some small changes

inputs/lecture_01.tex

@@ -114,8 +114,10 @@ The converse to this fact is also true:
             \begin{figure}[H]
                 \centering
                 \begin{tikzpicture}
-                    \begin{axis}[samples=1000, width=10cm, height=5cm]
+                    \begin{axis}[samples=1000, width=10cm, height=5cm, xmin=-2, xmax=2]
-                        \addplot[] {and(x >= -1,  x < 1) * 0.5 + (x >= 1)};
+                        \addplot[ domain=-2.5:-1]{ 0 };
+                        \addplot[ domain=-1:1] { 1 / 2 };
+                        \addplot[ domain=1:2.5] { 1 };
                     \end{axis}
                 \end{tikzpicture}
             \end{figure}

inputs/lecture_10.tex

@@ -33,7 +33,7 @@ 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:intsinxx},}\\\text{dominated convergence}}}{=}&
+        &\overset{\substack{\text{\autoref{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))\\
@@ -42,7 +42,7 @@ where $\mu = \bP X^{-1}$.
 \end{refproof}
 
 \begin{fact}
-    \label{fact:intsinxx}
+    \label{fact:sincint}
     \[
     \int_0^\infty \frac{\sin x}{x} \dif x = \frac{\pi}{2}
     \]
@@ -259,6 +259,73 @@ for all $f \in  C_b(\R)$.
     $X_n \xrightarrow{\text{dist}} X$ iff
     $F_n(t) \to  F(t)$ for all continuity points $t$ of $F$.
 \end{theorem}
+% \begin{proof}\footnote{This proof was not done in the lecture,
+%     but can be found in the official notes from lecture 13}
+%     ``$\implies$''
+%     Suppose $\mu_n \implies \mu$.
+%     Let $F_n$ and $F$ denote the respective density functions.
+%     Fix a continuity point $x_0 \in \R$ of $F$.
+%     We'll show
+%     \[
+%     \limsup_{n \to \infty} F_n(x_0) \le  F(x_0) + \epsilon
+%     \]
+%     and
+%     \[
+%     \liminf_{ \to \infty} F_n(x_0) \ge  F(x_0) - \epsilon
+%     \]
+%     for all $\epsilon > 0$.
+%     Fix some $\epsilon > 0$.
+%     Choose $\delta > 0$ such that $F(x_0 + \delta) < F(x_0) + \epsilon$
+%     and define
+%     \[
+%     g(x) \coloneqq \begin{cases}
+%         1 &\text{if }  x \le x_0,\\
+%         1 - \frac{1}{\delta}(x - x_0)&
+%             \text{if } x \in (x_0, x_0 + \delta],\\
+%         0 &\text{if } x \ge x_0 + \delta.
+%     \end{cases}
+%     \] 
+%     Since $g$ is continuous and bounded, we have
+%     \[
+%     \int g \dif \mu_n \to  \int g \dif \mu.
+%     \] 
+%     It is clear that $\One_{(-\infty, x_0]} \le g$.
+%     Hence
+%     \[
+%         F_n(x_0) = \int \One_{(-\infty, x_0]} \dif \mu_n \le \int g \dif \mu_n.
+%     \] 
+%     It follows that
+%     \begin{IEEEeqnarray*}{rCl}
+%     \limsup_{n} F_n(x_0)
+%     &\le& \limsup_n \int g \dif \mu_n\\
+%     &=& \lim_n \int g \dif \mu_n\\
+%     &=& \int g \dif \mu\\
+%     &\overset{g \le  \One_{(-\infty, x + \delta]}}{=}& F(x + \delta)\\
+%     &=& F(x) + \epsilon.
+%     \end{IEEEeqnarray*}
+%     The assertion about $\liminf_{n \to \infty} F_n(x_0)$ 
+%     follows by a similar argument.
+% 
+%     ``$\impliedby$''
+%     Assume that $F_n(x) \to  F(x)$ at all continuity points of $F$.
+%     We need to show
+%     \[
+%     \fgrall g \in C_b(\R) .~\int g \dif \mu_n \to \int g \dif \mu.
+%     \] 
+%     Let $C$ denote the set of continuity points of $f$.
+%     We apply measure theoretic induction:
+%     \begin{itemize}
+%         \item For $g = \One_{(a,b]}$, $a< b \in C$,
+%         we have
+%         \[\int g \dif \mu_n = F_n(b) - F_n(a) \to F(b) - F(a) = \int g  \dif \mu.\]
+%         \item For $g = \sum_{i} \alpha_i \One_{(a_i, b_i]}$,
+%             $a_i < b_i \in C$,
+%             we get $\int g \dif \mu_n \to  \int g \dif \mu$ 
+%             by the same argument.
+%         \item % TODO continue from Lec13 page 21 (iii)
+%     \end{itemize}
+% 
+% \end{proof}
 \begin{theorem}[Levy's continuity theorem]\label{levycontinuity}
     % Theorem 2
     $X_n \xrightarrow{\text{dist}} X$ iff

