lecture 14 -> 15

inputs/lecture_1.tex

@@ -46,7 +46,7 @@ First, let us recall some basic definitions:
 \end{fact}
 The converse to this fact is also true:
 \begin{theorem}[Kolmogorov's existence theorem / basic existence theorem of probability theory]
-    \label{kolmogorovxistence}
+    \label{kolmogorovexistence}
     Let $\cF(\R)$ be the set of all distribution functions on $\R$
     and let $\cM(\R)$ be the set of all probability measures on $\R$.
     Then there is a one-to-one correspondence between $\cF(\R)$ and $\cM(\R)$

inputs/lecture_14.tex

@@ -1,255 +1,162 @@
-\lecture{14}{2023-06-06}{}
+\lecture{14}{2023-05-25}{Conditional expectation}
 
-We want to derive some properties of conditional expectation.
+\section{Conditional expectation}
 
-\begin{theorem}[Law of total expectation] % Thm 1
+\subsection{Introduction}
-    \label{ceprop1}
-    \label{totalexpectation}
-    \[
-        \bE[\bE[X | \cG ]] = \bE[X].
-    \]
-\end{theorem}
-\begin{proof}
-    Apply (b) from the definition for $G = \Omega  \in \cG$.
-\end{proof}
-\begin{theorem} % Thm 2
-    \label{ceprop2}
-    If $X$ is $\cG$-measurable, then $X = \bE[X | \cG]$ a.s..
-\end{theorem}
-\begin{proof}
-    Suppose $\bP[X \neq Y] > 0$.
-    Without loss of generality $\bP[X > Y] > 0$.
-    Hence $\bP[ X > Y + \frac{1}{n}]> 0$ for some $n \in \N$.
-    Let $A \coloneqq \{X > Y + \frac{1}{n}\}$.
-    % TODO
-\end{proof}
 
