some diagrams

inputs/lecture_1.tex

@@ -72,6 +72,16 @@ The converse to this fact is also true:
                     1 & x \in (1,\infty).\\
                 \end{cases}
             \]
+
+            \begin{figure}[H]
+                \centering
+                \begin{tikzpicture}
+                    \begin{axis}[samples=1000, xmin=-1, xmax=2, width=10cm, height=5cm]
+                        \addplot[] {and(x>0,x<=1) * x + (x>1)};
+                    \end{axis}
+                \end{tikzpicture}
+            \end{figure}
+
         \item \vocab{Exponential distribution}:
             \[
             F(x) = \begin{cases}
@@ -79,10 +89,19 @@ The converse to this fact is also true:
                 0 & x < 0.
             \end{cases}
             \]
+            \begin{figure}[H]
+                \centering
+                \begin{tikzpicture}
+                    \begin{axis}[samples=1000, smooth, width=10cm, height=5cm, xmin=-2, xmax=5]
+                        \addplot[] {(x > 0) * (1 - exp( - 5 * x))};
+                    \end{axis}
+                \end{tikzpicture}
+            \end{figure}
         \item \vocab{Gaussian distribution}:
             \[
             \Phi(x) \coloneqq \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-\frac{y^2}{2}} dy.
             \]
+
         \item $\bP[X = 1] = \bP[X = -1] =  \frac{1}{2}$ :
             \[
             F(x) = \begin{cases}
@@ -91,5 +110,14 @@ The converse to this fact is also true:
                 1 & x \in [1, \infty).
             \end{cases}
             \]
+            \begin{figure}[H]
+                \centering
+                \begin{tikzpicture}
+                    \begin{axis}[samples=1000, width=10cm, height=5cm]
+                        \addplot[] {and(x >= -1,  x < 1) * 0.5 + (x >= 1)};
+                    \end{axis}
+                \end{tikzpicture}
+            \end{figure}
+
     \end{enumerate}
 \end{example}

inputs/lecture_12.tex

@@ -70,8 +70,8 @@ First, we need to prove some properties of characteristic functions.
             \begin{subproof}
                 For $y \ge 0$, we have
                 \begin{IEEEeqnarray*}{rCl}
-                    |e^{\i y} - 1| &=& |\int_0^y \cos(s) \d s + \i \int_0^y \sin(s) \d s|\\
+                    |e^{\i y} - 1| &=& |\int_0^y \cos(s) \dif s + \i \int_0^y \sin(s) \dif s|\\
-                                   &=& |\int_0^y e^{\i s} \d s|\\
+                                   &=& |\int_0^y e^{\i s} \dif s|\\
                                    &\overset{\text{Jensen}}{\le}& \int_0^y |e^{\i s}| ds = y.
                 \end{IEEEeqnarray*}
                 For $y < 0$, we have $|e^{\i y} - 1| = |e^{-\i y} - 1|$
@@ -122,21 +122,21 @@ First, we need to prove some properties of characteristic functions.
 \begin{refproof}{lec12_2}
     We have
     \begin{IEEEeqnarray*}{rCl}
-        \phi_X(t) &=& \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^\infty e^{\i t x} e^{-\frac{x^2}{2}} \d x\\
+        \phi_X(t) &=& \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^\infty e^{\i t x} e^{-\frac{x^2}{2}} \dif x\\
-                  &=& \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^\infty (\cos(tx) + \i \sin(tx)) e^{-\frac{x^2}{2}} \d x\\
+                  &=& \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^\infty (\cos(tx) + \i \sin(tx)) e^{-\frac{x^2}{2}} \dif x\\
-                  &=& \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^\infty \cos(t x) e^{-\frac{x^2}{2}} \d x,\\
+                  &=& \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^\infty \cos(t x) e^{-\frac{x^2}{2}} \dif x,\\
     \end{IEEEeqnarray*}
     since, as $x \mapsto \sin(tx)$ is odd and $x \mapsto e^{-\frac{x^2}{2}}$
     is even, their product is odd, wich gives that the integral is $0$.
 
