Fix factorial symbol and change variable k to n

.latexmkrc

@@ -1,4 +1,3 @@
 $pdflatex="pdflatex -shell-escape -interaction=nonstopmode -synctex=1 %O %S";
 $out_dir = 'build';
 $pdf_mode = 1;
-

Makefile

@@ -1,19 +1,25 @@
 PRESENTATIONS := $(patsubst src/%/presentation.tex,build/presentation_%.pdf,$(wildcard src/*/presentation.tex))
 HANDOUTS := $(patsubst build/presentation_%.pdf,build/presentation_%_handout.pdf,$(PRESENTATIONS))
 
+RC_PDFLATEX := $(shell grep '$$pdflatex' .latexmkrc \
+	       | sed -e 's/.*"\(.*\)".*/\1/' -e 's/%S//' -e 's/%O//')
+
 .PHONY: all
 all: $(PRESENTATIONS) $(HANDOUTS)
 
 build/presentation_%.pdf: src/%/presentation.tex build/prepared
-	TEXINPUTS=./lib/cel-slides-template-2025:$$TEXINPUTS latexmk $<
-	mv build/presentation.pdf $@
+	TEXINPUTS=./lib/cel-slides-template-2025:$(dir $<):$$TEXINPUTS \
+		latexmk -outdir=build/$* $<
+	cp build/$*/presentation.pdf $@
 
 build/presentation_%_handout.pdf: src/%/presentation.tex build/prepared
-	TEXINPUTS=./lib/cel-slides-template-2025:$$TEXINPUTS latexmk -pdflatex='pdflatex %O "\def\ishandout{1}\input{%S}"' $<
-	mv build/presentation.pdf $@
+	TEXINPUTS=./lib/cel-slides-template-2025:$(dir $<):$$TEXINPUTS \
+		latexmk -outdir=build/$*_handout \
+		-pdflatex='$(RC_PDFLATEX) %O "\def\ishandout{1}\input{%S}"' $<
+	cp build/$*_handout/presentation.pdf $@
 
 build/prepared:
-	mkdir -p build
+	mkdir build
 	touch build/prepared
 
 .PHONY: clean

src/2026-01-16/presentation.tex

@@ -30,11 +30,15 @@
 \usepackage{tikz}
 \usepackage{tikz-3dplot}
 \usetikzlibrary{spy, external, intersections, positioning}
-%\tikzexternalize[prefix=build/]
+
+\ifdefined\ishandout\else
+\tikzexternalize
+\fi
 
 \usepackage{pgfplots}
 \pgfplotsset{compat=newest}
 \usepgfplotslibrary{fillbetween}
+\usepgfplotslibrary{groupplots}
 
 \usepackage{enumerate}
 \usepackage{listings}
@@ -45,6 +49,7 @@
 \usepackage{amsmath}
 \usepackage{graphicx}
 \usepackage{calc}
+\usepackage{amssymb}
 
 \title{WT Tutorium 5}
 \author[Tsouchlos]{Andreas Tsouchlos}
@@ -64,7 +69,6 @@
 \newlength{\depthofsumsign}
 \setlength{\depthofsumsign}{\depthof{$\sum$}}
 \newlength{\totalheightofsumsign}
-\newlength{\heightanddepthofargument}
 \newcommand{\nsum}[1][1.4]{
     \mathop{
         \raisebox
@@ -97,7 +101,96 @@
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \subsection{Theorie Wiederholung}
 
