removed whitespace

inputs/a_1_counterexamples.tex

@@ -3,4 +3,3 @@
 Exercise 4.3
 10.2
 
-

inputs/lecture_10.tex

@@ -21,7 +21,7 @@ where $\mu = \bP X^{-1}$.
     Note that the term on the RHS is integrable, as
     \[
     \lim_{t \to 0} \frac{e^{-\i t b} - e^{-\i t a}}{- \i t} \pi(t) = a - b
-    \] 
+    \]
     and note that $\phi(0) = 1$ and $|\phi(t)| \le 1$.
     % TODO think about this
 
@@ -45,7 +45,7 @@ where $\mu = \bP X^{-1}$.
     \label{fact:intsinxx}
     \[
     \int_0^\infty \frac{\sin x}{x} dx = \frac{\pi}{2}
-    \] 
+    \]
     where the LHS is an improper Riemann-integral.
     Note that the LHS is not Lebesgue-integrable.
     It follows that
@@ -65,7 +65,7 @@ where $\mu = \bP X^{-1}$.
     Then $\bP$ has a continuous probability density given by
     \[
     f(x) = \frac{1}{2 \pi} \int_{\R} e^{-\i t x} \phi_{\R}(t) dt.
-    \] 
+    \]
 \end{theorem}
 
 \begin{example}
@@ -74,12 +74,12 @@ where $\mu = \bP X^{-1}$.
           Then
           \[
           \phi_{\R}(t) = \int e^{\i t x} d \delta_0(x) = e^{\i t 0 } = 1
-          \] 
+          \]
         \item Let $\bP = \frac{1}{2} \delta_1 + \frac{1}{2} \delta_{-1}$.
             Then
             \[
             \phi_{\R}(t) = \frac{1}{2} e^{\i t} + \frac{1}{2} e^{- \i t} = \cos(t)
-            \] 
+            \]
     \end{itemize}
 \end{example}
 \begin{refproof}{thm:lec10_3}
@@ -93,16 +93,16 @@ where $\mu = \bP X^{-1}$.
         Then
         \[
         |e^{-\i t x} \phi(t)| \le  |\phi(t)|
-        \] 
+        \]
         and $\phi \in L^1$, hence $f(x_n) \to  f(x)$
         by the dominated convergence theorem.
-        
+
     \end{subproof}
 
     We'll show that for all  $a < b$ we have
     \[
     \bP\left( (a,b] \right)  = \int_a^b (x) dx.\label{thm10_3eq1}
-    \] 
+    \]
     Let $F$ be the distribution function of $\bP$.
     It is enough to prove \autoref{thm10_3eq1}
     for all continuity points $a $ and $ b$ of $F$.
@@ -123,8 +123,8 @@ However, Fourier analysis is not only useful for continuous probability density
     Then
     \[
     \forall x \in  \R ~ \bP\left( \{x\}  \right) = \lim_{T \to \infty} \frac{1}{2 T} \int_{-T}^T e^{-\i t x } \phi(t) dt.
-    \] 
+    \]
-    
+
 \end{theorem}
 \begin{refproof}{bochnersformula}
     We have
@@ -140,10 +140,10 @@ However, Fourier analysis is not only useful for continuous probability density
         1, &y = x,\\
         0, &y \neq  x.
     \end{cases}
-    \] 
+    \]
     Hence
     \begin{IEEEeqnarray*}{rCl}
-            \lim_{T \to \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T  (y-x)}{T (y-x)} d \bP(y) &=& \bP\left( \{x\}\right) 
+            \lim_{T \to \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T  (y-x)}{T (y-x)} d \bP(y) &=& \bP\left( \{x\}\right)
     \end{IEEEeqnarray*}
     % TODO by dominated convergence?
 \end{refproof}
@@ -177,7 +177,7 @@ However, Fourier analysis is not only useful for continuous probability density
 \begin{theorem}[Bochner's theorem]\label{bochnersthm}
     The converse to \autoref{thm:lec_10thm5} holds, i.e.~ any
     $\phi: \R \to  \C$ satisfying (a) and (b) of \autoref{thm:lec_10thm5}
-    must be the Fourier transform of a probability measure $\bP$ 
+    must be the Fourier transform of a probability measure $\bP$
     on $(\R, \cB(\R))$.
 \end{theorem}
 Unfortunately, we won't prove \autoref{bochnersthm} in this lecture.

