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Finish solution for exercise 2a

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Andreas Tsouchlos 2026-01-14 01:30:12 +01:00
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src/2026-01-16/presentation.tex
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@ -249,7 +249,16 @@
X = h_1(U,V) = U \\ X = h_1(U,V) = U \\
Y = h_2(U,V) = \frac{V}{U} Y = h_2(U,V) = \frac{V}{U}
\end{array} \end{array}
\right. \\[4mm] \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} \mathcal{J} = \begin{pmatrix}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\[2mm] \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\[2mm]
\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}
@ -258,32 +267,37 @@
1 & 0 \\ 1 & 0 \\
- \frac{v}{u^2} & \frac{1}{u} - \frac{v}{u^2} & \frac{1}{u}
\end{pmatrix} \end{pmatrix}
\end{gather*} \end{gather}
\vspace{3mm}
\begin{align*} \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) f_{U,V}(u,v) &= \lvert \text{det}(\mathcal{J}) \rvert
= \left\{ \begin{array}{ll} \cdot f_{X,Y} \big(h_1(u,v),h_2(u,v)\big)
\frac{1}{u} \cdot \left(u + \frac{v}{u}\right) &,\hspace{2mm} 0 < v \le u \le 1 \\ = \frac{1}{u} \cdot \left(u + \frac{v}{u}\right)
0 &,\hspace{2mm} \text{sonst} = 1 + \frac{v}{u^2}, && \hspace*{-20mm} 0 < v \le u \le 1 \\[3mm]
\end{array} \right.
= \left\{ \begin{array}{ll}
1 + \frac{v}{u^2} &,\hspace{2mm} 0 < v \le u \le 1 \\
0 &,\hspace{2mm} \text{sonst}
\end{array} \right.
\end{align*} \end{align*}
\vspace{3mm} \vspace*{-22mm}
\begin{gather*} \pause\begin{align*}
\text{Für } 0 < v \le 1: \hspace{5mm} f_V(v) f_V(v) &= \int_{-\infty}^{\infty} f_{U,V}(u,v) du
= \int_{-\infty}^{\infty} f_{U,V}(u,v) du = \int_{v}^{1} 1 + \frac{v}{u^2} du
= \int_{v}^{1} 1 + \frac{v}{u^2} dx
= \left[ u - \frac{v}{u} \right]_v^1 = \left[ u - \frac{v}{u} \right]_v^1
= 2(1-v) \\[3mm] = 2(1-v), && \hspace*{-20mm} 0 < v \le 1
f_Z(z) = \left\{\begin{array}{ll} \end{align*}
2(1-z) \hspace{3mm}&,\hspace{3mm} 0 < z \le 1 \\ \vspace{5mm}
0 \hspace{3mm}&,\hspace{3mm} \text{sonst}\\ \pause\begin{gather*}
\end{array}\right. 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{gather*}
\pause \item Verwenden Sie einen alternativen Ansatz zur Berechnung der \end{enumerate}
% tex-fmt: on
\end{frame}
\begin{frame}
\frametitle{Aufgabe 2: Transformationssatz für 2D-Dichten}
\begin{enumerate}[a{)}]
\setcounter{enumi}{1}
\item Verwenden Sie einen alternativen Ansatz zur Berechnung der
Dichte. Hinweis: Beginnen Sie mit $P (Z \le z) = \ldots$ Dichte. Hinweis: Beginnen Sie mit $P (Z \le z) = \ldots$
\pause\begin{align*} \pause\begin{align*}
\end{align*} \end{align*}
@ -291,7 +305,6 @@
\pause\begin{align*} \pause\begin{align*}
\end{align*} \end{align*}
\end{enumerate} \end{enumerate}
% tex-fmt: on
\end{frame} \end{frame}
\end{document} \end{document}