Add solution for exercise 1

src/2026-01-30/presentation.tex

@@ -83,6 +83,19 @@
 % \tikzstyle{every node}=[font=\small]
 % \captionsetup[sub]{font=small}
 
+\newlength{\hght}
+\newlength{\wdth}
+
+\newcommand{\canceltotikz}[3][.5ex]{
+    \setlength{\hght}{\heightof{$#3$}}
+    \setlength{\wdth}{\widthof{$#3$}}
+    \makebox[0pt][l]{
+        \tikz[baseline]{\draw[-latex](0,-#1)--(\wdth,\hght+#1)
+            node[shift={(1mm,.5mm)}]{#2};
+        }
+    }#3
+}
+
 %
 %
 % Document body
@@ -121,6 +134,102 @@
     % tex-fmt: on
 \end{frame}
 
+\begin{frame}
+    \frametitle{Aufgabe 1: Korrelationskoeffizienten}
+
+    Es ist die Zufallsvariable $X \sim \mathcal{N}(0,1)$ gegeben. Berechnen Sie
+    jeweils den Korrelationskoeffizienten $\rho_{XY}$ für
+
+    % tex-fmt: off
+    \begin{enumerate}[a{)}]
+        \item $Y = aX + b \hspace{8mm}\text{mit } a, b \in R
+            \text{ und } a \neq 0$.
+            \pause \begin{gather*}
+                \rho_{XY} = \frac{\text{cov}(X,Y)}{\sqrt{V(X)V(Y)}}
+            \end{gather*}
+            \pause\begin{align*}
+                \text{cov}(X,Y) &= E(XY) - \canceltotikz[1ex]{0}{E(X)} E(Y)
+                    = E(XY) \\
+                    &= E(aX^2 + bX) = a\underbrace{E(X^2)}_{= V(X) = 1}
+                    + b\canceltotikz[1ex]{0}{E(X)} = a
+            \end{align*}
+            \pause\begin{gather*}
+                V(Y) = E\big( (Y - E(Y))^2 \big) = E\big( (aX)^2 \big)
+                    = a^2 \underbrace{E(X^2)}_{= V(X) = 1} = a^2
+            \end{gather*}
+            \pause\begin{align*}
+                \rho_{XY} = \frac{a}{\sqrt{a^2}} = \frac{a}{\lvert a \rvert}
+                = \left\{ \begin{array}{c}
+                    +1, \hspace{5mm} a > 0 \\
+                    -1, \hspace{5mm} a < 0
+                \end{array}
+                \right.
+            \end{align*}
+    \end{enumerate}
+    % tex-fmt: on
+\end{frame}
+
+\begin{frame}
+    \frametitle{Aufgabe 1: Korrelationskoeffizienten}
+
+    Es ist die Zufallsvariable $X \sim \mathcal{N}(0,1)$ gegeben. Berechnen Sie
+    jeweils den Korrelationskoeffizienten $\rho_{XY}$ für
+
+    % tex-fmt: off
+    \begin{enumerate}[a{)}]
+        \setcounter{enumi}{1}
+        \item $Y = X^2$.
+            \pause \begin{gather*}
+                \rho_{XY} = \frac{\text{cov}(X,Y)}{\sqrt{V(X)V(Y)}}
+            \end{gather*}
+            \pause\begin{columns}
+                \column{\kitfourcolumns}
+                \centering
+                \begin{gather*}
+                    \text{cov}(X,Y) = E(XY) - \canceltotikz[1ex]{0}{E(X)} E(Y)
+                        = E(XY) = E(X^3)
+                \end{gather*}
+                \vspace*{-12mm}
+                \pause\begin{gather*}
+                    \hspace*{-18mm} = \int_{-\infty}^{\infty}
+                        \underbrace{x^3}_\text{ungerade}
+                        \cdot\underbrace{f_X(x)}_\text{gerade} dx = 0 \\[7mm]
+                    \rho_{XY} = 0
+                \end{gather*}
+                \column{\kittwocolumns}
+                \centering
+                \begin{figure}[H]
+                    \centering
+                    \begin{tikzpicture}
+                        \begin{axis}[
+                            domain=-3:3,
+                            width=10cm,
+                            height=6.5cm,
+                            samples=100,
+                            xtick={0},
+                            ytick={0},
+                            legend pos = south east,
+                            legend cell align = left,
+                        ]
+                            \addplot+[scol1, mark=none, line width=1pt]
+                                {1 / sqrt(2*pi) * exp(-x^2)};
+                            \addlegendentry{$f_X(x)$}
+                            \addplot+[scol2, mark=none, line width=1pt]
+                                {0.01 * x^3};
+                            \addlegendentry{$x^3$}
+                        \end{axis}
+
+                        \node at (8.7, 4.7) {\footnotemark};
+                    \end{tikzpicture}
+                \end{figure}
+            \end{columns}
+    \end{enumerate}
+    % tex-fmt: on
+
+    \footnotetext{Die zwei Kurven sind bezüglich der $y$-Achse
+    unterschiedlich skaliert.}
+\end{frame}
+
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \section{Aufgabe 2}