Add solution for 2b

src/2026-01-16/presentation.tex

@@ -239,7 +239,8 @@
                     \partial u}y & \frac{\displaystyle
                     \partial}{\displaystyle \partial v}y
                 \end{pmatrix}
-                = \begin{pmatrix}
+                =
+                \begin{pmatrix}
                     \frac{\displaystyle \partial}{\displaystyle
                     \partial u}h_1(u,v) & \frac{\displaystyle
                     \partial}{\displaystyle \partial v}h_1(u,v)  \\[2mm]
@@ -357,20 +358,115 @@
 \begin{frame}
     \frametitle{Aufgabe 2: Transformationssatz für 2D-Dichten}
 
-    % tex-fmt: off
+    \begin{minipage}[c]{0.64\textwidth}
-    \begin{enumerate}[a{)}]
+        % tex-fmt: off
-        \setcounter{enumi}{1}
+        \begin{enumerate}[a{)}]
-        \item Verwenden Sie einen alternativen Ansatz zur Berechnung der
+            \setcounter{enumi}{1}
-            Dichte.\\
+            \item Verwenden Sie einen alternativen Ansatz zur Berechnung der
-            \textit{Hinweis}: Beginnen Sie mit $P (Z \le z) = \ldots$
+                Dichte.\\
-            \pause\begin{align*}
+                \textit{Hinweis}: Beginnen Sie mit $P (Z \le z) = \ldots$
-                P(Z \le z) = P(XZ \le z) &= \int_{-\infty}^{\infty}
+        \end{enumerate}
-                    \int_{-\infty}^{z/x} f_{X,Y}(x,y) dy dx \\[1mm]
+        % tex-fmt: on
-                &= \int_{-\infty}^{\infty} \int_{-\infty}^{z}
+    \end{minipage}%
-                    f_{X,Y}\left(x, \frac{u}{x}\right) \frac{1}{x} \; du dx
+    \begin{minipage}[c]{0.35\textwidth}
-            \end{align*}
+        \begin{lightgrayhighlightbox}
-    \end{enumerate}
+            \vspace*{-8mm}
-    % tex-fmt: on
+            % 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}