lecture 16

inputs/lecture_15.tex

@@ -229,7 +229,7 @@ Assume $Y = \One_B$, then $Y$ simple, then take the limit (using that $Y$ is bou
     of $\sigma(\sigma(X), \cG)$, then
      \[
          \bE[X | \sigma(\cG, \cH)] \overset{\text{a.s.}}{=} \bE[X | \cG].
-    \] 
+    \]
 \end{theorem}
 \begin{example}
     If $X$ is independent of $\cG$,
@@ -249,7 +249,6 @@ Assume $Y = \One_B$, then $Y$ simple, then take the limit (using that $Y$ is bou
         \bE[S_n | \cF_n] + \bE[X_{n+1} | \cF_n]\\
                              &\overset{\text{a.s.}}{=}& S_n + \bE[X_{n+1} | \cF_n]\\
                              &\overset{\text{\autoref{ceprop12}}}{=}& S_{n} + \bE[X_n]\\
-                             &=& S_n
+                             &=& S_n.
     \end{IEEEeqnarray*}
-    
 \end{example}

inputs/lecture_16.tex

@@ -0,0 +1,247 @@
+\lecture{16}{2023-06-13}{}
+
+\subsection{Conditional expectation}
+
+\begin{theorem}
+    \label{ceprop11}
+    \label{ceroleofindependence}
+    Let $X$ be a random variable,
+    and let $\cG, \cH$ be $\sigma$-algebras.
+
+    If $\cH$ is independent of $\sigma\left( \sigma(X), \cG \right)$,
+    then
+    \[
+    \bE[X | \sigma(\cG, \cH)] \overset{\text{a.s.}}{=} \bE[X | \cG].
+    \] 
+
+    In particular, if $X$ is independent of $\cG$,
+    then
+    \[
+    \bE[X | \cG] \overset{\text{a.s.}}{=} \bE[X].
+    \] 
+\end{theorem}
+
+\todo{Definition of independence wrt a $\sigma$-algebra}
+
+\begin{proof}
+    Let $\cH$ be independent of $\sigma(\sigma(X), \cG)$.
+    Then for all $H \in \cH$, we have that $\One_H$
+    and any random variable measurable with respect to either $\sigma(X)$ 
+    or $\cG$ must be independent.
+
+    It suffices to consider the case of $X \ge 0$.
+    Let $G \in \cG$ and $H \in \cH$.
+    By assumption, $X \One_G$ and $\One_H$ are independent.
+    Let $Z \coloneqq  \bE[X | \cG]$.
+    Then
+    \begin{IEEEeqnarray*}{rCl}
+        \underbrace{\bE[X;G \cap H]}_{\coloneqq \int_{G \cap H} X \dif \bP} &=& \bE[(X \One_G) \One_H]\\
+                                                                                                  &=& \bE[X \One_G] \bE[\One_H]\\
+                                                                                                  &=& \bE[Z \One_G] \bP(H)\\
+                                                                                                  &=& \bE[Z; G \cap H]
+    \end{IEEEeqnarray*}
+
+    The identity above means, that the measures $A \mapsto \bE[X; A]$
+    and  $A \mapsto \bE[Z; A]$
+    agree on the  $\sigma$-algebra $\sigma(\cG, \cH)$ for events
+    of the form $G \cap H$.
+    Since sets of this form generate $\sigma(\cG, \cH)$,
+    these two measures must agree on  $\sigma(\cG, \cH)$.
+    The claim of the theorem follows by the uniqueness of conditional expectation.
+
+    To deduce the second statement, choose $\cG = \{\emptyset, \Omega\}$.
+\end{proof}
+
+
+\subsection{The Radon Nikodym theorem}
+
+First, let us recall some basic facts:
+\begin{fact}
+    Let $(\Omega, \cF, \mu)$ be a \vocab[Measure space!$\sigma$-finite]{$\sigma$-finite
+    measure space},
+    i.e.~$\Omega$ can be decomposed into countably many subsets of finite measure.
+    Let $f: \Omega \to [0, \infty)$ be measurable.
+    Define $\nu(A) \coloneqq \int_A f \dif \mu$.
+    Then $\nu$ is also a $\sigma$-finite measure on $(\Omega, \cF)$.\todo{Application of mct}
