diff --git a/inputs/lecture_1.tex b/inputs/lecture_1.tex
index 10f049f..3de936a 100644
--- a/inputs/lecture_1.tex
+++ b/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 $\cM(\R)$
diff --git a/inputs/lecture_14.tex b/inputs/lecture_14.tex
new file mode 100644
index 0000000..a5b9a4a
--- /dev/null
+++ b/inputs/lecture_14.tex
@@ -0,0 +1,173 @@
+\lecture{14}{2023-05-25}{Conditional expectation}
+
+\section{Conditional expectation}
+
+\subsection{Introduction}
+
+Consider a probability space $(\Omega, \cF, \bP)$
+and two events $A, B \in \cF$ with $\bP(B) > 0$.
+
+\begin{definition}
+    The \vocab{conditional probability}  of $A$ given $B$ is defined as
+    \[
+    \bP(A | B) \coloneqq \frac{\bP(A \cap B)}{\bP(B)}.
+    \]
+\end{definition}
+
+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
+\[
+\bE[X | Y = y_j] \coloneqq  \sum_{i=1}^m x_i \bP[X=x_i | Y = y_j].
+\]
+
+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}
+    \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 $G \in \cG$.
+
+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}
+
+
+
+
+
+
+
+
+
+
+
diff --git a/inputs/lecture_2.tex b/inputs/lecture_2.tex
index 5360dc4..1c38d36 100644
--- a/inputs/lecture_2.tex
+++ b/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
diff --git a/inputs/lecture_3.tex b/inputs/lecture_3.tex
index 2e8c762..8708d78 100644
--- a/inputs/lecture_3.tex
+++ b/inputs/lecture_3.tex
@@ -1,3 +1,4 @@
+\lecture{3}{}{}
 \todo{Lecture 3 needs to be finished}
 \begin{notation}
     Let $\cB_n$ denote $\cB(\R^n)$.
diff --git a/inputs/lecture_5.tex b/inputs/lecture_5.tex
index 5364619..b2f53e1 100644
--- a/inputs/lecture_5.tex
+++ b/inputs/lecture_5.tex
@@ -1,4 +1,4 @@
-% Lecture 5 2023-04-21
+\lecture{5}{2023-04-21}{}
 \subsection{The laws of large numbers}
 
 
diff --git a/probability_theory.tex b/probability_theory.tex
index 9ab7263..0b93608 100644
--- a/probability_theory.tex
+++ b/probability_theory.tex
@@ -37,6 +37,7 @@
 \input{inputs/lecture_11.tex}
 \input{inputs/lecture_12.tex}
 \input{inputs/lecture_13.tex}
+\input{inputs/lecture_14.tex}
 
 \cleardoublepage
 
diff --git a/wtheo.sty b/wtheo.sty
index e42ca30..1aab83a 100644
--- a/wtheo.sty
+++ b/wtheo.sty
@@ -103,3 +103,4 @@
 \DeclareSimpleMathOperator{Exp}
 
 \newcommand*\dif{\mathop{}\!\mathrm{d}}
+\newcommand\lecture[3]{{\color{gray}\hfill Lecture #1 (#2)}}