1016 lines
35 KiB
TeX
1016 lines
35 KiB
TeX
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\groupname{Communication Engineering Lab (CEL)}
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\usepackage{amsmath}
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\usepackage{graphicx}
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\usepackage{calc}
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\usepackage{amssymb}
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\title{WT Tutorium 5}
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\author[Tsouchlos]{Andreas Tsouchlos}
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\date[]{16. Januar 2026}
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\newcommand{\res}{src/2026-01-16/res}
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\newlength{\depthofsumsign}
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\setlength{\depthofsumsign}{\depthof{$\sum$}}
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\newlength{\totalheightofsumsign}
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\newcommand{\nsum}[1][1.4]{
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\mathop{
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\raisebox
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{-#1\depthofsumsign+1\depthofsumsign}
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{\scalebox
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{#1}
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{$\displaystyle\sum$}%
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}
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}
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}
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% \tikzstyle{every node}=[font=\small]
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% Document body
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\begin{document}
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\begin{frame}[title white vertical, picture=images/IMG_7801-cut]
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\titlepage
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\end{frame}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\section{Aufgabe 1}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\subsection{Theorie Wiederholung}
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\begin{frame}
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\frametitle{Summen Unabhängiger Zufallsvariablen}
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\begin{gather*}
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Z = X + Y, \hspace{10mm}X,Y \text{ unabhängig} \\[4mm]
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\begin{array}{rl}
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\text{Faltungssatz (diskret): } & P_Z(n) =
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\nsum_{k=0}^{n} P_X(k)P_Y(n-k) \\[2mm]
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\text{Charakteristische Funktion: } & \phi_Z(s) = \phi_X(s)
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\cdot \phi_Y(s)
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\end{array}
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\end{gather*}
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\end{frame}
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\begin{frame}
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\frametitle{Poisson-Verteilung}
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\vspace*{-10mm}
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\begin{itemize}
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\item Binomialverteilung für $N\rightarrow \infty$ mit
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$pN=\text{const.}=: \lambda$ \\
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\begin{itemize}
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\item ``Übergang von diskreter auf stetige
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Zeitachse bei fester mittlerer Rate'' \\
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\item $\lambda \equiv$ ``mittlere Rate an Treffern
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pro Zeitabschnitt''
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\end{itemize}
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\item Beispiele
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\begin{itemize}
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\item Sternschnuppen pro Stunde
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\item Anzahl an Websitebesuchern pro Minute
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\end{itemize}
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\end{itemize}
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\pause
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\begin{gather*}
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X \sim \text{Poisson}(\lambda)
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\end{gather*}
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\vspace*{-2mm}
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\begin{gather*}
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P_X(k) = \frac{\lambda^k}{k!}e^{-\lambda} \\[2mm]
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\phi_X(s) = \text{exp}\left(\lambda (e^{js} -1)\right)
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\end{gather*}
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\vspace*{-2mm}
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\begin{align*}
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E(X) &= \lambda\\
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V(X) &= \lambda
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\end{align*}
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\end{frame}
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\begin{frame}
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\frametitle{Zusammenfassung}
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\vspace*{-20mm}
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\begin{columns}[t]
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\column{\kitthreecolumns}
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\begin{greenblock}{Poisson-Verteilung}
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\vspace*{-6mm}
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\begin{gather*}
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X \sim \text{Poisson}(\lambda) \\[3mm]
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P_X(k) = \frac{\lambda^k \cdot e^{-\lambda}}{k!} \\[4mm]
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\phi_X(s) = \text{exp}\left(\lambda (e^{js} -1)\right)
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\end{gather*}
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\end{greenblock}
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\begin{greenblock}{Binomialentwicklung}
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\vspace*{-6mm}
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\begin{gather*}
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\nsum_{k=0}^{n} \binom{n}{k}a^k b^{n-k} = (a+b)^n, \hspace{15mm}
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\binom{n}{k} = \frac{n!}{(n-k!)k!}
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\end{gather*}
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\end{greenblock}
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\column{\kitthreecolumns}
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\begin{greenblock}{Faltungssatz (diskrete ZV)}
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\vspace*{-6mm}
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\begin{gather*}
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Z = X + Y, \hspace{10mm}X,Y \text{ unabhängig} \\[3mm]
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P_Z(n) = \nsum_{k=0}^{n} P_X(k)P_Y(n-k)
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\end{gather*}
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\end{greenblock}
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\begin{greenblock}{Charakteristische Funktion einer Summe von ZV}
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\vspace*{-6mm}
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\begin{gather*}
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Z = X + Y, \hspace{10mm}X,Y \text{ unabhängig} \\[3mm]
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\phi_Z(s) = \phi_X(s) \cdot \phi_Y(s)
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\end{gather*}
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\end{greenblock}
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\end{columns}
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\end{frame}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\subsection{Aufgabe}
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\begin{frame}
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\frametitle{Aufgabe 1:\\Faltungssatz \& Charakteristische Funktion}
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Es seien zwei unabhängige poissonverteilte Zufallsvariablen $X$ und
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$Y$ mit den Parametern $\lambda_1$
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bzw. $\lambda_2$ gegeben.
