lecture 17

inputs/lecture_16.tex

@@ -221,8 +221,7 @@ Typically $\cF_n = \sigma(X_1, \ldots, X_n)$ for a sequence of random variables.
 
 
 \begin{example}
-    \begin{itemize}
+    The simple random walk:
-        \item The simple random walk:
 
             Let $\xi_1, \xi_2, ..$ iid,
             $\bP[\xi_i = 1] = \bP[\xi_i = -1] = \frac{1}{2}$,
@@ -231,7 +230,8 @@ Typically $\cF_n = \sigma(X_1, \ldots, X_n)$ for a sequence of random variables.
             Then $X_n$ is $\cF_n$-measurable.
             Showing that $(X_n)_n$ is a martingale
              is left as an exercise.
-        \item See exercise sheet 9.
+\end{example}
-        \item The branching process (next lecture).
+\begin{example}
-    \end{itemize}
+    See exercise sheet 9.
+    \todo{Copy}
 \end{example}

inputs/lecture_17.tex

@@ -0,0 +1,166 @@
+\lecture{17}{2023-06-15}{}
+
+\begin{definition}[Stochastic process]
+    % TODO
+\end{definition}
+
+\begin{goal}
+    What about a ``gambling strategy''?
+    
+    Consider a stochastic process $(X_n)_{n \in \N}$.
+
+    Note that the increments $X_{n+1} - X_n$ can be thought of as the win
+    or loss per round of a game.
+    Suppose that there is another stochastic process
+    $(C_n)_{n \ge 1}$  such that $C_n$ is determined
+    by the information gathered up until time $n$,
+    i.e.~$C_n$ is measurable with respect to $\cF_{n-1}$.
+    If such process $C_n$ exists, we say that $ C_n$ is
+    \vocab[Stochastic process!previsible]{previsible}.
+    Think of $C_n$ as our strategy of playing the game.
+    Then $C_n(X_n - X_{n-1})$ defines the win in the $n$-th game,
+    while
+    \[
+    Y_n \coloneqq  \sum_{j=1}^n C_j(X_j - X_{j-1})
+    \] 
+    defines the cumulative win process.
+\end{goal}
+\begin{lemma}
+    If $(C_n)_{n \ge 1}$ is previsible and $(X_n)_{n \ge 0}$
+    is a (sub/super-) martingale
+    and there exists a constant $K_n$ such that $|C_n(\omega)| \le  K_n$.
+    Then $(Y_n)_{n \ge 1}$ is also a (sub/super-) martingale.
+\end{lemma}
+\begin{proof}
+    Exercise. \todo{Copy}
+\end{proof}
+\begin{remark}
+    The assumption of $K_n$ being constant can be weakened to
+    $C_n \in  L^p(\bP)$, $X_n \in  ^q(\bP)$ with $\frac{1}{p} + \frac{1}{q} = 1$.
+\end{remark}
+
+Suppose we have $(X_n)$ adapted, $X_n \in  L^1(\bP)$,
+$(C_n)_{n \ge 1}$ previsible.
+We play according to the following principle:
+Pick two real numbers $a < b$.
+Wait until $X_n \le  a$, then start playing.
+Stop playing when $X_n \ge b$.
+I.e.~define
+\begin{itemize}
+    \item $C_1 \coloneqq  0$,
+    \item $C_n \coloneqq  \One_{\{C_{n-1} = 1\}} \cdot \One_{\{X_{n-1} \le b\}} + \One_{\{C_{n-1} = 0\} } \One_{\{X_{n-1}\}  < a}$.
+\end{itemize}
+
+\begin{definition}
+Fix $N \in \N$ and let
+\[U_n^X([a,b]) \coloneqq \# \{\text{Upcrossings of $[a,b]$ made by $n \mapsto X_n(\omega)$ by time $n$}\},\]
+i.e.~$U_n([a,b])(\omega)$ is the largest $k \in \N_0$ such that we can find a
+sequence
+$0 \le  s_1 < t_1 < s_2 < t_2 < \ldots < s_k < t_k \le N$
+such that $X_{s_j}(\omega) < a$ and $X_{t_j}(\omega) > b$ for all $1 \le  j \le  k$.
+\end{definition}
+Clearly $U_N^X([a,b]) \uparrow$ as $N$ increases.
+It follows that the monotonic limit $U_\infty([a,b]) \coloneqq  \lim_{N \to \infty} U_N([a,b])$ exists pointwise.
+\begin{lemma} % Lemma 1
+    \[
+    \{\omega | \liminf_{N \to  \infty} Z_N(\omega) < a < b <
+    \limsup_{N \to \infty} Z_N(\omega)\} \subseteq 
+    \{\omega: U^{Z}_\infty([a,b])(\omega) = \infty\}
+    \] 
+    for every sequence of measurable functions $(Z_n)_{n \ge 1}$.
+    
