,
@@ -193,22 +193,22 @@ we need the following theorem, which we won't prove here:
     If we look at a process
     \[
     S_n = \sum_{i=1}^{n}  X_i
-    \] 
+    \]
     for some $(X_n)_n$, then
     \[
     S^T = (\sum_{i=1}^{T \wedge n} X_i)_n
-    \] 
+    \]
     and
     \[
     S_T = \sum_{i=1}^{T}  X_i.
-    \] 
+    \]
 \end{example}
 
 \begin{theorem}
     If $(X_n)_n$ is a supermartingale and $T$ is a stopping time,
     then $X^T$ is also a supermartingale,
     and we have $\bE[X_{T \wedge n}] \le  \bE[X_0]$ for all $n$.
-    If $(X_n)_n$ is a martingale, then so is $X^T$ 
+    If $(X_n)_n$ is a martingale, then so is $X^T$
     and $\bE[X_{T \wedge n}] \le  \bE[X_0]$.
 \end{theorem}
 \begin{proof}
@@ -222,7 +222,7 @@ we need the following theorem, which we won't prove here:
     It is also clear that $X^T_n$ is integrable since
     \[
     \bE[|X^T_n|] \le  \sum_{k=1}^{n} \bE[|X_k|] < \infty.
-    \] 
+    \]
 
     We have
     \begin{IEEEeqnarray*}{rCl}
@@ -259,7 +259,7 @@ we need the following theorem, which we won't prove here:
     Then $\bP[T < \infty] = 1$, but
     \[
     1 = \bE[S_T] \neq  \bE[S_0] = 0.
-    \] 
+    \]
 \end{example}
 
 \begin{theorem}[Optional Stopping]

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