From 876bbad136e2bd72fc801f52e01b37e37e490b39 Mon Sep 17 00:00:00 2001 From: Andreas Tsouchlos Date: Thu, 15 Jan 2026 03:26:33 +0100 Subject: [PATCH] Add solution for 2b --- src/2026-01-16/presentation.tex | 126 ++++++++++++++++++++++++++++---- 1 file changed, 111 insertions(+), 15 deletions(-) diff --git a/src/2026-01-16/presentation.tex b/src/2026-01-16/presentation.tex index 4a390aa..b6019e5 100644 --- a/src/2026-01-16/presentation.tex +++ b/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{enumerate}[a{)}] - \setcounter{enumi}{1} - \item Verwenden Sie einen alternativen Ansatz zur Berechnung der - Dichte.\\ - \textit{Hinweis}: Beginnen Sie mit $P (Z \le z) = \ldots$ - \pause\begin{align*} - P(Z \le z) = P(XZ \le z) &= \int_{-\infty}^{\infty} - \int_{-\infty}^{z/x} f_{X,Y}(x,y) dy dx \\[1mm] - &= \int_{-\infty}^{\infty} \int_{-\infty}^{z} - f_{X,Y}\left(x, \frac{u}{x}\right) \frac{1}{x} \; du dx - \end{align*} - \end{enumerate} - % tex-fmt: on + \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}