inputs/lecture_11.tex

@@ -21,7 +21,7 @@ For intuition, watch \url{https://3blue1brown.com/lessons/clt}.
 
 \begin{example}
     We throw a fair die $n = 100$ times and denote the sum of the faces
-    by $S_n \coloneqq  X_1 + \ldots + X_n$, where $X_1,\ldots, X_n$ 
+    by $S_n \coloneqq  X_1 + \ldots + X_n$, where $X_1,\ldots, X_n$
     are i.i.d.~and uniformly distributed on $\{1,\ldots,6\}$.
     Then $\bE[S_n] = 350$ and $\sqrt{\Var(S_n)} = \sigma \approx 17.07$.
     \todo{Missing pictures}
@@ -32,7 +32,7 @@ For intuition, watch \url{https://3blue1brown.com/lessons/clt}.
 \end{question}
 By definition, $\Var(X) = \bE[(X- \bE(X))^2]$, hence $\sqrt{\Var(X)}$
 can be interpreted as a distance.
-One could also define $\Var(X)$ to be $\bE[|X - \bE(X)|]$ but this is not 
+One could also define $\Var(X)$ to be $\bE[|X - \bE(X)|]$ but this is not
 well behaved.
 
 
@@ -95,7 +95,7 @@ If $S_n \sim \Bin(n,p)$ and $[a,b] \subseteq  \R$, we have
     With this in mind, a better approximation is
     \[
     \bP[S \le  25] = \bP[S \le  25.5] \approx \Phi\left( \frac{5.5}{\sqrt{10} } \right) \approx 0.9541.
-    \] 
+    \]
 \end{example}
 
 \begin{example}

inputs/lecture_13.tex

@@ -23,7 +23,7 @@ if $X_1, X_2,\ldots$ are i.i.d.~with $ \mu = \bE[X_1]$,
     Then the CLT holds, i.e.~
     \[
     \frac{\sum_{i=1}^n (X_i - \mu_i)}{S_n} \xrightarrow{(d)}  \cN(0,1).
-    \] 
+    \]
 \end{theorem}
 
 \begin{theorem}[Lyapunov condition]
@@ -34,7 +34,7 @@ if $X_1, X_2,\ldots$ are i.i.d.~with $ \mu = \bE[X_1]$,
     Then, assume that, for some $\delta > 0$,
     \[
     \lim_{n \to \infty} \sum_{i=1}^{n} \bE[(X_i - \mu_i)^{2 + \delta}] = 0
-    \] 
+    \]
     (\vocab{Lyapunov condition}).
     Then the CLT holds.
 \end{theorem}
@@ -54,7 +54,7 @@ details can be found in the notes.\notes
     if
     \[
         \lim_{a \to \infty} \sup_{n \in \N} \bP[|X_n| > a] = 0.
-    \] 
+    \]
 \end{definition}
 \begin{example}+[Exercise 8.1]
     \todo{Copy}
@@ -194,7 +194,7 @@ for all $f \in C_b$ and $x \to  e^{\i t x}$ is continuous and bounded.
     It suffices to show that
     \[
         \frac{A}{2} \left| \int_{-\frac{2}{A}}^{\frac{2}{A}} \phi_n(t) dt\right| - 1 \ge  1 - 2\epsilon
-    \] 
+    \]
     or
     \[
     1 - \frac{A}{4} \left|\int_{-\frac{2}{A}}^{\frac{2}{A}} \phi_n(t) dt \right| \le \epsilon,

inputs/lecture_14.tex

@@ -74,7 +74,7 @@ We now want to generalize this to arbitrary random variables.
          then
          \[
              \bE[X | \cG] = (\omega \mapsto \bE[X])
-         \] 
+         \]
          is a constant random variable.
 \end{remark}
 
@@ -128,13 +128,13 @@ and then do the harder proof.
     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 $H = L^2(\Omega, \cF, \bP)$ 
+    Let $H = L^2(\Omega, \cF, \bP)$
     and $K = L^2(\Omega, \cG, \bP)$.
 