     \begin{IEEEeqnarray*}{rCl}
         \phi'_X(t) &=& \bE[\i X e^{\i t X}] \\
-        &=& \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty \i x \left( \cos(t x) + \i \sin(tx) \right) e^{-\frac{x^2}{2}} \d x\\
+        &=& \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty \i x \left( \cos(t x) + \i \sin(tx) \right) e^{-\frac{x^2}{2}} \dif x\\
-        &=& \frac{1}{\sqrt{2 \pi}} \left( \i \int_{-\infty}^\infty x \cos(tx) \right) e^{-\frac{x^2}{2}} \d x\\
+        &=& \frac{1}{\sqrt{2 \pi}} \left( \i \int_{-\infty}^\infty x \cos(tx) \right) e^{-\frac{x^2}{2}} \dif x\\
-        &=& \frac{1}{\sqrt{2 \pi} } \left(\underbrace{\i \int_{-\infty}^\infty x \cos(tx) e^{-\frac{x^2}{2}}  \d x}_{= 0} + \int_{-\infty}^\infty - \sin(t x) e^{-\frac{x^2}{2}} \d x\right)\\
+        &=& \frac{1}{\sqrt{2 \pi} } \left(\underbrace{\i \int_{-\infty}^\infty x \cos(tx) e^{-\frac{x^2}{2}}  \dif x}_{= 0} + \int_{-\infty}^\infty - \sin(t x) e^{-\frac{x^2}{2}} \dif x\right)\\
-        &=& \int_{-\infty}^\infty \underbrace{\sin(tx)}_{y(x)} \underbrace{ \frac{1}{\sqrt{2 \pi} }(-x) e^{\i\frac{x^2}{2}}}_{f'(x)} \d x\\
+        &=& \int_{-\infty}^\infty \underbrace{\sin(tx)}_{y(x)} \underbrace{ \frac{1}{\sqrt{2 \pi} }(-x) e^{\i\frac{x^2}{2}}}_{f'(x)} \dif x\\
         &=& \underbrace{[ \sin(tx) \frac{1}{\sqrt{2 \pi}  e^{-\frac{x^2}{2}}}]_{x=-\infty}^\infty}_{=0}
-        - \int_{-\infty}^\infty t \cos(tx) \frac{1}{\sqrt{2 \pi} } e^{-\frac{x^2}{2}} \d x\\
+        - \int_{-\infty}^\infty t \cos(tx) \frac{1}{\sqrt{2 \pi} } e^{-\frac{x^2}{2}} \dif x\\
         &=& -t \phi_X(t)
     \end{IEEEeqnarray*}
     Thus, for all $t \in \R$
@@ -172,18 +172,20 @@ Now, we can finally prove the CLT:
     Let $t \in \R$.
     Then
     \begin{IEEEeqnarray*}{rCl}
-    \phi_{V_n}(t) = \bE[e^{\i t Y_n}] = \bE[e^{\i t \left( \frac{Y_1 + \ldots + Y_n}{\sqrt{n} } \right) }] \\
+        \phi_{V_n}(t) &=& \bE[e^{\i t Y_n}]\\
-    &=& \bE[e^{\i t \frac{Y_1}{\sqrt{n}}}] \cdot  \ldots \cdot  \bE[e^{\i t \frac{Y_n}{\sqrt{n} }}]\\
+                      &=& \bE[e^{\i t \left( \frac{Y_1 + \ldots + Y_n}{\sqrt{n} } \right) }] \\
-    &=& \left( \phi(\frac{t}{\sqrt{n} } \right)^n.
+    &=& \bE\left[e^{\i t \frac{Y_1}{\sqrt{n}}}\right] \cdot  \ldots \cdot  \bE\left[e^{\i t \frac{Y_n}{\sqrt{n} }}\right]\\
+    &=& \left( \phi\left(\frac{t}{\sqrt{n} }\right) \right)^n.
     \end{IEEEeqnarray*}
     where $\phi(t) \coloneqq  \phi_{Y_1}(t)$.
 
     We have
     \begin{IEEEeqnarray*}{rCl}
-        \phi(s) &=& \phi(0) + \phi'(0) s + \frac{\phi''(0)}{2} s^2 + o(s^2), \text{as $s \to 0$}\\
+        \phi(s) &=& \phi(0) + \phi'(0) s + \frac{\phi''(0)}{2} s^2 + o(s^2), \text{ as $s \to 0$}\\
                 &=& 1 - \underbrace{\i \bE[Y_1] s}_{=0}
-                - \bE[Y_1^2] \frac{s^2}{2} + o(s^2)\\
+                - \bE[Y_1^2] \frac{s^2}{2} + o(s^2), \text{ as $s \to 0$}
-                &=& 1 - \frac{s^2}{2} + o(s^2), \text{as $s \to $}
+\\
+                &=& 1 - \frac{s^2}{2} + o(s^2), \text{ as $s \to 0$}
     \end{IEEEeqnarray*}
 
