some cleanup

inputs/lecture_13.tex

@@ -46,7 +46,7 @@ We will not prove the \autoref{lindebergclt} or \autoref{lyapunovclt}
 in this lecture. However, they are quite important.
 
 We will now sketch the proof of \autoref{levycontinuity},
-details can be found in the notes.\todo{Complete this}
+details can be found in the notes.\notes
 A generalized version of \autoref{levycontinuity} is the following:
 \begin{theorem}[A generalized version of Levy's continuity \autoref{levycontinuity}]
     \label{genlevycontinuity}
@@ -120,7 +120,7 @@ A generalized version of \autoref{levycontinuity} is the following:
 
 We will prove \autoref{levycontinuity} assuming
 \autoref{lec10_thm1}.
-\autoref{lec10_thm1} will be shown in the notes.\todo{TODO}
+\autoref{lec10_thm1} will be shown in the notes.\notes
 We will need the following:
 \begin{lemma}
     \label{lec13_lem1}
@@ -239,7 +239,7 @@ We still need to show that $\mu_n \implies \mu$.
     Then $a_n \to  a$.
 \end{fact}
 \begin{subproof}
-    \todo{in the notes}
+    \notes
 \end{subproof}
 Assume that $\mu_n$ does not converge to $\mu$.
 By \autoref{lec10_thm1}, pick a continuity point $x_0$ of $F$,

inputs/lecture_14.tex

@@ -87,7 +87,8 @@ We now want to generalize this to arbitrary random variables.
     \]
 \end{definition}
 
-\paragraph{Plan}
+\subsection{Existence of conditional probability}
+
 We will give two different proves of \autoref{conditionalexpectation}.
 The first one will use orthogonal projections.
 The second will use the Radon-Nikodym theorem.
@@ -108,7 +109,7 @@ and then do the harder proof.
     \end{enumerate}
 \end{lemma}
 \begin{proof}
-    \todo{Notes}
+    \notes
 \end{proof}
 
 \begin{refproof}{conditionalexpectation}
@@ -162,7 +163,7 @@ and then do the harder proof.
         $0 \overset{\text{a.s.}}{\le} Z_n \uparrow$.
     \end{claim}
     \begin{subproof}
-        \todo{Notes}
+        \notes
     \end{subproof}
 
     Define $Z(\omega) \coloneqq  \limsup_{n \to \infty} Z_n(\omega)$.

inputs/lecture_15.tex

@@ -1,4 +1,5 @@
 \lecture{15}{2023-06-06}{}
+\subsection{Properties of conditional expectation}
 
 We want to derive some properties of conditional expectation.
 
@@ -12,16 +13,20 @@ We want to derive some properties of conditional expectation.
 \begin{proof}
     Apply (b) from the definition for $G = \Omega  \in \cG$.
 \end{proof}
-\begin{theorem} % Thm 2
+\begin{theorem}
     \label{ceprop2}
-    If $X$ is $\cG$-measurable, then $X = \bE[X | \cG]$ a.s..
+    If $X$ is $\cG$-measurable, then $X \overset{\text{a.s.}}{=}\bE[X | \cG]$.
 \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
+    Then
+    \[
+    \int_A X \dif \bP \ge \frac{1}{n}\bP(A) + \int_A Y \dif \bP,
+    \] 
+    contradicting property (b) from \autoref{conditionalexpectation}.
 \end{proof}
 
 \begin{example}
@@ -40,7 +45,7 @@ We want to derive some properties of conditional expectation.
     \]
 \end{theorem}
 \begin{proof}
-    Trivial % TODO
+    trivial\todo{add details}
 \end{proof}
 
 \begin{theorem}[Positivity]
@@ -52,7 +57,7 @@ We want to derive some properties of conditional expectation.
 \begin{proof}
     Let $W $ be a version of $\bE[X | \cG]$.
     Suppose $\bP[ W < 0] > 0$.
-    Then $G \coloneqq  \{W < -\frac{1}{n}\} \in \cG$
+    Then \[G \coloneqq  \{W < -\frac{1}{n}\} \in \cG.\]
     For some $n \in \N$, we have $\bP[G] > 0$.
     However it follows that
     \[
@@ -106,7 +111,7 @@ We want to derive some properties of conditional expectation.
     \] 
 \end{theorem}
 \begin{proof}
-    \todo{in the notes}
+    \notes
 \end{proof}
 \begin{theorem}[Conditional dominated convergence theorem]
     \label{ceprop7}
@@ -118,14 +123,14 @@ We want to derive some properties of conditional expectation.
     
