Add complete theory for exercise 1
This commit is contained in:
@@ -80,6 +80,20 @@
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}
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}
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\newlength{\depthofprodsign}
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\setlength{\depthofprodsign}{\depthof{$\prod$}}
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\newlength{\totalheightofprodsign}
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\newcommand{\nprod}[1][1.4]{
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\mathop{
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\raisebox
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{-#1\depthofprodsign+1\depthofprodsign}
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{\scalebox
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{#1}
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{$\displaystyle\prod$}%
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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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% \captionsetup[sub]{font=small}
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@@ -122,7 +136,8 @@
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\begin{itemize}
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\item Einfache Stichprobe
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\begin{gather*}
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X_1, \ldots, X_N \hspace{2mm}\overbrace{\text{unabhäng und haben
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X_1, \ldots, X_N
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\hspace{2mm}\overbrace{\text{unabhängig und haben
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dieselbe Verteilung}}^{\text{``iid.''}}
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\hspace*{5mm} \rightarrow\hspace*{5mm}
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\bm{X} :=
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@@ -157,7 +172,7 @@
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X_1 \\
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\vdots \\
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X_N
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\end{pmatrix}\sim f_{\bm{X}}$
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\end{pmatrix}\sim P_{\bm{X}}$
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};
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\node[right=of model] (x) {
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@@ -246,7 +261,7 @@
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X_1 \\
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\vdots \\
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X_N
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\end{pmatrix}\sim f_{\bm{X}}$
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\end{pmatrix}\sim P_{\bm{X}}$
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};
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\draw[
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@@ -266,7 +281,7 @@
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\end{frame}
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\begin{frame}
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\frametitle{Punktschätzer I}
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\frametitle{Punktschätzer}
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\vspace*{-10mm}
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@@ -276,6 +291,113 @@
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\begin{figure}[H]
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\centering
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\only<1>{
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\begin{tikzpicture}
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\node[
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rectangle,
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densely dashed,
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draw,
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inner sep=5mm,
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] (x) {
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$
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\bm{x} =
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\begin{pmatrix}
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26{,}2 \\
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27{,}8 \\
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25{,}7 \\
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\vdots
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\end{pmatrix}
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$
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};
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\node[
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draw opacity=0,
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fill opacity=0,
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rectangle,
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right=of x,
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minimum width=5cm, minimum height=2cm,
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draw=kit-green, fill=kit-green!20,
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line width=1pt,
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align=center,
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inner sep=3mm
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] (est) {Schätzer\\[5mm] $T_N(\bm{x}) =
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\displaystyle\frac{1}{N}
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\nsum_{i=0}^{N} x_i$};
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\node[
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draw opacity=0,
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fill opacity=0,
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above=of est,
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rectangle,
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densely dashed,
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draw,
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inner sep=5mm,
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] (model) {
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$X_i \sim \mathcal{N}(\mu = \vartheta,
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\sigma^2 = 1)$
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};
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\node[right=of est, draw opacity=0, fill
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opacity=0] (theta) {$\hat{\vartheta} = 26{,}0$};
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\node[below] at (x.south) {Beobachtung};
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\node[above, draw opacity=0, fill opacity=0]
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at (model.north) {Parametrisiertes Modell};
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\end{tikzpicture}
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}%
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\only<2>{
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\begin{tikzpicture}
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\node[
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rectangle,
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densely dashed,
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draw,
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inner sep=5mm,
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] (x) {
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$
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\bm{x} =
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\begin{pmatrix}
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26{,}2 \\
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27{,}8 \\
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25{,}7 \\
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\vdots
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\end{pmatrix}
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$
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};
