yaref

inputs/lecture_03.tex

@@ -66,7 +66,7 @@ Define $\cF = \bigcup_{n \in \N} \cF_n$. Then $\cF$ is an algebra.
 We'll show that if we define $\lambda: \cF \to  [0,1]$ with
 $\lambda(A) = \lambda_n(A)$ for any $n$ where this is well defined,
 then $\lambda$ is countably additive on $\cF$.
-Using \autoref{caratheodory},  $\lambda$ will extend uniquely to a probability measure on $\sigma(\cF)$.
+Using \yaref{caratheodory},  $\lambda$ will extend uniquely to a probability measure on $\sigma(\cF)$.
 
 We want to prove:
 \begin{claim}
@@ -107,7 +107,7 @@ We want to prove:
     thus $\cF \subseteq  \cB_\infty$. Since $\cB_\infty$ is a $\sigma$-algebra,
     $\sigma(\cF) \subseteq  \cB_\infty$.
 \end{refproof}
-For the proof of \autoref{claim:lambdacountadd},
+For the proof of \yaref{claim:lambdacountadd},
 we are going to use the following:
 \begin{fact}
     \label{fact:finaddtocountadd}

inputs/lecture_04.tex

@@ -1,6 +1,6 @@
 \lecture{4}{}{End of proof of Kolmogorov's consistency theorem}
 
-To finish the proof of \autoref{claim:lambdacountadd},
+To finish the proof of \yaref{claim:lambdacountadd},
 we need the following:
 \begin{fact}
     \label{lec4fact1}
@@ -91,7 +91,7 @@ we need the following:
             so $\{x_k^{(n)}\}_n$ is bounded.
     \end{itemize}
 
-    By \autoref{lec4fact1},
+    By \yaref{lec4fact1},
     there is an infinite set $S \subseteq  \N$,
     such that $\{x_k^{(n)}\}_{n \in S}$
     converges for every $k$.
@@ -115,7 +115,7 @@ we need the following:
 \end{refproof}
 
 \begin{refproof}{claim:lambdacountadd}
-    In order to apply \autoref{fact:finaddtocountadd},
+    In order to apply \yaref{fact:finaddtocountadd},
     we need the following:
     \begin{claim}
         For any sequence $B_n \in \cF$
@@ -163,7 +163,7 @@ we need the following:
         \[
         \bigcap_{k=1}^n L_k^\ast \neq \emptyset.
         \]
-        By \autoref{lem:intersectioncompactsets},
+        By \yaref{lem:intersectioncompactsets},
         it follows that
         \[
         \bigcap_{k \in \N} L_k^\ast \neq \emptyset.
@@ -190,7 +190,7 @@ hence
 \end{IEEEeqnarray*}
 
 For the definition of $\lambda$
-as well as the proof of \autoref{claim:lambdacountadd}
+as well as the proof of \yaref{claim:lambdacountadd}
 we have only used that $(\lambda_n)_{n \in \N}$
 is a consistent family.
-Hence we have in fact shown \autoref{thm:kolmogorovconsistency}.
+Hence we have in fact shown \yaref{thm:kolmogorovconsistency}.

inputs/lecture_05.tex

@@ -9,7 +9,7 @@ The RHS is constant, which we can explicitly compute from the distribution of th
 We fix a probability space $(\Omega, \cF, \bP)$ once and for all.
 
 \begin{theorem}
-    \label{lln}
+    \label{lln} % TODO - yaref
     Let $X_1, X_2,\ldots$ be i.i.d.~random variables on $(\R, \cB(\R))$
     and $m = \bE[X_i] < \infty$
     and $\sigma^{2} = \Var(X_i) = \bE[ (X_i - \bE(X_i))^2] = \bE[X_i^2] - \bE[X_i]^2 < \infty$.
@@ -37,7 +37,7 @@ We fix a probability space $(\Omega, \cF, \bP)$ once and for all.
             \begin{IEEEeqnarray*}{rCl}
                 \bP\left[ \left| \frac{X_1 + \ldots + X_n}{n} - m\right| > \epsilon\right]
                 &=& \bP\left[\left|\frac{S_n}{n}-m\right| > \epsilon\right]\\
-                &\overset{\text{Chebyshev}}{\le }&
+                &\overset{\yaref{thm:chebyshev}}{\le }&
                     \frac{\Var\left( \frac{S_n}{n} \right) }{\epsilon^2}
                     = \frac{1}{n} \frac{\Var(X_1)}{\epsilon^2}
                     \xrightarrow{n \to \infty} 0
@@ -69,7 +69,7 @@ Consider the following:
     where $X_n$ has distribution
     $\frac{1}{n^2} \delta_n + \frac{1}{n^2} \delta_{-n} + (1-\frac{2}{n^2}) \delta_0$.
     We have $\bP[X_n \neq 0] = \frac{2}{n^2}$.
-    Since this is summable, Borel-Cantelli yields
+    Since this is summable, \yaref{thm:borelcantelli} yields
     \[
         \bP[X_{n} \neq 0 \text{ for infinitely many $n$}] = 0.
     \]

inputs/lecture_06.tex

@@ -1,6 +1,6 @@
 \lecture{6}{}{Proof of SLLN}
 \begin{refproof}{lln}
-    We want to deduce the SLLN (\autoref{lln}) from \autoref{thm2}.
+    We want to deduce the SLLN (\yaref{lln}) from \yaref{thm2}.
     W.l.o.g.~let us assume that $\bE[X_i] = 0$
     (otherwise define $X'_i \coloneqq  X_i - \bE[X_i]$).
     We will show that $\frac{S_n}{n} \xrightarrow{a.s.} 0$.
@@ -8,7 +8,7 @@
     Then the $Y_i$ are independent and we have $\bE[Y_i] = 0$
     and $\Var(Y_i) = \frac{\sigma^2}{i^2}$.
     Thus $\sum_{i=1}^\infty \Var(Y_i) < \infty$.
-    From \autoref{thm2} we obtain that $\sum_{i=1}^\infty Y_i$ converges a.s.
+    From \yaref{thm2} we obtain that $\sum_{i=1}^\infty Y_i$ converges a.s.
     \begin{claim}
         Let $(a_n)$ be a sequence in $\R$
         such that $\sum_{n=1}^{\infty} \frac{a_n}{n}$ converges,
@@ -45,9 +45,9 @@
     The SLLN follows from the claim.
 \end{refproof}
 
-In order to prove \autoref{thm2}, we need the following:
+In order to prove \yaref{thm2}, we need the following:
 \begin{theorem}[Kolmogorov's inequality]
-    \label{thm:kolmogorovineq}
+    \yalabel{Kolmogorov's Inequality}{Kolmogorov}{thm:kolmogorovineq}
     If $X_1,\ldots, X_n$ are independent with $\bE[X_i] = 0$
     and $\Var(X_i) = \sigma_i^2$, then
     \[
@@ -139,7 +139,7 @@ In order to prove \autoref{thm2}, we need the following:
     \begin{IEEEeqnarray*}{rCl}
         &&\max \{|S_{m+1} - S_m|, |S_{m+2} - S_m|, \ldots, |S_{m+l} - S_m|\}\\
         &=& \max \{|X_{m+1}|, |X_{m+1} + X_{m+2}|, \ldots, |X_{m+1} + X_{m+2} + \ldots + X_{m+l}|\}\\
-        &\overset{\text{\autoref{thm:kolmogorovineq}}}{\le}&
+        &\overset{\yaref{thm:kolmogorovineq}}{\le}&
         \frac{1}{\epsilon^2} \sum_{i=m}^{l} \Var(X_i)\\
         &\le & \frac{1}{\epsilon^2} \sum_{i=m}^\infty \Var(X_i)
         \xrightarrow{m \to \infty} 0,
@@ -200,6 +200,3 @@ In order to prove \autoref{thm2}, we need the following:
     Hence $\frac{t}{N_t} \to  m$.
 
 \end{proof}
-
-
-

inputs/lecture_07.tex

@@ -5,7 +5,7 @@
     when the $X_n$ are independent.
 \end{goal}
 \begin{theorem}[Kolmogorov's three-series theorem] % Theorem 3
-    \label{thm:kolmogorovthreeseries}
+    \yalabel{Kolmogorov's Three-Series Theorem}{3 Series}{thm:kolmogorovthreeseries}
     \label{thm3}
     Let $X_n$ be a family of independent random variables.
     \begin{enumerate}[(a)]
@@ -21,7 +21,7 @@
             Then all three series above converge for every $C > 0$.
     \end{enumerate}
 \end{theorem}
-For the proof we'll need a slight generalization of \autoref{thm2}:
+For the proof we'll need a slight generalization of \yaref{thm2}:
 \begin{theorem} %[Theorem 4]
     \label{thm4}
     Let $\{X_n\}_n$ be independent and \vocab{uniformly bounded}
@@ -31,7 +31,7 @@ For the proof we'll need a slight generalization of \autoref{thm2}:
     converge.
 \end{theorem}
 \begin{refproof}{thm3}
-    Assume, that we have already proved \autoref{thm4}.
+    Assume, that we have already proved \yaref{thm4}.
     We prove part (a) first.
     Put $Y_n = X_n \cdot \One_{\{|X_n| \le  C\}}$.
     Since the $X_n$ are independent, the $Y_n$ are independent as well.
@@ -40,11 +40,11 @@ For the proof we'll need a slight generalization of \autoref{thm2}:
     $\sum_{n \ge  1}  \int_{|X_n| \le  C} X_n \dif\bP = \sum_{n \ge 1}  \bE[Y_n]$
     and $\sum_{n \ge 1} \int_{|X_n| \le C} X_n^2 \dif\bP - \left( \int_{|X_n| \le  C} X_n \dif\bP \right)^2 = \sum_{n \ge 1}  \Var(Y_n)$
     converges.
-    By \autoref{thm4} it follows that $\sum_{n \ge 1}  Y_n < \infty$
+    By \yaref{thm4} it follows that $\sum_{n \ge 1}  Y_n < \infty$
     almost surely.
     Let $A_n \coloneqq \{\omega : |X_n(\omega)| > C\}$.
     Since $\sum_{n \ge 1}  \bP(A_n) < \infty$ by assumption,
-    Borel-Cantelli yields $\bP[\text{infinitely many $A_n$ occur}] = 0$.
+    \yaref{thm:borelcantelli} yields $\bP[\text{infinitely many $A_n$ occur}] = 0$.
 