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% tex-fmt: off
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\begin{enumerate}[a{)}]
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\item Zeigen Sie, dass die Summe $Z = X + Y$ ebenfalls
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Poisson-verteilt ist mit dem Parameter $\lambda = \lambda_1 +
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\lambda_2$. Nutzen Sie dazu den Faltungssatz für die Addition
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zweier Zufallsvariablen.
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\item Erbringen Sie denselben Nachweis mithilfe der
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charakteristischen Funktion.
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\end{enumerate}
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% tex-fmt: on
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\end{frame}
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\begin{frame}[fragile]
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\frametitle{Aufgabe 1:\\Faltungssatz \& Charakteristische Funktion}
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\vspace*{-3mm}
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Es seien zwei unabhängige poissonverteilte Zufallsvariablen $X$ und
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$Y$ mit den Parametern $\lambda_1$
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bzw. $\lambda_2$ gegeben.
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% tex-fmt: off
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\begin{enumerate}[a{)}]
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\item Zeigen Sie, dass die Summe $Z = X + Y$ ebenfalls
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Poisson-verteilt ist mit dem Parameter $\lambda = \lambda_1 +
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\lambda_2$. Nutzen Sie dazu den Faltungssatz für die Addition
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zweier Zufallsvariablen.
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\pause\begin{gather*}
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X \sim \text{Poisson}(\lambda_1) \hspace{3mm}
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\Leftrightarrow \hspace{3mm} P_X(k)
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= \frac{\lambda_1^k \cdot e^{-\lambda_1}}{k!} \hspace{30mm}
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Y \sim \text{Poisson}(\lambda_2) \hspace{3mm}
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\Leftrightarrow \hspace{3mm} P_Y(k)
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= \frac{\lambda_2^k \cdot e^{-\lambda_2}}{k!}
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\end{gather*}
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\pause\begin{align*}
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P_Z(n) &= P_{X+Y}(n) = \nsum_{k=0}^{n} P_X(k)P_Y(n-k)
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= \nsum_{k=0}^{n} \frac{\lambda_1^k \cdot e^{-\lambda_1}}{k!}
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\cdot \frac{\lambda_2^{n-k} \cdot e^{-\lambda_2}}{(n-k)!}
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\end{align*}
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\vspace*{-4mm}
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\pause\begin{align*}
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&= e^{-(\lambda_1 + \lambda_2)} \nsum_{k=0}^{n}
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\frac{1}{k! (n-k)!} \lambda_1^k \lambda_2^{n-k} \\[3mm]
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&= \frac{e^{-(\lambda_1 + \lambda_2)}}{n!} \nsum_{k=0}^{n}
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\frac{n!}{k! (n-k)!} \lambda_1^k \lambda_2^{n-k} \\[3mm]
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&= \frac{e^{-(\lambda_1 + \lambda_2)}}{n!} \nsum_{k=0}^{n}
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\binom{n}{k} \lambda_1^k \lambda_2^{n-k}
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= \frac{e^{-(\lambda_1 + \lambda_2)}}{n!}
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( \lambda_1 + \lambda_2 )^n
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=: \frac{\lambda^n e^{-\lambda}}{n!} \\[6mm]
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& \hspace*{-15mm}\Rightarrow Z \sim \text{Poisson}(\lambda_1 + \lambda_2)
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\end{align*}
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\end{enumerate}
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% tex-fmt: on
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\end{frame}
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\begin{frame}
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\frametitle{Aufgabe 1:\\Faltungssatz \& Charakteristische Funktion}
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% tex-fmt: off
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\begin{enumerate}[a{)}]
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\setcounter{enumi}{1}
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\item Erbringen Sie denselben Nachweis mithilfe der
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charakteristischen Funktion.