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\node[
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draw opacity=0,
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fill opacity=0,
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rectangle,
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right=of x,
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minimum width=5cm, minimum height=2cm,
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draw=kit-green, fill=kit-green!20,
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line width=1pt,
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align=center,
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inner sep=3mm
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] (est) {Schätzer\\[5mm] $T_N(\bm{x}) =
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\displaystyle\frac{1}{N}
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\nsum_{i=0}^{N} x_i$};
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\node[
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above=of est,
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rectangle,
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densely dashed,
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draw,
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inner sep=5mm,
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] (model) {
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$X_i \sim \mathcal{N}(\mu = \vartheta,
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\sigma^2 = 1)$
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};
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\node[right=of est, draw opacity=0, fill
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opacity=0] (theta) {$\hat{\vartheta}
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= 26{,}0$};
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\node[below] at (x.south) {Beobachtung};
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\node[above] at (model.north) {Parametrisiertes Modell};
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\end{tikzpicture}
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}%
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\only<3->{
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\begin{tikzpicture}
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\node[
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rectangle,
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@@ -302,9 +424,9 @@
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line width=1pt,
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align=center,
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inner sep=3mm
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] (est) {Schätzer\\[5mm] $T(\bm{x}) =
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] (est) {Schätzer\\[5mm] $T_N(\bm{x}) =
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\displaystyle\frac{1}{N}
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\sum_{i=0}^{N} x_i$};
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\nsum_{i=0}^{N} x_i$};
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\node[
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above=of est,
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@@ -313,7 +435,8 @@
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draw,
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inner sep=5mm,
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] (model) {
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$X_i \sim \mathcal{N}(\mu = \vartheta, \sigma^2 = 1)$
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$X_i \sim \mathcal{N}(\mu = \vartheta,
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\sigma^2 = 1)$
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};
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\node[right=of est] (theta) {$\hat{\vartheta}
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@@ -327,26 +450,113 @@
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\draw[-{Latex}, line width=1pt] (model) -- (est);
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\draw[-{Latex}, line width=1pt] (est) -- (theta);
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\end{tikzpicture}
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}
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\end{figure}
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\pause
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\pause
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\item Punktschätzer: Rechenvorschrift zur Berechnung von
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Parametern aus Beobachtungen \\
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\pause
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$\rightarrow$ Schätzer hängen von den Realisierungen ab
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und sind damit selbst auch zufällig \\
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$\rightarrow$ Schätzer haben selbst einen Erwartungswert
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und eine Varianz
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$\rightarrow$ Schätzer haben einen Erwartungswert und eine Varianz
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\end{itemize}
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\end{frame}
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\begin{frame}
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\frametitle{Punktschätzer II}
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\frametitle{Likelihood und Log-Likelihood (Diskret)}
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\vspace*{-10mm}
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\begin{itemize}
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\item Maximum Likelihood (ML) Schätzer\\
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\begin{minipage}{0.21\textwidth}
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\phantom{a}
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\end{minipage}
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\begin{minipage}{0.16\textwidth}
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\centering
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\begin{align*}
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\hat{\vartheta}_\text{ML}
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= \argmax_\vartheta \hspace{2mm} P(\bm{X} = \bm{x}
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\vert \vartheta)
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\end{align*}
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\end{minipage}%
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\visible<2->{
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\begin{minipage}{0.15\textwidth}
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\centering
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\begin{align*}
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\hspace*{-3mm} = \argmax_\vartheta
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\hspace{2mm} L_{\bm{x}} (\vartheta)
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\end{align*}
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\end{minipage}%
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}
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\visible<3->{
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\begin{minipage}{0.13\textwidth}
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\centering
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\begin{align*}
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\hspace*{-10mm} = \argmax_\vartheta
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\hspace{2mm} l_{\bm{x}} (\vartheta)
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\end{align*}
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\end{minipage}%
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}
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\begin{figure}[H]
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\centering
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``Welches $\vartheta$ maximiert die
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Wahrscheinlichkeit die beobachtete Realisierung zu bekommen?''