     $K$ is closed, since a pointwise limit of $\cG$-measurable

inputs/lecture_15.tex

@@ -25,7 +25,7 @@ We want to derive some properties of conditional expectation.
     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}
 
@@ -61,7 +61,7 @@ We want to derive some properties of conditional expectation.
     However it follows that
     \[
         \int_G \bP[X | \cG] \dif \bP \le  -\frac{1}{n} \bP[G] < 0 \le  \int_G X \dif \bP.
-    \] 
+    \]
 \end{proof}
 \begin{theorem}[Conditional monotone convergence theorem]
     \label{ceprop5}
@@ -77,11 +77,11 @@ We want to derive some properties of conditional expectation.
     we have
     \[
         \bE[X_n | \cG] \overset{\text{a.s.}}{\ge } 0
-    \] 
+    \]
     and
     \[
         \bE[X_n | \cG] \uparrow \text{a.s.}
-    \] 
+    \]
     (consider $X_{n+1} - X_n$ ).
 
     Define $Z \coloneqq  \limsup_{n \to \infty} Z_n$.
@@ -91,7 +91,7 @@ We want to derive some properties of conditional expectation.
     Take some $G \in \cG$.
     We know by (b) % TODO REF
     that $\bE[Z_n \One_G] = \bE[X_n \One_G]$.
-    The LHS increases to  $\bE[Z \One_G]$ by the monotone 
+    The LHS increases to  $\bE[Z \One_G]$ by the monotone
     convergence theorem.
     Again by MCT, $\bE[X_n \One_G]$ increases to
     $\bE[X \One_G]$.
@@ -105,7 +105,7 @@ We want to derive some properties of conditional expectation.
     Then
     \[
         \bE[ \liminf_{n \to \infty} X_n | \cG] \le  \liminf_{n \to \infty} \bE[X_n | \cG].
-    \] 
+    \]
 \end{theorem}
 \begin{proof}
     \notes
@@ -117,7 +117,7 @@ We want to derive some properties of conditional expectation.
     Suppose $|X_n(\omega)| < X(\omega)$ a.e.~
     and  $\int |X| \dif \bP < \infty$.
     Then $X_n(\omega) \to  X\left( \omega \right) \implies \bE[ X_n | \cG] \to \bE[X | \cG]$.
-    
+
 \end{theorem}
 \begin{proof}
     \notes
@@ -144,7 +144,7 @@ For conditional expectation, we have
     such that
     \[
     c(x) = \sup_n(a_n x + b_n).
-    \] 
+    \]
 \end{fact}
 \begin{refproof}{cjensen}
     By \autoref{convapprox}, $c(x) \ge  a_n X + b_n$
@@ -153,14 +153,14 @@ For conditional expectation, we have
     \[
         \bE[c(X) | \cG] \ge a_n \bE[X | \cG] + \bE[b_n | \cG]
         = a_n \bE[X | \cG] + b_n \text{ a.s.}
-    \] 
+    \]
     for all $n$.
     Using that  a countable union of sets o f measure zero has measure zero,
     we conclude that a.s~this happens simultaneously for all $n$.
     Hence
     \[
         \bE[c(X) | \cG] \ge  \sup_n (a_n \bE[X | \cG] + b_n) \overset{\text{\autoref{convapprox}}}{=} c(\bE(X | \cG)).
-    \] 
+    \]
 \end{refproof}
 
 Recall
@@ -170,7 +170,7 @@ Recall
     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{fact}
 