-\begin{example}
+Consider a probability space $(\Omega, \cF, \bP)$
-    Suppose $X \in L^1(\bP)$, $\cG \coloneqq  \sigma(X)$.
+and two events $A, B \in \cF$ with $\bP(B) > 0$.
-    Then $X$ is measurable with respect to $\cG$.
-    Hence $\bE[X | \cG] = X$.
-\end{example}
-
-\begin{theorem}[Linearity]
-    \label{ceprop3}
-    \label{celinearity}
-    For all $a,b \in \R$
-    we have
-    \[
-        \bE[a X_1 + bX_2 | \cG] = a \bE[X_1 | \cG] + b \bE[X_2|\cG].
-    \]
-\end{theorem}
-\begin{proof}
-    Trivial % TODO
-\end{proof}
-
-\begin{theorem}[Positivity]
-    \label{ceprop4}
-    % 4
-    \label{cpositivity}
-    If $X \ge  0$, then $\bE[X | \cG] \ge 0$ a.s.
-\end{theorem}
-\begin{proof}
-    Let $W $ be a version of $\E[X | \cG]$.
-    Suppose $\bP[ W < 0] > 0$.
-    Then $G \coloneqq  \{W < -\frac{1}{n}\} \in \cG$
-    For some $n \in \N$, we have $\bP[G] > 0$.
-    However it follows that
-    \[
-        \int_G \bP[X | \cG] \dif \bP \le  -\frac{1}{n} \bP[G] < 0 \le  \int_G X \dif \bP.
-    \] 
-\end{proof}
-\begin{theorem}[Conditional monotone convergence theorem]
-    \label{ceprop5}
-    % 5
-    \label{mcmt}
-    Let $X_n,X \in L^1(\Omega, \cF, \bP)$.
-    Suppose $X_n \ge 0$ with $X_n \uparrow X$.
-    Then $\bE[X_n|\cG] \uparrow \bE[X|\cG]$.
-    
-\end{theorem}
-\begin{proof}
-    Let $Z_n$ be a version of $\bE[X_n | Y]$.
-    Since  $X_n \ge 0$ and $X_n \uparrow$,
-    by \autoref{cpositivity},
-    we have
-    \[
-        \bE[X_n | \cG] \overset{\text{a.s.}}{\ge } 0
-    \] 
-    and
-    \[
-        \bE[X_n | \cG] \uparrow \text{a.s.}
-    \] 
-    (consider $X_{n+1} - X_n$ ).
-
-    Define $Z \coloneqq  \limsup_{n \to \infty} Z_n$.
-    Then $Z$ is $\cG$-measurable
-    and $Z_n \uparrow Z$ a.s.
-
-    Take some $G \in \cG$.
-    We know by (b) % TODO REF
-    that $\be[Z_n \One_G] = \bE[X_n \One_G]$.
-    The LHS increases to  $\bE[Z \One_G]$ by the monotone 
-    convergence theorem.
-    Again by MCT, $\bE[X_n \One_G]$ increases to
-    $\bE[X \One_G]$.
-    Hence $Z$ is a version of $\bE[X | \cG]$.
-\end{proof}
-
-\begin{theorem}[Conditional Fatou]
-    \label{ceprop6}
-    \label{cfatou}
-    Let $X_n \in L^1(\Omega, \cF, \bP)$, $X_n \ge 0$.
-    Then
-    \[
-        \bE[ \liminf_{n \to \infty} X_n | \cG] \le  \liminf_{n \to \infty} \bE[X_n | \cG].
-    \] 
-\end{theorem}
-\begin{proof}
-    \todo{in the notes}
-\end{proof}
-\begin{theorem}[Conditional dominated convergence theorem]
-    \label{ceprop7}
-    \label{cdct}
-    Let $X_n,X \in L^1(\Omega, \cF, \bP)$.
-    Suppose $|X_n(\omega)| < X(\omega)$ a.e.~
-    and  $\int |X| \dif \bP < \infty$.
-    Then $X_n(\omega) \to  X\left( \omega \right) \implies \bE[ X_n | \cG] \to \bE[X | \cG]$.
-    
-\end{theorem}
-\begin{proof}
-    \todo{in the notes}