-% TODO:
+\begin{frame}
+    \frametitle{Summen Unabhängiger Zufallsvariablen}
+
+    \begin{gather*}
+        Z = X + Y, \hspace{10mm}X,Y \text{ unabhängig} \\[4mm]
+        \begin{array}{rl}
+            \text{Faltungssatz (diskret): } & P_Z(n) =
+            \nsum_{k=0}^{n} P_X(k)P_Y(n-k) \\[2mm]
+            \text{Charakteristische Funktion: } & \phi_Z(s) = \phi_X(s)
+            \cdot \phi_Y(s)
+        \end{array}
+    \end{gather*}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Poisson-Verteilung}
+
+    \vspace*{-10mm}
+
+    \begin{itemize}
+        \item Binomialverteilung für $N\rightarrow \infty$ mit
+            $pN=\text{const.}=: \lambda$ \\
+            \begin{itemize}
+                \item ``Übergang von diskreter auf stetige
+                    Zeitachse bei fester mittlerer Rate'' \\
+                \item $\lambda \equiv$ ``mittlere Rate an Treffern
+                    pro Zeitabschnitt''
+            \end{itemize}
+        \item Beispiele
+            \begin{itemize}
+                \item Sternschnuppen pro Stunde
+                \item Anzahl an Websitebesuchern pro Minute
+            \end{itemize}
+    \end{itemize}
+
+    \pause
+    \begin{gather*}
+        X \sim \text{Poisson}(\lambda)
+    \end{gather*}
+    \vspace*{-2mm}
+    \begin{gather*}
+        P_X(n) = \frac{\lambda^n}{n!}e^{-\lambda} \\[2mm]
+        \phi_X(s) = \text{exp}\left(\lambda (e^{js} -1)\right)
+    \end{gather*}
+    \vspace*{-2mm}
+    \begin{align*}
+        E(X) &= \lambda\\
+        V(X) &= \lambda
+    \end{align*}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Zusammenfassung}
+
+    \vspace*{-20mm}
+
+    \begin{columns}[t]
+        \column{\kitthreecolumns}
+        \begin{greenblock}{Poisson-Verteilung}
+            \vspace*{-6mm}
+            \begin{gather*}
+                X \sim \text{Poisson}(\lambda) \\[3mm]
+                P_X(n) = \frac{\lambda^n \cdot e^{-\lambda}}{n!} \\[4mm]
+                \phi_X(s) = \text{exp}\left(\lambda (e^{js} -1)\right)
+            \end{gather*}
+        \end{greenblock}
+        \begin{greenblock}{Binomialentwicklung}
+            \vspace*{-6mm}
+            \begin{gather*}
+                \nsum_{k=0}^{n} \binom{n}{k}a^k b^{n-k} = (a+b)^n, \hspace{15mm}
+                \binom{n}{k} = \frac{n!}{(n-k)!k!}
+            \end{gather*}
+        \end{greenblock}
+        \column{\kitthreecolumns}
+        \begin{greenblock}{Faltungssatz (diskrete ZV)}
+            \vspace*{-6mm}
+            \begin{gather*}
+                Z = X + Y, \hspace{10mm}X,Y \text{ unabhängig} \\[3mm]
+                P_Z(n) = \nsum_{k=0}^{n} P_X(k)P_Y(n-k)
+            \end{gather*}
+        \end{greenblock}
+        \begin{greenblock}{Charakteristische Funktion einer Summe von ZV}
+            \vspace*{-6mm}
+            \begin{gather*}
+                Z = X + Y, \hspace{10mm}X,Y \text{ unabhängig} \\[3mm]
+                \phi_Z(s) = \phi_X(s) \cdot \phi_Y(s)
+            \end{gather*}
+        \end{greenblock}
+    \end{columns}
+\end{frame}
 
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \subsection{Aufgabe}
@@ -124,6 +217,8 @@
 \begin{frame}[fragile]
     \frametitle{Aufgabe 1:\\Faltungssatz \& Charakteristische Funktion}
 
+    \vspace*{-3mm}
+
     Es seien zwei unabhängige poissonverteilte Zufallsvariablen $X$ und
     $Y$ mit den Parametern $\lambda_1$
     bzw. $\lambda_2$ gegeben.