 \begin{theorem}[Conditional Hölder's inequality]
@@ -181,7 +181,7 @@ Recall
     Then
     \[
         \bE(X Y | \cG) \le \bE(|X|^p | \cG)^{\frac{1}{p}} \bE(|Y|^q | \cG)^{\frac{1}{q}}.
-    \] 
+    \]
 \end{theorem}
 \begin{proof}
     \todo{Exercise}
@@ -194,7 +194,7 @@ Recall
     Then
     \[
         \bE\left[\bE[X | \cG] \mid \cH\right] \overset{\text{a.s.}}{=} \bE[X | \cH].
-    \] 
+    \]
 \end{theorem}
 \begin{proof}
     By definition, $\bE[\bE[X | \cG] | \cH]$ is  $\cH$-measurable.
@@ -214,7 +214,7 @@ Recall
     If $Y$ is $\cG$-measurable and bounded, then
     \[
         \bE[YX| \cG] \overset{\text{a.s.}}{=} Y \bE[X | \cG].
-    \] 
+    \]
 \end{theorem}
 \begin{proof}
     Assume w.l.o.g.~$X \ge 0$.
@@ -239,13 +239,13 @@ Assume $Y = \One_B$, then $Y$ simple, then take the limit (using that $Y$ is bou
     then
     \[
     \bE[X | \sigma(\cG, \cH)] \overset{\text{a.s.}}{=} \bE[X | \cG].
-    \] 
+    \]
 
     In particular, if $X$ is independent of $\cG$,
     then
     \[
     \bE[X | \cG] \overset{\text{a.s.}}{=} \bE[X].
-    \] 
+    \]
 \end{theorem}
 
 \begin{example}[Martingale property of the simple random walk]

inputs/lecture_16.tex

@@ -5,7 +5,7 @@
 \begin{refproof}{ceroleofindependence}
     Let $\cH$ be independent of $\sigma(\sigma(X), \cG)$.
     Then for all $H \in \cH$, we have that $\One_H$
-    and any random variable measurable with respect to either $\sigma(X)$ 
+    and any random variable measurable with respect to either $\sigma(X)$
     or $\cG$ must be independent.
 
     It suffices to consider the case of $X \ge 0$.
@@ -55,7 +55,7 @@ The Radon Nikodym theorem is the converse of that:
     Suppose
     \[
     \forall A \in  \cF . ~ \mu(A) = 0 \implies  \nu(A) = 0.
-    \] 
+    \]
     Then
     \begin{enumerate}[(1)]
         \item there exists $Z: \Omega \to [0, \infty)$ measurable,
@@ -70,7 +70,7 @@ The Radon Nikodym theorem is the converse of that:
 \begin{definition}
     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} 
+    we say that $\nu$ is \vocab{absolutely continuous}
     w.r.t.~$\mu$.
     This is written as $\nu \ll \mu$.
 \end{definition}
@@ -81,7 +81,7 @@ of conditional expectation:
     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 $\cG$. 
+    Obviously, $\nu \ll \bP$ on $\cG$.
     Then apply \autoref{radonnikodym}.
 \end{proof}
 
@@ -89,7 +89,7 @@ of conditional expectation:
 \begin{refproof}{radonnikodym}
     We will only sketch the proof.
     A full proof can be found in the official notes.
-    
+
     \paragraph{Step 1: Uniqueness} \notes
     \paragraph{Step 2: Reduction to the finite measure case}
         \notes
@@ -114,7 +114,7 @@ of conditional expectation:
         \item For all $f \in \cC$, we have
             \[
             \int_\Omega f \dif \mu \le \nu(\Omega) < \infty.
-            \] 
+            \]
     \end{enumerate}
 
     Define $\alpha \coloneqq  \sup \{ \int f \dif \mu : f \in \cC\} \le \nu(\Omega) < \infty$.
@@ -142,7 +142,7 @@ of conditional expectation:
         Then $\lambda(A) = 0$ since $\mu$ is finite.
 
         Assume the claim does not hold.
-        Then there must be some $k \in \N$, $A \in \cF$ 
+        Then there must be some $k \in \N$, $A \in \cF$
         such that $\lambda(A) - \frac{1}{k} \mu(A) > 0$.
         Fix this $A$ and $k$.
         Then $A$ satisfies condition (i) of being good,
@@ -187,8 +187,8 @@ Typically $\cF_n = \sigma(X_1, \ldots, X_n)$ for a sequence of random variables.
     if it is adapted to $\cF_n$ but
     \[
     \bE[X_{n+1} | \cF_n] \overset{\text{a.s.}}{\ge} X_n.
-    \] 
+    \]
-    It is called a \vocab{super-martingale} 
+    It is called a \vocab{super-martingale}
     if it is adapted but $\bE[X_{n+1} | \cF_n] \overset{\text{a.s.}}{\le} X_n$.
 \end{definition}
 \begin{corollary}

inputs/lecture_17.tex

@@ -11,7 +11,7 @@
 
 \begin{goal}
     What about a ``gambling strategy''?
-    
+
     Consider a stochastic process $(X_n)_{n \in \N}$.
 