 
     For the proof of (b), suppose $\sum_{n\ge 1} X_n(\omega) < \infty$
@@ -59,7 +59,7 @@ For the proof we'll need a slight generalization of \autoref{thm2}:
     \]
     Then the $Y_n$ are independent and $\sum_{n \ge 1} Y_n(\omega) < \infty$
     almost surely and the $Y_n$ are uniformly bounded.
-    By \autoref{thm4} $\sum_{n \ge 1} \bE[Y_n]$ and $\sum_{n \ge 1} \Var(Y_n)$
+    By \yaref{thm4} $\sum_{n \ge 1} \bE[Y_n]$ and $\sum_{n \ge 1} \Var(Y_n)$
     converge.
     Define
     \[
@@ -70,7 +70,7 @@ For the proof we'll need a slight generalization of \autoref{thm2}:
     \]
     Then the $Z_n$ are independent, uniformly bounded and  $\sum_{n \ge 1}  Z_n(\omega) < \infty$
     almost surely.
-    By \autoref{thm4} we have
+    By \yaref{thm4} we have
     $\sum_{n \ge 1} \bE(Z_n) < \infty$
     and $\sum_{n \ge 1} \Var(Z_n) < \infty$.
 
@@ -88,8 +88,8 @@ For the proof we'll need a slight generalization of \autoref{thm2}:
     $\sum_{n \ge 1} \Var(Z_n)$ to conclude that this series converges
     as well.
 \end{refproof}
-Recall \autoref{thm2}.
+Recall \yaref{thm2}.
-We will see, that the converse of \autoref{thm2} is true if the $X_n$ are uniformly bounded.
+We will see, that the converse of \yaref{thm2} is true if the $X_n$ are uniformly bounded.
 More formally:
 \begin{theorem}[Theorem 5]
     \label{thm5}
@@ -99,14 +99,14 @@ More formally:
     then $\sum_{n \ge 1} \Var(X_n) < \infty$.
 \end{theorem}
 \begin{refproof}{thm4}
-    Assume we have proven \autoref{thm5}.
+    Assume we have proven \yaref{thm5}.
 
     ``$\impliedby$'' Assume $\{X_n\} $ are independent, uniformly bounded
     and $\sum_{n \ge 1}  \bE(X_n) < \infty$ as well as $\sum_{n \ge 1}  \Var(X_n) < \infty$.
     We need to show that $\sum_{n \ge 1} X_n < \infty$ a.s.
     Let $Y_n \coloneqq  X_n - \bE(X_n)$.
     Then the $Y_n$ are independent, $\bE(Y_n) = 0$ and $\Var(Y_n) = \Var(X_n)$.
-    By \autoref{thm2} $\sum_{n \ge 1} Y_n < \infty$ a.s.
+    By \yaref{thm2} $\sum_{n \ge 1} Y_n < \infty$ a.s.
     Thus $\sum_{n \ge 1}  X_n < \infty$ a.s.
 
     ``$\implies$'' We assume that $\{X_n\}$ are independent, uniformly bounded
@@ -145,23 +145,23 @@ More formally:
         $\sum_{n \ge 1} \left(Y_n(\omega, \omega') - Z_n(\omega, \omega') \right)= \sum_{n \ge 1} \left(X_n(\omega) - X_n(\omega')\right)$.
             Thus $\sum_{n \ge 1} \left( Y_n(\omega, \omega') - Z_n(\omega, \omega') \right) < \infty$ a.s.~on $\Omega_0\otimes\Omega_0$.
     \end{subproof}
-    By \autoref{thm5}, $\sum_{n} \Var(X_n) = \frac{1}{2}\sum_{n \ge 1}  \Var(Y_n - Z_n) < \infty$ a.s.
+    By \yaref{thm5}, $\sum_{n} \Var(X_n) = \frac{1}{2}\sum_{n \ge 1}  \Var(Y_n - Z_n) < \infty$ a.s.
     Define $U_n \coloneqq  X_n - \bE(X_n)$.
     Then $\bE(U_n) = 0$ and the $U_n$ are independent
     and uniformly bounded.
     We have $\sum_{n} \Var(U_n) = \sum_{n}  \Var(X_n) < \infty$.
-    Thus $\sum_{n} U_n$ converges a.s.~by \autoref{thm2}.
+    Thus $\sum_{n} U_n$ converges a.s.~by \yaref{thm2}.
     Since by assumption $\sum_{n} X_n < \infty$ a.s.,
     it follows that $\sum_{n} \bE(X_n) < \infty$.
 \end{refproof}
 \begin{remark}
-    In the proof of \autoref{thm4}
+    In the proof of \yaref{thm4}
-    ``$\impliedby$'' is just a trivial application of \autoref{thm2}
+    ``$\impliedby$'' is just a trivial application of \yaref{thm2}
     and uniform boundedness was not used.
     The idea of `` $\implies$ '' will lead to coupling. % TODO ?
 \end{remark}
-A proof of \autoref{thm5} can be found in the notes.\notes
+A proof of \yaref{thm5} can be found in the notes.\notes
-\begin{example}[Application of \autoref{thm4}]
+\begin{example}[Application of \yaref{thm4}]
     The series $\sum_{n}  \frac{1}{n^{\frac{1}{2} + \epsilon}}$
     does not converge for $\epsilon < \frac{1}{2}$.
     However

inputs/lecture_08.tex

@@ -59,7 +59,7 @@ which does not depend on the realisation of the first $k$ random variables
 for any $k \in \N$.
 
 \begin{theorem}[Kolmogorov's 0-1 law]
-    \label{kolmogorov01}
+    \yalabel{Kolmogorov's 0-1 Law}{0-1 Law}{kolmogorov01}
     Let $X_n, n \in \N$ be a sequence of independent random variables
     and let $\cT$ denote their tail-$\sigma$-algebra.
     Then  $\cT$ is \vocab{$\bP$-trivial}, i.e.~$\bP[A] \in \{0,1\}$

inputs/lecture_09.tex

@@ -1,7 +1,7 @@
 \lecture{9}{}{Percolation, Introduction to characteristic functions}
 \subsubsection{Application: Percolation}
 
-We will now discuss another application of Kolmogorov's $0-1$-law, percolation.
+We will now discuss another application of \yaref{kolmogorov01}, percolation.
 
 \begin{definition}[\vocab{Percolation}]
     Consider the graph with nodes $\Z^d$, $d \ge 2$, where edges from the lattice are added with probability $p$. The added edges are called \vocab[Percolation!Edge!open]{open};
@@ -178,7 +178,7 @@ We have
 \end{remark}
 
 \begin{theorem}[Inversion formula] % thm1
-    \label{inversionformula}
+    \yalabel{Inversion Formula}{Inversion Formula}{inversionformula}
     Let $(\Omega, \cB(\R), \bP)$ be a probability space.
     Let $F$ be the distribution function of  $\bP$
     (i.e.~$F(x) = \bP((-\infty, x])$ for all $x \in \R$ ).
@@ -193,7 +193,7 @@ We have
 We will prove this later.
 
 \begin{theorem}[Uniqueness theorem] % thm2
-    \label{charfuncuniqueness}
+    \yalabel{Uniqueness Theorem}{Uniqueness}{charfuncuniqueness}
     Let $\bP$ and $\Q$ be two probability measures on $(\R, \cB(\R))$.
     Then $\phi_\bP = \phi_\Q \implies \bP = \Q$.
 
@@ -202,20 +202,20 @@ We will prove this later.
     from $\phi$.
 \end{theorem}
 \begin{refproof}{charfuncuniqueness}
-    Assume that we have already shown \autoref{inversionformula}.
+    Assume that we have already shown the \yaref{inversionformula}.
     Suppose that $F$ and $G$ are the distribution functions of $\bP$ and $\Q$.
     Let $a,b \in \R$ with $a < b$.
     Assume that $a $ and $b$ are continuity points of both $F$ and $G$.
-    By \autoref{inversionformula} we have
+    By the \yaref{inversionformula} we have
     \begin{IEEEeqnarray*}{rCl}
         F(b) - F(a) = G(b) - G(a) \label{eq:charfuncuniquefg}
     \end{IEEEeqnarray*}
 
-    Since $F$ and $G$ are monotonic, \autoref{eq:charfuncuniquefg}
+    Since $F$ and $G$ are monotonic, \yaref{eq:charfuncuniquefg}
     holds for all $a < b$ outside a countable set.
 
     Take $a_n$ outside this countable set, such that $a_n \ssearrow -\infty$.
-    Then, \autoref{eq:charfuncuniquefg} implies that
+    Then, \yaref{eq:charfuncuniquefg} implies that
     $F(b) - F(a_n) = G(b) - G(a_n)$ hence  $F(b) = G(b)$.
     Since $F$ and $G$ are right-continuous, it follows that $F = G$.
 \end{refproof}

inputs/lecture_10.tex

@@ -14,7 +14,7 @@ where $\mu = \bP X^{-1}$.
 