@@ -136,17 +231,19 @@
             zweier Zufallsvariablen.
             \pause\begin{gather*}
                 X \sim \text{Poisson}(\lambda_1) \hspace{3mm}
-                    \Leftrightarrow \hspace{3mm} P_X(k)
-                    = \frac{\lambda_1^k \cdot e^{-\lambda_1}}{k!} \hspace{30mm}
+                    \Leftrightarrow \hspace{3mm} P_X(n)
+                    = \frac{\lambda_1^n \cdot e^{-\lambda_1}}{n!} \hspace{30mm}
                 Y \sim \text{Poisson}(\lambda_2) \hspace{3mm}
-                    \Leftrightarrow \hspace{3mm} P_Y(k)
-                    = \frac{\lambda_2^k \cdot e^{-\lambda_2}}{k!}
+                    \Leftrightarrow \hspace{3mm} P_Y(n)
+                    = \frac{\lambda_2^n \cdot e^{-\lambda_2}}{n!}
             \end{gather*}
-            \vspace{2mm}
             \pause\begin{align*}
                 P_Z(n) &= P_{X+Y}(n) = \nsum_{k=0}^{n} P_X(k)P_Y(n-k)
                 = \nsum_{k=0}^{n} \frac{\lambda_1^k \cdot e^{-\lambda_1}}{k!}
-                    \cdot \frac{\lambda_2^{n-k} \cdot e^{-\lambda_2}}{(n-k)!} \\[3mm]
+                    \cdot \frac{\lambda_2^{n-k} \cdot e^{-\lambda_2}}{(n-k)!}
+            \end{align*}
+            \vspace*{-4mm}
+            \pause\begin{align*}
                 &= e^{-(\lambda_1 + \lambda_2)} \nsum_{k=0}^{n}
                     \frac{1}{k! (n-k)!} \lambda_1^k \lambda_2^{n-k} \\[3mm]
                 &= \frac{e^{-(\lambda_1 + \lambda_2)}}{n!} \nsum_{k=0}^{n}
@@ -155,12 +252,37 @@
                     \binom{n}{k} \lambda_1^k \lambda_2^{n-k}
                 = \frac{e^{-(\lambda_1 + \lambda_2)}}{n!}
                     ( \lambda_1 + \lambda_2 )^n
-                =: \frac{\lambda^n e^{-\lambda}}{n!}
+                    =: \frac{\lambda^n e^{-\lambda}}{n!} \\[6mm]
+                & \hspace*{-15mm}\Rightarrow Z \sim \text{Poisson}(\lambda_1 + \lambda_2)
             \end{align*}
-        \pause\item Erbringen Sie denselben Nachweis mithilfe der
+    \end{enumerate}
+    % tex-fmt: on
+\end{frame}
+
+\begin{frame}
+    \frametitle{Aufgabe 1:\\Faltungssatz \& Charakteristische Funktion}
+
+    % tex-fmt: off
+    \begin{enumerate}[a{)}]
+        \setcounter{enumi}{1}
+        \item Erbringen Sie denselben Nachweis mithilfe der
             charakteristischen Funktion.
+            \pause\begin{gather*}
+                X \sim \text{Poisson}(\lambda_1) \hspace{3mm}
+                    \Leftrightarrow \hspace{3mm}
+                    \phi_X(s) = \text{exp}\left(\lambda_1 (e^{js} -1)\right)
+                \hspace{30mm}
+                Y \sim \text{Poisson}(\lambda_1) \hspace{3mm}
+                    \Leftrightarrow \hspace{3mm}
+                    \phi_Y(s) = \text{exp}\left(\lambda_2 (e^{js} -1)\right)
+            \end{gather*}
+            \vspace*{-5mm}
             \pause\begin{align*}
-                % TODO: Write solution
+                \phi_Z(s) &= \phi_X(s) \cdot \phi_Y(s) \\
+                &= \text{exp}\left(\lambda_2 (e^{js} -1)\right) \cdot
+                    \text{exp}\left(\lambda_2 (e^{js} -1)\right) \\
+                &= \text{exp}\left((\lambda_1 + \lambda_2) (e^{js} -1)\right) \\[4mm]
+                & \hspace*{-15mm}\Rightarrow Z \sim \text{Poisson}(\lambda_1 + \lambda_2)
             \end{align*}
     \end{enumerate}
     % tex-fmt: on
@@ -172,7 +294,429 @@
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \subsection{Theorie Wiederholung}
 