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\pause\begin{gather*}
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X \sim \text{Poisson}(\lambda_1) \hspace{3mm}
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\Leftrightarrow \hspace{3mm}
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\phi_X(s) = \text{exp}\left(\lambda_1 (e^{js} -1)\right)
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\hspace{30mm}
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Y \sim \text{Poisson}(\lambda_1) \hspace{3mm}
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\Leftrightarrow \hspace{3mm}
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\phi_Y(s) = \text{exp}\left(\lambda_2 (e^{js} -1)\right)
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\end{gather*}
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\vspace*{-5mm}
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\pause\begin{align*}
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\phi_Z(s) &= \phi_X(s) \cdot \phi_Y(s) \\
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&= \text{exp}\left(\lambda_2 (e^{js} -1)\right) \cdot
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\text{exp}\left(\lambda_2 (e^{js} -1)\right) \\
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&= \text{exp}\left((\lambda_1 + \lambda_2) (e^{js} -1)\right) \\[4mm]
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& \hspace*{-15mm}\Rightarrow Z \sim \text{Poisson}(\lambda_1 + \lambda_2)
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\end{align*}
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\end{enumerate}
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% tex-fmt: on
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\end{frame}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\section{Aufgabe 2}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\subsection{Theorie Wiederholung}
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\begin{frame}
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\frametitle{Mehrdimensionale Zufallsvariablen}
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\vspace*{-20mm}
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\begin{columns}[t]
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\column{\kitfourcolumns}
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\begin{itemize}
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\item Randdichte
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\begin{align*}
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f_X(x) = \int_{-\infty}^{\infty} f_{X,Y}(x,y) dy
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\end{align*}
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\end{itemize}
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\column{\kittwocolumns}
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\begin{figure}[H]
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\centering
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\begin{tikzpicture}[
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/pgfplots/scale only axis,
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/pgfplots/width=5cm,
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/pgfplots/height=5cm
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]
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\begin{axis}[
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name=main axis,
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view={0}{90},
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ticks=none,
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xlabel={$x$},ylabel={$y$},
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]
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\addplot3[
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surf, shader=interp,
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samples=40,
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domain=-3:3, y domain=-3:3
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]
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{1/(2*pi*sqrt(0.5)) * exp(-1/(2*(1 -
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sqrt(0.5))) * (x^2 -2*sqrt(0.5)*x*y + y^2) )};
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\end{axis}
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\node[below] at
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($(main axis.south west) + (-.5, -.5)$) {$f_{X,Y}(x,y)$};
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\begin{axis}[
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anchor=south west,
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at=(main axis.north west),
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height=2cm,
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ticks=none,
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ylabel={$f_X(x)$},
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samples=50,
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domain=-3:3,
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xmin=-3,xmax=3,
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]
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\addplot[line width=1pt] {1/sqrt(2*pi) *
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exp(-x^2/2)};
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\end{axis}
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\begin{axis}[
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anchor=north west,
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at=(main axis.north east),
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width=2cm,
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ticks=none,
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xlabel={$f_Y(y)$},
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samples=50,
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domain=-3:3,
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ymin=-3,ymax=3,
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]
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\addplot[line width=1pt] ( {1/sqrt(2*pi)
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* exp(-x^2/2)}, {x} );
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\end{axis}
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\end{tikzpicture}
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\end{figure}
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\end{columns}
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\pause
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\vspace*{-45mm}
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\begin{columns}
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\column{\kitfourcolumns}
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\begin{itemize}
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\item Umrechnung von Dichten mit dem Transformationssatz
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\begin{gather*}
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X = h_1(U,V), \hspace{5mm} Y = h_2(U,V) \\[5mm]
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\mathcal{J} =
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\begin{pmatrix}
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\frac{\displaystyle \partial}{\displaystyle
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\partial u}x & \frac{\displaystyle
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\partial}{\displaystyle \partial v}x \\[3mm]
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\frac{\displaystyle \partial}{\displaystyle
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\partial u}y & \frac{\displaystyle
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\partial}{\displaystyle \partial v}y
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\end{pmatrix}
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=
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\begin{pmatrix}
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\frac{\displaystyle \partial}{\displaystyle
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\partial u}h_1(u,v) & \frac{\displaystyle