 
 \begin{refproof}{inversionformula}
-    We will prove that the limit in the RHS of \autoref{invf}
+    We will prove that the limit in the RHS of \yaref{invf}
     exists and is equal to the LHS.
     Note that the term on the RHS is integrable, as
     \[
@@ -31,7 +31,7 @@ where $\mu = \bP X^{-1}$.
         &=& \lim_{T \to \infty} \frac{1}{2 \pi} \int_{\R} \underbrace{\int_{-T}^T \left[ \frac{\cos(t (x-b)) - \cos(t(x-a))}{-\i t}\right] \dif t}_{=0 \text{, as the function is odd}} \bP(\dif x) \\
         &&  + \lim_{T \to \infty} \frac{1}{2\pi} \int_{\R}\int_{-T}^T \frac{\sin(t ( x - b)) - \sin(t(x-a))}{-t} \dif t \bP(\dif x)\\
         &=& \lim_{T \to \infty} \frac{1}{\pi} \int_\R \int_{0}^T \frac{\sin(t(x-a)) - \sin(t(x-b))}{t} \dif t \bP(\dif x)\\
-        &\overset{\substack{\text{\autoref{fact:sincint},}\\\text{dominated convergence}}}{=}&
+        &\overset{\substack{\yaref{fact:sincint},\text{dominated convergence}}}{=}&
           \frac{1}{\pi} \int -\frac{\pi}{2} \One_{x < a} + \frac{\pi}{2} \One_{x > a}
           - (- \frac{\pi}{2} \One_{x < b} + \frac{\pi}{2} \One_{x > b}) \bP(\dif x)\\
         &=& \frac{1}{2} \bP(\{a\} ) + \frac{1}{2} \bP(\{b\}) + \bP((a,b))\\
@@ -103,7 +103,7 @@ where $\mu = \bP X^{-1}$.
     \bP\left( (a,b] \right)  = \int_a^b f(x) \dif x.\label{thm10_3eq1}
     \]
     Let $F$ be the distribution function of $\bP$.
-    It is enough to prove \autoref{thm10_3eq1}
+    It is enough to prove \yaref{thm10_3eq1}
     for all continuity points $a $ and $ b$ of $F$.
     We have
     \begin{IEEEeqnarray*}{rCl}
@@ -112,12 +112,14 @@ where $\mu = \bP X^{-1}$.
             &=& \frac{1}{2\pi} \int_{\R} \phi(t) \left( \frac{e^{-\i t b} - e^{-\i t a}}{- \i t} \right) \dif t\\
             &\overset{\text{dominated convergence}}{=}& \lim_{T \to \infty} \frac{1}{2\pi} \int_{-T}^{T} \phi(t) \left( \frac{e^{-\i t b} - e^{- \i t a}}{- \i t} \right) \dif t
     \end{IEEEeqnarray*}
-    By \autoref{inversionformula}, the RHS is equal to $F(b) - F(a) = \bP\left( (a,b] \right)$.
+    By the \yaref{inversionformula},
+    the RHS is equal to $F(b) - F(a) = \bP\left( (a,b] \right)$.
 \end{refproof}
 
 However, Fourier analysis is not only useful for continuous probability density functions:
 
-\begin{theorem}[Bochner's formula for the mass at a point]\label{bochnersformula} % Theorem 4
+\begin{theorem}[Bochner's formula for the mass at a point]
+    \yalabel{Bochner's Formula for the Mass at a Point}{Bochner}{bochnersformula} % Theorem 4
     Let $\bP \in M_1(\lambda)$.
     Then
     \[
@@ -174,13 +176,14 @@ However, Fourier analysis is not only useful for continuous probability density
                                                        &=& \int_{\R} \left| \sum_{l}  c_l e^{\i t_l x}\right|^2 \ge  0
     \end{IEEEeqnarray*}
 \end{refproof}
-\begin{theorem}[Bochner's theorem]\label{thm:bochner}
+\begin{theorem}[Bochner's theorem]
-    The converse to \autoref{thm:lec_10thm5} holds, i.e.~any
+    \yalabel{Bochner's Theorem for Positive Definite Functions}{Bochner's Theorem}{thm:bochner}%
-    $\phi: \R \to  \C$ satisfying (a) and (b) of \autoref{thm:lec_10thm5}
+    The converse to \yaref{thm:lec_10thm5} holds, i.e.~any
+    $\phi: \R \to  \C$ satisfying (a) and (b) of \yaref{thm:lec_10thm5}
     must be the Fourier transform of a probability measure $\bP$
     on $(\R, \cB(\R))$.
 \end{theorem}
-Unfortunately, we won't prove \autoref{thm:bochner} in this lecture.
+Unfortunately, we won't prove \yaref{thm:bochner} in this lecture.
 
 
 \begin{definition}[Convergence in distribution / weak convergence]
@@ -325,7 +328,8 @@ for all $f \in  C_b(\R)$.
 %     \end{itemize}
 % 
 % \end{proof}
-\begin{theorem}[Levy's continuity theorem]\label{levycontinuity}
+\begin{theorem}[Levy's continuity theorem]
+    \yalabel{Levy's Continuity Theorem}{Levy}{levycontinuity}
     % Theorem 2
     $X_n \xrightarrow{\text{d}} X$ iff
     $\phi_{X_n}(t) \to \phi(t)$ for all $t \in \R$.

inputs/lecture_11.tex

@@ -52,7 +52,8 @@ In order to make things nicer, we do the following:
 Then $\bE[\frac{S_n - \bE[S_n]}{\sqrt{\Var(S_n)}}] = 0$
 and $\Var(\frac{S_n - \bE[S_n]}{\sqrt{\Var(S_n)}}) = 1$.
 
-\begin{theorem}[Central limit theorem, 1920s, Lindeberg and Levy]\label{clt}
+\begin{theorem}[Central limit theorem, 1920s, Lindeberg and Levy]%
+    \yalabel{Central Limit Theorem}{CLT}{clt}
     Let $X_1,X_2,\ldots$ be i.i.d.~random variables
     with $\bE[X_1] = \mu$ and $\Var(X_1) = \sigma^2 \in (0, \infty)$.
 
@@ -81,9 +82,9 @@ There exists a special case of this theorem, which was proved earlier:
     Let $X_1, X_2,\ldots$ i.i.d.~with $X_1 \sim \Ber(p)$.
     Then $\bE[X_1] = p$ and $\Var(X_1) = p(1-p )$.
     Furthermore $\sum_{i=1}^n X_i \sim \Bin(n,p)$,
-    and the special case follows from \autoref{clt}.
+    and the special case follows from \yaref{clt}.
 \end{proof}
-\autoref{preclt} is a useful tool for approximating the Binomial distribution with the normal distribution.
+\yaref{preclt} is a useful tool for approximating the Binomial distribution with the normal distribution.
 If $S_n \sim \Bin(n,p)$ and $[a,b] \subseteq  \R$, we have
 \[\bP[a \le S_n \le b] = \bP\left[\frac{a - np}{\sqrt{np(1-p)}}  \le \frac{S_n -np}{\sqrt{n p (1-p)}} \le  \frac{b - np}{\sqrt{n p (1-p)} }\right] \approx \Phi(b') - \Phi(a').\]
 
@@ -105,7 +106,7 @@ If $S_n \sim \Bin(n,p)$ and $[a,b] \subseteq  \R$, we have
 
     More formally: Let $X_1,X_2,\ldots$ be i.i.d.~with $\bP[X_1=1] = \bP[X_1=-1] = \frac{1}{2}$ and consider $S_n \coloneqq \sum_{i=1}^n X_i$.
 
-    Then \autoref{clt} states, that $S_n \approx \cN(0,n)$.
+    Then the \yaref{clt} states, that $S_n \approx \cN(0,n)$.
 \end{example}
 
 \begin{example}
@@ -129,4 +130,3 @@ If $S_n \sim \Bin(n,p)$ and $[a,b] \subseteq  \R$, we have
     We have  $p\cdot (1-p) \le \frac{1}{4}$,
     thus $n \approx (1.96)^2 \cdot 100^2 \cdot \frac{1}{4} = 9600$ suffices.
 \end{example}
-

inputs/lecture_12.tex

@@ -1,12 +1,12 @@
 \lecture{12}{2023-05-16}{Proof of the CLT}
 
-We now want to prove \autoref{clt}.
+We now want to prove the \yaref{clt}.
 The plan is to do the following:
 \begin{enumerate}[1.]
     \item Identify the characteristic function of a standard normal
     \item Show that the characteristic functions of the $V_n$ converge pointwise
         to that of $\cN$.
-    \item Apply  \autoref{levycontinuity}
+    \item Apply  \yaref{levycontinuity}
 \end{enumerate}
 
 First, we need to prove some properties of characteristic functions.
@@ -94,7 +94,7 @@ First, we need to prove some properties of characteristic functions.
             For arbitrary $h \in \R$, we have
             \begin{IEEEeqnarray*}{rCl}
                 |e^{\i t X} \frac{e^{\i h X}}{h}| &\le & \left| \frac{1}{h} \left( e^{\i h X} - 1 \right)\right|\\
-                                                  &\overset{\text{\autoref{charfprop:c1}}}{\le}& \left|\frac{1}{h} \i h X\right| = |X|.
+                                                  &\overset{\yaref{charfprop:c1}}{\le}& \left|\frac{1}{h} \i h X\right| = |X|.
             \end{IEEEeqnarray*}
             Thus the dominated convergence theorem can be applied and we obtain
             \[
@@ -156,7 +156,7 @@ First, we need to prove some properties of characteristic functions.
 \end{refproof}
 
 
-Now, we can finally prove the CLT:
+Now, we can finally prove the \yaref{clt}:
 \begin{refproof}{clt}
     Let $X_1,X_2,\ldots$ be i.i.d.~random variables
     with $\bE[X_1] = \mu_1$, $\Var(X_1) = \sigma^2$.
@@ -212,7 +212,7 @@ Now, we can finally prove the CLT:
     \[
     \phi_n(t) \xrightarrow{n \to \infty} e^{-\frac{t^2}{2}} = \phi_{\cN(0,1)}(t).
     \]
-    Using \autoref{levycontinuity}, we obtain \autoref{clt}.
+    Using \yaref{levycontinuity}, we obtain the \yaref{clt}.
 \end{refproof}
 
 \begin{remark}

inputs/lecture_13.tex

@@ -1,5 +1,5 @@
 \lecture{13}{2023-05}{}
-%The difficult part is to show \autoref{levycontinuity}.
+%The difficult part is to show \yaref{levycontinuity}.
 %This is the last lecture, where we will deal with independent random variables.
 
