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+\lecture{14}{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}