-\end{proof}
-
-Recall
-\begin{theorem}[Jensen's inequality]
-    If $c : \R \to  \R$ is convex and $\bE[|c \circ X|] < \infty$,
-    then $\bE[c \circ X] \ge c(\bE[X])$.
-\end{theorem}
-
-For conditional expectation, we have
-\begin{theorem}[Conditional Jensen's inequality]
-    \label{ceprop8}
-    \label{cjensen}
-    Let $X \in L^1(\Omega, \cF, \bP)$.
-    If $c : \R \to  \R$ is convex and $\bE[|c \circ X|] < \infty$,
-    then $\bE[c \circ X | \cG] \ge c(\bE[X | \cG])$ a.s.
-\end{theorem}
-\begin{fact}
-    \label{convapprox}
-    If $c$ is convex, then there are two sequences of real numbers
-    $a_n, b_n \in \R$
-    such that
-    \[
-    c(x) = \sup_n(a_n x + b_n).
-    \] 
-\end{fact}
-\begin{refproof}{cjensen}
-    By \autoref{convapprox}, $c(x) \ge  a_n X + b_n$
-    for all $n$.
-    Hence
-    \[
-        \bE[c(X) | \cG] \ge a_n \bE[X | \cG] + \bE[b_n | \cG]
-        = a_n \bE[X | \cG] + b_n \text{a.s.}
-    \] 
-    for all $n$.
-    Using that  a countable union of sets o f measure zero has measure zero,
-    we conclude that a.s~this happens simultaneously for all $n$.
-    Hence
-    \[
-        \bE[c(X) | \cG] \ge  \sup_n (a_n \bE[X | \cG] + b_n) \overset{\text{\autoref{convapprox}}}{=} c(\bE(X | \cG)).
-    \] 
-\end{refproof}
-
-Recall
-\begin{theorem}[Hölder's inequality]
-    Let $p,q \ge  1$ such that $\frac{1}{p} + \frac{1}{q} = 1$.
-    Suppose $X \in L^p(\bP)$ and $Y \in L^q(\bP)$.
-    Then
-    \[
-        \bE(X Y) \le  \underbrace{\bE(|X|^p)^{\frac{1}{p}}}_{\text{\reflectbox{$\coloneqq$}} \|X\|_{L^p}}  \bE(|Y|^q)^{\frac{1}{q}}.
-    \] 
-\end{theorem}
-
-\begin{theorem}[Conditional Hölder's inequality]
-    \label{ceprop9}
-    \label{choelder}
-    Let $p,q \ge  1$ such that $\frac{1}{p} + \frac{1}{q} = 1$.
-    Suppose $X \in L^p(\bP)$ and $Y \in L^q(\bP)$.
-    Then
-    \[
-        \bE(X Y | \cG) \le \bE(|X|^p | \cG)^{\frac{1}{p}} \bE(|Y|^q | \cG)^{\frac{1}{q}}.
-    \] 
-\end{theorem}
-\begin{proof}
-    Similar to the proof of Hölder's inequality.
-    \todo{Exercise}
-\end{proof}
-
-\begin{theorem}[Tower property]
-    % 10
-    \label{ceprop10}
-    \label{ctower}
-    Suppose $\cF \supset \cG \supset \cH$ are sub-$\sigma$-algebras.
-    Then
-    \[
-        \bE\left[\bE[X | \cG] \mid \cH\right] = \bE[X | \cH].
-    \] 
-\end{theorem}
-\begin{proof}
-    \todo{Exercise}
-\end{proof}
-
-\begin{theorem}[Taking out what is known]
-    % 11
-    \label{ceprop11}
-    \label{takingoutwhatisknown}
-
-    If $Y$ is $\cG$-measurable and bounded, then
-    \[
-        \bE[YX| \cG] \overset{\text{a.s.}}{=} Y \bE[X | \cG].
-    \] 
-\end{theorem}
-\begin{proof}
-    Assume w.l.o.g.~$X \ge 0$.
-Assume $Y = \One_B$, then $Y$ simple, then take the limit (using that $Y$ is bounded).
-    \todo{Exercise}
-\end{proof}
 