 We have seen, that
@@ -12,7 +12,7 @@ if $X_1, X_2,\ldots$ are i.i.d.~with $ \mu = \bE[X_1]$,
 \end{question}
 
 \begin{theorem}[Lindeberg CLT]
-    \label{lindebergclt}
+    \yalabel{Lindeberg's CLT}{Lindeberg CLT}{lindebergclt}
     Assume $X_1, X_2, \ldots,$ are independent (but not necessarily identically distributed) with $\mu_i = \bE[X_i] < \infty$ and $\sigma_i^2 = \Var(X_i) < \infty$.
     Let $S_n = \sqrt{\sum_{i=1}^{n} \sigma_i^2}$
     and assume that
@@ -29,7 +29,7 @@ if $X_1, X_2,\ldots$ are i.i.d.~with $ \mu = \bE[X_1]$,
 \end{theorem}
 
 \begin{theorem}[Lyapunov condition]
-    \label{lyapunovclt}
+    \yalabel{Lyapunov's CLT}{Lyapunov CLT}{lyapunovclt}
     Let $X_1, X_2,\ldots$ be independent, $\mu_i = \bE[X_i] < \infty$,
     $\sigma_i^2 = \Var(X_i) < \infty$
     and $S_n \coloneqq  \sqrt{\sum_{i=1}^n \sigma_i^2}$.
@@ -45,10 +45,10 @@ if $X_1, X_2,\ldots$ are i.i.d.~with $ \mu = \bE[X_1]$,
     The Lyapunov condition implies the Lindeberg condition.
     (Exercise).
 \end{remark}
-We will not prove the \autoref{lindebergclt} or \autoref{lyapunovclt}
+We will not prove \yaref{lindebergclt} or \yaref{lyapunovclt}
 in this lecture. However, they are quite important.
 
-We will now sketch the proof of \autoref{levycontinuity},
+We will now sketch the proof of \yaref{levycontinuity},
 details can be found in the notes.\notes
 \begin{definition}
     Let $(X_n)_n$ be a sequence of random variables.
@@ -62,8 +62,8 @@ details can be found in the notes.\notes
 \begin{example}+[Exercise 8.1]
     \todo{Copy}
 \end{example}
-A generalized version of \autoref{levycontinuity} is the following:
+A generalized version of \yaref{levycontinuity} is the following:
-\begin{theorem}[A generalized version of Levy's continuity \autoref{levycontinuity}]
+\begin{theorem}[A generalized version of \yaref{levycontinuity}]
     \label{genlevycontinuity}
     Suppose we have random variables $(X_n)_n$ such that
     $\bE[e^{\i t X_n}] \xrightarrow{n \to \infty}  \phi(t)$ for all $t \in \R$
@@ -77,12 +77,12 @@ A generalized version of \autoref{levycontinuity} is the following:
         \item $\phi$ is continuous at $0$.
     \end{enumerate}
 \end{theorem}
-\todo{Proof of \autoref{genlevycontinuity} (Exercise 8.2)}
+\todo{Proof of \yaref{genlevycontinuity} (Exercise 8.2)}
 \begin{example}
     Let $Z \sim \cN(0,1)$ and $X_n \coloneqq  n Z$.
     We have $\phi_{X_n}(t) = \bE[[e^{\i t X_n}] = e^{-\frac{1}{2} t^2 n^2} \xrightarrow{n \to \infty} \One_{\{t = 0\} }$.
     $\One_{\{t = 0\}}$ is not continuous at $0$.
-    By \autoref{genlevycontinuity}, $X_n$ can not converge to a real-valued
+    By \yaref{genlevycontinuity}, $X_n$ can not converge to a real-valued
     random variable.
 
     Exercise: $X_n \xrightarrow{(d)} \overline{X}$,
@@ -102,12 +102,12 @@ A generalized version of \autoref{levycontinuity} is the following:
         \frac{1}{n} \bE[ (X_1+ \ldots + X_n)^2]\\
                                                           &=& \sigma^2
     \end{IEEEeqnarray*}
-    For $a > 0$, by Chebyshev's inequality, % TODO
+    For $a > 0$, by \yaref{thm:chebyshev},
     we have
     \[
     \bP\left[ \left| \frac{S_n}{\sqrt{n}} \right| > a \right] \leq \frac{\sigma^2}{a^2} \xrightarrow{a \to \infty} 0.
     \]
-    verifying \autoref{genlevycontinuity}.
+    verifying \yaref{genlevycontinuity}.
 \end{example}
 
 \begin{example}
@@ -133,9 +133,9 @@ A generalized version of \autoref{levycontinuity} is the following:
     Exercise: $\phi_{\frac{S_n}{n}}(t) = e^{-|t|} = \phi_{C_1}(t)$, thus $\frac{S_n}{n} \sim  C$.
 \end{example}
 
-We will prove \autoref{levycontinuity} assuming
+We will prove \yaref{levycontinuity} assuming
-\autoref{lec10_thm1}.
+\yaref{lec10_thm1}.
-\autoref{lec10_thm1} will be shown in the notes.\notes
+\yaref{lec10_thm1} will be shown in the notes.\notes
 We will need the following:
 \begin{lemma}
     \label{lec13_lem1}
@@ -217,7 +217,7 @@ for all $t \in \R$.
         Apply dominated convergence.
     \end{subproof}
     So to prove $\mu_n\left( (-A,A) \right) \ge 1 - 2 \epsilon$,
-    apply \autoref{s7e1}.
+    apply \yaref{s7e1}.
     It suffices to show that
     \[
         \frac{A}{2} \left| \int_{-\frac{2}{A}}^{\frac{2}{A}} \phi_n(t) dt\right| - 1 \ge  1 - 2\epsilon
@@ -226,11 +226,11 @@ for all $t \in \R$.
     \[
     1 - \frac{A}{4} \left|\int_{-\frac{2}{A}}^{\frac{2}{A}} \phi_n(t) dt \right| \le \epsilon,
     \]
-    which follows from \autoref{levyproofc1eqn2}.
+    which follows from \yaref{levyproofc1eqn2}.
 \end{refproof}
 
 % Step 2
-By \autoref{lec13_lem1}
+By \yaref{lec13_lem1}
 there exists a right continuous, non-decreasing $F $
 and a subsequence $(F_{n_k})_k$ of $(F_n)_n$ where $F_n$ is
 the probability distribution function of $\mu_n$,
@@ -251,7 +251,7 @@ such that $F_{n_k}(x) \to F(x)$ for all $x$ where $F$ is continuous.
     \mu_{n_k}\left( (- \infty, x] \right) = F_{n_k}(x) \to F(x).
     \]
     Again, given $\epsilon > 0$, there exists $A > 0$, such that
-    $\mu_{n_k}\left( (-A,A) \right) > 1 - 2 \epsilon$ (\autoref{levyproofc1}).
+    $\mu_{n_k}\left( (-A,A) \right) > 1 - 2 \epsilon$ (\yaref{levyproofc1}).
 
     Hence $F(x) \ge  1 - 2 \epsilon$ for $x > A $
     and $F(x) \le  2\epsilon$ for $x < -A$.
@@ -262,13 +262,13 @@ Since $F$ is a probability distribution function, there exists
 a probability measure $\nu$ on $\R$ such that $F$ is the distribution
 function of $\nu$.
 Since $F_{n_k}(x) \to F_n(x)$ at all continuity points $x$ of $F$,
-by \autoref{lec10_thm1} we obtain that
+by \yaref{lec10_thm1} we obtain that
 $\mu_{n_k} \overset{k \to \infty}{\implies} \nu$.
 Hence
 $\phi_{\mu_{n_k}}(t) \to \phi_\nu(t)$, by the other direction of that theorem.
 But by assumption,
 $\phi_{\mu_{n_k}}(\cdot ) \to  \phi_n(\cdot )$ so $\phi_{\mu}(\cdot) = \phi_{\nu}(\cdot )$.
-By \autoref{charfuncuniqueness}, we get $\mu = \nu$.
+By the \yaref{charfuncuniqueness}, we get $\mu = \nu$.
 
 We have shown, that $\mu_{n_k} \implies \mu$ along a subsequence.
 We still need to show that $\mu_n \implies \mu$.
@@ -281,7 +281,7 @@ We still need to show that $\mu_n \implies \mu$.
 %     \notes
 % \end{subproof}
 Assume that $\mu_n$ does not converge to $\mu$.
-By \autoref{lec10_thm1}, pick a continuity point $x_0$ of $F$,
+By \yaref{lec10_thm1}, pick a continuity point $x_0$ of $F$,
 such that $F_n(x_0) \not\to F(x_0)$.
 Pick $\delta > 0$ and a subsequence $F_{n_1}(x_0), F_{n_2}(x_0), \ldots$
 which are all outside $(F(x_0) - \delta, F(x_0) + \delta)$.

inputs/lecture_14.tex

@@ -89,7 +89,7 @@ We now want to generalize this to arbitrary random variables.
 
 \subsection{Existence of Conditional Probability}
 
-We will give two different proves of \autoref{conditionalexpectation}.
+We will give two different proves of \yaref{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
@@ -139,7 +139,7 @@ and then do the harder proof.
 
     $K$ is closed, since a pointwise limit of $\cG$-measurable
     functions is $\cG$ measurable (if it exists).
-    By \autoref{orthproj},
+    By \yaref{orthproj},
     there exists $z \in K$ such that
     \[\bE[(X - Z)^2] = \inf \{ \bE[(X- W)^2] ~|~ W \in L^2(\cG)\}\]
     and
@@ -168,5 +168,6 @@ and then do the harder proof.
 