-% TODO:
+\begin{frame}
+    \frametitle{Mehrdimensionale Zufallsvariablen}
+
+    \vspace*{-20mm}
+
+    \begin{columns}[t]
+        \column{\kitfourcolumns}
+        \begin{itemize}
+            \item Randdichte
+                \begin{align*}
+                    f_X(x) = \int_{-\infty}^{\infty} f_{X,Y}(x,y) dy
+                \end{align*}
+        \end{itemize}
+        \column{\kittwocolumns}
+        \begin{figure}[H]
+            \centering
+
+            \begin{tikzpicture}[
+                    /pgfplots/scale only axis,
+                    /pgfplots/width=5cm,
+                    /pgfplots/height=5cm
+                ]
+
+                \begin{axis}[
+                        name=main axis,
+                        view={0}{90},
+                        ticks=none,
+                        xlabel={$x$},ylabel={$y$},
+                    ]
+                    \addplot3[
+                        surf, shader=interp,
+                        samples=40,
+                        domain=-3:3, y domain=-3:3
+                    ]
+                    {1/(2*pi*sqrt(0.5)) * exp(-1/(2*(1 -
+                    sqrt(0.5))) * (x^2 -2*sqrt(0.5)*x*y + y^2) )};
+                \end{axis}
+
+                \node[below] at
+                ($(main axis.south west) + (-.5, -.5)$) {$f_{X,Y}(x,y)$};
+
+                \begin{axis}[
+                        anchor=south west,
+                        at=(main axis.north west),
+                        height=2cm,
+                        ticks=none,
+                        ylabel={$f_X(x)$},
+                        samples=50,
+                        domain=-3:3,
+                        xmin=-3,xmax=3,
+                    ]
+                    \addplot[line width=1pt] {1/sqrt(2*pi) *
+                    exp(-x^2/2)};
+                \end{axis}
+
+                \begin{axis}[
+                        anchor=north west,
+                        at=(main axis.north east),
+                        width=2cm,
+                        ticks=none,
+                        xlabel={$f_Y(y)$},
+                        samples=50,
+                        domain=-3:3,
+                        ymin=-3,ymax=3,
+                    ]
+                    \addplot[line width=1pt] ( {1/sqrt(2*pi)
+                    * exp(-x^2/2)}, {x} );
+                \end{axis}
+            \end{tikzpicture}
+        \end{figure}
+    \end{columns}
+
+    \pause
+    \vspace*{-45mm}
+    \begin{columns}
+        \column{\kitfourcolumns}
+        \begin{itemize}
+            \item Umrechnung von Dichten mit dem Transformationssatz
+                \begin{gather*}
+                    X = h_1(U,V), \hspace{5mm} Y = h_2(U,V) \\[5mm]
+                    \mathcal{J} =
+                    \begin{pmatrix}
+                        \frac{\displaystyle \partial}{\displaystyle
+                        \partial u}x & \frac{\displaystyle
+                        \partial}{\displaystyle \partial v}x  \\[3mm]
+                        \frac{\displaystyle \partial}{\displaystyle
+                        \partial u}y & \frac{\displaystyle
+                        \partial}{\displaystyle \partial v}y
+                    \end{pmatrix}
+                    =
+                    \begin{pmatrix}
+                        \frac{\displaystyle \partial}{\displaystyle
+                        \partial u}h_1(u,v) & \frac{\displaystyle
+                        \partial}{\displaystyle \partial v}h_1(u,v)  \\[3mm]
+                        \frac{\displaystyle \partial}{\displaystyle
+                        \partial u}h_2(u,v) & \frac{\displaystyle
+                        \partial}{\displaystyle \partial v}h_2(u,v)
+                    \end{pmatrix} \\[5mm]
+                    f_{U,V}(u,v) = \lvert
+                    \text{det}(\mathcal{J}) \rvert
+                    \cdot f_{X,Y} \big(h_1(u,v),h_2(u,v)\big)
+                \end{gather*}
+        \end{itemize}
+        \column{\kittwocolumns}
+    \end{columns}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Unabhängigkeit \& Korrelation I}
+
+    \vspace*{-10mm}
+
+    \begin{itemize}
+        \item Unabhängige ZV (stetig)
+            \begin{columns}
+                \column{\kitthreecolumns}
+                \begin{align*}
+                    X,Y \text{ unabhängig}
+                    \hspace{5mm} \Leftrightarrow \hspace{5mm}
+                    f_{X,Y}(x,y) = f_X(x)f_Y(y)
+                \end{align*}
+                \column{\kitthreecolumns}
+                \begin{lightgrayhighlightbox}
+                    Erinnerung: Unabhängige Ereignisse
+                    \begin{align*}
+                        A,B \text{ \normalfont unabhängig}
+                        \hspace{5mm} \Leftrightarrow \hspace{5mm}
+                        P(AB) = P(A)P(B)
+                    \end{align*}
+                    \vspace*{-13mm}
+                \end{lightgrayhighlightbox}
+            \end{columns}
+            \pause
+        \item Kovarianz
+            \begin{columns}
+                \column{\kitthreecolumns}
+                \begin{align*}
+                    \text{cov}(X,Y) &= E\bigg( \big(X - E(X)\big) \big(Y
+                    - E(Y)\big) \bigg) \\
+                    &= E(XY) - E(X)E(Y)
+                \end{align*}
+                \column{\kitthreecolumns}
+                \begin{lightgrayhighlightbox}
+                    Erinnerung: Varianz
+                    \begin{align*}
+                        V(X) = E\big( \left(X - E(X)\right)^2 \big) =
+                        E(X^2) - E^2(X)
+                    \end{align*}
+                    \vspace*{-13mm}
+                \end{lightgrayhighlightbox}