     Setting $s \coloneqq  \frac{t}{\sqrt{n}}$ we obtain
@@ -194,7 +196,7 @@ Now, we can finally prove the CLT:
 
     \[
     \phi_{V_n}(t) = \left( \phi\left( \frac{t}{\sqrt{n} } \right) \right)^n =
-    (1 - \frac{t^2}{2 n } + o\left( \frac{t^2}{n} \right)^n \xrightarrow{n \to \infty} e^{-\frac{t^2}{2}},
+    1 - \frac{t^2}{2 n } + o\left( \frac{t^2}{n} \right)^n \xrightarrow{n \to \infty} e^{-\frac{t^2}{2}},
     \]
     where we have used the following:
 

inputs/lecture_13.tex

@@ -15,8 +15,10 @@ if $X_1, X_2,\ldots$ are i.i.d.~with $ \mu = \bE[X_1]$,
     \label{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 $\lim_{n \to \infty} \frac{1}{S_n^2} \bE\left[(X_i - \mu_i)^2 \One_{|X_i - \mu_i| > \epsilon \S_n}\right] = 0$ for all $\epsilon > 0$
+    and assume that
-    (\vocab{Lindeberg condition}, ``The truncated variance is negligible compared to the variance.'').
+    \[\lim_{n \to \infty} \frac{1}{S_n^2} \bE\left[(X_i - \mu_i)^2 \One_{|X_i - \mu_i| > \epsilon \S_n}\right] = 0\]
+    for all $\epsilon > 0$
+    (\vocab{Lindeberg condition}\footnote{``The truncated variance is negligible compared to the variance.''}).
 
     Then the CLT holds, i.e.~
     \[
@@ -94,16 +96,26 @@ A generalized version of \autoref{levycontinuity} is the following:
 \end{example}
 
 \begin{example}
-    Suppose $C$ is a random variable which is Cauchy distributed, i.e.~$C$
+    Suppose $C$ is a random variable which is \vocab[Cauchy distribution]{Cauchy distributed}, i.e.~$C$
     has probability distribution $f_C(x) = \frac{1}{\pi} \frac{1}{1 + x^2}$.
 
+    \begin{figure}[H]
+        \centering
+        \begin{tikzpicture}
+            \begin{axis}[samples=100, smooth]
+                \addplot[] { (1/3.14159265358979323846) * (1 / ( 1 + x * x))};
+            \end{axis}
+        \end{tikzpicture}
+        \caption{Probability density function of $C$}
+    \end{figure}
+
     We know that $\bE[|C|] = \infty$.
 
     We have $\phi_C(t) = \bE[e^{\i t C}] = e^{-|t|}$.
     Suppose $C_1, C_2, \ldots, C_n$ are i.i.d.~Cauchy distributed
     and let $S_n \coloneqq  C_1 + \ldots + C_n$.
 
-    Exercise: $\phi_{S_n}(t) = e^{-|t|} = \phi_{C_1}(t)$, thus $S_n \sim  C$.
+    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
@@ -229,7 +241,7 @@ We still need to show that $\mu_n \implies \mu$.
 \begin{subproof}
     \todo{in the notes}
 \end{subproof}
-Assume $\mu_n$ does not converge to $\mu$.
+Assume that $\mu_n$ does not converge to $\mu$.
 By \autoref{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 $F_{n_1}(x_0), F_{n_2}(x_0), \ldots$

inputs/lecture_6.tex

@@ -25,7 +25,7 @@
     The SLLN follows from the claim.
 \end{refproof}
 
-We need the following inequality:
+We need the fol]
 \begin{theorem}[Kolmogorov's inequality]
     If $X_1,\ldots, X_n$ are independent with $\bE[X_i] = 0$
     and $\Var(X_i) = \sigma_i^2$, then
@@ -41,7 +41,8 @@ We need the following inequality:
 
     We have
     \begin{IEEEeqnarray*}{rCl}
-        \int_{A_i} (\underbrace{X_1 + \ldots + X_i}_C + \underbrace{X_{i+1} + \ldots + X_n}_D)^2 d \bP &=& \int_{A_i} C^2 d\bP + \underbrace{\int_{A_i} D^2 d \bP}_{\ge  0} + 2 \int_{A_i} CD d\bP\\
+        &&\int_{A_i} (\underbrace{X_1 + \ldots + X_i}_C + \underbrace{X_{i+1} + \ldots + X_n}_D)^2 d \bP\\
+        &=& \int_{A_i} C^2 d\bP + \underbrace{\int_{A_i} D^2 d \bP}_{\ge  0} + 2 \int_{A_i} CD d\bP\\
                                                                                                        &\ge & \int_{A_i}  \underbrace{C^2}_{\ge \epsilon^2} d \bP + 2 \int \underbrace{\One_{A_i} (X_1 + \ldots + X_i)}_E \underbrace{(X_{i+1} + \ldots + X_n)}_D d \bP\\
                                                                                                        &\ge& \int_{A_i} \epsilon^2 d\bP
     \end{IEEEeqnarray*}