     Note that the increments $X_{n+1} - X_n$ can be thought of as the win
@@ -27,7 +27,7 @@
     while
     \[
     Y_n \coloneqq  \sum_{j=1}^n C_j(X_j - X_{j-1})
-    \] 
+    \]
     defines the cumulative win process.
 \end{goal}
 \begin{lemma}
@@ -70,11 +70,11 @@ It follows that the monotonic limit $U_\infty([a,b]) \coloneqq  \lim_{N \to \inf
     \label{lec17l1}
     \[
     \{\omega | \liminf_{N \to  \infty} Z_N(\omega) < a < b <
-    \limsup_{N \to \infty} Z_N(\omega)\} \subseteq 
+    \limsup_{N \to \infty} Z_N(\omega)\} \subseteq
     \{\omega: U^{Z}_\infty([a,b])(\omega) = \infty\}
-    \] 
+    \]
     for every sequence of measurable functions $(Z_n)_{n \ge 1}$.
-    
+
 \end{lemma}
 \begin{lemma} % 2
     \label{lec17l2}
@@ -111,7 +111,7 @@ It follows that the monotonic limit $U_\infty([a,b]) \coloneqq  \lim_{N \to \inf
     by the monotone convergence theorem
     \[
     \bE(U_n([a,b])] \uparrow \bE[U_\infty([a,b])].
-    \] 
+    \]
 \end{proof}
 
 Assume now, that our process $(X_n)_{n \ge 1}$ is a supermartingale
@@ -119,7 +119,7 @@ bounded in $L^1(\bP)$.
 Let
 \[
 \Lambda \coloneqq  \{\omega | X_n(\omega) \text{ does not converge to anything in $[-\infty,\infty]$}\}.
-\] 
+\]
 We have
 \begin{IEEEeqnarray*}{rCl}
     \Lambda &=& \{\omega | \liminf_N X_N(\omega) < \limsup_N X_N(\omega)\}\\
@@ -160,7 +160,7 @@ The second part follows from
 \end{claim}
 \begin{subproof}
     We need to show $\sup_n \bE(|X_n|) < \infty$.
-    Since the supermartingale is non-negative, we have $\bE[|X_n|] = \bE[X_n]$ 
+    Since the supermartingale is non-negative, we have $\bE[|X_n|] = \bE[X_n]$
     and since it is a supermartingale $\bE[X_n]  \le \bE[X_0]$.
 \end{subproof}
 

inputs/lecture_18.tex

@@ -3,7 +3,7 @@
 Recall our key lemma \ref{lec17l3} for supermartingales from last time:
 \[
     (b-a) \bE[U_N([a,b])] \le  \bE[(X_n - a)^-].
-\] 
+\]
 
 What happens for submartingales?
 If $(X_n)_{n \in \N}$ is a submartingale, then $(-X_n)_{n \in \N}$ is a supermartingale.
@@ -40,12 +40,12 @@ Hence the same holds for submartingales, i.e.
     By the SLLN, we have
     \[
     \frac{1}{n} \sum_{k=1}^{n}  Z_k \xrightarrow{a.s.} \bE[Z_1] = zp - 1.
-    \] 
+    \]
     Hence
     \[
         \left(\frac{X_n}{x}\right)^{\frac{1}{n}} = u^{\frac{1}{n} \sum_{k=1}^n Z_k}
         \xrightarrow{a.s.} u^{zp -1}.
-    \] 
+    \]
     Since $(X_n)_{n \ge 0}$ is a martingale, we must have $\bE[u^{Z_1}] = 1$.
     Hence $2p - 1 < 0$, because $u > 1$.
 