+            \end{columns}
+        \item Korrelation
+            \begin{align*}
+                E(XY)
+            \end{align*}
+            \pause
+        \item Korrelationskoeffizient
+            \begin{align*}
+                \rho_{XY} = \frac{\text{cov}(X,Y)}{\sqrt{V(X)V(Y)}}
+                \hspace{25mm} \rho_{XY} = 0
+                \hspace{2mm}\Leftrightarrow\hspace{2mm}
+                E(XY) = E(X)E(Y)
+            \end{align*}
+    \end{itemize}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Unabhängigkeit \& Korrelation II}
+
+    \vspace*{-15mm}
+
+    \begin{itemize}
+        \item Korrelation misst einen linearen Zusammenhang zwischen zwei ZV.\\
+            Unabhängigkeit gibt an ob zwei ZV ``überhaupt zusammenhängen''
+            \begin{align*}
+                \hspace{5mm} X,Y \text{ unabhängig}
+                \hspace{5mm}\Rightarrow\hspace{5mm}
+                X,Y \text{ unkorreliert}
+            \end{align*}
+        \item Bei gemeinsam normalverteilten ZV gilt zusätzlich
+            \begin{align*}
+                \hspace{5mm} X,Y \text{ unkorreliert}
+                \hspace{5mm}\Rightarrow\hspace{5mm}
+                X,Y \text{ unabhängig}
+            \end{align*}
+            \vspace*{5mm}
+            \pause
+        \item Korrelation und Unabhängigkeit haben nichts mit den
+            Einzelverteilungen zu tun. Sie sind ``eine Ebene höher''
+            \begin{figure}[H]
+                \centering
+
+                \begin{subfigure}{0.32\textwidth}
+                    \begin{tikzpicture}[
+                            /pgfplots/scale only axis,
+                            /pgfplots/width=3.5cm,
+                            /pgfplots/height=3.5cm
+                        ]
+
+                        \begin{axis}[
+                                name=main axis,
+                                view={0}{90},
+                                ticks=none,
+                                xlabel={$x$},ylabel={$y$},
+                            ]
+                            \addplot3[
+                                surf, shader=interp,
+                                samples=40,
+                                domain=-3:3, y domain=-3:3
+                            ]
+                            {1/(2*pi*sqrt(0.5)) * exp(-1/(2*(1 -
+                            sqrt(0.5))) * (x^2 -2*sqrt(0.5)*x*y + y^2) )};
+                        \end{axis}
+
+                        \node[below] at
+                        ($(main axis.south west) + (-.5, -.5)$)
+                        {$f_{X,Y}(x,y)$};
+
+                        \begin{axis}[
+                                anchor=south west,
+                                at=(main axis.north west),
+                                height=2cm,
+                                ticks=none,
+                                ylabel={$f_X(x)$},
+                                samples=50,
+                                domain=-3:3,
+                                xmin=-3,xmax=3,
+                            ]
+                            \addplot[line width=1pt] {1/sqrt(2*pi) *
+                            exp(-x^2/2)};
+                        \end{axis}
+
+                        \begin{axis}[
+                                anchor=north west,
+                                at=(main axis.north east),
+                                width=2cm,
+                                ticks=none,
+                                xlabel={$f_Y(y)$},
+                                samples=50,
+                                domain=-3:3,
+                                ymin=-3,ymax=3,
+                            ]
+                            \addplot[line width=1pt] ( {1/sqrt(2*pi)
+                            * exp(-x^2/2)}, {x} );
+                        \end{axis}
+                    \end{tikzpicture}
+                \end{subfigure}%
+                \begin{subfigure}{0.32\textwidth}
+                    \begin{tikzpicture}[
+                            /pgfplots/scale only axis,
+                            /pgfplots/width=3.5cm,
+                            /pgfplots/height=3.5cm
+                        ]
+
+                        \begin{axis}[
+                                name=main axis,
+                                view={0}{90},
+                                ticks=none,
+                                xlabel={$x$},ylabel={$y$},
+                            ]
+                            \addplot3[
+                                surf, shader=interp,
+                                samples=40,
+                                domain=-3:3, y domain=-3:3
+                            ]
+                            {1/(2*pi) * exp(-1/2 * (x^2 + y^2) )};
+                        \end{axis}
+
+                        \node[below] at
+                        ($(main axis.south west) + (-.5, -.5)$)
+                        {$f_{X,Y}(x,y)$};
+
+                        \begin{axis}[
+                                anchor=south west,
+                                at=(main axis.north west),
+                                height=2cm,
+                                ticks=none,
+                                ylabel={$f_X(x)$},
+                                samples=50,
+                                domain=-3:3,
+                                xmin=-3,xmax=3,
+                            ]
+                            \addplot[line width=1pt] {1/sqrt(2*pi) *
+                            exp(-x^2/2)};
+                        \end{axis}
+
+                        \begin{axis}[
+                                anchor=north west,