 \end{theorem}
 \begin{proof}
-    \todo{in the notes}
+    \notes
 \end{proof}
 
 Recall
-\begin{theorem}[Jensen's inequality]
+\begin{fact}[Jensen's inequality]
     If $c : \R \to  \R$ is convex and $\bE[|c \circ X|] < \infty$,
-    then $\bE[c \circ X] \ge c(\bE[X])$.
+    then $\bE[c \circ X] \overset{\text{a.s.}}{\ge} c(\bE[X])$.
-\end{theorem}
+\end{fact}
 
 For conditional expectation, we have
 \begin{theorem}[Conditional Jensen's inequality]
@@ -150,7 +155,7 @@ For conditional expectation, we have
     Hence
     \[
         \bE[c(X) | \cG] \ge a_n \bE[X | \cG] + \bE[b_n | \cG]
-        = a_n \bE[X | \cG] + b_n \text{a.s.}
+        = a_n \bE[X | \cG] + b_n \text{ a.s.}
     \] 
     for all $n$.
     Using that  a countable union of sets o f measure zero has measure zero,
@@ -162,14 +167,14 @@ For conditional expectation, we have
 \end{refproof}
 
 Recall
-\begin{theorem}[Hölder's inequality]
+\begin{fact}[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}
+\end{fact}
 
 \begin{theorem}[Conditional Hölder's inequality]
     \label{ceprop9}
@@ -193,7 +198,7 @@ Recall
     Suppose $\cF \supset \cG \supset \cH$ are sub-$\sigma$-algebras.
     Then
     \[
-        \bE\left[\bE[X | \cG] \mid \cH\right] = \bE[X | \cH].
+        \bE\left[\bE[X | \cG] \mid \cH\right] \overset{\text{a.s.}}{=} \bE[X | \cH].
     \] 
 \end{theorem}
 \begin{proof}
@@ -219,7 +224,8 @@ Assume $Y = \One_B$, then $Y$ simple, then take the limit (using that $Y$ is bou
 \begin{definition}
     Let $\cG$ and $\cH$ be $\sigma$-algebras.
     We call $\cG$ and $\cH$ \vocab[$\sigma$-algebra!independent]{independent},
-    \todo{TODO}
+    if $\bP(G \cap H) = \bP(G) \bP(H)$
+    for all events $G \in \cG$, $H \in \cH$.
 \end{definition}
 
 \begin{theorem}[Role of independence]
@@ -251,7 +257,7 @@ Assume $Y = \One_B$, then $Y$ simple, then take the limit (using that $Y$ is bou
 
     For $\bE[S_{n+1} | \cF_n]$ we obtain
     \begin{IEEEeqnarray*}{rCl}
-        \bE[S_{n+1} | \cF_n] &\overset{\autoref{celinearity}}{=}&
+        \bE[S_{n+1} | \cF_n] &\overset{\text{\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]\\

inputs/lecture_16.tex

@@ -68,7 +68,8 @@ The Radon Nikodym theorem is the converse of that:
     Such a $Z$ is called the \vocab{Radon-Nikodym derivative}.
 \end{theorem}
 \begin{definition}
-    Whenever the property $\forall  A \in  \cF, \mu(A)= 0 \implies \nu(A) = 0$,
+    Whenever the property $\forall  A \in  \cF, \mu(A) = 0 \implies \nu(A) = 0$
+    holds for two measures $\mu$ and $\nu$,
     we say that $\nu$ is \vocab{absolutely continuous} 
     w.r.t.~$\mu$.
     This is written as $\nu \ll \mu$.
@@ -86,16 +87,19 @@ of conditional expectation:
 
 
 \begin{refproof}{radonnikodym}
-    We will only sketch the proof. A full proof can be found in the notes.
+    We will only sketch the proof.
+    A full proof can be found in the official notes.
     
