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inputs/lecture_1.tex
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@ -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 $\cM(\R)$

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\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
\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}
\subsection{Introduction}
\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}
Consider a probability space $(\Omega, \cF, \bP)$
and two events $A, B \in \cF$ with $\bP(B) > 0$.
\begin{definition}
Let $\cG$ and $\cH$ be $\sigma$-algebras.
We call $\cG$ and $\cH$ \vocab[$\sigma$-algebra!independent]{independent},
if % TODO
The \vocab{conditional probability} of $A$ given $B$ is defined as
\[
\bP(A | B) \coloneqq \frac{\bP(A \cap B)}{\bP(B)}.
\]
\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
Suppose we have two random variables $X$ and $Y$ on $\Omega$,
such that $X$ takes distinct values $x_1, x_2,\ldots, x_{m}$
and $Y$ takes distinct values $y_1,\ldots, y_n$.
Then for this case, define the \vocab{conditional expectation}
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}}{=}&
\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
\int_{\{Y = y_j\} } Z \dif \bP &=& z_j \int_{\{Y = y_j\}} \dif \bP\\
&=& z_j \bP[Y=y_j]\\
&=&\sum_{i=1}^m x_i \bP[X = x_i | Y = y_j] \bP[Y = y_j]\\
&=&\sum_{i=1}^m x_i \bP[X = x_i, Y = y_j]\\
&=& \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}

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\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}

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inputs/lecture_2.tex
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\lecture{2}{}{}
\section{Independence and product measures}
In order to define the notion of independence, we first need to construct

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inputs/lecture_3.tex
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\lecture{3}{}{}
\todo{Lecture 3 needs to be finished}
\begin{notation}
Let $\cB_n$ denote $\cB(\R^n)$.

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inputs/lecture_5.tex
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% Lecture 5 2023-04-21
\lecture{5}{2023-04-21}{}
\subsection{The laws of large numbers}

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wtheo.sty
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\DeclareSimpleMathOperator{Exp}
\newcommand*\dif{\mathop{}\!\mathrm{d}}
\newcommand\lecture[3]{{\color{gray}\hfill Lecture #1 (#2)}}