inputs/lecture_13.tex

@@ -152,15 +152,39 @@ We will need the following:
     Then $\mu\left( (-A,A) \right) \ge  \frac{A}{2} \left| \int_{-\frac{2}{A}}^{\frac{2}{A}} \phi(t) d t \right| - 1$.
 \end{lemma}
 \begin{refproof}{s7e1}
-    Exercise.\todo{TODO}
+    We have
+    \begin{IEEEeqnarray*}{rCl}
+        \int_{-\frac{2}{A}}^{\frac{2}{A}} \phi(t) \dif t
+        &=& \int_{-\frac{2}{A}}^{\frac{2}{A}} \int_{\R} e^{\i t x} \mu(\dif x) \dif t\\
+        &=& \int_{\R} \int_{-\frac{2}{A}}^{\frac{2}{A}} e^{\i t x} \dif t \mu(\dif x)\\
+        &=& \int_{\R} \int_{-\frac{2}{A}}^{\frac{2}{A}} \cos(t x) \dif t \mu(\dif x)\\
+        &=& \int_{\R} \frac{2 \sin\left( \frac{2x}{A}\right) }{x} \mu(\dif x).\\
+    \end{IEEEeqnarray*}
+    Hence
+    \begin{IEEEeqnarray*}{rCl}
+        \frac{A}{2}\left|\int_{-\frac{2}{A}}^{\frac{2}{A}} \phi(t) \dif t\right|
+        &=& \left| A \int_{\R} \frac{\sin\left( \frac{2x}{A} \right) }{x} \mu(\dif t)\right|\\
+        &=& 2\left| \int_{\R} \sinc\left( \frac{2x}{A} \right) \mu(\dif t)\right|\\
+        &\le& 2 \left[ \int_{|x| < A} \underbrace{\left|\sinc\left( \frac{2x}{A} \right) \right|}_{\le 1} \mu(\dif x)
+            + \int_{|x| \ge A} \left|\sinc\left( \frac{2x}{A} \right)\right| \mu(\dif x) \right]\\
+        &\le& 2 \left[ \mu\left( (-A,A) \right)
+            + \frac{A}{2} \int_{|x| \ge A} \frac{\sin(2x / A)|}{|x|} \mu(\dif x) \right]\\
+        &\le& 2 \left[ \mu\left( (-A,A) \right)
+            + \frac{A}{2} \int_{|x| \ge A} \frac{1}{A} \mu(\dif x) \right]\\
+        &\le& 2 \mu((-A,A)) + \mu((-A,A)^c)\\
+        &=& 1 + \mu((-A,A)).
+    \end{IEEEeqnarray*}
 \end{refproof}
 
 
-
 \begin{refproof}{levycontinuity}
-``$\implies$ '' If $\mu_n \implies \mu$, then
+``$\implies$ '' If $\mu_n \implies \mu$,
-$\int f d \mu_n \to  \int f d \mu$
+then by definition
-for all $f \in C_b$ and $x \to  e^{\i t x}$ is continuous and bounded.
+$\int f \dif \mu_n \to  \int f \dif \mu$
+for all $f \in C_b$.
+Since $x \to  e^{\i t x}$ is continuous and bounded,
+it follows that $\phi_n(t) \to \phi(t)$
+for all $t \in \R$.
 
 
 ``$ \impliedby$''
@@ -237,8 +261,8 @@ such that $F_{n_k}(x) \to F(x)$ for all $x$ where $F$ is continuous.
 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 $F$.
+Since $F_{n_k}(x) \to F_n(x)$ at all continuity points $x$ of $F$,
-By \autoref{lec10_thm1} we obtain that
+by \autoref{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.
@@ -269,8 +293,6 @@ However $G_1, G_2, \ldots$ is not converging to $F$,
 as this would fail at $x_0$. This is a contradiction.
 \end{refproof}
 
-
-
 % IID is over now
 \subsection{Summary}
 What did we learn:
@@ -281,5 +303,3 @@ What did we learn:
     \item Kolmogorov's three series theorem
     \item Fourier transform, weak convergence and CLT
 \end{itemize}
-
-

inputs/lecture_15.tex

@@ -125,6 +125,7 @@ We want to derive some properties of conditional expectation.
 