-    \paragraph{Step 1: Uniqueness}
+    \paragraph{Step 1: Uniqueness} \notes
-        See notes.
     \paragraph{Step 2: Reduction to the finite measure case}
-        See notes.
+        \notes
     \paragraph{Step 3: Getting hold of $Z$}
     Assume now that $\mu$ and $\nu$ are two finite measures.
     Let
-    \[\cC \coloneqq  \{f: \Omega \to [0,\infty] | \forall A \in \cF.~\int_A f \dif \mu \le  \nu(A)\}.\]
+    \[
+        \cC \coloneqq  \left\{f: \Omega \to [0,\infty] \middle| %
+        \forall A \in \cF.~\int_A f \dif \mu \le  \nu(A)\right\}.
+    \]
 
     We have $\cC \neq  \emptyset$ since $0 \in \cC$.
     The goal is to find a maximal function $Z$ in $\cC$.
@@ -145,7 +149,7 @@ of conditional expectation:
         but it need not satisfy (ii).
 
         The tricky part is to make $A$ smaller such that it also
-        satisfies (ii).\todo{Copy from notes}
+        satisfies (ii).\notes
     \end{subproof}
 \end{refproof}
 
@@ -153,6 +157,8 @@ of conditional expectation:
 
 \section{Martingales}
 
+\subsection{Definition}
+
 We have already worked with martingales, but we will define them
 rigorously now.
 
@@ -168,15 +174,16 @@ Typically $\cF_n = \sigma(X_1, \ldots, X_n)$ for a sequence of random variables.
     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
+    if the following hold:
     \begin{itemize}
-        \item $X_n$ is $\cF_n$-measurable for all $n$
+        \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},
+    $(X_n)_n$ 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.
@@ -185,12 +192,12 @@ Typically $\cF_n = \sigma(X_1, \ldots, X_n)$ for a sequence of random variables.
     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 $f: \R \to  \R$ is a convex function such that $f(X_n) \in L^1(\bP)$.
-    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}.
+    Suppose that $(X_n)_n$ is a martingale\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}.
+    Apply \autoref{cjensen}.
 \end{proof}
 
 \begin{corollary}

inputs/lecture_17.tex

@@ -1,5 +1,8 @@
 \lecture{17}{2023-06-15}{}
 
+\subsection{Doob's martingale convergence theorem}
+
+
 \begin{definition}[Stochastic process]
     A \vocab{stochastic process} is a collection of random
     variables $(X_t)_{t \in T}$ for some index set $T$.
@@ -64,6 +67,7 @@ such that $X_{s_j}(\omega) < a$ and $X_{t_j}(\omega) > b$ for all $1 \le  j \le
 Clearly $U_N^X([a,b]) \uparrow$ as $N$ increases.
 It follows that the monotonic limit $U_\infty([a,b]) \coloneqq  \lim_{N \to \infty} U_N([a,b])$ exists pointwise.
 \begin{lemma} % Lemma 1
+    \label{lec17l1}
     \[
     \{\omega | \liminf_{N \to  \infty} Z_N(\omega) < a < b <
     \limsup_{N \to \infty} Z_N(\omega)\} \subseteq 
@@ -73,8 +77,9 @@ It follows that the monotonic limit $U_\infty([a,b]) \coloneqq  \lim_{N \to \inf
     
 \end{lemma}
 \begin{lemma} % 2
+    \label{lec17l2}
     Let $Y_n(\omega) \coloneqq \sum_{j=1}^n C_j(X_j - X_{j-1})$.
-    Then $Y_N \ge  (b -a) U_N([a,b]) - (X_N - a)^{-}$.
+    Then \[Y_N \ge  (b -a) U_N([a,b]) - (X_N - a)^{-}.\]
 \end{lemma}
 \begin{proof}
     Every upcrossing of $[a,b]$ increases the  value of $Y$ by $(b-a)$,
@@ -82,21 +87,24 @@ It follows that the monotonic limit $U_\infty([a,b]) \coloneqq  \lim_{N \to \inf
 \end{proof}
 