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\partial}{\displaystyle \partial v}h_1(u,v) \\[3mm]
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\frac{\displaystyle \partial}{\displaystyle
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\partial u}h_2(u,v) & \frac{\displaystyle
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\partial}{\displaystyle \partial v}h_2(u,v)
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\end{pmatrix} \\[5mm]
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f_{U,V}(u,v) = \lvert
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\text{det}(\mathcal{J}) \rvert
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\cdot f_{X,Y} \big(h_1(u,v),h_2(u,v)\big)
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\end{gather*}
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\end{itemize}
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\column{\kittwocolumns}
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\end{columns}
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\end{frame}
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\begin{frame}
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\frametitle{Unabhängigkeit \& Korrelation I}
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\vspace*{-10mm}
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\begin{itemize}
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\item Unabhängige ZV (stetig)
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\begin{columns}
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\column{\kitthreecolumns}
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\begin{align*}
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X,Y \text{ unabhängig}
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\hspace{5mm} \Leftrightarrow \hspace{5mm}
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f_{X,Y}(x,y) = f_X(x)f_Y(y)
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\end{align*}
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\column{\kitthreecolumns}
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\begin{lightgrayhighlightbox}
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Erinnerung: Unabhängige Ereignisse
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\begin{align*}
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X,Y \text{ \normalfont unabhängig}
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\hspace{5mm} \Leftrightarrow \hspace{5mm}
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P(AB) = P(A)P(B)
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\end{align*}
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\vspace*{-13mm}
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\end{lightgrayhighlightbox}
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\end{columns}
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\pause
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\item Kovarianz
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\begin{columns}
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\column{\kitthreecolumns}
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\begin{align*}
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\text{cov}(X,Y) &= E\bigg( \big(X - E(X)\big) \big(Y
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- E(Y)\big) \bigg) \\
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&= E(XY) - E(X)E(Y)
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\end{align*}
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\column{\kitthreecolumns}
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\begin{lightgrayhighlightbox}
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Erinnerung: Varianz
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\begin{align*}
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V(X) = E\big( \left(X - E(X)\right)^2 \big) = E(X^2) - E^2(X)
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\end{align*}
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\vspace*{-13mm}
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\end{lightgrayhighlightbox}
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\end{columns}
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\item Korrelation
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\begin{align*}
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E(XY)
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\end{align*}
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\pause
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\item Korrelationskoeffizient
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\begin{align*}
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\rho_{XY} = \frac{\text{cov}(X,Y)}{\sqrt{V(X)V(Y)}}
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\hspace{25mm} \rho_{XY} = 0
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\hspace{2mm}\Leftrightarrow\hspace{2mm}
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E(XY) = E(X)E(Y)
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\end{align*}
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\end{itemize}
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\end{frame}
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\begin{frame}
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\frametitle{Unabhängigkeit \& Korrelation II}
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\vspace*{-15mm}
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\begin{itemize}
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\item Korrelation misst einen linearen Zusammenhang zwischen zwei ZV.\\
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Unabhängigkeit gibt an ob zwei ZV ``überhaupt zusammenhängen''
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\begin{align*}
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\hspace{5mm} X,Y \text{ unabhängig}
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\hspace{5mm}\Rightarrow\hspace{5mm}
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X,Y \text{ unkorreliert}
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\end{align*}
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\item Bei gemeinsam normalverteilten ZV gilt zusätzlich
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\begin{align*}
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\hspace{5mm} X,Y \text{ unkorreliert}
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\hspace{5mm}\Rightarrow\hspace{5mm}
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X,Y \text{ unabhängig}
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\end{align*}
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\vspace*{5mm}
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\pause
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\item Korrelation und Unabhängigkeit haben nichts mit den
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Einzelverteilungen zu tun. Sie sind ``eine Ebene höher''
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\begin{figure}[H]
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\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}
|
|
|
|
\begin{frame}
|
|
\frametitle{Aufgabe 2: Transformationssatz für 2D-Dichten}
|
|
|
|
Die Zufallsvariable $(X; Y)^T$ habe die gemeinsame
|
|
Wahrscheinlichkeitsdichte $f (x, y) = x + y$ für
|
|
$x, y \in (0; 1]$ und null sonst.
|
|
|
|
% tex-fmt: off
|
|
\begin{enumerate}[a{)}]
|
|
\item Berechnen Sie die Dichte von $(Z = X \cdot Y)$ mithilfe des
|
|
Transformationssatzes.
|
|
\item Verwenden Sie einen alternativen Ansatz zur Berechnung der
|
|
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.
|
|
|
|
% tex-fmt: off
|
|
\begin{enumerate}[a{)}]
|
|
\item Berechnen Sie die Dichte von $(Z = X \cdot Y)$ mithilfe des
|
|
Transformationssatzes.
|
|
\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*}
|
|
\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}
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\end{gather*}
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\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.
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|
\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}
|
|
|