+                                at=(main axis.north east),
+                                width=2cm,
+                                ticks=none,
+                                xlabel={$f_Y(y)$},
+                                samples=50,
+                                domain=-3:3,
+                                ymin=-3,ymax=3,
+                            ]
+                            \addplot[line width=1pt] ( {1/sqrt(2*pi)
+                            * exp(-x^2/2)}, {x} );
+                        \end{axis}
+                    \end{tikzpicture}
+                \end{subfigure}%
+                \begin{subfigure}{0.32\textwidth}
+                    \begin{tikzpicture}[
+                            /pgfplots/scale only axis,
+                            /pgfplots/width=3.5cm,
+                            /pgfplots/height=3.5cm
+                        ]
+
+                        \begin{axis}[
+                                name=main axis,
+                                view={0}{90},
+                                ticks=none,
+                                xlabel={$x$},ylabel={$y$},
+                            ]
+                            \addplot3[
+                                surf, shader=interp,
+                                samples=40,
+                                domain=-3:3, y domain=-3:3
+                            ]
+                            {1/(2*pi*sqrt(0.5)) * exp(-1/(2*(1 -
+                            sqrt(0.5))) * (x^2 +2*sqrt(0.5)*x*y + y^2) )};
+                        \end{axis}
+
+                        \node[below] at
+                        ($(main axis.south west) + (-.5, -.5)$)
+                        {$f_{X,Y}(x,y)$};
+
+                        \begin{axis}[
+                                anchor=south west,
+                                at=(main axis.north west),
+                                height=2cm,
+                                ticks=none,
+                                ylabel={$f_X(x)$},
+                                samples=50,
+                                domain=-3:3,
+                                xmin=-3,xmax=3,
+                            ]
+                            \addplot[line width=1pt] {1/sqrt(2*pi) *
+                            exp(-x^2/2)};
+                        \end{axis}
+
+                        \begin{axis}[
+                                anchor=north west,
+                                at=(main axis.north east),
+                                width=2cm,
+                                ticks=none,
+                                xlabel={$f_Y(y)$},
+                                samples=50,
+                                domain=-3:3,
+                                ymin=-3,ymax=3,
+                            ]
+                            \addplot[line width=1pt] ( {1/sqrt(2*pi)
+                            * exp(-x^2/2)}, {x} );
+                        \end{axis}
+                    \end{tikzpicture}
+                \end{subfigure}
+            \end{figure}
+    \end{itemize}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Zusammenfassung}
+
+    \vspace*{-20mm}
+
+    \begin{columns}[t]
+        \column{\kittwocolumns}
+        \begin{greenblock}{Korrelationskoeffizient}
+            \vspace*{-6mm}
+            \begin{gather*}
+                \rho_{XY} = \frac{\text{cov}(X,Y)}{\sqrt{V(X)V(Y)}}
+            \end{gather*}
+        \end{greenblock}
+        \begin{greenblock}{Kovarianz}
+            \vspace*{-6mm}
+            \begin{gather*}
+                \text{cov}(X,Y) = E(X Y) - E(X)E(Y)
+            \end{gather*}
+        \end{greenblock}
+        \begin{greenblock}{Randdichte}
+            \vspace*{-6mm}
+            \begin{gather*}
+                f_X(x) = \int_{-\infty}^{\infty} f_{X,Y}(x,y) dy
+            \end{gather*}
+        \end{greenblock}
+        \column{\kitfourcolumns}
+        \begin{greenblock}{Umrechnung von Dichten mit dem Transformationssatz}
+            \vspace*{-6mm}
+            \begin{gather*}
+                X = h_1(U,V), \hspace{5mm} Y = h_2(U,V) \\[5mm]
+                \mathcal{J} =
+                \begin{pmatrix}
+                    \frac{\displaystyle \partial}{\displaystyle
+                    \partial u}x & \frac{\displaystyle
+                    \partial}{\displaystyle \partial v}x  \\[3mm]
+                    \frac{\displaystyle \partial}{\displaystyle
+                    \partial u}y & \frac{\displaystyle
+                    \partial}{\displaystyle \partial v}y
+                \end{pmatrix}
+                =
+                \begin{pmatrix}
+                    \frac{\displaystyle \partial}{\displaystyle
+                    \partial u}h_1(u,v) & \frac{\displaystyle
+                    \partial}{\displaystyle \partial v}h_1(u,v)  \\[3mm]
+                    \frac{\displaystyle \partial}{\displaystyle
+                    \partial u}h_2(u,v) & \frac{\displaystyle
+                    \partial}{\displaystyle \partial v}h_2(u,v)
+                \end{pmatrix} \\[5mm]
+                f_{U,V}(u,v) = \lvert
+                \text{det}(\mathcal{J}) \rvert
+                \cdot f_{X,Y} \big(h_1(u,v),h_2(u,v)\big)
+            \end{gather*}
+        \end{greenblock}
+        \begin{greenblock}{Erwartungswert \& Varianz}
+            \vspace*{-6mm}
+            \begin{align*}
+                V(X) &= E\big( (X - E(X))^2 \big) = E(X^2) - E^2(X) \\
+                E(X) &= \int_{-\infty}^{\infty} x f_X(x) dx \\
+                E(g(X)) &= \int_{-\infty}^{\infty} g(x) f_X(x) dx
+            \end{align*}
+        \end{greenblock}
+    \end{columns}
+\end{frame}
 