 \begin{lemma} %3
+    \label{lec17l3}
     Suppose $(X_n)_n$ is a supermartingale.
-    Then in the above setup, $(b-a) \bE[U_N([a,b])] \le  \bE[(X_n - a)^-]$.
+    Then in the above setup
+    \[(b-a) \bE[U_N([a,b])] \le  \bE[(X_n - a)^-].\]
 \end{lemma}
 \begin{proof}
-    Obvious from lemma 2 % TODO REF
+    This is obvious from \autoref{lec17l2}
     and the supermartingale property.
 \end{proof}
 \begin{corollary}
-    Let $(X_n)_n$ be a \vocab[Supermartingale!bounded]{supermartingale bounded in $L^1(\bP)$ },
+    Let $(X_n)_n$ be a
-    i.e.~$\sup_n \bE[|X_n| ] < \infty$.
+    \vocab[Supermartingale!bounded]{supermartingale bounded in $L^1(\bP)$},
+    i.e.~$\sup_n \bE[|X_n|] < \infty$.
     Then $(b-a) \bE(U_\infty) \le  |a| + \sup_n \bE(|X_n|)$.
     In particular, $\bP[U_\infty = \infty] = 0$.
 \end{corollary}
 \begin{proof}
-    By lemma 3 % TODO REF
+    By \autoref{lec17l3}
     we have that
     \[(b-a) \bE[U_N([a,b])] \le  \bE[ | X_N| ] + |a| \le  \sup_n \bE[|X_n|] + |a|.\]
     Since $U_N(\cdot) \ge  0$ and $U_N(\cdot ) \uparrow U_\infty(\cdot )$,

inputs/lecture_18.tex

@@ -1,10 +1,9 @@
 \lecture{18}{2023-06-20}{}
 
-Recall our key lemma for supermartingales from last time:
+Recall our key lemma \ref{lec17l3} for supermartingales from last time:
 \[
     (b-a) \bE[U_N([a,b])] \le  \bE[(X_n - a)^-].
 \] 
-% TODO Ref
 
 What happens for submartingales?
 If $(X_n)_{n \in \N}$ is a submartingale, then $(-X_n)_{n \in \N}$ is a supermartingale.
@@ -14,11 +13,14 @@ Hence the same holds for submartingales, i.e.
     a.s.~to a finite limit, which is a.s.~finite.
 \end{lemma}
 
+\subsection{Doob's $L^p$ inequality}
+
+
 \begin{question}
     What about $L^p$ convergence of martingales?
 \end{question}
 
-\begin{example}[Branching process]
+\begin{example}[\vocab{Branching process}]
     Let $ (Z_n)_{n \ge 1}$ be i.i.d.~$\pm 1$ with
     $\bP[Z_n = 1] = p \in (0,1)$.
 
@@ -26,11 +28,11 @@ Hence the same holds for submartingales, i.e.
     Define $X_{n+1} \coloneqq  u^{Z_{n+1}} X_{n}$.
 
 
-    \paragraph{Exercise}
+    \begin{exercise}
         Given $u \ge 0$, find $p = p(u)$
         such that  $(X_n)_n$ is a martingale w.r.t.~the canonical filtration.
-
+    \end{exercise}
-    % TODO
+    \todo{TODO}
 
 
     By \autoref{doobmartingaleconvergence}, there is an
@@ -51,10 +53,11 @@ Hence the same holds for submartingales, i.e.
     $N_0(\epsilon)$ (possibly random)
     such that for all $n > N_0(\epsilon)$
     \[
-        \left( \frac{X_n}{x} \right)^{\frac{1}{n}} \le  u^{2p - 1}(1 + \epsilon) \implies x [\underbrace{u^{2p - 1} (1+\epsilon)}_{<1}]^n \xrightarrow{n \to \infty} 0.
+        \left( \frac{X_n}{x} \right)^{\frac{1}{n}} \le  u^{2p - 1}(1 + \epsilon) %
+        \implies x [\underbrace{u^{2p - 1} (1+\epsilon)}_{<1}]^n \xrightarrow{n \to \infty} 0.
     \]
-    Hence it can not converge in $L^1$.
+    Thus it can not converge in $L^1$.
-    % TODO Confusion
+    % TODO Make this less confusing
 
 \end{example}
 
@@ -75,16 +78,21 @@ consider $L^2$.
     Since $\bE[X_n | \cF_{n-1}] = X_{n-1}$ a.s.,
     by induction $\bE[X_n | \cF_{k}] = X_k$ a.s.~for all $k \le  n$.
     Play with conditional expectation.
-    % TODO Exercise
+    \todo{Exercise}
 \end{proof}
 
