lecture 20 part 1

inputs/lecture_15.tex

@@ -189,7 +189,7 @@ Recall
 \begin{theorem}[Tower property]
     % 10
     \label{ceprop10}
-    \label{ctower}
+    \label{cetower}
     Suppose $\cF \supset \cG \supset \cH$ are sub-$\sigma$-algebras.
     Then
     \[

inputs/lecture_19.tex

@@ -218,15 +218,14 @@ Let $(\Omega, \cF, \bP)$ as always and let $(\cF_n)_n$ always be a filtration.
 \end{proof}
 
 \begin{theorem}
+    \label{ceismartingale}
     Let $X \in L^p$ for some $p \ge 1$.
     Then $X_n \coloneqq  \bE[X | \cF_n]$ defines a martingale which converges
     to $X$ in $L^p$.
 \end{theorem}
-\begin{proof}
-    
-\end{proof}
 
 \begin{theorem}
+    \label{martingaleisce}
     Let $p > 1$.
     Let $(X_n)_n$ be a martingale bounded in $L^p$.
     Then there exists a random variable $X \in  L^p$, such that

inputs/lecture_20.tex

@@ -0,0 +1,118 @@
+\begin{refproof}{ceismartingale}
+    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}, $|X_n| \le  \bE[|X| | \cF_n]$,
+    hence $\sup_{\substack{n \in \N \\ \omega \in  \Omega}} | X_n(\omega)| < \infty$.
+    Thus $(X_n)_n$ is a martingale in $L^{\infty} \subseteq L^2$.
+    By the convergence theorem for martingales in $L^2$ % TODO REF
+    there exists a random variable $Y$,
+    such that $X_n \xrightarrow{L^2} Y$.
+
+    Fix $m \in \N$ and $A \in \cF_m$.
+    Then
+    \begin{IEEEeqnarray*}{rCl}
+        \int_A Y \dif \bP
+        &=& \lim_{n \to \infty} \int_A X_n \dif \bP\\
+        &=& \lim_{n \to \infty} \bE[X_n \One_A]\\
+        &=& \lim_{n \to \infty} \bE[\bE[X | \cF_n] \One_A]\\
+        &\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 $A \in  \cF$.
+    Hence $X = Y$ a.s., so $X_n \xrightarrow{L^2} X$.
+    Since $(X_n)_n$ is uniformly bounded, this also means
+    $X_n \xrightarrow{L^p}  X$.
+
+
+    Second step:
+    Now let $X \in  L^p$ be general and define
+    \[
+    X'(\omega) \coloneqq  \begin{cases}
+        X(\omega)& \text{ if }  |X(\omega)| \le  M,\\
+        0&\text{ otherwise}
+    \end{cases}
+    \]
+    for some $M  > 0$.
+    Then $X' \in  L^\infty$ and
+    \begin{IEEEeqnarray*}{rCl}
+        \int | X - X'|^p \dif \bP &=& \int_{\{|X| > M\} } |X|^p \dif \bP \xrightarrow{M \to \infty} 0
+    \end{IEEEeqnarray*}
+    as $\bP$ is \vocab{regular}, \todo{Definition?}
+    i.e.~$\forall  \epsilon > 0 \exists  k . \bP[|X|^p \in [-k,k] \ge  1-\epsilon$.
+
+    % Take some $\epsilon > 0$ and $M$ large enough such that
+    % \[
+    % \int |X - X'| \dif \bP < \epsilon.
+    % \]
+
+    % Let $(X_n')_n$ be the martingale given by $(\bE[X' | \cF_n])_n$.
+    % Then $X_n' \xrightarrow{L^p} X'$ by the first step.
+
+    % It is
+    % \begin{IEEEeqnarray*}{rCl}
+    %     \|X_n - X_n'\|_{L^p}^p  &=& \bE[\bE[X - X' | \cF_n]^{p}]\\
+    %                             &\overset{\text{Jensen}}{\le}& \bE[\bE[(X- X')^p | \cF_n]\\
+    %                             &=& \|X - X'\|_{L^p}^p\\
+    %                             &<& \epsilon.
+    % \end{IEEEeqnarray*}
+
+    Hence
+    \[
+    \|X_n - X\|_{L^p} \le |X_n - X_n'|_{L^p} + |X_n' - X'|_{L^p} +  | X - X'|_{L^p} \le 3 \epsilon.
+    \]
+    Thus $X_n \xrightarrow{L^p} X$.
+\end{refproof}
+
+For the proof of \autoref{martingaleisce},
+we need the following theorem, which we won't prove here:
+\begin{theorem}[Banach Alaoglu]
+    \label{banachalaoglu}
+    Let $X$ be a  normed vector space and $X^\ast$ its
+    continuous dual.
+    Then the closed unit ball in $X^\ast$ is compact
+    w.r.t.~the ${\text{weak}}^\ast$ topology.
+\end{theorem}
+\begin{fact}
+    We have $L^p \cong (L^q)^\ast$ for $\frac{1}{p} + \frac{1}{q} = 1$
+    via
+    \begin{IEEEeqnarray*}{rCl}
+        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}
+
+\begin{refproof}{martingaleisce}
+    Since $(X_n)_n$ is bounded in $L^p$, by \autoref{banachalaoglu},
+    there exists $X \in L^p$ and a subsequence
+    $(X_{n_k})_k$ 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 $(L^\infty)^\ast \not\cong  L^1$).
+
+    Let $A \in  \cF_m$ for some fixed $m$ and write
+    $Y = \One_A$.
+    Then
+    \begin{IEEEeqnarray*}{rCl}
+    \int_A X \dif \bP
+    &=& \lim_{k \to \infty} \int_A X_{n_k} \dif \bP\\
+    &=& \lim_{k \to \infty} \bE[X_{n_k} \One_A]\\
+    &\overset{\text{for }n_k \ge m}{=}& \int_{k \to \infty} \bE[X_m \One_A].
+    \end{IEEEeqnarray*}
+    Hence $X_n = \bE[X | \cF_m]$ by the uniqueness of conditional expectation
+    and by \autoref{ceismartingale},
+    we get the convergence.
+    
+\end{refproof}
+
+
+