 \begin{definition}
-    Let $\cG$ and $\cH$ be $\sigma$-algebras.
+    The \vocab{conditional probability}  of $A$ given $B$ is defined as
-    We call $\cG$ and $\cH$ \vocab[$\sigma$-algebra!independent]{independent},
+    \[
-    if % TODO
+    \bP(A | B) \coloneqq \frac{\bP(A \cap B)}{\bP(B)}.
+    \]
 \end{definition}
 
-\begin{theorem}[Role of independence]
+Suppose we have two random variables $X$ and $Y$ on $\Omega$,
-    \label{ceprop12}
+such that $X$ takes distinct values $x_1, x_2,\ldots, x_{m}$
-    \label{roleofindependence}
+and $Y$ takes distinct values $y_1,\ldots, y_n$.
-    If $\cH$ is a sub-$\sigma$-algebra of $\cF$ and  $\cH$ is independent
+Then for this case, define the \vocab{conditional expectation}
-    of $\sigma(\sigma(X), \cG)$, then
+of $X$ given  $Y = y_j$ as
 \[
-         \bE[X | \sigma(\cG, \cH)] \overset{\text{a.s.}}{=} \bE[X | \cG].
+\bE[X | Y = y_j] \coloneqq  \sum_{i=1}^m x_i \bP[X=x_i | Y = y_j].
 \]
-\end{theorem}
-\begin{example}
-    If $X$ is independent of $\cG$,
-    then $\bE[X | \cG] \overset{\text{a.s.}}{=} \bE[X]$.
-\end{example}
-\begin{example}[Martingale property of the simple random walk]
-    Suppose $X_1,X_2,\ldots$ are i.i.d.~with $\bP[X_i = 1] = \bP[X_i = -1] = \frac{1}{2}$.
-    Let $S_n \coloneqq  \sum_{i=1}^n X_i$ be the \vocab{simple random walk}.
-    Let $\cF$ denote the $\sigma$-algebra on the product space.
-    Define $\cF_n \coloneqq  \sigma(X_1,\ldots)$.
-    Intuitively, $\cF_n$ contains all the information gathered until time  $n$.
-    We have $\cF_1 \subset \cF_2 \subset \cF_3 \subset \ldots$
 