-\begin{fact}[Parallelogram identity]
+\begin{fact}[\vocab{Parallelogram identity}]
-    % TODO
+    Let $X, Y \in L^2$.
+    Then
+    \[
+        2 \bE[X^2] + 2 \bE[Y^2] = \bE[(X+Y)^2] + \bE[(X-Y)^2].
+    \] 
 \end{fact}
 
-\begin{theorem}
+\begin{theorem}\label{martingaleconvergencel2}
     Suppose that $(X_n)_n$ is a martingale bounded in
-    $L^2$, i.e.~$\sup_n \bE[X_n^2] < \infty$.
+    $L^2$,\\
+    i.e.~$\sup_n \bE[X_n^2] < \infty$.
     Then there is a random variable $X_\infty$ such that
     \[
     X_n \xrightarrow{L^2}  X_\infty.
@@ -100,7 +108,7 @@ consider $L^2$.
     \bE[X_n^2] = \bE[X_0^2] + \sum_{j=1}^{n} \bE[Y_j^2]
     \] 
     by \autoref{martingaleincrementsorthogonal}
-    (this is known as the \vocab{parallelogram identity}). % TODO Move
+    % (this is known as the \vocab{parallelogram identity}). % TODO how exactly is this used here?
     In particular,
     \[
     \sup_n \bE[X_n^2] < \infty \iff \sum_{j=1}^{\infty} \bE[Y_j^2] < \infty.
@@ -141,7 +149,7 @@ First, we need a very important inequality:
     \end{enumerate}
 \end{theorem}
 
-We first need
+In order to prove \autoref{dooblp}, we first need
 \begin{lemma}
     \label{dooplplemma}
     Let $p > 1$ and $X,Y$ non-negative random variable
@@ -176,7 +184,7 @@ We first need
 
     Suppose now $Y  \not\in L^p$.
     Then look at $Y_M = Y \wedge M$.
-    Apply the case of $Y \in L^p$ and use the monotone convergence theorem.
+    Apply the above to $Y_M \in L^p$ and use the monotone convergence theorem.
 \end{proof}
 
 \begin{refproof}{dooblp}
@@ -185,9 +193,13 @@ We first need
     \[
     E_j = \{|X_1| \le \ell, |X_2| \le \ell, \ldots, |X_{j-1}| \le \ell, |X_j| \ge \ell\}.
     \] 
-    Then $\bP[E_j] \overset{\text{Markov}}{\le } \frac{1}{\ell} \int_{E_j} |X_j| \dif \bP (\ast\ast)$.
+    Then 
+    \begin{equation}
+        \bP[E_j] \overset{\text{Markov}}{\le } \frac{1}{\ell} \int_{E_j} |X_j| \dif \bP
+        \label{lec18eq2star}
+    \end{equation}
     Since $(X_n)_n$ is a sub-martingale, $(|X_n|)_n$ is also a sub-martingale
-    (by \autoref{jensen}).
+    (by \autoref{cjensen}).
     Hence
     \begin{IEEEeqnarray*}{rCl}
         \bE[\One_{E_j}(|X_n| - |X_{j}|) | \cF_j]
@@ -196,14 +208,14 @@ We first need
     \end{IEEEeqnarray*}
     By the law of total expectation, \autoref{totalexpectation},
     it follows that
-    \[
+    \begin{equation}
-    \bE[\One_{E_j} (|X_n| - |X_j|)] \ge 0 (\ast\ast\ast).
+        \bE[\One_{E_j} (|X_n| - |X_j|)] \ge 0. \label{lec18eq3star}
-    \] 
+    \end{equation}
     
     Now
     \begin{IEEEeqnarray*}{rCl}
         \bP(E) &=& \sum_{j=1}^n \bP(E_j)\\
-               &\overset{(\ast\ast) (\ast\ast\ast)}{\le }& \frac{1}{\ell} \left( \int_{E_1} |X_n| \dif \bP + \ldots + \int_{E_n} |X_n| \dif \bP \right)\\
+               &\overset{\eqref{lec18eq2star}, \eqref{lec18eq3star}}{\le }& \frac{1}{\ell} \left( \int_{E_1} |X_n| \dif \bP + \ldots + \int_{E_n} |X_n| \dif \bP \right)\\
                &=& \frac{1}{\ell} \int_E |X_n| \dif \bP
     \end{IEEEeqnarray*}
 

inputs/lecture_19.tex

@@ -54,7 +54,7 @@ However, some subsets can be easily described, e.g.
     where we have applied Markov's inequality. % TODO REF
 