     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  $G \in \cG$.
+    by the \yaref{cmct},
+    $\bE(Z \One_G) = \bE(X \One_G)$ for all  $G \in \cG$.
 \end{refproof}

inputs/lecture_15.tex

@@ -5,7 +5,7 @@ We want to derive some properties of conditional expectation.
 
 \begin{theorem}[Law of total expectation]
     \label{ceprop1}
-    \label{totalexpectation}
+    \yalabel{Law of Total Expectation}{Total Expectation}{totalexpectation}
     \[
         \bE[\bE[X | \cG ]] = \bE[X].
     \]
@@ -26,7 +26,7 @@ We want to derive some properties of conditional expectation.
     \[
     \int_A X \dif \bP \ge \frac{1}{n}\bP(A) + \int_A Y \dif \bP,
     \]
-    contradicting property (b) from \autoref{conditionalexpectation}.
+    contradicting property (b) from \yaref{conditionalexpectation}.
 \end{proof}
 
 \begin{example}
@@ -37,7 +37,7 @@ We want to derive some properties of conditional expectation.
 
 \begin{theorem}[Linearity]
     \label{ceprop3}
-    \label{celinearity}
+    \yalabel{Linearity of Conditional Expectation}{Linearity}{celinearity}
     For all $a,b \in \R$
     we have
     \[
@@ -50,7 +50,7 @@ We want to derive some properties of conditional expectation.
 
 \begin{theorem}[Positivity]
     \label{ceprop4}
-    \label{cpositivity}
+    \yalabel{Positivity of Conditional Expectation}{Positivity}{cpositivity}
     If $X \ge  0$, then $\bE[X | \cG] \ge 0$ a.s.
 \end{theorem}
 \begin{proof}
@@ -65,7 +65,7 @@ We want to derive some properties of conditional expectation.
 \end{proof}
 \begin{theorem}[Conditional monotone convergence theorem]
     \label{ceprop5}
-    \label{mcmt}
+    \yalabel{Conditional Monotone Converence Theorem}{MCT}{cmct}
     Let $X_n,X \in L^1(\Omega, \cF, \bP)$.
     Suppose $X_n \ge 0$ with $X_n \uparrow X$.
     Then $\bE[X_n|\cG] \uparrow \bE[X|\cG]$.
@@ -73,7 +73,7 @@ We want to derive some properties of conditional expectation.
 \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},
+    by the \yaref{cpositivity},
     we have
     \[
         \bE[X_n | \cG] \overset{\text{a.s.}}{\ge } 0
@@ -100,7 +100,7 @@ We want to derive some properties of conditional expectation.
 
 \begin{theorem}[Conditional Fatou]
     \label{ceprop6}
-    \label{cfatou}
+    \yalabel{Conditional Fatou's Lemma}{Fatou}{cfatou}
     Let $X_n \in L^1(\Omega, \cF, \bP)$, $X_n \ge 0$.
     Then
     \[
@@ -112,7 +112,7 @@ We want to derive some properties of conditional expectation.
 \end{proof}
 \begin{theorem}[Conditional dominated convergence theorem]
     \label{ceprop7}
-    \label{cdct}
+    \yalabel{Conditional Dominated Convergence Theorem}{DCT}{cdct}
     Let $X_n,Y \in L^1(\Omega, \cF, \bP)$.
     Suppose that $\sup_n |X_n(\omega)| < Y(\omega)$ a.e.~
     and that $X_n$ converges to a pointwise limit $X$.
@@ -124,7 +124,7 @@ We want to derive some properties of conditional expectation.
 
 Recall
 \begin{fact}[Jensen's inequality]
-    \label{jensen}
+    \yalabel{Jensen's Inequality}{Jensen}{jensen}
     If $c : \R \to  \R$ is convex and $\bE[|c \circ X|] < \infty$,
     then $\bE[c \circ X] \overset{\text{a.s.}}{\ge} c(\bE[X])$.
 \end{fact}
@@ -132,7 +132,7 @@ Recall
 For conditional expectation, we have
 \begin{theorem}[Conditional Jensen's inequality]
     \label{ceprop8}
-    \label{cjensen}
+    \yalabel{Jensen's Inequality}{Jensen}{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.
@@ -147,7 +147,7 @@ For conditional expectation, we have
     \]
 \end{fact}
 \begin{refproof}{cjensen}
-    By \autoref{convapprox}, $c(x) \ge  a_n X + b_n$
+    By \yaref{convapprox}, $c(x) \ge  a_n X + b_n$
     for all $n$.
     Hence
     \[
@@ -159,12 +159,13 @@ For conditional expectation, we have
     we conclude that a.s~this happens simultaneously for all $n$.
     Hence
     \[
-        \bE[c(X) | \cG] \ge  \sup_n (a_n \bE[X | \cG] + b_n) \overset{\text{\autoref{convapprox}}}{=} c(\bE(X | \cG)).
+        \bE[c(X) | \cG] \ge  \sup_n (a_n \bE[X | \cG] + b_n) \overset{\yaref{convapprox}}{=} c(\bE(X | \cG)).
     \]
 \end{refproof}
 
 Recall
 \begin{fact}[Hölder's inequality]
+    \yalabel{Hölder's Inequality}{Hölder}{thm:hoelder}
     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
@@ -175,7 +176,7 @@ Recall
 
 \begin{theorem}[Conditional Hölder's inequality]
     \label{ceprop9}
-    \label{choelder}
+    \yalabel{Hölder's Inequality}{Hölder}{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
@@ -203,7 +204,7 @@ Recall
 
 \begin{theorem}[Tower property]
     \label{ceprop10}
-    \label{cetower}
+    \yalabel{Tower Property}{Tower}{cetower}
     Suppose $\cF \supset \cG \supset \cH$ are sub-$\sigma$-algebras.
     Then
     \[
@@ -245,7 +246,7 @@ Assume $Y = \One_B$, then $Y$ simple, then take the limit (using that $Y$ is bou
 
 \begin{theorem}[Role of independence]
     \label{ceprop12}
-    \label{ceroleofindependence}
+    \yalabel{Role of Independence}{Independence}{ceroleofindependence}
     Let $X$ be a random variable,
     and let $\cG, \cH$ be $\sigma$-algebras.
 
@@ -272,10 +273,10 @@ 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{\text{\autoref{celinearity}}}{=}&
+        \bE[S_{n+1} | \cF_n] &\overset{\yaref{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]\\
+                             &\overset{\yaref{ceroleofindependence}}{=}& S_{n} + \bE[X_n]\\
                              &=& S_n.
     \end{IEEEeqnarray*}
 \end{example}

inputs/lecture_16.tex

@@ -48,7 +48,7 @@ Note that in this setting, if $\mu(A) = 0$ it follows that $\nu(A) = 0$.
 
 The Radon Nikodym theorem is the converse of that:
 \begin{theorem}[Radon-Nikodym]
-    \label{radonnikodym}
+    \yalabel{Radon-Nikodym Theorem}{Radon-Nikodym}{radonnikodym}
 
     Let $\mu$ and $\nu$ be two $\sigma$-finite measures
     on $(\Omega, \cF)$.
@@ -85,14 +85,14 @@ The Radon Nikodym theorem is the converse of that:
 \end{definition}
 
 
-With \autoref{radonnikodym} we get a very short proof of the existence
+With the \yaref{radonnikodym} we get a very short proof of the existence
 of conditional expectation:
-\begin{proof}[Second proof of \autoref{conditionalexpectation}]
+\begin{proof}[Second proof of \yaref{conditionalexpectation}]
     Let $(\Omega, \cF, \bP)$ as always, $X \in L^1(\bP)$ and $\cG \subseteq  \cF$.
     It suffices to consider the case of $X \ge 0$.
     For all $G \in \cG$, define $\nu(G) \coloneqq  \int_G X \dif \bP$.
     Obviously, $\nu \ll \bP$ on $\cG$.
-    Then apply \autoref{radonnikodym}.
+    Then apply the \yaref{radonnikodym}.
 \end{proof}
 
 
@@ -212,7 +212,7 @@ Typically $\cF_n = \sigma(X_1, \ldots, X_n)$ for a sequence of random variables.
     Likewise, if $f$ is concave, then $((f(X_n))_n$ is a supermartingale.
 \end{corollary}
 \begin{proof}
-    Apply \autoref{cjensen}.
+    Apply \yaref{cjensen}.
 \end{proof}
 
 \begin{corollary}

inputs/lecture_17.tex

@@ -130,9 +130,9 @@ exists pointwise.
 \end{lemma}
 \begin{proof}
     Since $C_n \ge 0$,
-    by \autoref{lem:gambling-strategy} we have that $Y_n$ is a supermartingale.
+    by \yaref{lem:gambling-strategy} we have that $Y_n$ is a supermartingale.
     Hence $\bE[Y_N] \le \bE[Y_1] = 0$.
-    From \autoref{lec17l2} it follows that
+    From \yaref{lec17l2} it follows that
     \[
         (b-a) \bE[U_N([a,b])] \le \bE[Y_n] + \bE[(X_N-a)^-] \le \bE[(X_N-a)^-].
     \] 
@@ -146,7 +146,7 @@ exists pointwise.
     In particular, $\bP[U_\infty = \infty] = 0$.
 \end{corollary}
 \begin{proof}
-    By \autoref{lec17l3}
+    By \yaref{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 $U_N(\cdot ) \uparrow U_\infty(\cdot )$,
@@ -160,8 +160,8 @@ Let us now consider the case that our process $(X_n)_{n \ge 1}$ is a supermartin
 bounded in $L^1(\bP)$.
 
 \begin{theorem}[Doob's martingale convergence theorem]
-    \label{doobmartingaleconvergence}
+    \yalabel{Doob's Martingale Convergence Theorem}{Doob}{doobmartingaleconvergence}
-    \label{doob}
+    \yalabel{Doob's Martingale Convergence Theorem}{Doob}{doob}
     Any supermartingale bounded in $L^1$ converges almost surely to a
     random variable, which is almost surely finite.
     In particular, any non-negative supermartingale converges a.s.~to a finite random variable.
@@ -179,8 +179,8 @@ We have
 \end{IEEEeqnarray*}
 
 We have $\Lambda_{a,b} \subseteq  \{\omega : U_{\infty}([a,b])(\omega) = \infty\}$
-by \autoref{lec17l1}.
+by \yaref{lec17l1}.
-By \autoref{lec17l3} we have $\bP(\Lambda_{a,b}) = 0$,
+By \yaref{lec17l3} we have $\bP(\Lambda_{a,b}) = 0$,
 hence $\bP(\Lambda) = 0$.
 Thus there exists a random variable $X_\infty$ such that $X_n \xrightarrow{a.s.}  X_\infty$.
 