+    Moreover, $\nu$ is finite iff $f$ is integrable.
+\end{fact}
+Note that in this setting, if $\mu(A) = 0$ it follows that $\nu(A) = 0$.
+
+The Radon Nikodym theorem is the converse of that:
+\begin{theorem}[Radon-Nikodym]
+    \label{radonnikodym}
+
+    Let $\mu$ and $\nu$ be two $\sigma$-finite measures
+    on $(\Omega, \cF)$.
+    Suppose
+    \[
+    \forall A \in  \cF . ~ \mu(A) = 0 \implies  \nu(A) = 0.
+    \] 
+    Then
+    \begin{enumerate}[(1)]
+        \item there exists $Z: \Omega \to [0, \infty)$ measurable,
+            such that
+            \[\forall  A \in  \cF . ~ \nu(A) = \int_A Z \dif \mu.\]
+        \item Such a $Z$ is unique up to equality a.e.~(w.r.t.  $\mu$).
+        \item $Z$ is integrable w.r.t.~$ \mu$ iff $\nu$ is a finite measure.
+    \end{enumerate}
+
+    Such a $Z$ is called the \vocab{Radon-Nikodym derivative}.
+\end{theorem}
+\begin{definition}
+    Whenever the property $\forall  A \in  \cF, \mu(A)= 0 \implies \nu(A) = 0$,
+    we say that $\nu$ is \vocab{absolutely continuous} 
+    w.r.t.~$\mu$.
+    This is written as $\nu \ll \mu$.
+\end{definition}
+
+With \autoref{radonnikodym} we get a very short proof of the existence
+of conditional expectation:
+\begin{proof}[Second proof of \autoref{conditionalexpectation}]
+    Let $(\Omega, \cF, \bP)$ as always, $X \in L^1(\bP)$ and $\cG \subseteq  \cF$.
+    It suffices to consider the case of $X \ge 0$.
+    For all $G \in \cG$, define $\nu(G) \coloneqq  \int_G X \dif \bP$.
+    Obviously, $\nu \ll \bP$ on $\cG$. 
+    Then apply \autoref{radonnikodym}.
+\end{proof}
+
+
+\begin{refproof}{radonnikodym}
+    We will only sketch the proof. A full proof can be found in the notes.
+    
+    \paragraph{Step 1: Uniqueness}
+        See notes.
+    \paragraph{Step 2: Reduction to the finite measure case}
+        See notes.
+    \paragraph{Step 3: Getting hold of $Z$}
+    Assume now that $\mu$ and $\nu$ are two finite measures.
+    Let
+    \[\cC \coloneqq  \{f: \Omega \to [0,\infty] | \forall A \in \cF.~\int_A f \dif \mu \le  \nu(A)\}.\]
+
+    We have $\cC \neq  \emptyset$ since $0 \in \cC$.
+    The goal is to find a maximal function $Z$ in $\cC$.
+    Obviously its integral will also be maximal.
+
+    \begin{enumerate}[(a)]
+        \item If $f,g \in \cC$, than $f \lor g$ (the pointwise maximum)
+            s also in $\cC$.
+        \item Suppose $\{f_n\}_{n \ge 1}$ is an increasing sequence in $\cC$.
+            Let $f$ be the pointwise limit.
+            Then $f \in \cC$.
+        \item For all $f \in \cC$, we have
+            \[
+            \int_\Omega f \dif \mu \le \nu(\Omega) < \infty.
+            \] 
+    \end{enumerate}
+
+    Define $\alpha \coloneqq  \sup \{ \int f \dif \mu : f \in \cC\} \le \nu(\Omega) < \infty$.
+    Let $f_n \in \cC, n\in \N$ be a sequence
+    with $\int f_n \dif \mu \to \alpha$.
+    Define $g_n \coloneqq \max \{f_1,\ldots,f_n\} \in \cC$.
+    Applying (b), we get that the pointwise limit, $Z$,
+    is an element of $\cC$.
+
+    \paragraph{Step 4: Showing that our choice of $Z$ works}
+    Define $\lambda(A) \coloneqq  \nu(A) - \int_A Z \dif \mu \ge 0$.
+    $\lambda$ is a measure.
+    \begin{claim}
+        $\lambda = 0$.
+    \end{claim}
+    \begin{subproof}
+        Call $G \in \cF$ \emph{good} if the following hold:
+        \begin{enumerate}[(i)]