     Since $\sup_n \bE[|X_n|^{1+\delta}] < \infty$,
-    we have that $\sup_n \bE[|X_n|] < \infty$ by Jensen. % TODO REF
+    we have that $\sup_n \bE[|X_n|] < \infty$ by Jensen (\autoref{cjensen}).
     Hence, choose $k$ large enough to make the relevant
     term less than $\epsilon$.
 \end{proof}
@@ -94,7 +94,7 @@ However, some subsets can be easily described, e.g.
                                      &\ge & \epsilon
             \end{IEEEeqnarray*}
             where the assumption that $X$ is in $L^1$ was used to apply
-            the reverse of Fatou's lemma.
+            the reverse of Fatou's lemma. % TODO reverse fatou
 
             This yields a contradiction since $\bP(F) = 0$.
         \item We want to apply part (a) to $F = \{ |X| > k\}$.
@@ -120,7 +120,10 @@ However, some subsets can be easily described, e.g.
 \begin{proof}
     Fix $\epsilon > 0$.
     Choose $\delta > 0$ such that
-    \[\forall  F \in \cF.~ \bP(F) < \delta \implies \bE[|X| \One_F] <\epsilon. (\ast)\]
+    \begin{equation}
+        \forall  F \in \cF.~ \bP(F) < \delta \implies \bE[|X| \One_F] <\epsilon.
+        \label{lec19eqstar}
+    \end{equation}
     Let $Y = \bE[X | \cG]$ for some sub-$\sigma$-algebra $\cG$.
     Then, by \autoref{condjensen}, $|Y| \le  \bE[ |X| | \cG]$.
     Hence $\bE[|Y|] \le  \bE[|X|]$.
@@ -132,7 +135,7 @@ However, some subsets can be easily described, e.g.
     \begin{IEEEeqnarray*}{rCl}
         \bE[|Y| \One_{\{|Y| > k\} }] &<& \epsilon
     \end{IEEEeqnarray*}
-    by $(\ast)$, since $\bP[|Y| > k] < \delta$.
+    by \eqref{lec19eqstar}, since $\bP[|Y| > k] < \delta$.
 \end{proof}
 
 \begin{theorem}
@@ -155,7 +158,7 @@ However, some subsets can be easily described, e.g.
         \end{cases}
     \end{IEEEeqnarray*}
 
-    $\phi$ is $1$-Lipshitz. % TODO
+    $\phi$ is $1$-Lipschitz. % TODO
 
     We have
     \begin{IEEEeqnarray*}{rCl}
@@ -168,7 +171,7 @@ However, some subsets can be easily described, e.g.
     \autoref{lec19f4} part (b).
     Similarly $\int_{|X| > k} |X - \phi(X)| \dif \bP < \epsilon$.
 
-    Since $\phi$ is Lipshitz,
+    Since $\phi$ is Lipschitz,
     $ X_n \xrightarrow{\bP} X \implies \phi(X_n) \xrightarrow{\bP} \phi(X)$.
     By the bounded convergence theorem % TODO
     $|\phi(X_n)| \le k \implies \int |  \phi(X_n) - \phi(X)| \dif \bP \to  0$.
@@ -232,6 +235,3 @@ Let $(\Omega, \cF, \bP)$ as always and let $(\cF_n)_n$ always be a filtration.
     Then there exists a random variable $X \in  L^p$, such that
     $X_n = \bE[X | \cF_n]$ for all $n$.
 \end{theorem}
-
-
-

inputs/lecture_20.tex

@@ -1,3 +1,4 @@
+\lecture{20}{2023-06-27}{}
 \begin{refproof}{ceismartingale}
     By the tower property (\autoref{cetower})
     it is clear that $(\bE[X | \cF_n])_n$
@@ -5,10 +6,11 @@
 
     First step:
     Assume that $X$ is bounded.
-    Then, by \autoref{cejensen}, $|X_n| \le  \bE[|X| | \cF_n]$,
+    Then, by \autoref{cjensen}, $|X_n| \le  \bE[|X| | \cF_n]$,
     hence $\sup_{\substack{n \in \N \\ \omega \in  \Omega}} | X_n(\omega)| < \infty$.
     Thus $(X_n)_n$ is a martingale in $L^{\infty} \subseteq L^2$.
-    By the convergence theorem for martingales in $L^2$ % TODO REF
+    By the convergence theorem for martingales in $L^2$
+    (\autoref{martingaleconvergencel2})
     there exists a random variable $Y$,
     such that $X_n \xrightarrow{L^2} Y$.
 