-    For $\bE[S_{n+1} | \cF_n]$ we obtain
+The random variable $Z = \bE[X | Y]$
+is defined as follows:
+If $Y(\omega) = y_j$ then
+\[
+Z(\omega) \coloneqq  \underbrace{\bE[X | Y = y_j]}_{\text{\reflectbox{$\coloneqq$}} z_j}.
+\]
+Note that $\Omega_j \coloneqq  \{\omega : Y(\omega) = y_j\}$
+defines a partition of $\Omega$ and on each $\Omega_j$
+(``the  $j^{\text{th}}$ $Y$-atom'')
+$ Z$ is constant.
+
+
+Let $\cG \coloneqq  \sigma(Y)$.
+Then $Z$ is measurable with respect to $\cG$.
+Furthermore
 \begin{IEEEeqnarray*}{rCl}
-        \bE[S_{n+1} | \cF_n] &\overset{\autoref{celinearity}}{=}&
+    \int_{\{Y = y_j\} } Z  \dif \bP &=& z_j \int_{\{Y = y_j\}} \dif \bP\\
-        \bE[S_n | \cF_n] + \bE[X_{n+1} | \cF_n]\\
+                                    &=& z_j \bP[Y=y_j]\\
-                             &\overset{\text{a.s.}}{=}& S_n + \bE[X_{n+1} | \cF_n]\\
+                                    &=&\sum_{i=1}^m  x_i \bP[X = x_i | Y = y_j] \bP[Y = y_j]\\
-                             &\overset{\text{\autoref{ceprop12}}}{=}& S_{n} + \bE[X_n]\\
+                                    &=&\sum_{i=1}^m x_i \bP[X = x_i, Y = y_j]\\
-                             &=& S_n
+                                    &=& \int_{\{Y = y_j\}} X \dif \bP.
 \end{IEEEeqnarray*}
+Hence
+\[
+\int_{G} Z \dif \bP = \int_{G} X \dif \bP
+\]
+for all $G \in \cG$.
 
