lecture 20

inputs/lecture_19.tex

@@ -219,7 +219,8 @@ Let $(\Omega, \cF, \bP)$ as always and let $(\cF_n)_n$ always be a filtration.
 
 \begin{theorem}
     \label{ceismartingale}
-    Let $X \in L^p$ for some $p \ge 1$.
+    Let $X \in L^p$ for some $p \ge 1$
+    and $\bigcup_n \cF_n \to \cF$.
     Then $X_n \coloneqq  \bE[X | \cF_n]$ defines a martingale which converges
     to $X$ in $L^p$.
 \end{theorem}

inputs/lecture_20.tex

@@ -22,8 +22,8 @@
         &\overset{A \in \cF_n}{=}& \lim_{\substack{n \to \infty\\n \ge m}} \bE[X \One_A]\\
     \end{IEEEeqnarray*}
     Hence $\int_A Y \dif \bP = \int_A X \dif \bP$ for all $m \in \N, A \in  \cF_m$.
-    Since $\cF = \sigma\left( \bigcup \cF_n \right)$
-    this holds for all $A \in  \cF$.
+    Since $\sigma(X) = \bigcup \cF_n$
+    this holds for all $A \in  \sigma(X)$.
     Hence $X = Y$ a.s., so $X_n \xrightarrow{L^2} X$.
     Since $(X_n)_n$ is uniformly bounded, this also means
     $X_n \xrightarrow{L^p}  X$.
@@ -111,8 +111,203 @@ we need the following theorem, which we won't prove here:
     Hence $X_n = \bE[X | \cF_m]$ by the uniqueness of conditional expectation
     and by \autoref{ceismartingale},
     we get the convergence.
-    
 \end{refproof}
 
+\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},
+    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}
+    A constant random variable $T = c$ is a stopping time.
+\end{example}
+
+\begin{example}[Hitting times]
+    For an adapted process $(X_n)_n$ 
+    with values in $\R$ and $A \in  \cB(\R)$, the \vocab{hitting time}
+    \[
+    T \coloneqq \inf \{n \in \N : X_n \in A\} 
+    \]
+    is a stopping time,
+    as
+    \[
+    \{T \le  n \} = \bigcup_{k=1}^n \{X_k \in A\} \in \cF_n.
+    \] 
+
+    However, the last exit time
+    \[
+    T \coloneqq  \sup \{n \in \N : X_n \in  A\} 
+    \] 
+    is not a stopping time.
+
+\end{example}
 