@@ -42,28 +44,31 @@
     \begin{IEEEeqnarray*}{rCl}
         \int | X - X'|^p \dif \bP &=& \int_{\{|X| > M\} } |X|^p \dif \bP \xrightarrow{M \to \infty} 0
     \end{IEEEeqnarray*}
-    as $\bP$ is \vocab{regular}, \todo{Definition?}
+    as $\bP$ is \vocab[Measure!regular]{regular}, \todo{Make this a definition?}
-    i.e.~$\forall  \epsilon > 0 \exists  k . \bP[|X|^p \in [-k,k] \ge  1-\epsilon$.
+    i.e.~$\forall \epsilon > 0 . ~\exists  k . ~\bP[|X|^p \in [-k,k] \ge  1-\epsilon$.
 
-    % Take some $\epsilon > 0$ and $M$ large enough such that
+    Take some $\epsilon > 0$ and $M$ large enough such that
-    % \[
+    \[
-    % \int |X - X'| \dif \bP < \epsilon.
+    \int |X - X'| \dif \bP < \epsilon.
-    % \]
+    \]
 
-    % Let $(X_n')_n$ be the martingale given by $(\bE[X' | \cF_n])_n$.
+    Let $(X_n')_n$ be the martingale given by $(\bE[X' | \cF_n])_n$.
-    % Then $X_n' \xrightarrow{L^p} X'$ by the first step.
+    Then $X_n' \xrightarrow{L^p} X'$ by the first step.
 
-    % It is
+    It is
-    % \begin{IEEEeqnarray*}{rCl}
+    \begin{IEEEeqnarray*}{rCl}
-    %     \|X_n - X_n'\|_{L^p}^p  &=& \bE[\bE[X - X' | \cF_n]^{p}]\\
+         \|X_n - X_n'\|_{L^p}^p
-    %                             &\overset{\text{Jensen}}{\le}& \bE[\bE[(X- X')^p | \cF_n]\\
+         &=& \bE[\bE[X - X' | \cF_n]^{p}]\\
-    %                             &=& \|X - X'\|_{L^p}^p\\
+         &\overset{\text{Jensen}}{\le}& \bE[\bE[(X - X')^p | \cF_n]]\\
-    %                             &<& \epsilon.
+         &=& \|X - X'\|_{L^p}^p\\
-    % \end{IEEEeqnarray*}
+         &<& \epsilon.
+    \end{IEEEeqnarray*}
 
     Hence
     \[
-    \|X_n - X\|_{L^p} \le |X_n - X_n'|_{L^p} + |X_n' - X'|_{L^p} +  | X - X'|_{L^p} \le 3 \epsilon.
+        \|X_n - X\|_{L^p} % 
+        \le \|X_n - X_n'\|_{L^p} + \|X_n' - X'\|_{L^p} +  \|X - X'\|_{L^p} %
+        \le 3 \epsilon.
     \]
     Thus $X_n \xrightarrow{L^p} X$.
 \end{refproof}
@@ -226,15 +231,16 @@ we need the following theorem, which we won't prove here:
 