@@ -67,7 +67,7 @@ consider $L^2$.
 \begin{fact}[Martingale increments are orthogonal in $L^2$ ]
     \label{martingaleincrementsorthogonal}
     Let $(X_n)_n$ be a martingale
-    and let $Y_n \coloneqq  X_n - X_{n-1}$ 
+    and let $Y_n \coloneqq  X_n - X_{n-1}$
     denote the \vocab{martingale increments}.
     Then for all $m \neq n$ we have that
     \[
@@ -86,7 +86,7 @@ consider $L^2$.
     Then
     \[
         2 \bE[X^2] + 2 \bE[Y^2] = \bE[(X+Y)^2] + \bE[(X-Y)^2].
-    \] 
+    \]
 \end{fact}
 
 \begin{theorem}\label{martingaleconvergencel2}
@@ -96,23 +96,23 @@ consider $L^2$.
     Then there is a random variable $X_\infty$ such that
     \[
     X_n \xrightarrow{L^2}  X_\infty.
-    \] 
+    \]
 \end{theorem}
 \begin{proof}
     Let $Y_n \coloneqq  X_n - X_{n-1}$ and write
     \[
     X_n = \sum_{j=1}^{n} Y_j.
-    \] 
+    \]
     We have
     \[
     \bE[X_n^2] = \bE[X_0^2] + \sum_{j=1}^{n} \bE[Y_j^2]
-    \] 
+    \]
     by \autoref{martingaleincrementsorthogonal}
     % (this is known as the \vocab{parallelogram identity}). % TODO how exactly is this used here?
     In particular,
     \[
     \sup_n \bE[X_n^2] < \infty \iff \sum_{j=1}^{\infty} \bE[Y_j^2] < \infty.
-    \] 
+    \]
 
     Since $(X_n)_n$ is bounded in  $L^2$,
     there exists $X_\infty$ such that $X_n \xrightarrow{\text{a.s.}} X_\infty$
@@ -128,7 +128,7 @@ consider $L^2$.
     limit
     \[
     \sum_{j \ge n + 1}  \xrightarrow{n\to \infty} 0
-    \] 
+    \]
     we get $\bE[(X_\infty - X_n)^2] \xrightarrow{n\to \infty} 0$.
 \end{proof}
 
@@ -192,8 +192,8 @@ In order to prove \autoref{dooblp}, we first need
     where
     \[
     E_j = \{|X_1| \le \ell, |X_2| \le \ell, \ldots, |X_{j-1}| \le \ell, |X_j| \ge \ell\}.
-    \] 
+    \]
-    Then 
+    Then
     \begin{equation}
         \bP[E_j] \overset{\text{Markov}}{\le } \frac{1}{\ell} \int_{E_j} |X_j| \dif \bP
         \label{lec18eq2star}
@@ -211,7 +211,7 @@ In order to prove \autoref{dooblp}, we first need
     \begin{equation}
         \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)\\
@@ -221,6 +221,6 @@ In order to prove \autoref{dooblp}, we first need
 
     This proves the first part.
 
-    For the second part, we apply the first part and 
+    For the second part, we apply the first part and
     \autoref{dooplplemma} (choose $Y \coloneqq  X_n^\ast$).
 \end{refproof}

inputs/lecture_19.tex

@@ -179,7 +179,7 @@ However, some subsets can be easily described, e.g.
 
     (1) $\implies$ (2)
 
-    $X_n \xrightarrow{L^1}  X \implies X_n \xrightarrow{\bP} X$ 
+    $X_n \xrightarrow{L^1}  X \implies X_n \xrightarrow{\bP} X$
     by Markov's inequality.
 
     Fix $\epsilon > 0$.
@@ -195,7 +195,7 @@ However, some subsets can be easily described, e.g.
     Then by \autoref{lec19f4} part (a) it follows that
     \[
     \int_{|X_n| > k} |X_n| \dif \bP \le  \underbrace{\int |X - X_n| \dif \bP}_{< \epsilon} + \int_{|X_n| > k} |X| \dif \bP \le  2 \epsilon.
-    \] 
+    \]
 \end{proof}
 
 \subsection{Martingale Convergence Theorems in \texorpdfstring{$L^p, p \ge 1$}{$Lp, p >= 1$}}
@@ -211,7 +211,7 @@ Let $(\Omega, \cF, \bP)$ as always and let $(\cF_n)_n$ always be a filtration.
     It is clear that $(\bE[X | \cF_n])_n$ is adapted to $(\cF_n)_n$.
 