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\end{figure}
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\pause
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\item Likelihoodfunktion
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\end{itemize}
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\vspace*{5mm}
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\begin{minipage}{0.5\textwidth}
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\centering
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\begin{align*}
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L_{\bm{x}}(\vartheta) = P(\bm{X} = \bm{x} \vert
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\vartheta) \overset{X_i \text{
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iid.}}{=\joinrel=\joinrel=} \nprod_{i=1}^{N}
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P(X_i = x_i \vert \vartheta)
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\end{align*}
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\end{minipage}%
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\begin{minipage}{0.5\textwidth}
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\centering
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\begin{lightgrayhighlightbox}
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\vspace*{-3mm}
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Beispiel
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\vspace*{-10mm}
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\begin{gather*}
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X_i \sim \text{\normalfont Binomial} (p = \vartheta, K) \\
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L_{\bm{x}}(\vartheta) = P(\bm{X}=\bm{x} \vert \vartheta) =
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\nprod_{i=1}^{N}
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\binom{K}{x_i}\vartheta^{x_i}(1-\vartheta)^{K-x_i}
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\end{gather*}
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\vspace*{-10mm}
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\end{lightgrayhighlightbox}
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\end{minipage}%
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\vspace*{5mm}
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\begin{itemize}
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\pause
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\item Log-Likelihoodfunktion
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\begin{align*}
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l_{\bm{x}}(\vartheta) = \ln \left( L_{\bm{x}}(\vartheta) \right)
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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{Eigenschaften von Punktschätzern}
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\vspace*{-10mm}
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\begin{itemize}
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\item Erwartungtreue
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\begin{gather*}
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E(\hat{\vartheta}) = E\big( T(\bm{X}) \big) = \vartheta
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E(\hat{\vartheta}) = E\big( T_N(\bm{X}) \big) = \vartheta
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\end{gather*}
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\begin{figure}[H]
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@@ -359,7 +569,7 @@
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\item Konsistenz
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\begin{gather*}
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\lim_{N\rightarrow \infty} P_\vartheta \big( \lvert
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T_N - \vartheta \rvert \ge \varepsilon \big) = 0
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\hat{\vartheta} - \vartheta \rvert \ge \varepsilon \big) = 0
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\end{gather*}
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\begin{figure}[H]
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@@ -371,22 +581,108 @@
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\vspace*{10mm}
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\pause
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\item Effizienz (für erwartungtreue Schätzer)
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\begin{minipage}{0.68\textwidth}
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\begin{gather*}
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V(\hat{\vartheta}) = \frac{1}{J(\vartheta)},
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\hspace*{5mm} J(\vartheta) = - E\left(
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\frac{\partial^2}{\partial \vartheta^2}
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\ln \mleft( f_\vartheta (\bm{X}) \mright)
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l_{\bm{X}}(\vartheta)
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\right)
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\end{gather*}
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\begin{figure}[H]
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\centering
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``Für jedes N hat der Schätzer jeweils die
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``Für jedes fixe N hat der Schätzer jeweils die
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kleinstmögliche Varianz''
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\end{figure}
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\end{minipage}%
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\begin{minipage}{0.3\textwidth}
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\begin{lightgrayhighlightbox}