-\end{example}
+We now want to generalize this to arbitrary random variables.
+\begin{theorem}
+    \label{conditionalexpectation}
+    Let $(\Omega, \cF, \bP)$ be a probability space, $X \in L^1(\bP)$
+    and $\cG \subseteq  \cF$ a sub-$\sigma$-algebra.
+    Then there exists a random variable $Z$
+    such that
+    \begin{enumerate}[(a)]
+        \item $Z$ is $\cG$-measurable and $Z \in L^1(\bP)$,
+        \item $\int_G Z \dif \bP = \int_G X \dif \bP$
+            for all $G \in \cG$.
+    \end{enumerate}
+
+    Such a $Z$ is unique up to sets of measure $0$ and is
+    called the \vocab{conditional expectation}  of $X$ given
+    the $\sigma$-algebra $\cG$ and written
+    $Z = \bE[X | \cG]$.
+\end{theorem}
+\begin{remark}
+     Suppose $\cG = \{\emptyset, \Omega\}$,
+         then
+         \[
+             \bE[X | \cG] = (\omega \mapsto \bE[X])
+         \] 
+         is a constant random variable.
+\end{remark}
+
+\paragraph{Plan}
+We will give two different proves of \autoref{conditionalexpectation}.
+The first one will use orthogonal projections.
+The second will use the Radon-Nikodym theorem.
+We'll first do the easy proof, derive some properties
+and then do the harder proof.
+
+\begin{lemma}
+    \label{orthproj}
+    Suppose $H$ is a \vocab{Hilbert space},
+    i.e.~$H$ is a vector space with an inner product $\langle \cdot, \cdot \rangle_H$ which defines a norm by $\|x\|_H^2 = \langle x, x\rangle_H$
+    making $H$ a complete metric space.
+
+    For any $x \in H$ and $K \subseteq H$ closed,
+    there exists a unique $z \in K$ such that the following equivalent conditions hold:
+    \begin{enumerate}[(a)]
+        \item  $\forall  y \in K : \langle x-z, y\rangle_H = 0$,
+        \item $\forall y \in K: \|z-x\|_H \le \|z-x\|_H$.
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    \todo{Notes}
+\end{proof}
+
+\begin{refproof}{conditionalexpectation}
+
+    Almost sure uniqueness of $Z$:
+
+    Suppose $X \in L^1$ and $Z$ and $Z'$ satisfy (a) and (b).
+    We need to show that $\bP[Z \neq Z'] = 0$.
+    By (a), we have $Z, Z' \in  L^1(\Omega, \cG, \bP)$.
+    By (b), $\bE[(Z - Z') \One_G] = 0$ for all $G \in \cG$.
+
+    Assume that $\bP[Z > Z'] > 0$.
+    Since $\{Z > Z' + \frac{1}{n}\} \uparrow \{Z > Z'\}$,
+    we see that $\bP[Z > Z' + \frac{1}{n}] > 0$ for some $n$.
+    However $\{Z > Z' + \frac{1}{n}\} \in \cG$,
+    which is a contradiction, since
+    \[
+    \bE[(Z - Z') \One_{Z - Z' > \frac{1}{n}}] \ge \frac{1}{n} \bP[ Z - Z' > \frac{1}{n}] > 0.
+    \] 
+
+
+    \bigskip
+    Existence of $\bE(X | \cG)$ for $X \in L^2$:
+
+    Let $H = L^2(\Omega, \cF, \bP)$ 
+    and $K = L^2(\Omega, \cG, \bP)$.
+
+    $K$ is closed, since a pointwise limit of $\cG$-measurable
+    functions is $\cG$ measurable (if it exists).
+    By \autoref{orthproj},
+    there exists $z \in K$ such that
+    \[\bE[(X - Z)^2] = \inf \{ \bE[(X- W)^2] ~|~ W \in L^2(\cG)\}\]
+    and
+    \begin{equation}
+    \forall  Y \in L^2(\cG) : \langle X - Z, Y\rangle = 0.
+    \label{lec13_boxcond}
+    \end{equation}
+    Now, if $G \in \cG$, then $Y \coloneqq  \One_G \in L^2(\cG)$
+    and by \eqref{lec13_boxcond} $\bE[Z \One_G] = \bE[X \One_G]$.
+
+
+    \bigskip
+    Existence of  $\bE(X | \cG)$ for $X \in L^1$ :
+
+    Let $X = X^+ - X^-$.
+    It suffices to show (a) and (b) for  $X^+$.
+    Choose bounded random variables $X_n \ge 0$ such that $X_n \uparrow X$.
+    Since each $X_n \in L^2$, we can choose a version $Z_n$ of $\bE(X_n | \cG)$.
+
+    \begin{claim}
+        $0 \overset{\text{a.s.}}{\le} Z_n \uparrow$.
+    \end{claim}
+    \begin{subproof}
+        \todo{Notes}
+    \end{subproof}
+
+    Define $Z(\omega) \coloneqq  \limsup_{n \to \infty} Z_n(\omega)$.
+    Then $Z$ is $\cG$-measurable and since $Z_n \uparrow Z$,
+    by MCT, $\bE(Z \One_G) = \bE(X \One_G)$ for all  $G \in \cG$.
+\end{refproof}