@@ -192,7 +192,7 @@ Thus there exists a random variable $X_\infty$ such that $X_n \xrightarrow{a.s.}
     We have.
     \begin{IEEEeqnarray*}{rCl}
         \bE[|X_\infty|] &=& \bE[\liminf_{n \to \infty} |X_n|]\\
-                        &\overset{\text{Fatou}}{\le }& \liminf_n \bE[|X_n|]\\
+                        &\overset{\yaref{cfatou}}{\le }& \liminf_n \bE[|X_n|]\\
                         &\le & \sup_n \bE[|X_n|]\\
                         &<& \infty.
     \end{IEEEeqnarray*}

inputs/lecture_18.tex

@@ -37,7 +37,7 @@ Hence the same holds for submartingales, i.e.
                             &=& X_n.
     \end{IEEEeqnarray*}
 
-    By \autoref{doobmartingaleconvergence},
+    By \yaref{doobmartingaleconvergence},
     there exists an a.s.~limit $X_\infty$.
     By the SLLN, we have almost surely
     \[
@@ -116,7 +116,7 @@ consider $L^2$.
     \[
     \bE[X_n^2] = \bE[X_0^2] + \sum_{j=1}^{n} \bE[Y_j^2]
     \]
-    by \autoref{martingaleincrementsorthogonal}.
+    by \yaref{martingaleincrementsorthogonal}.
     In particular,
     \[
     \sup_n \bE[X_n^2] < \infty \iff \sum_{j=1}^{\infty} \bE[Y_j^2] < \infty.
@@ -124,7 +124,7 @@ consider $L^2$.
 
     Since $(X_n)_n$ is bounded in  $L^2$,
     there exists $X_\infty$ such that $X_n \xrightarrow{\text{a.s.}} X_\infty$
-    by \autoref{doob}.
+    by \yaref{doob}.
 
     It remains to show $X_n \xrightarrow{L^2} X_\infty$.
     For any $r \in \N$, consider
@@ -143,7 +143,7 @@ consider $L^2$.
 Now let $p \ge 1$ be not necessarily $2$.
 First, we need a very important inequality:
 \begin{theorem}[Doob's $L^p$ inequality]
-    \label{dooblp}
+    \yalabel{Doob's Martingale Inequalities}{Doob}{dooblp}
     Suppose that $(X_n)_n$ is a martingale
     or a non-negative submartingale.
     Let $X_n^\ast \coloneqq \max \{|X_1|, |X_2|, \ldots, |X_n|\}$
@@ -158,7 +158,7 @@ First, we need a very important inequality:
     \end{enumerate}
 \end{theorem}
 
-In order to prove \autoref{dooblp}, we first need
+In order to prove \yaref{dooblp}, we first need
 \begin{lemma}
     \label{dooplplemma}
     Let $p > 1$ and $X,Y$ non-negative random variables
@@ -182,7 +182,7 @@ In order to prove \autoref{dooblp}, we first need
          &=& \int Y(\omega)^p \dif \bP(\omega)\\
          &=&\int_{\Omega} \left( \int_0^{Y(\omega)} p \ell^{p-1} \dif \ell
           \right) \dif \bP(\omega)\\
-         &\overset{\text{Fubini}}{=}&
+         &\overset{\yaref{thm:fubini}}{=}&
            \int_0^\infty p \ell^{p-1}\underbrace{\int_\Omega \One_{Y \ge \ell}\dif \bP}_%
              {\bP[Y \ge \ell]} \dif\ell. \label{l18star}
     \end{IEEEeqnarray}
@@ -192,7 +192,7 @@ In order to prove \autoref{dooblp}, we first need
         \eqref{l18star}
          &\le& \int_0^\infty p \ell^{p-2}
          \int_{\{Y(\omega) \ge \ell\}} X(\omega) \bP(\dif \omega)\dif \ell\\
-        &\overset{\text{Fubini}}{=}&
+         &\overset{\yaref{thm:fubini}}{=}&
           \int_\Omega  X(\omega) \int_{0}^{Y(\omega)} p \ell^{p-2} \dif \ell\bP(\dif \omega)\\
         &=& \frac{p}{p-1} \int_{\omega} X(\omega) Y (\omega)^{p-1} \bP(\dif \omega)\\
         &\overset{\text{Hölder}}{\le}& \frac{p}{p-1} \|X\|_{L^p} \|Y\|_{p}^{p-1},
@@ -212,11 +212,11 @@ In order to prove \autoref{dooblp}, we first need
     \]
     Then
     \begin{equation}
-        \bP[E_j] \overset{\text{Markov}}{\le } \frac{1}{\ell} \int_{E_j} |X_j| \dif \bP
+        \bP[E_j] \overset{\yaref{thm:markov}}{\le } \frac{1}{\ell} \int_{E_j} |X_j| \dif \bP
         \label{lec18eq2star}
     \end{equation}
     We have that $(|X_n|)_n$ is a submartingale,
-    by \autoref{cor:convexmartingale}
+    by \yaref{cor:convexmartingale}
     in the case of $X_n$ being a martingale
     and trivially if $X_n$ is non-negative.
     Hence
@@ -225,7 +225,7 @@ In order to prove \autoref{dooblp}, we first need
         &=& \One_{E_j} \bE[(|X_n| - |X_{j}|)|\cF_j]\\
         &\overset{\text{a.s.}}{\ge }& 0.
     \end{IEEEeqnarray*}
-    By the law of total expectation, \autoref{totalexpectation},
+    By the \yaref{totalexpectation},
     it follows that
     \begin{equation}
         \bE[\One_{E_j} (|X_n| - |X_j|)] \ge 0. \label{lec18eq3star}
@@ -241,6 +241,6 @@ In order to prove \autoref{dooblp}, we first need
     This proves the first part.
 
     For the second part, we apply the first part and
-    \autoref{dooplplemma} (choose $Y \coloneqq  X_n^\ast$).
+    \yaref{dooplplemma} (choose $Y \coloneqq  X_n^\ast$).
 \end{refproof}
 \todo{Branching process}

inputs/lecture_19.tex

@@ -57,10 +57,10 @@ However, some subsets can be easily described, e.g.
                 \sup_n \underbrace{\bP[|X_n| > K]^{\frac{1}{q}}}_%
                   {\le K^{-\frac{1}{q}} \bE[|X_n|]^{\frac{1}{q}}}\\
     \end{IEEEeqnarray*}
-    where we have applied Markov's inequality. % TODO REF
+    where we have applied \yaref{thm:markov}.
 
     Since $\sup_n \bE[|X_n|^{1+\delta}] < \infty$,
-    we have that $\sup_n \bE[|X_n|] < \infty$ by Jensen (\autoref{jensen}).
+    we have that $\sup_n \bE[|X_n|] < \infty$ by \yaref{jensen}.
     Hence for $K$ large enough relevant term is less than $\epsilon$.
 \end{proof}
 
@@ -93,7 +93,7 @@ However, some subsets can be easily described, e.g.
             but $\int_{F_n} |X| \dif \bP \ge  \epsilon$.
 
             Since $\sum_{n}  \bP(F_n) < \infty$,
-            by \autoref{borelcantelli},
+            by \yaref{thm:borelcantelli},
             \[\bP[\underbrace{\limsup_n F_n}_{\text{\reflectbox{$\coloneqq$}}F}] = 0.\]
             We have
             \begin{IEEEeqnarray*}{rCl}
@@ -108,7 +108,7 @@ However, some subsets can be easily described, e.g.
 
             This yields a contradiction since $\bP(F) = 0$.
         \item We want to apply part (a) to $F = \{ |X| > k\}$.
-            By Markov, $\bP(F) \le  \frac{1}{k} \bE[|X|]$.
+            By \yaref{thm:markov}, $\bP(F) \le  \frac{1}{k} \bE[|X|]$.
             Since $\bE[|X|] < \infty$, we can choose $k$ large enough
             to get $\bP(F) \le \delta$.
     \end{enumerate}
@@ -120,7 +120,7 @@ However, some subsets can be easily described, e.g.
     \[
     \bE[|X_n| \One_{|X_n| > k}] \le  \bE[|Y| \One_{|Y| > k}] < \epsilon
     \]
-    for $k$ large enough by \autoref{lec19f4} (b).
+    for $k$ large enough by \yaref{lec19f4} (b).
 \end{refproof}
 
 \begin{fact}\label{lec19f5}
@@ -135,9 +135,9 @@ However, some subsets can be easily described, e.g.
         \label{lec19eqstar}
     \end{equation}
     Let $Y = \bE[X | \cG]$ for some sub-$\sigma$-algebra $\cG$.
-    Then, by \autoref{cjensen}, $|Y| \le  \bE[ |X| | \cG]$.
+    Then, by \yaref{cjensen}, $|Y| \le  \bE[ |X| | \cG]$.
     Hence $\bE[|Y|] \le  \bE[|X|]$.
-    By Markov's inequality,
+    By \yaref{thm:markov},
     it follows that $\bP[|Y| > k] < \delta$
     for $k > \frac{\bE[|X|]}{\delta}$.
     Note that $\{|Y| > k\}  \in \cG$.
@@ -179,19 +179,19 @@ However, some subsets can be easily described, e.g.
         + \int |\phi(X_n) - \phi(X)| \dif \bP\\
     \end{IEEEeqnarray*}
     We have $\int_{|X_n| > k} \underbrace{|X_n - \phi(X_n)|}_{\le  |X_n| + | \phi(X_n)| \le  2 |X_n|}  \dif \bP\le  \epsilon$ by uniform integrability and
-    \autoref{lec19f4} part (b).
+    \yaref{lec19f4} part (b).
     Similarly $\int_{|X| > k} |X - \phi(X)| \dif \bP < \epsilon$.
 
     Since $\phi$ is Lipschitz,
     $ X_n \xrightarrow{\bP} X \implies \phi(X_n) \xrightarrow{\bP} \phi(X)$.
-    By the bounded convergence theorem, \autoref{thm:boundedconvergence},
+    By the \yaref{thm:boundedconvergence}
     $|\phi(X_n)| \le k \implies \int |  \phi(X_n) - \phi(X)| \dif \bP \to  0$.
 
 
     (1) $\implies$ (2)
 
     $X_n \xrightarrow{L^1}  X \implies X_n \xrightarrow{\bP} X$
-    by Markov's inequality (see \autoref{claim:convimpll1p}).
+    by \yaref{thm:markov} (see \yaref{claim:convimpll1p}).
 
     Fix $\epsilon > 0$.
     We have
@@ -202,8 +202,8 @@ However, some subsets can be easily described, e.g.
     \end{IEEEeqnarray*}
     for all $\delta > 0$ and suitable $k$.
 