     Let  $X_n \coloneqq  \bE[X | \cF_n]$.
-    Consider 
+    Consider
     \begin{IEEEeqnarray*}{rCl}
         \bE[X_n - X_{n-1} | \cF_{n-1}]
         &=& \bE[\bE[X | \cF_n] - \bE[X | \cF_{n-1}] | \cF_{n-1}]\\

inputs/lecture_21.tex

@@ -108,7 +108,7 @@ is the unique solution to this problem.
     \begin{subproof}
         \todo{TODO}
         % We have $\sigma(\One_{X_{n+1} \in B}) \subseteq \sigma(X_{n}, \xi_{n+1})$.
-        % $\sigma(X_1,\ldots,X_{n-1})$ 
+        % $\sigma(X_1,\ldots,X_{n-1})$
         % is independent of $\sigma( \sigma(\One_{X_{n+1} \in B}), X_n)$.
         % Hence the claim follows from \autoref{ceroleofindependence}.
     \end{subproof}

inputs/lecture_5.tex

@@ -49,7 +49,7 @@ For the proof of (b) we need the following general result:
 \end{theorem}
 
 \begin{proof}
-    
+
 
 \end{proof}
 \begin{question}
@@ -60,7 +60,7 @@ This does not hold. Consider for example $X_n = \frac{1}{n^2} \delta_n + \frac{1
 
 \begin{refproof}{lln}
     \begin{enumerate}
-        \item[(b)] 
+        \item[(b)]
     \end{enumerate}
 \end{refproof}
 

inputs/lecture_6.tex

@@ -21,7 +21,7 @@
         Then $a_1 + \ldots + a_n = (S_1 - S_0) + 2(S_2 - S_1) + 3(S_3 - S_2) +
         \ldots + n (S_n - S_{n-1})$.
         Thus $a_1 + \ldots + a_n = n S_n - (S1 $ % TODO
-        
+   
     \end{subproof}
     The SLLN follows from the claim.
 \end{refproof}
@@ -50,12 +50,12 @@ We need the fol]
     (By the independence of $X_1,\ldots, X_n$ and therefore that of $E$ and $D$ and $\bE(X_{i+1}) = \ldots = \bE(X_n) = 0$ we have $\int D E d\bP = 0$.)
 
     % TODO
-    
+
 \end{proof}
 
 \begin{refproof}{thm2}
     % TODO
-    
+
 \end{refproof}
 
 
@@ -67,7 +67,7 @@ We need the fol]
     Let $S_n \coloneqq  \sum_{i=1}^n X_i$.
     For all $t > 0$ let \[
     N_t \coloneqq  \sup \{n : S_n \le  t\}.
-    \] 
+    \]
     Then $\frac{N_t}{t} \xrightarrow{a.s.} \frac{1}{m}$ as $t \to \infty$.
 \end{theorem}
 
@@ -92,7 +92,7 @@ We need the fol]
     By definition, we have $S_{N_t} \le  t \le  S_{N_t + t}$.
     Then $\frac{S_{N_t}}{N_t} \le  \frac{t}{N_t} \le S_{N_t + 1}{N_t} \le  \frac{S_{N_t + 1}}{N_t + 1} \cdot  \frac{N_t + 1}{N_t}$.
     Hence $\frac{t}{N_t} \to  m$.
-    
+
 \end{proof}
 
 

inputs/prerequisites.tex

@@ -98,7 +98,7 @@ from the lecture on stochastic.
     \end{claim}
     \begin{subproof}
     We can use the same counterexample as in c).
-    
+
     $\bP[\lim_{n \to \infty} X_n = 0] \ge \bP[X_n = 0] = 1 - \frac{1}{n} \to 0$.
     We have already seen, that $X_n$ does not converge in $L_1$.
     \end{subproof}
@@ -166,7 +166,7 @@ We used Chebyshev's inequality. Linearity of $\bE$, $\Var(cX) = c^2\Var(X)$ and
     Then
     \[
     \lim_{n \to \infty} \int_{\R} f(x) \cos(n x) \lambda(\dif x) = 0.
-    \] 
+    \]
 \end{theorem}