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \subsection{Aufgabe}
@@ -189,16 +733,18 @@
         \item Berechnen Sie die Dichte von $(Z = X \cdot Y)$ mithilfe des
             Transformationssatzes.
         \item Verwenden Sie einen alternativen Ansatz zur Berechnung der
-            Dichte. Hinweis: Beginnen Sie mit $P (Z \le z) = \ldots$
+            Dichte.\\
+            \textit{Hinweis}: Beginnen Sie mit $P (Z \le z) = \ldots$
         \item Berechnen Sie den Korrelationskoeffizienten $\rho_{XY}$ .
     \end{enumerate}
     % tex-fmt: on
-
 \end{frame}
 
 \begin{frame}
     \frametitle{Aufgabe 2: Transformationssatz für 2D-Dichten}
 
+    \vspace*{-15mm}
+
     Die Zufallsvariable $(X; Y)^T$ habe die gemeinsame
     Wahrscheinlichkeitsdichte $f (x, y) = x + y$ für
     $x, y \in (0; 1]$ und null sonst.
@@ -207,22 +753,263 @@
     \begin{enumerate}[a{)}]
         \item Berechnen Sie die Dichte von $(Z = X \cdot Y)$ mithilfe des
             Transformationssatzes.
-            \pause\begin{align*}
-                f(x) = \displaystyle\int_{-\infty}^{\infty} f(x,y) dy
-                    = x + 0{,}5 \\
-                f(y) = \displaystyle\int_{-\infty}^{\infty} f(x,y) dx
-                    = y + 0{,}5
+            \pause\begin{gather*}
+                \left.
+                \begin{array}{l}
+                    U := X \\
+                    V := Z = X \cdot Y
+                \end{array}
+                \right\}
+                \Rightarrow
+                \left\{
+                \begin{array}{l}
+                    X = h_1(U,V) = U \\
+                    Y = h_2(U,V) = \frac{V}{U}
+                \end{array}
+                \hspace{20mm}
+                \left(\begin{array}{l}
+                    0 < x \le 1 \Rightarrow 0 < u \le 1 \\
+                    0 < y \le 1 \Rightarrow 0 < v \le u \le 1
+                \end{array}
+                \right)
+                \right.
+            \end{gather*}
+            \vspace*{5mm}
+            \pause\begin{gather*}
+                \mathcal{J} = \begin{pmatrix}
+                    \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}  \\[2mm]
+                    \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}
+                \end{pmatrix}
+                 = \begin{pmatrix}
+                     1 & 0 \\
+                     - \frac{v}{u^2} & \frac{1}{u}
+                \end{pmatrix}
+            \end{gather*}
+            \begin{align*}
+                f_{U,V}(u,v) &= \lvert \text{det}(\mathcal{J}) \rvert
+                    \cdot f_{X,Y} \big(h_1(u,v),h_2(u,v)\big)
+                    = \frac{1}{u} \cdot \left(u + \frac{v}{u}\right)
+                = 1 + \frac{v}{u^2}, && \hspace*{-20mm} 0 < v \le u \le 1 \\[3mm]
             \end{align*}
-        \pause \item Verwenden Sie einen alternativen Ansatz zur Berechnung der
-            Dichte. Hinweis: Beginnen Sie mit $P (Z \le z) = \ldots$
-            \pause\begin{align*}
-            \end{align*}
-        \pause \item Berechnen Sie den Korrelationskoeffizienten $\rho_{XY}$ .
+            \vspace*{-22mm}
             \pause\begin{align*}
+                f_V(v) &= \int_{-\infty}^{\infty} f_{U,V}(u,v) du
+                    = \int_{v}^{1} 1 + \frac{v}{u^2} du
+                    = \left[ u - \frac{v}{u} \right]_v^1
+                    = 2(1-v), && \hspace*{-20mm} 0 < v \le 1
             \end{align*}
+            \vspace{5mm}
+            \pause\begin{gather*}
+                f_Z(z) = \left\{\begin{array}{ll}
+                    2(1-z) \hspace{3mm}&,\hspace{3mm} 0 < z \le 1 \\
+                    0 \hspace{3mm}&,\hspace{3mm} \text{sonst}\\
+                \end{array}\right.
+            \end{gather*}
     \end{enumerate}
     % tex-fmt: on
+\end{frame}
 