-    Hence $\bP[|X_n| > k] < \delta$ by Markov's inequality.
+    Hence $\bP[|X_n| > k] < \delta$ by \yaref{thm:markov}.
-    Then by \autoref{lec19f4} part (a) it follows that
+    Then by \yaref{lec19f4} part (a) it follows that
     \[
     \int_{|X_n| > k} |X_n| \dif \bP \le  \underbrace{\int |X - X_n| \dif \bP}_{< \epsilon} + \int_{|X_n| > k} |X| \dif \bP \le  2 \epsilon.
     \]

inputs/lecture_20.tex

@@ -1,16 +1,16 @@
 \lecture{20}{2023-06-27}{}
 \begin{refproof}{ceismartingale}
-    By the tower property (\autoref{cetower})
+    By the \yaref{cetower}
     it is clear that $(\bE[X | \cF_n])_n$
     is a martingale.
 
     First step:
     Assume that $X$ is bounded.
-    Then, by \autoref{cjensen}, $|X_n| \le  \bE[|X| | \cF_n]$,
+    Then, by \yaref{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 $L^2$
-    (\autoref{martingaleconvergencel2})
+    (\yaref{martingaleconvergencel2})
     there exists a random variable $Y$,
     such that $X_n \xrightarrow{L^2} Y$.
 
@@ -74,10 +74,10 @@
     Thus $X_n \xrightarrow{L^p} X$.
 \end{refproof}
 
-For the proof of \autoref{martingaleisce},
+For the proof of \yaref{martingaleisce},
 we need the following theorem, which we won't prove here:
 \begin{theorem}[Banach Alaoglu]
-    \label{banachalaoglu}
+    \yalabel{Banach Alaoglu}{Banach Alaoglu}{banachalaoglu}
     Let $X$ be a  normed vector space and $X^\ast$ its
     continuous dual.
     Then the closed unit ball in $X^\ast$ is compact
@@ -96,7 +96,7 @@ we need the following theorem, which we won't prove here:
 \end{fact}
 
 \begin{refproof}{martingaleisce}
-    Since $(X_n)_n$ is bounded in $L^p$, by \autoref{banachalaoglu},
+    Since $(X_n)_n$ is bounded in $L^p$, by \yaref{banachalaoglu},
     there exists $X \in L^p$ and a subsequence
     $(X_{n_k})_k$ such that for all $Y \in L^q$ ($\frac{1}{p} + \frac{1}{q} = 1$ )
     \[
@@ -115,7 +115,7 @@ we need the following theorem, which we won't prove here:
     &\overset{\text{for }n_k \ge m}{=}& \bE[X_m \One_A].
     \end{IEEEeqnarray*}
     Hence $X_n = \bE[X | \cF_m]$ by the uniqueness of conditional expectation
-    and by \autoref{ceismartingale},
+    and by \yaref{ceismartingale},
     we get the convergence.
 \end{refproof}
 
@@ -298,7 +298,7 @@ we need the following theorem, which we won't prove here:
 \end{example}
 
 \begin{theorem}[Optional Stopping]
-    \label{optionalstopping}
+    \yalabel{Optional Stopping Theorem}{Optional Stopping}{optionalstopping}
     Let $(X_n)_n$ be a supermartingale
     and let $T$ be a stopping time
     taking values in $\N$.
@@ -320,7 +320,7 @@ we need the following theorem, which we won't prove here:
     $\bE[X_T] = \bE[X_0]$.
 \end{theorem}
 \begin{proof}
-    (i) was already done in \autoref{roptionalstoppingi}.
+    (i) was already done in \yaref{roptionalstoppingi}.
 
     (ii): Since $(X_n)_n$ is bounded, we get that
     \begin{IEEEeqnarray*}{rCl}
@@ -345,3 +345,17 @@ we need the following theorem, which we won't prove here:
     applying this to $(X_n)_n$ and $(-X_n)_n$,
     which are both supermartingales.
 \end{proof}
+\begin{remark}+
+    Let $(X_n)_n$ be a supermartingale and $T$ a stopping time.
+    If $(X_n)_n$ itself is not bounded,
+    but  $T$ ensures boundedness,
+    i.e. $T < \infty$ a.s.~and $(X_{T \wedge n})_n$
+    is uniformly bounded,
+    the \yaref{optionalstopping} can still be applied, as
+    \[
+        \bE[X_T] = \bE[X_{T \wedge T}]
+        \overset{\yaref{optionalstopping}}{\le} \bE[X_{T \wedge 0}]
+            = \bE[X_0].
+    \]
+\end{remark}
+

inputs/lecture_21.tex

@@ -6,7 +6,7 @@ This is the last lecture relevant for the exam.
 
 \begin{goal}
     We want to see an application of the
-    optional stopping theorem \ref{optionalstopping}.
+    \ref{optionalstopping}.
 \end{goal}
 
 \begin{notation}
@@ -110,7 +110,7 @@ is the unique solution to this problem.
         % 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}.
+        % Hence the claim follows from \yaref{ceroleofindependence}.
     \end{subproof}
 \end{example}
 

inputs/lecture_23.tex

@@ -4,7 +4,7 @@
 \subsubsection{Construction of iid random variables.}
 
 \begin{itemize}
-    \item Definition of a consistent family (\autoref{def:consistentfamily})
+    \item Definition of a consistent family (\yaref{def:consistentfamily})
     \item Important construction:
 
             Consider a distribution function $F$ and define
@@ -16,22 +16,22 @@
     \item Examples of consistent and inconsistent families
         \todo{Exercises}
     \item Kolmogorov's consistency theorem
-        (\autoref{thm:kolmogorovconsistency})
+        (\yaref{thm:kolmogorovconsistency})
 \end{itemize}
 
 \subsubsection{Limit theorems}
 \begin{itemize}
     \item Work with iid.~random variables.
-    \item Notions of convergence (\autoref{def:convergence})
+    \item Notions of convergence (\yaref{def:convergence})
     \item Implications between different notions of convergence (very important) and counter examples.
-        (\autoref{thm:convergenceimplications})
+        (\yaref{thm:convergenceimplications})
 
-    \item Laws of large numbers: (\autoref{lln})
+    \item Laws of large numbers: (\yaref{lln})
             \begin{itemize}
                 \item WLLN: convergence in probability
                 \item SLLN: weak convergence
             \end{itemize}
-    \item \autoref{thm2} (building block for SLLN):
+    \item \yaref{thm2} (building block for SLLN):
         Let $(X_n)$ be independent with mean $0$ and $\sum \sigma_n^2 < \infty$,
         then $ \sum X_n $ converges a.s.
         \begin{itemize}
@@ -43,12 +43,12 @@
 
                 $\sum \frac{\pm 1}{ n^{\frac{1}{2} -\epsilon}}$ does not converge a.s.~for any $\epsilon > 0$.
         \end{itemize}
-    \item Kolmogorov's inequality (\autoref{thm:kolmogorovineq})
+    \item \yaref{thm:kolmogorovineq}
-    \item Kolmogorov's $0-1$-law. (\autoref{kolmogorov01})
+    \item \yaref{kolmogorov01}
 
         In particular, a series of independent random variables converges with probability $0$ or $1$.
 
-    \item Kolmogorov's 3 series theorem. (\autoref{thm:kolmogorovthreeseries})
+    \item \yaref{thm:kolmogorovthreeseries}
         \begin{itemize}
             \item What are those $3$ series?
             \item Applications
@@ -59,15 +59,15 @@
 
 \begin{itemize}
     \item Definition of Fourier transform
-        (\autoref{def:characteristicfunction})
+        (\yaref{def:characteristicfunction})
     \item The Fourier transform uniquely determines the probability distribution.
         It is bounded, so many theorems are easily applicable.
-    \item Uniqueness theorem (\autoref{charfuncuniqueness}),
+    \item \yaref{charfuncuniqueness},
-        inversion formula (\autoref{inversionformula}), ...
+        \yaref{inversionformula}, ...
-    \item Levy's continuity theorem (\autoref{levycontinuity}),
+    \item \yaref{levycontinuity},
-        (\autoref{genlevycontinuity})
+        \yaref{genlevycontinuity}
-    \item Bochner's theorem for positive definite function (\autoref{thm:bochner})
+    \item \yaref{thm:bochner}
-    \item Bochner's theorem for the mass at a point (\autoref{bochnersformula})
+    \item \yaref{bochnersformula}
     \item Related notions
         \todo{TODO}
         \begin{itemize}
@@ -81,7 +81,7 @@
 \paragraph{Weak convergence}
 \begin{itemize}
     \item Definition of weak convergence % ( test against continuous, bounded functions).
-        (\autoref{def:weakconvergence})
+        (\yaref{def:weakconvergence})
     \item Examples:
         \begin{itemize}
             \item $(\delta_{\frac{1}{n}})_n$,
@@ -93,9 +93,8 @@
     \item Non-examples: $(\delta_n)_n$
     \item How does one prove weak convergence? How does one write this down in a clear way?
         \begin{itemize}
-            \item \autoref{lec10_thm1},
+            \item \yaref{lec10_thm1},
-            \item Levy's continuity theorem
+            \item \yaref{levycontinuity},
-                \ref{levycontinuity},
             \item Generalization of Levy's continuity theorem
                 \ref{genlevycontinuity}
         \end{itemize}
@@ -111,12 +110,12 @@
 