inputs/lecture_8.tex

@@ -85,7 +85,7 @@ for any $k \in \N$.
         This follows from the independence of the $X_i$.
         It is
         \[
-        \sigma\left( X_1,\ldots,X_n \right) = \sigma\left(\underbrace{\{X_{1}^{-1}(B_1) \cap \ldots \cap X_n^{-1}(B_n)\} | B_1,\ldots,B_n \in \cB(\R)\}}_{\text{\reflectbox{$\coloneqq$}}\cA} \right).
+        \sigma\left( X_1,\ldots,X_n \right) = \sigma\left(\underbrace{\{X_{1}^{-1}(B_1) \cap \ldots \cap X_n^{-1}(B_n) | B_1,\ldots,B_n \in \cB(\R)\}}_{\text{\reflectbox{$\coloneqq$}}\cA} \right).
         \]
         $\cA$ is a semi-algebra, since
         \begin{enumerate}[(i)]

inputs/lecture_9.tex

@@ -64,9 +64,11 @@ Why is $\sqrt{n}$ the right order? (Handwavey argument)
 
 Suppose $X_1, X_2,\ldots$ are i.i.d. $\cN(0,1)$.
 The mean of the l.h.s.~is $0$ and for the variance we get
-\[
+\begin{IEEEeqnarray*}{rCl}
-\Var(\frac{X_1 + \ldots + X_n - n \bE(X_1)}{\sqrt{n} }) = \Var\left( \frac{X_1+ \ldots + X_n}{\sqrt{n} } \right) = \frac{1}{n} \left( \Var(X_1) + \ldots + \Var(X_n) \right) = 1
+    \Var(\frac{X_1 + \ldots + X_n - n \bE(X_1)}{\sqrt{n} }) &=& \Var\left( \frac{X_1+ \ldots + X_n}{\sqrt{n} } \right)\\
-\]
+                                                            &=& \frac{1}{n} \left( \Var(X_1) + \ldots + \Var(X_n) \right) = 1
+\end{IEEEeqnarray*}
+
 For the r.h.s.~we get a mean of  $0$ and a variance of $1$.
 So, to determine what $(\ast)$ could mean, it is necessary that $\sqrt{n}$
 is the right scaling.
@@ -77,8 +79,17 @@ This notion of convergence will be defined in terms of characteristic functions
 
 \subsection{Characteristic functions and Fourier transform}
 
-Consider $(\R, \cB(\R), \bP)$.
+\begin{definition}
-For every $t \in \R$ define a function $\phi(t) \coloneqq \phi_\bP(t) \coloneqq \int_{\R} e^{\i t x} \bP(dx)$.
+    Consider $(\R, \cB(\R), \bP)$.
+    The \vocab{characteristic function} of $\bP$ is defined as
+    \begin{IEEEeqnarray*}{rCl}
+        \phi_{\bP}: \R &\longrightarrow & \C \\
+        t  &\longmapsto & \int_{\R} e^{\i t x} \bP(\dif x).
+    \end{IEEEeqnarray*}
+\end{definition}
+\begin{abuse}
+    $\phi_\bP(t)$ will often be abbreviated as $\phi(t)$.
+\end{abuse}
 We have
 \[
 \phi(t) = \int_{\R} \cos(tx) \bP(dx) + \i \int_{\R} \sin(tx) \bP(dx).
@@ -88,7 +99,6 @@ We have
     \item We have $\phi(0) = 1$.
     \item $|\phi(t)| \le  \int_{\R} |e^{\i t x} | \bP(dx) = 1$.
 \end{itemize}
-We call $\phi_{\bP}$ the \vocab{characteristic function} of $\bP$.
 
 \begin{remark}
     Suppose $(\Omega, \cF, \bP)$ is an arbitrary probability space and

wtheo.sty

@@ -23,6 +23,10 @@
 \usepackage{float}
 %\usepackage{algorithmicx}
 
+\usepackage{pgfplots}
+\pgfplotsset{compat = newest}
+
+
 \newcounter{subsubsubsection}[subsubsection]
 \renewcommand\thesubsubsubsection{\thesubsubsection.\arabic{subsubsubsection}}
 \newcommand\subsubsubsection[1]
@@ -97,3 +101,5 @@
 \DeclareSimpleMathOperator{Bin}
 \DeclareSimpleMathOperator{Ber}
 \DeclareSimpleMathOperator{Exp}
+
+\newcommand*\dif{\mathop{}\!\mathrm{d}}