+\begin{frame}
+    \frametitle{Aufgabe 2: Transformationssatz für 2D-Dichten}
+
+    \begin{minipage}[c]{0.64\textwidth}
+        % tex-fmt: off
+        \begin{enumerate}[a{)}]
+            \setcounter{enumi}{1}
+            \item Verwenden Sie einen alternativen Ansatz zur Berechnung der
+                Dichte.\\
+                \textit{Hinweis}: Beginnen Sie mit $P (Z \le z) = \ldots$
+        \end{enumerate}
+        % tex-fmt: on
+    \end{minipage}%
+    \begin{minipage}[c]{0.35\textwidth}
+        \begin{lightgrayhighlightbox}
+            \vspace*{-8mm}
+            % tex-fmt: off
+            \begin{gather*}
+                \text{Bekannt: } \hspace{10mm} f_{X,Y}(x,y) = x + y
+            \end{gather*}
+            % tex-fmt: on
+            \vspace*{-12mm}
+        \end{lightgrayhighlightbox}
+    \end{minipage}
+
+    \pause
+    \begin{align*}
+        P(Z \le z) = \int_{-\infty}^{z} f_Z(t) dt
+    \end{align*}
+
+    \begin{minipage}{0.4\textwidth}
+        \pause
+        \begin{figure}[H]
+            \centering
+
+            \begin{tikzpicture}
+                \begin{axis}[
+                        view={20}{30},
+                        xlabel=$x$, ylabel=$y$, zlabel={$f_{X,Y}(x,y)$},
+                        xmin=0, xmax=1, ymin=0, ymax=1, zmin=0, zmax=2,
+                        xtick={0,0.5,1},ytick={0,0.5,1},ztick={0,1,2},
+                        point meta min=0, point meta max=2,
+                        declare function={cutoff(\x) = 0.3/\x;},
+                        legend,
+                    ]
+                    \addplot3[
+                        surf, shader=interp,
+                        samples=40,
+                        domain=0:1, y domain=0:1
+                    ] (
+                        x,
+                        {y * min(1, cutoff(x))},
+                        {x + (y * min(1, cutoff(x)))}
+                    );
+                    \addlegendentry{$x\cdot y \le z$}
+
+                    \addplot3[
+                        surf, shader=interp,
+                        samples=40,
+                        domain=0.3:1, y domain=0:1,
+                        fill=gray,
+                        draw=none,
+                        point meta=1.1,
+                        colormap name=cividis,
+                    ] (
+                        x,
+                        {cutoff(x) + y*(1 - cutoff(x))},
+                        {x + (cutoff(x) + y*(1 - cutoff(x)))}
+                    );
+
+                    \addplot3[
+                        mesh,
+                        samples=15,
+                        domain=0:1, y domain=0:1,
+                        draw=black,
+                        opacity=0.3
+                    ] {x + y};
+                \end{axis}
+            \end{tikzpicture}
+        \end{figure}
+    \end{minipage}%
+    \begin{minipage}{0.58\textwidth}
+        \pause
+        \begin{align*}
+            P(Z \le z) &= P(XY \le z) = \int_{-\infty}^{\infty}
+            \int_{-\infty}^{z/x} f_{X,Y}(x,y) dy dx
+        \end{align*}
+        \vspace*{-10mm}
+        \pause
+        \begin{align*}
+            \overset{
+                \begin{subarray}{l}
+                    u = xy \\
+                    du = xdy
+            \end{subarray}}{=}
+            &\int_{-\infty}^{\infty} \int_{-\infty}^{z} f_{X,Y}(x,
+            \frac{u}{x})\frac{1}{x}\; du dx \\[2mm]
+            = &\int_{-\infty}^{z}
+            \underbrace{\int_{-\infty}^{\infty} f_{X,Y}(x,
+            \frac{u}{x})\frac{1}{x}\; dx}_{f_Z(u)}du \\
+        \end{align*}
+    \end{minipage}
+
+    \pause
+    \begin{gather*}
+        0 < y \le 1 \hspace{5mm} \Rightarrow\hspace{5mm} 0 <
+        \frac{u}{x} \le 1 \hspace{5mm}\Rightarrow\hspace{5mm} 0 <
+        u \le x \le 1 \\
+        f_Z(u) = \int_{-\infty}^{\infty} f_{X,Y}(x,
+        \frac{u}{x})\frac{1}{x}\; dx
+        = \int_{z}^{1} 1 + \frac{u}{x^2} dx = 2(1-u), \hspace{5mm} 0 < u \le 1
+    \end{gather*}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Aufgabe 2: Transformationssatz für 2D-Dichten}
+
+    \vspace*{-15mm}
+
+    \begin{minipage}[c]{0.5\textwidth}
+        % tex-fmt: off
+        \begin{enumerate}[a{)}]
+            \setcounter{enumi}{2}
+            \item Berechnen Sie den Korrelationskoeffizienten $\rho_{XY}$.
+        \end{enumerate}
+        % tex-fmt: on
+    \end{minipage}%
+    \begin{minipage}[c]{0.5\textwidth}
+        \begin{lightgrayhighlightbox}
+            \vspace*{-8mm}
+            % tex-fmt: off
+            \begin{gather*}
+                \text{Bekannt: } \hspace{10mm}
+                \left\{\hspace{2mm}
+                    \begin{array}{l}
+                        f_{X,Y}(x,y) = x + y \\
+                        f_{Z}(z) = 2(1-z), \hspace{10mm} Z = X\cdot Y
+                    \end{array}
+                    \right.
+            \end{gather*}
+            % tex-fmt: on
+            \vspace*{-10mm}
+        \end{lightgrayhighlightbox}
+    \end{minipage}
+    \vspace*{2mm}
+    \pause
+    \begin{gather*}
+        \rho_{XY} = \frac{\text{cov}(X,Y)}{\sqrt{V(X)V(Y)}},
+        \hspace{15mm}
+        \begin{array}{l}
+            \text{cov}(X,Y) = \overbrace{E(XY)}^{E(Z)} - E(X)E(Y) \\
+            V(X) = E(X^2) - E^2(X)
+        \end{array},
+        \hspace*{15mm}
+        E(X) = \int_{-\infty}^{\infty} xf_X(x) dx
+    \end{gather*}
+    \vspace*{5mm}
+    \pause
+    \begin{gather*}
+        f_X(x) = \int_{-\infty}^{\infty} f_{X,Y}(x,y) dy
+        = \int_{0}^{1} x + y dy
+        = \left[ xy + \frac{y^2}{2} \right]_0^1 = x + \frac{1}{2}
+    \end{gather*}
+    \vspace*{-3mm}
+    \pause
+    \begin{gather*}
+        f(x,y) = f(y,x) \Rightarrow
+        \left\{
+            \begin{array}{l}
+                E(X) = E(Y) \\
+                V(X) = V(Y)
+            \end{array}
+            \right.
+            \hspace{5mm} \Rightarrow \hspace{5mm}
+            \rho_{XY} = \frac{E(Z) - E^2(X)}{E(X^2) - E^2(X)}
+        \end{gather*}
+        \vspace*{5mm}
+        \pause
+        \begin{gather*}
+            \left.
+            \begin{array}{rl}
+                E(X) &= \displaystyle \int_{-\infty}^{\infty} x f_X(x) dx
+                = \int_{0}^{1} x(x+ \frac{1}{2}) dx
+                = \left[\frac{x^3}{3} + \frac{x^2}{4} \right]_0^1
+                = \frac{7}{12} \\
+                E(X^2) &= \displaystyle \int_{-\infty}^{\infty}
+                x^2 f_X(x) dx
+                = \int_{0}^{1} x^2 (x + \frac{1}{2} ) dx
+                = \left[\frac{x^4}{4} + \frac{x^3}{6} \right]_0^1
+                = \frac{5}{12} \\
+                E(Z) &= \displaystyle \int_{-\infty}^{\infty} z f_Z(z) dz
+                = \int_{0}^{1} z \cdot 2(1-z) dz
+                = 2 \left[ \frac{z^2}{2} - \frac{z^3}{3} \right]_0^1
+                = \frac{1}{3}
+            \end{array}
+            \hspace{3mm}
+        \right\}
+        \hspace{5mm} \Rightarrow \hspace{5mm}
+        \rho_{XY} = \frac{\frac{1}{3} - (\frac{7}{12})^2}{\frac{5}{12}
+        - (\frac{7}{12})^2} = -\frac{1}{11}
+    \end{gather*}
 \end{frame}
 
 \end{document}