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Cramér-Rao Ungleichung \\
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\vspace*{-6mm}
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\begin{gather*}
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V(\hat{\vartheta}) \le \frac{1}{J(\vartheta)}
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\end{gather*}
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\vspace*{-10mm}
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\end{lightgrayhighlightbox}
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\end{minipage}
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\end{itemize}
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\end{frame}
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\begin{frame}
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\frametitle{Zusammenfassung}
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\vspace*{-10mm}
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\begin{columns}
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\column{\kitthreecolumns}
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% \begin{greenblock}{Einfache Stichprobe}
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% \vspace*{-8mm}
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% \begin{gather*}
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% \bm{X} =
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% \begin{pmatrix}
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% X_1 \\
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% \vdots \\
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% X_N
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% \end{pmatrix},\hspace{5mm}
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% X_1, \ldots, X_N \text{ iid.}
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% \end{gather*}
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% \vspace*{-3mm}
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% \end{greenblock}
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\begin{greenblock}{Likelihood und co. (diskret)}
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\vspace*{-10mm}
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\begin{align*}
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\text{Likelihoodfunktion: } &L_{\bm{x}} (\vartheta) = P\left(
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\bm{X} = \bm{x}
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\vert \vartheta \right) \\[3mm]
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\text{Log-Likelihoodfunktion: } &l_{\bm{x}}
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(\vartheta) = \ln \left( L_{\bm{x}}
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(\vartheta) \right) \\[3mm]
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\text{ML-Schätzer: } &\hat{\vartheta}_\text{ML} =
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\argmax_\vartheta
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\hspace{2mm} l_{\bm{x}} (\vartheta)
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\end{align*}
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\vspace*{-6mm}
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\end{greenblock}
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\begin{greenblock}{Eigenschaften von Schätzern}
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\vspace*{-10mm}
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\begin{align*}
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\text{Erwartungtreue: } & E\left( \hat{\vartheta}
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\right) = \vartheta \\
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\text{Konsistenz: } & \lim_{N\rightarrow \infty}
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P\left( \lvert \hat{\vartheta}
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- \vartheta \rvert \ge \varepsilon
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\right) = 0 \\
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\text{Effizienz: } & V(\hat{\vartheta}) =
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\frac{1}{J(\vartheta)},\hspace{5mm} J(\vartheta) = - E\left(
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\frac{\partial^2}{\partial \vartheta^2}
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l_{\bm{x}}(\vartheta) \right)
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\end{align*}
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\vspace*{-3mm}
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\end{greenblock}
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\column{\kitthreecolumns}
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\begin{greenblock}{Erwartungswert \& Varianz Rechenregeln}
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\vspace*{-10mm}
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\begin{align*}
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E(aX) &= aE(X) \\
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E(X + b) &= E(X) + b \\
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E(X + Y) &= E(X) + E(Y) \\[5mm]
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V(aX) &= a^2V(X) \\
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V(X + b) &= E(X) \\
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V(X + Y) &= V(X) + V(Y)
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\end{align*}
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\vspace*{-8mm}
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\end{greenblock}
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\begin{greenblock}{Tschebyscheff Ungleichung}
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\vspace*{-8mm}
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\begin{align*}
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P\left( \lvert X - E(X) \rvert \ge \varepsilon \right) \le