 Recall
 \begin{fact}[Jensen's inequality]
+    \label{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}

inputs/lecture_19.tex

@@ -25,7 +25,8 @@
 \begin{definition}
     A sequence of random variables $(X_n)_n$ is called \vocab{uniformly integrable} (UI),
     if
-    \[\forall  \epsilon > 0 .~\exists  k > 0 .~ \forall n.~\bE[|X_n| \One_{\{|X_n > k\} }] < \epsilon.\]
+    \[\forall  \epsilon > 0 .~\exists  K > 0 .~ \forall n.~
+        \bE[|X_n| \One_{\{|X_n| > K\} }] < \epsilon.\]
 
     Similarly, we define uniformly integrable for sets of random variables.
 \end{definition}
@@ -45,24 +46,32 @@ However, some subsets can be easily described, e.g.
     Choose $q$ such that $\frac{1}{p} + \frac{1}{q} = 1$.
     Then
     \begin{IEEEeqnarray*}{rCl}
-        \bE[|X_n| \One_{|X_n| > k}] &\le& \bE[|X_n|^p]^{\frac{1}{p}} \bP[|X_n| > k]^{\frac{1}{q}}\\
+        \bE[|X_n| \One_{|X_n| > K}]
+          &\le& \bE[|X_n|^p]^{\frac{1}{p}} \bP[|X_n| > k]^{\frac{1}{q}},\\
     \end{IEEEeqnarray*}
     i.e.
     \begin{IEEEeqnarray*}{rCl}
-        \sup_n\bE[|X_n| \One_{|X_n| > k}] &\le& \underbrace{\sup_n\bE[|X_n|^p]^{\frac{1}{p}}}_{< \infty} \sup_n \underbrace{\bP[|X_n| > k]^{\frac{1}{q}}}_{\le k^{\frac{1}{q}} \bE[|X_n|]^{\frac{1}{q}}}\\
+        \sup_n\bE[|X_n| \One_{|X_n| > k}]
+            &\le& \underbrace{\sup_n\bE[|X_n|^p]^{\frac{1}{p}}}_{< \infty}
+                \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
 
     Since $\sup_n \bE[|X_n|^{1+\delta}] < \infty$,
-    we have that $\sup_n \bE[|X_n|] < \infty$ by Jensen (\autoref{cjensen}).
+    we have that $\sup_n \bE[|X_n|] < \infty$ by Jensen (\autoref{jensen}).
-    Hence, choose $k$ large enough to make the relevant
+    Hence for $K$ large enough relevant term is less than $\epsilon$.
-    term less than $\epsilon$.
 \end{proof}
 
 \begin{fact}\label{lec19f2}
     If $(X_n)_n$ is uniformly integrable,
     then $(X_n)_n$ is bounded in $L^1$.
 \end{fact}
+\begin{proof}
+    Take some $\epsilon > 0$ and $K$ such that
+    $\sup_n\bE[|X_n| \One_{|X_n| > K}] < \epsilon$.
+    Then $\sup_n\|X_n\|_{L^1} \le  K + \epsilon$.
+\end{proof}
 
 \begin{fact}\label{lec19f3}
     Suppose $Y \in  L^1(\bP)$ and $\sup_n |X_n(\cdot )| \le  Y(\cdot )$.
@@ -125,7 +134,7 @@ 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{condjensen}, $|Y| \le  \bE[ |X| | \cG]$.
+    Then, by \autoref{cjensen}, $|Y| \le  \bE[ |X| | \cG]$.
     Hence $\bE[|Y|] \le  \bE[|X|]$.
     It follows that $\bP[|Y| > k] < \delta$
     for $k$ suitably large,
@@ -191,7 +200,7 @@ 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 $\bP[|X_n| \ge k] < \delta$ by Markov's inequality.
     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.
@@ -227,6 +236,7 @@ Let $(\Omega, \cF, \bP)$ as always and let $(\cF_n)_n$ always be a filtration.
     Then $X_n \coloneqq  \bE[X | \cF_n]$ defines a martingale which converges
     to $X$ in $L^p$.
 \end{theorem}
+% \todo{Proof ?}
 
 \begin{theorem}
     \label{martingaleisce}

inputs/lecture_20.tex

@@ -44,7 +44,7 @@
     \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[Measure!regular]{regular}, \todo{Make this a definition?}
+    as $\bP$ is regular,
     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