 \subsubsubsection{CLT}
 \begin{itemize}
-    \item Statement of the CLT
+    \item Statement of the \yaref{clt}
     \item Several versions:
         \begin{itemize}
-            \item iid (\autoref{clt}),
+            \item iid,
-            \item Lindeberg (\autoref{lindebergclt}),
+            \item \yaref{lindebergclt},
-            \item Lyapanov (\autoref{lyapunovclt})
+            \item \yaref{lyapunovclt}
         \end{itemize}
     \item How to apply this? Exercises!
 \end{itemize}
@@ -124,11 +123,11 @@
 \subsubsection{Conditional expectation}
 \begin{itemize}
     \item Definition and existence of conditional expectation for $X \in L^1(\Omega, \cF, \bP)$
-        (\autoref{conditionalexpectation})
+        (\yaref{conditionalexpectation})
     \item If $H = L^2(\Omega, \cF, \bP)$, then $\bE[ \cdot | \cG]$
         is the (unique) projection on the closed subspace $L^2(\Omega, \cG, \bP)$.
         Why is this a closed subspace? Why is the projection orthogonal?
-    \item Radon-Nikodym Theorem \ref{radonnikodym}
+    \item \yaref{radonnikodym}
         (Proof not relevant for the exam)
     \item (Non-)examples of mutually absolutely continuous measures
         Singularity in this context? % TODO
@@ -137,30 +136,30 @@
 \subsubsection{Martingales}
 
 \begin{itemize}
-    \item Definition of Martingales (\autoref{def:martingale})
+    \item Definition of Martingales (\yaref{def:martingale})
-    \item Doob's convergence theorem (\autoref{doobmartingaleconvergence}),
+    \item Doob's convergence theorem (\yaref{doobmartingaleconvergence}),
-        Upcrossing inequality (\autoref{lec17l1}, \autoref{lec17l2}, \autoref{lec17l3})
+        Upcrossing inequality (\yaref{lec17l1}, \yaref{lec17l2}, \yaref{lec17l3})
         (downcrossings for submartingales)
     \item Examples of Martingales converging a.s.~but not in $L^1$
-        (\autoref{ex:martingale-not-converging-in-l1})
+        (\yaref{ex:martingale-not-converging-in-l1})
     \item Bounded in $L^2$ $\implies$ convergence in $L^2$
-        (\autoref{martingaleconvergencel2}).
+        (\yaref{martingaleconvergencel2}).
     \item Martingale increments are orthogonal in $L^2$!
-        (\autoref{martingaleincrementsorthogonal})
+        (\yaref{martingaleincrementsorthogonal})
     \item Doob's (sub-)martingale inequalities
-        (\autoref{dooblp}),
+        (\yaref{dooblp}),
     \item $\bP[\sup_{k \le n} M_k \ge x]$ $\leadsto$ Look at martingale inequalities!
         Estimates might come from Doob's inequalities if $(M_k)_k$ is a (sub-)martingale.
     \item Doob's $L^p$ convergence theorem
-        (\autoref{ceismartingale}, \autoref{martingaleisce}).
+        (\yaref{ceismartingale}, \yaref{martingaleisce}).
         \begin{itemize}
-            \item Why is $p > 1$ important? \textbf{Role of Banach-Alaoglu}
+            \item Why is $p > 1$ important? \textbf{Role of \yaref{banachalaoglu}}
             \item This is an important proof.
         \end{itemize}
-    \item Uniform integrability (\autoref{def:ui})
+    \item Uniform integrability (\yaref{def:ui})
-    \item What are stopping times? (\autoref{def:stopping-time})
+    \item What are stopping times? (\yaref{def:stopping-time})
     \item (Non-)examples of stopping times
-    \item \textbf{Optional stopping theorem} (\autoref{optionalstopping})
+    \item \textbf{\yaref{optionalstopping}}
         - be really comfortable with this.
 \end{itemize}
 

inputs/prerequisites.tex

@@ -35,7 +35,7 @@ from the lecture on stochastic.
                 This notion of convergence was actually
                 defined during the course of the lecture,
                 but has been added here for completeness;
-                see \autoref{def:weakconvergence}.
+                see \yaref{def:weakconvergence}.
             }
             ($X_n \xrightarrow{\text{d}} X$)
             iff for every continuous, bounded $f: \R \to  \R$
@@ -128,7 +128,7 @@ from the lecture on stochastic.
     \begin{subproof}
         Let $F$ be the distribution function of $X$
         and $(F_n)_n$ the distribution functions of $(X_n)_n$.
-        By \autoref{lec10_thm1}
+        By \yaref{lec10_thm1}
         it suffices to show that $F_n(t) \to F(t)$ for all continuity
         points $t$ of $F$.
         Let $t$ be a continuity point of $F$.
@@ -155,7 +155,7 @@ from the lecture on stochastic.
         \label{claim:convimplpl1}
         $X_n \xrightarrow{\bP} X \notimplies X_n\xrightarrow{L^1} X$.%
         \footnote{Note that the implication holds under certain assumptions,
-            see \autoref{thm:l1iffuip}.}
+            see \yaref{thm:l1iffuip}.}
     \end{claim}
     \begin{subproof}
     Take $([0,1], \cB([0,1 ]), \lambda)$
@@ -170,7 +170,7 @@ from the lecture on stochastic.
     $X_n \xrightarrow{\text{a.s.}} X \notimplies X_n\xrightarrow{L^1} X$.
     \end{claim}
     \begin{subproof}
-        We can use the same counterexample as in \autoref{claim:convimplpl1}
+        We can use the same counterexample as in \yaref{claim:convimplpl1}
 
     $\bP[\lim_{n \to \infty} X_n = 0] \ge \bP[X_n = 0] = 1 - \frac{1}{n} \to 0$.
     We have already seen, that $X_n$ does not converge in $L_1$.
@@ -220,7 +220,7 @@ from the lecture on stochastic.
 \end{refproof}
 
 \begin{theorem}[Bounded convergence theorem]
-    \label{thm:boundedconvergence}
+    \yalabel{Bounded Convergence Theorem}{Bounded convergence}{thm:boundedconvergence}
     Suppose that $X_n \xrightarrow{\bP} X$ and there exists
     some $K$ such that $|X_n| \le K$ for all $n$.
     Then $X_n \xrightarrow{L^1} X$.
@@ -262,7 +262,7 @@ from the lecture on stochastic.
 
 \begin{theorem}+[Riemann-Lebesgue]
     %\footnote{see exercise 3.3}
-    \label{riemann-lebesgue}
+    \yalabel{Riemann-Lebesgue}{Riemann-Lebesgue}{riemann-lebesgue}
     Let $f: \R \to  \R$ be integrable.
     Then
     \[
@@ -271,6 +271,7 @@ from the lecture on stochastic.
 \end{theorem}
 
 \begin{theorem}+[Fubini-Tonelli]
+    \yalabel{Fubuni-Tonelli}{Fubini}{thm:fubini}
     %\footnote{exercise sheet 1}
     Let $(\Omega_{i}, \cF_i, \bP_i), i \in \{0,1\}$
     be probability spaces and $\Omega \coloneqq \Omega_0 \otimes \Omega_1$,
@@ -296,6 +297,7 @@ from the lecture on stochastic.
 This is taken from section 6.1 of the notes on Stochastik.
 
 \begin{theorem}[Markov's inequality]
+    \yalabel{Markov's Inequality}{Markov}{thm:markov}
     Let $X$ be a random variable and $a > 0$.
     Then
     \[
@@ -311,6 +313,7 @@ This is taken from section 6.1 of the notes on Stochastik.
 \end{proof}
 
 \begin{theorem}[Chebyshev's inequality]
+    \yalabel{Chebyshev's Inequality}{Chebyshev}{thm:chebyshev}
     Let $X$ be a random variable and $a > 0$.
     Then
     \[
@@ -322,14 +325,14 @@ This is taken from section 6.1 of the notes on Stochastik.
     \begin{IEEEeqnarray*}{rCl}
         \bP[|X-\bE(X)| \ge a]
         &=& \bP[|X - \bE(X)|^2 \ge a^2]\\
-        &\overset{\text{Markov}}{\le}& \frac{\bE[|X - \bE(X)|^2]}{a^2}.
+        &\overset{\yaref{thm:markov}}{\le}& \frac{\bE[|X - \bE(X)|^2]}{a^2}.
     \end{IEEEeqnarray*}
 \end{proof}
 
 How do we prove that something happens almost surely?
 The first thing that should come to mind is:
 \begin{lemma}[Borel-Cantelli]
-    \label{borelcantelli}
+    \yalabel{Borel-Cantelli}{Borel-Cantelli}{thm: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$

jrpie-yaref.sty

@@ -0,0 +1,49 @@
+\NeedsTeXFormat{LaTeX2e}
+\ProvidesPackage{jrpie-yaref}[2023/07/28 - yet another ref]
+
+\RequirePackage{hyperref}
+\RequirePackage{amstext}
+
+\newcommand{\yaref@text@large}[1]{%
+    \ifcsname yaref@longlabel@#1\endcsname%
+        \hyperref[#1]{\csname yaref@longlabel@#1\endcsname\ \ref*{#1}}%
+    \else%
+        \autoref{#1}%
+    \fi%
+}
+\newcommand{\yaref@text@small}[1]{%
+    \ifcsname yaref@shortlabel@#1\endcsname%
+        \hyperref[#1]{\csname yaref@shortlabel@#1\endcsname}%
+    \else%
+        (\ref{#1})%
+    \fi%
+}
+\newcommand{\yaref@math@large}[1]{%
+    \text{\yaref@text@large{#1}}%
+}
+\newcommand{\yaref@math@small}[1]{%
+    \text{\yaref@text@small{#1}}%
+}
+\newcommand{\yaref@math@verysmall}[1]{%
+    \yaref@math@small{#1}%
+}
+
+\newcommand{\yalabel}[3]{%
+    \write\@auxout{\noexpand\expandafter\noexpand\gdef\noexpand\csname yaref@longlabel@#3\noexpand\endcsname{#1}}%
+    \write\@auxout{\noexpand\expandafter\noexpand\gdef\noexpand\csname yaref@shortlabel@#3\noexpand\endcsname{#2}}%
+    \expandafter\gdef\csname yaref@longlabel@#3\endcsname{#1}%
+    \expandafter\gdef\csname yaref@shortlabel@#3\endcsname{#2}%
+    \label{#3}%
+}
+
+\newcommand{\yaref}[1]{%
+    \relax\ifmmode%
+        \mathchoice
+            {\yaref@math@large{#1}}      % display style
+            {\yaref@math@large{#1}}      % text style
+            {\yaref@math@small{#1}}      % script style
+            {\yaref@math@verysmall{#1}}  % scriptscript style
+    \else%
+        \yaref@text@large{#1}%
+    \fi%
+}

wtheo.sty

@@ -8,6 +8,7 @@
 \usepackage[index]{mkessler-vocab}
 \usepackage{mkessler-code}
 \usepackage{jrpie-math}
+\usepackage{jrpie-yaref}
 \usepackage[normalem]{ulem}
 \usepackage{pdflscape}
 \usepackage{longtable}