     We have
     \begin{IEEEeqnarray*}{rCl}
-        \bE[X^T_n - X^T_{n-1} | \cF_{n-1}]
+        &&\bE[X^T_n - X^T_{n-1} | \cF_{n-1}]\\
-        &=& \bE[X_n \One_{\{T \ge n\}} + \sum_{k=1}^{n-1}  X_k \One_{\{ T = k\} } - X_{n-1}(\One_{T \ge  n}  + \One_{\{T = n-1\}})
+        &=& \bE\left[X_n \One_{\{T \ge n\}} + \sum_{k=1}^{n-1}  X_k \One_{\{ T = k\} }
-        + \sum_{k=1}^{n-2}  X_k \One_{\{T = k\} }  | \cF_{n-1}]\\
+        - X_{n-1}(\One_{T \ge  n}  + \One_{\{T = n-1\}})\right.\\
+        &&\left.+ \sum_{k=1}^{n-2}  X_k \One_{\{T = k\} }  \middle| \cF_{n-1}\right]\\
         &=&  \bE[(X_n - X_{n-1}) \One_{\{ T \ge  n\} } | \cF_{n-1}]\\
-        &=& \One_{\{ T \ge  n\}} (\bE[X_n | \cF_{n-1}] - X_{n-1})\\
+        &=& \One_{\{ T \ge  n\}} (\bE[X_n | \cF_{n-1}] - X_{n-1})
-        && \begin{cases}
+        \begin{cases}
             \le 0\\
             = 0 \text{ if $(X_n)_n$ is a martingale}.
-        \end{cases}.
+        \end{cases}
     \end{IEEEeqnarray*}
 
 \end{proof}

inputs/lecture_21.tex

@@ -1,4 +1,5 @@
 \lecture{21}{2023-06-29}{}
+\subsection{An Application of the Optional Stopping Theorem}
 
 This is the last lecture relevant for the exam.
 (Apart from lecture 22 which will be a repetion).
@@ -60,7 +61,7 @@ Suppose that $A^c \subseteq \R$ is a bounded domain.
 Given a bounded function $f$ on $\R$,
 we want a function $u$ which is bounded,
 such that
-$Lu = 0$ on $A^c$ and $u = f$ on $A$.
+$\mathbf{L}u = 0$ on $A^c$ and $u = f$ on $A$.
 \end{goal}
 
 We will show that $u(x) = \bE_x[f(X_{T_A})]$
@@ -98,12 +99,18 @@ is the unique solution to this problem.
     Intuitively, conditioned on $X_n$, $X_{n+1}$ should
     be independent of $\sigma(X_1,\ldots, X_{n-1})$.
 
-    \begin{claim}
+    \begin{claim*}
         For a set $B$, we have
-        $\bE[\One_{X_{n+1} \in B} | \sigma(X_1,\ldots, X_n)] = \bE[\One_{X_{n+1} \in B} | \sigma(X_n)]$.
+        \[
-    \end{claim}
+        \bE[\One_{X_{n+1} \in B} | \sigma(X_1,\ldots, X_n)] %
+        = \bE[\One_{X_{n+1} \in B} | \sigma(X_n)].\]
+    \end{claim*}
     \begin{subproof}
-    The rest of the lecture was very chaotic...
+        \todo{TODO}
+        % We have $\sigma(\One_{X_{n+1} \in B}) \subseteq \sigma(X_{n}, \xi_{n+1})$.
+        % $\sigma(X_1,\ldots,X_{n-1})$ 
+        % is independent of $\sigma( \sigma(\One_{X_{n+1} \in B}), X_n)$.
+        % Hence the claim follows from \autoref{ceroleofindependence}.
     \end{subproof}
 \end{example}
 

inputs/lecture_3.tex

@@ -1,5 +1,5 @@
 \lecture{3}{}{}
-\todo{Lecture 3 needs to be finished}
+\todo{My battery died during this lecture, so this still needs to be finished}
 \begin{notation}
     Let $\cB_n$ denote $\cB(\R^n)$.
 \end{notation}

inputs/prerequisites.tex

@@ -121,6 +121,7 @@
 How do we prove that something happens almost surely?
 The first thing that should come to mind is:
 \begin{lemma}[Borel-Cantelli]
+    \label{borelcantelli}
     If we have a sequence of events  $(A_n)_{n \ge 1}$
     such that $\sum_{n \ge 1}  \bP(A_n) < \infty$,
     then $\bP[ A_n \text{for infinitely many $n$}] = 0$

wtheo.sty

@@ -104,3 +104,4 @@
 
 \newcommand*\dif{\mathop{}\!\mathrm{d}}
 \newcommand\lecture[3]{\hrule{\color{darkgray}\hfill{\tiny[Lecture #1, #2]}}}
+\newcommand\notes{\todo{TODO: copy from official notes}}