 
+\begin{example}
+    Consider the simple random walk, i.e.
+    $X_n$ i.i.d.~with $\bP[X_n = 1] = \bP[X_n = -1] = \frac{1}{2}$.
+    Set $S_n \coloneqq \sum_{i=1}^{n}  X_n$.
+    Then
+    \[
+    T \coloneqq \inf \{n \in \N : S_n \ge A \lor S_n \le  B\}
+    \] 
+    is a stopping time.
+\end{example}
+
+\begin{example}
+    If $T_1, T_2$ are stopping times with respect to the same filtration,
+    then
+    \begin{itemize}
+        \item $T_1 + T_2$,
+        \item $\min \{T_1, T_2\}$ and
+        \item $\max \{T_1, T_2\}$
+    \end{itemize}
+    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 $T$ 
+    and a stochastic process $(X_n)_n$.
+    \begin{itemize}
+        \item The behaviour of $ X_n$ until $T$,
+            i.e.~looking at the \vocab{stopped process}
+            \[
+            X^T \coloneqq  \left(X_{T \wedge n}\right)_{n \in \N}
+            \].
+        \item  The value of $(X_n)_n)$ at time $T$,
+            i.e.~looking at $X_T$.
+    \end{itemize}
+\end{remark}
+\begin{example}
+    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 $X^T$ 
+    and $\bE[X_{T \wedge n}] \le  \bE[X_0]$.
+\end{theorem}
+\begin{proof}
+    First, we need to show that $X^T$ is adapted.
+    This is clear since
+    \begin{IEEEeqnarray*}{rCl}
+        X^T_n &=& X_T \One_{T < n} + X_n \One_{T \ge n}\\
+              &=& \sum_{k=1}^{n-1}  X_k \One_{T = k} + X_n \One_{T \ge n}.
+    \end{IEEEeqnarray*}
+
+    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}
+        \bE[X^T_n - X^T_{n-1} | \cF_{n-1}]
+        &=& \bE[X_n \One_{\{T \ge n\}} + \sum_{k=1}^{n-1}  X_k \One_{\{ T = k\} } - X_{n-1}(\One_{T \ge  n}  + \One_{\{T = n-1\}})
+        + \sum_{k=1}^{n-2}  X_k \One_{\{T = k\} }  | \cF_{n-1}]\\
+        &=&  \bE[(X_n - X_{n-1}) \One_{\{ T \ge  n\} } | \cF_{n-1}]\\
+        &=& \One_{\{ T \ge  n\}} (\bE[X_n | \cF_{n-1}] - X_{n-1})\\
+        && \begin{cases}
+            \le 0\\
+            = 0 \text{ if $(X_n)_n$ is a martingale}.
+        \end{cases}.
+    \end{IEEEeqnarray*}
+
+\end{proof}
+
+\begin{remark}
+    \label{roptionalstoppingi}
+    We now want a similar statement for $X_T$.
+    In the case that $T \le M$ is bounded,
+    we get from the above that
+    \[
+    \bE[X_T] \overset{n \ge M}{=} \bE[X^T_n]  \begin{cases}
+        \le \bE[X_0] & \text{ supermartingale},
+        = \bE[X_0] & \text{ martingale}.
+    \end{cases}
+    \]
+
+    However if $T$ is not bounded, this does not hold in general.
+\end{remark}
+\begin{example}
+    Let $(S_n)_n$ be the simple random walk
+    and take $T \coloneqq  \inf \{n : S_n = 1\}$.
+    Then $\bP[T < \infty] = 1$, but
+    \[
+    1 = \bE[S_T] \neq  \bE[S_0] = 0.
+    \] 
+\end{example}
+
+\begin{theorem}[Optional Stopping]
+    \label{optionalstopping}
+    Let $(X_n)_n$ be a supermartingale
+    and let $T$ be a stopping time
+    taking values in $\N$.
+
+    If one of the following holds
+    \begin{itemize}[(i)]
+        \item $T \le M$ is bounded,
+        \item $(X_n)_n$ is uniformly bounded
+            and $T < \infty$  a.s.,
+        \item $\bE[T] < \infty$
+            and $|X_n(\omega) - X_{n-1}(\omega)| \le K$
+            for all $n \in \N, \omega \in \Omega$ and
+            some $K > 0$,
+    \end{itemize}
+    then $\bE[X_T] \le  \bE[X_0]$.
+
+    If $(X_n)_n$ even is a martingale, then
+    under the same conditions
+    $\bE[X_T] = \bE[X_0]$.
+\end{theorem}
+\begin{proof}
+    (i) was dealt with in \autoref{roptionalstoppingi}.
+
+    (ii): Since $(X_n)_n$ is bounded, we get that
+    \begin{IEEEeqnarray*}{rCl}
+        \bE[|X_T - X_0|] &\overset{\text{dominated convergence}}{=}& \lim_{n \to \infty} \bE[|X_{T \wedge n} - X_0|]\\
+                         &\overset{\text{part (i)}}{\le}& 0.
+    \end{IEEEeqnarray*}
+
+    (iii): It is
+    \begin{IEEEeqnarray*}{rCl}
+        |X_{T \wedge n}- X_0| &\le& | \sum_{k=1}^{T \wedge n}  X_k - X_{k-1}|\\
+                              &\le & (T \wedge n) \cdot  K\\
+                              &\le & T \cdot K < \infty.
+    \end{IEEEeqnarray*}
+
+    Hence, we can apply dominated convergence and obtain
+    \begin{IEEEeqnarray*}{rCl}
+        \bE[X_T - X_0] &=& \lim_{n \to \infty} \bE[X_{T \wedge n} - X_0].
+    \end{IEEEeqnarray*}
+    Thus, we can apply (ii).
+
+
+    The statement about martingales follows from
+    applying this to $(X_n)_n$ and $(-X_n)_n$,
+    which are both supermartingales.
+\end{proof}

probability_theory.tex

@@ -43,6 +43,7 @@
 \input{inputs/lecture_17.tex}
 \input{inputs/lecture_18.tex}
 \input{inputs/lecture_19.tex}
+\input{inputs/lecture_20.tex}
 
 \cleardoublepage