inputs/lecture_15.tex

@@ -0,0 +1,255 @@
+\lecture{15}{2023-06-06}{}
+
+We want to derive some properties of conditional expectation.
+
+\begin{theorem}[Law of total expectation] % Thm 1
+    \label{ceprop1}
+    \label{totalexpectation}
+    \[
+        \bE[\bE[X | \cG ]] = \bE[X].
+    \]
+\end{theorem}
+\begin{proof}
+    Apply (b) from the definition for $G = \Omega  \in \cG$.
+\end{proof}
+\begin{theorem} % Thm 2
+    \label{ceprop2}
+    If $X$ is $\cG$-measurable, then $X = \bE[X | \cG]$ a.s..
+\end{theorem}
+\begin{proof}
+    Suppose $\bP[X \neq Y] > 0$.
+    Without loss of generality $\bP[X > Y] > 0$.
+    Hence $\bP[ X > Y + \frac{1}{n}]> 0$ for some $n \in \N$.
+    Let $A \coloneqq \{X > Y + \frac{1}{n}\}$.
+    % TODO
+\end{proof}
+
+\begin{example}
+    Suppose $X \in L^1(\bP)$, $\cG \coloneqq  \sigma(X)$.
+    Then $X$ is measurable with respect to $\cG$.
+    Hence $\bE[X | \cG] = X$.
+\end{example}
+
+\begin{theorem}[Linearity]
+    \label{ceprop3}
+    \label{celinearity}
+    For all $a,b \in \R$
+    we have
+    \[
+        \bE[a X_1 + bX_2 | \cG] = a \bE[X_1 | \cG] + b \bE[X_2|\cG].
+    \]
+\end{theorem}
+\begin{proof}
+    Trivial % TODO
+\end{proof}
+
+\begin{theorem}[Positivity]
+    \label{ceprop4}
+    % 4
+    \label{cpositivity}
+    If $X \ge  0$, then $\bE[X | \cG] \ge 0$ a.s.
+\end{theorem}
+\begin{proof}
+    Let $W $ be a version of $\E[X | \cG]$.
+    Suppose $\bP[ W < 0] > 0$.
+    Then $G \coloneqq  \{W < -\frac{1}{n}\} \in \cG$
+    For some $n \in \N$, we have $\bP[G] > 0$.
+    However it follows that
+    \[
+        \int_G \bP[X | \cG] \dif \bP \le  -\frac{1}{n} \bP[G] < 0 \le  \int_G X \dif \bP.
+    \] 
+\end{proof}
+\begin{theorem}[Conditional monotone convergence theorem]
+    \label{ceprop5}
+    % 5
+    \label{mcmt}
+    Let $X_n,X \in L^1(\Omega, \cF, \bP)$.
+    Suppose $X_n \ge 0$ with $X_n \uparrow X$.
+    Then $\bE[X_n|\cG] \uparrow \bE[X|\cG]$.
+    
+\end{theorem}
+\begin{proof}
+    Let $Z_n$ be a version of $\bE[X_n | Y]$.
+    Since  $X_n \ge 0$ and $X_n \uparrow$,
+    by \autoref{cpositivity},
+    we have
+    \[
+        \bE[X_n | \cG] \overset{\text{a.s.}}{\ge } 0
+    \] 
+    and
+    \[
+        \bE[X_n | \cG] \uparrow \text{a.s.}
+    \] 
+    (consider $X_{n+1} - X_n$ ).
+
+    Define $Z \coloneqq  \limsup_{n \to \infty} Z_n$.
+    Then $Z$ is $\cG$-measurable
+    and $Z_n \uparrow Z$ a.s.
+
+    Take some $G \in \cG$.
+    We know by (b) % TODO REF
+    that $\be[Z_n \One_G] = \bE[X_n \One_G]$.
+    The LHS increases to  $\bE[Z \One_G]$ by the monotone 
+    convergence theorem.
+    Again by MCT, $\bE[X_n \One_G]$ increases to
+    $\bE[X \One_G]$.
+    Hence $Z$ is a version of $\bE[X | \cG]$.
+\end{proof}
+
+\begin{theorem}[Conditional Fatou]
+    \label{ceprop6}
+    \label{cfatou}
+    Let $X_n \in L^1(\Omega, \cF, \bP)$, $X_n \ge 0$.
+    Then
+    \[
+        \bE[ \liminf_{n \to \infty} X_n | \cG] \le  \liminf_{n \to \infty} \bE[X_n | \cG].
+    \] 
+\end{theorem}
+\begin{proof}
+    \todo{in the notes}
+\end{proof}
+\begin{theorem}[Conditional dominated convergence theorem]
+    \label{ceprop7}
+    \label{cdct}
+    Let $X_n,X \in L^1(\Omega, \cF, \bP)$.
+    Suppose $|X_n(\omega)| < X(\omega)$ a.e.~
+    and  $\int |X| \dif \bP < \infty$.
+    Then $X_n(\omega) \to  X\left( \omega \right) \implies \bE[ X_n | \cG] \to \bE[X | \cG]$.
+    
+\end{theorem}
+\begin{proof}
+    \todo{in the notes}
+\end{proof}
+
+Recall
+\begin{theorem}[Jensen's inequality]
+    If $c : \R \to  \R$ is convex and $\bE[|c \circ X|] < \infty$,
+    then $\bE[c \circ X] \ge c(\bE[X])$.
+\end{theorem}
+
+For conditional expectation, we have
+\begin{theorem}[Conditional Jensen's inequality]
+    \label{ceprop8}
+    \label{cjensen}
+    Let $X \in L^1(\Omega, \cF, \bP)$.
+    If $c : \R \to  \R$ is convex and $\bE[|c \circ X|] < \infty$,
+    then $\bE[c \circ X | \cG] \ge c(\bE[X | \cG])$ a.s.
+\end{theorem}
+\begin{fact}
+    \label{convapprox}
+    If $c$ is convex, then there are two sequences of real numbers