+\end{lemma}
+\begin{lemma} % 2
+    Let $Y_n(\omega) \coloneqq \sum_{j=1}^n C_j(X_j - X_{j-1})$.
+    Then $Y_N \ge  (b -a) U_N([a,b]) - (X_N - a)^{-}$.
+\end{lemma}
+\begin{proof}
+    Every upcrossing of $[a,b]$ increases the  value of $Y$ by $(b-a)$,
+    while the last intverval of play  $(X_n -a)^{-}$ overemphasizes the loss.
+\end{proof}
+
+\begin{lemma} %3
+    Suppose $(X_n)_n$ is a supermartingale.
+    Then in the above setup, $(b-a) \bE[U_N([a,b])] \le  \bE[(X_n - a)^-]$.
+\end{lemma}
+\begin{proof}
+    Obvious from lemma 2 % TODO REF
+    and the supermartingale property.
+\end{proof}
+\begin{corollary}
+    Let $(X_n)_n$ be a \vocab[Supermartingale!bounded]{supermartingale bounded in $L^1(\bP)$ },
+    i.e.~$\sup_n \bE[|X_n| ] < \infty$.
+    Then $(b-a) \bE(U_\infty) \le  |a| + \sup_n \bE(|X_n|)$.
+    In particular, $\bP[U_\infty = \infty] = 0$.
+\end{corollary}
+\begin{proof}
+    By lemma 3 % TODO REF
+    we have that
+    \[(b-a) \bE[U_N([a,b])] \le  \bE[ | X_N| ] + |a| \le  \sup_n \bE[|X_n|] + |a|.\]
+    Since $U_N(\cdot) \ge  0$ and $U_N(\cdot ) \uparrow U_\infty(\cdot )$,
+    by the monotone convergence theorem
+    \[
+    \bE(U_n([a,b])] \uparrow \bE[U_\infty([a,b])].
+    \] 
+\end{proof}
+
+Assume now, that our process $(X_n)_{n \ge 1}$ is a supermartingale
+bounded in $L^1(\bP)$.
+Let
+\[
+\Lambda \coloneqq  \{\omega | X_n(\omega) \text{ does not converge to anything in $[-\infty,\infty]$}\}.
+\] 
+We have
+\begin{IEEEeqnarray*}{rCl}
+    \Lambda &=& \{\omega | \liminf_N X_N(\omega) < \limsup_N X_N(\omega)\}\\
+            &=& \{\omega | \liminf_N X_N(\omega) < a < b \limsup_N X_N(\omega)\} \\
+            &=& \bigcup_{a,b \in \Q} \underbrace{\{\omega | \liminf_N X_N(\omega) < a < b < \limsup_N X_N(\omega)\}}_{\Lambda_{a,b}} \\
+\end{IEEEeqnarray*}
+
+We have $\Lambda_{a,b} \subseteq  \{\omega : U_{\infty}([a,b])(\omega) = \infty\} $ by lemma 1. % TODO REF
+By lemma 3 % TODO REF
+we have $\bP(\Lambda_{a,b}) = 0$, hence $\bP(\Lambda) = 0$.
+Hence there exists a random variable $X_\infty$ such that $X_n \xrightarrow{a.s.}  X_\infty$.
+
+\begin{claim}
+    $\bP[X_\infty \in  \{\pm \infty\}] = 0$.
+\end{claim}
+\begin{subproof}
+    It suffices to show that $\bE[|X_\infty|] < \infty$.
+    We have.
+    \begin{IEEEeqnarray*}{rCl}
+        \bE[|X_\infty|] &=& \bE[\liminf_{n \to \infty} |X_n|]\\
+                        &\overset{\text{Fatou}}{\le }& \liminf_n \bE[|X_n|]\\
+                        &\le & \sup_n \bE[|X_n|]\\
+                        &<& \infty.
+    \end{IEEEeqnarray*}
+\end{subproof}
+
+We have thus shown
+\begin{theorem}[Doob's martingale convergence theorem]
+    \label{doobmartingaleconvergence}
+    Any supermartingale bounded in $L^1$ converges almost surely to a
+    random variable, which is almost surely finite.
+    In particular, any non-negative supermartingale converges a.s.~to a finite random variable.
+\end{theorem}
+The second part follows from
+\begin{claim}
+    Any non-negative supermartingale is bounded in $L^1$.
+\end{claim}
+\begin{subproof}
+    We need to show $\sup_n \bE(|X_n|) < \infty$.
+    Since the supermartingale is non-negative, we have $\bE[|X_n|] = \bE[X_n]$ 
+    and since it is a supermartingale $\bE[X_n]  \le \bE[X_0]$.
+    
+\end{subproof}
+
+\todo{rearrange proof}
+
+
+
+\begin{example}[Branching process]
+    % TODO
+
+\end{example}
+
+

probability_theory.tex

@@ -40,6 +40,7 @@
 \input{inputs/lecture_14.tex}
 \input{inputs/lecture_15.tex}
 \input{inputs/lecture_16.tex}
+\input{inputs/lecture_17.tex}
 
 \cleardoublepage