+            \item $\lambda(G) - \frac{1}{k}\mu(G) > 0$.
+            \item $\forall  B \subseteq G, B \in \cF. ~ \lambda(B) - \frac{1}{k}\mu(B) \ge 0$.
+        \end{enumerate}
+        Suppose we know that for all $A \in  \cF, k \in \N$
+        we have
+        $\lambda(A) \le  \frac{1}{k} \mu(A)$.
+        Then $\lambda(A) = 0$ since $\mu$ is finite.
+
+        Assume the claim does not hold.
+        Then there must be some $k \in \N$, $A \in \cF$ 
+        such that $\lambda(A) - \frac{1}{k} \mu(A) > 0$.
+        Fix this $A$ and $k$.
+        Then $A$ satisfies condition (i) of being good,
+        but it need not satisfy (ii).
+
+        The tricky part is to make $A$ smaller such that it also
+        satisfies (ii).\todo{Copy from notes}
+    \end{subproof}
+\end{refproof}
+
+
+
+\section{Martingales}
+
+We have already worked with martingales, but we will define them
+rigorously now.
+
+\begin{definition}[Filtration]
+    A \vocab{filtration} is a sequence $(\cF_n)$ of $\sigma$-algebras
+    such that $\cF_n \subseteq \cF_{n+1}$ for all $n \ge 1$.
+\end{definition}
+Intuitively, we can think of a $\cF_n$ as the set of information
+we have gathered up to time $n$.
+Typically $\cF_n = \sigma(X_1, \ldots, X_n)$ for a sequence of random variables.
+
+\begin{definition}
+    Let $(\cF_n)$ be a filtration and
+    $X_1,\ldots,X_n$ be random variables such that $X_i \in L^1(\bP)$.
+    Then we say that $(X_n)_{n \ge 1}$ is an $(\cF_n)_n$-\vocab{martingale}
+    if
+    \begin{itemize}
+        \item $X_n$ is $\cF_n$-measurable for all $n$
+            ($X_n$ is \vocab[Sequence!adapted to a filtration]{adapted to the filtration} $\cF_n$ ).
+        \item $\bE[X_{n+1} | \cF_n] \overset{\text{a.s.}}{=} X_n$
+            for all $n$.
+    \end{itemize}
+
+    $(X_n)$ is called a \vocab{sub-martingale},
+    if it is adapted to $\cF_n$ but
+    \[
+    \bE[X_{n+1} | \cF_n] \overset{\text{a.s.}}{\ge} X_n.
+    \] 
+    It is called a \vocab{super-martingale} 
+    if it is adapted but $\bE[X_{n+1} | \cF_n] \overset{\text{a.s.}}{\le} X_n$.
+\end{definition}
+\begin{corollary}
+    Suppose that $f: \R \to  \R$ is a convex function such that $f(xn) \in L^1(\bP)$.
+    Suppose that $(X_n)_n$ is a martingal\footnote{In this form it means, that there is some filtration, that we don't explicitly specify}.
+    Then $(f(X_n))_n$ is a sub-martingale.
+\end{corollary}
+\begin{proof}
+    Apply \autoref{cejensensinequality}.
+\end{proof}
+
+\begin{corollary}
+    If $(X_n)_n$ is a martingale,
+    then $\bE[X_n] = \bE[X_0]$.
+\end{corollary}
+
+
+\begin{example}
+    \begin{itemize}
+        \item The simple random walk:
+
+            Let $\xi_1, \xi_2, ..$ iid,
+            $\bP[\xi_i = 1] = \bP[\xi_i = -1] = \frac{1}{2}$,
+            $X_n \coloneqq  \xi_1 + \ldots + \xi_n$
+            and $\cF_n \coloneqq  \sigma(\xi_1, \ldots, \xi_n) = \sigma(X_1, \ldots, X_n)$.
+            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.
+        \item The branching process (next lecture).
+    \end{itemize}
+\end{example}
+
+
+
+
+
+
+
+
+
+

probability_theory.tex

@@ -39,6 +39,7 @@
 \input{inputs/lecture_13.tex}
 \input{inputs/lecture_14.tex}
 \input{inputs/lecture_15.tex}
+\input{inputs/lecture_16.tex}
 
 \cleardoublepage
 

temp_lenovo

@@ -0,0 +1,28 @@
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