+    $a_n, b_n \in \R$
+    such that
+    \[
+    c(x) = \sup_n(a_n x + b_n).
+    \] 
+\end{fact}
+\begin{refproof}{cjensen}
+    By \autoref{convapprox}, $c(x) \ge  a_n X + b_n$
+    for all $n$.
+    Hence
+    \[
+        \bE[c(X) | \cG] \ge a_n \bE[X | \cG] + \bE[b_n | \cG]
+        = a_n \bE[X | \cG] + b_n \text{a.s.}
+    \] 
+    for all $n$.
+    Using that  a countable union of sets o f measure zero has measure zero,
+    we conclude that a.s~this happens simultaneously for all $n$.
+    Hence
+    \[
+        \bE[c(X) | \cG] \ge  \sup_n (a_n \bE[X | \cG] + b_n) \overset{\text{\autoref{convapprox}}}{=} c(\bE(X | \cG)).
+    \] 
+\end{refproof}
+
+Recall
+\begin{theorem}[Hölder's inequality]
+    Let $p,q \ge  1$ such that $\frac{1}{p} + \frac{1}{q} = 1$.
+    Suppose $X \in L^p(\bP)$ and $Y \in L^q(\bP)$.
+    Then
+    \[
+        \bE(X Y) \le  \underbrace{\bE(|X|^p)^{\frac{1}{p}}}_{\text{\reflectbox{$\coloneqq$}} \|X\|_{L^p}}  \bE(|Y|^q)^{\frac{1}{q}}.
+    \] 
+\end{theorem}
+
+\begin{theorem}[Conditional Hölder's inequality]
+    \label{ceprop9}
+    \label{choelder}
+    Let $p,q \ge  1$ such that $\frac{1}{p} + \frac{1}{q} = 1$.
+    Suppose $X \in L^p(\bP)$ and $Y \in L^q(\bP)$.
+    Then
+    \[
+        \bE(X Y | \cG) \le \bE(|X|^p | \cG)^{\frac{1}{p}} \bE(|Y|^q | \cG)^{\frac{1}{q}}.
+    \] 
+\end{theorem}
+\begin{proof}
+    Similar to the proof of Hölder's inequality.
+    \todo{Exercise}
+\end{proof}
+
+\begin{theorem}[Tower property]
+    % 10
+    \label{ceprop10}
+    \label{ctower}
+    Suppose $\cF \supset \cG \supset \cH$ are sub-$\sigma$-algebras.
+    Then
+    \[
+        \bE\left[\bE[X | \cG] \mid \cH\right] = \bE[X | \cH].
+    \] 
+\end{theorem}
+\begin{proof}
+    \todo{Exercise}
+\end{proof}
+
+\begin{theorem}[Taking out what is known]
+    % 11
+    \label{ceprop11}
+    \label{takingoutwhatisknown}
+
+    If $Y$ is $\cG$-measurable and bounded, then
+    \[
+        \bE[YX| \cG] \overset{\text{a.s.}}{=} Y \bE[X | \cG].
+    \] 
+\end{theorem}
+\begin{proof}
+    Assume w.l.o.g.~$X \ge 0$.
+Assume $Y = \One_B$, then $Y$ simple, then take the limit (using that $Y$ is bounded).
+    \todo{Exercise}
+\end{proof}
+
+\begin{definition}
+    Let $\cG$ and $\cH$ be $\sigma$-algebras.
+    We call $\cG$ and $\cH$ \vocab[$\sigma$-algebra!independent]{independent},
+    if % TODO
+\end{definition}
+
+\begin{theorem}[Role of independence]
+    \label{ceprop12}
+    \label{roleofindependence}
+    If $\cH$ is a sub-$\sigma$-algebra of $\cF$ and  $\cH$ is independent
+    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 $\cG$,
+    then $\bE[X | \cG] \overset{\text{a.s.}}{=} \bE[X]$.
+\end{example}
+\begin{example}[Martingale property of the simple random walk]
+    Suppose $X_1,X_2,\ldots$ are i.i.d.~with $\bP[X_i = 1] = \bP[X_i = -1] = \frac{1}{2}$.
+    Let $S_n \coloneqq  \sum_{i=1}^n X_i$ be the \vocab{simple random walk}.
+    Let $\cF$ denote the $\sigma$-algebra on the product space.
+    Define $\cF_n \coloneqq  \sigma(X_1,\ldots)$.
+    Intuitively, $\cF_n$ contains all the information gathered until time  $n$.
+    We have $\cF_1 \subset \cF_2 \subset \cF_3 \subset \ldots$
+
+    For $\bE[S_{n+1} | \cF_n]$ we obtain
+    \begin{IEEEeqnarray*}{rCl}
+        \bE[S_{n+1} | \cF_n] &\overset{\autoref{celinearity}}{=}&
+        \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
+    \end{IEEEeqnarray*}
+    
+\end{example}

inputs/lecture_2.tex

@@ -1,3 +1,4 @@
+\lecture{2}{}{}
 \section{Independence and product measures}
 
 In order to define the notion of independence, we first need to construct

inputs/lecture_3.tex

@@ -1,3 +1,4 @@
+\lecture{3}{}{}
 \todo{Lecture 3 needs to be finished}
 \begin{notation}
     Let $\cB_n$ denote $\cB(\R^n)$.

inputs/lecture_5.tex

@@ -1,4 +1,4 @@
-% Lecture 5 2023-04-21
+\lecture{5}{2023-04-21}{}
 \subsection{The laws of large numbers}
 
 

wtheo.sty

@@ -103,3 +103,4 @@
 \DeclareSimpleMathOperator{Exp}
 
 \newcommand*\dif{\mathop{}\!\mathrm{d}}
+\newcommand\lecture[3]{{\color{gray}\hfill Lecture #1 (#2)}}