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\frac{V(X)}{\varepsilon^2}
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\end{align*}
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\vspace*{-6mm}
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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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@@ -447,22 +743,22 @@
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\begin{align*}
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\hspace*{-77mm}
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L_{\bm{x}}(\lambda) &= P(\bm{X} = \bm{x} | \lambda) =
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\prod_{i=1}^{N} P(X_i=x_i | \lambda) =
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\prod_{i=1}^{N} \frac{\lambda^{x_i}}{x_i!} e^{-\lambda}
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\nprod_{i=1}^{N} P(X_i=x_i | \lambda) =
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\nprod_{i=1}^{N} \frac{\lambda^{x_i}}{x_i!} e^{-\lambda}
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\end{align*}
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\vspace*{-3mm}
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\pause
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\begin{align*}
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l_{\bm{x}}(\lambda) &= \ln \left(
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L_{\bm{x}}(\lambda) \right) = \ln \left(
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\prod_{i=1}^{N} \frac{\lambda^{x_i}}{x_i!}
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\nprod_{i=1}^{N} \frac{\lambda^{x_i}}{x_i!}
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e^{-\lambda} \right)
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=
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\sum_{i=1}^{N}\left[\ln \left( e^{-\lambda} \right) +
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\nsum_{i=1}^{N}\left[\ln \left( e^{-\lambda} \right) +
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||||
\ln \left( \lambda^{x_i} \right)
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- \ln \left( x_i! \right)\right]
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= - N \lambda + \sum_{i=1}^{N} \left[ x_i \ln \left(
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||||
\lambda \right) - \sum_{n=1}^{x_i} \ln \left( n
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= - N \lambda + \nsum_{i=1}^{N} \left[ x_i \ln \left(
|
||||
\lambda \right) - \nsum_{n=1}^{x_i} \ln \left( n
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||||
\right) \right]
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||||
\end{align*}
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||||
\vspace*{5mm}
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@@ -472,18 +768,19 @@
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\begin{array}{l}
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||||
\displaystyle\frac{\partial
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l_{\bm{x}}(\lambda)}{\lambda} = -N +
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||||
\frac{1}{\lambda} \sum_{i=1}^{N} x_i \overset{!}{=} 0
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||||
\Rightarrow \lambda = \frac{\sum_{i=1}^{N} x_i}{N} \\[7mm]
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||||
\frac{1}{\lambda} \nsum_{i=1}^{N} x_i \overset{!}{=} 0
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||||
\Rightarrow \lambda = \frac{1}{N} \nsum_{i=1}^{N}
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x_i \\[7mm]
|
||||
\displaystyle\frac{\partial^2
|
||||
l_{\bm{x}}(\lambda)}{\partial
|
||||
\lambda^2} = - \frac{1}{\lambda^2} \sum_{i=1}^{N} x_i < 0
|
||||
\lambda^2} = - \frac{1}{\lambda^2} \nsum_{i=1}^{N} x_i < 0
|
||||
\end{array}
|
||||
% tex-fmt: off
|
||||
\right\}
|
||||
% tex-fmt: on
|
||||
\Rightarrow \hat{\lambda}_\text{ML} =
|
||||
\argmax_\lambda \hspace{2mm} l_{\bm{x}}(\lambda) =
|
||||
\frac{\sum_{i=1}^{N} x_i}{N}
|
||||
\frac{1}{N} \nsum_{i=1}^{N} x_i
|
||||
%
|
||||
% \hat{\lambda}_\text{ML} = \argmax_\lambda
|
||||
% \hspace{2mm} \ln \left( l_{\bm{x}} (\lambda) \right)
|
||||
@@ -507,8 +804,8 @@
|
||||
\pause
|
||||
\begin{gather*}
|
||||
E(\hat{\lambda}_\text{ML}) = E \left(\frac{1}{N}
|
||||
\sum_{i=1}^{N} X_i \right)
|
||||
= \frac{1}{N} \sum_{i=1}^{N} E(X_i) = \frac{1}{N}
|
||||
\nsum_{i=1}^{N} X_i \right)
|
||||
= \frac{1}{N} \nsum_{i=1}^{N} E(X_i) = \frac{1}{N}
|
||||
\cdot N \lambda = \lambda
|
||||
\hspace{7mm}\Rightarrow\hspace{7mm} \text{Schätzer
|
||||
ist erwartungstreu}
|
||||
@@ -523,7 +820,7 @@
|
||||
\begin{minipage}{0.16\textwidth}
|
||||
\begin{gather*}
|
||||
E\left( \lvert \hat{\lambda}_\text{ML} - \lambda
|
||||
\rvert > \varepsilon
|
||||
\rvert \ge \varepsilon
|
||||
\right)
|
||||
\end{gather*}
|
||||
\end{minipage}%
|
||||
@@ -532,7 +829,7 @@
|
||||
\begin{gather*}
|
||||
= E\left( \lvert \hat{\lambda}_\text{ML} -
|
||||
E\left(\hat{\lambda}_\text{ML}\right) \rvert
|
||||
> \varepsilon
|
||||
\ge \varepsilon
|
||||
\right)
|
||||
\le
|
||||
\frac{V\left(\hat{\lambda}_\text{ML}\right)}{\varepsilon^2}
|
||||
@@ -542,8 +839,8 @@
|
||||
\pause
|
||||
\begin{gather*}
|
||||
V\left(\hat{\lambda}_\text{ML}\right) = V \left(
|
||||
\frac{1}{N} \sum_{i=1}^{N} X_i \right) =
|
||||
\frac{1}{N^2} \sum_{i=1}^{N} V(X_i) =
|
||||
\frac{1}{N} \nsum_{i=1}^{N} X_i \right) =
|
||||
\frac{1}{N^2} \nsum_{i=1}^{N} V(X_i) =
|
||||
\frac{N\lambda}{N^2} = \frac{\lambda}{N}
|
||||
\end{gather*}
|
||||
\pause
|
||||
@@ -564,10 +861,10 @@
|
||||
\begin{gather*}
|
||||
J\left( \lambda \right) = - E
|
||||
\left(
|
||||
\frac{\partial^2}{\partial \lambda^2} l_{\bm{x}}
|
||||
\frac{\partial^2}{\partial \lambda^2} l_{\bm{X}}
|
||||
(\lambda) \right)
|
||||
= - E \left( \frac{1}{\lambda^2} \sum_{i=1}^{N} X_i \right)
|
||||
= \frac{1}{\lambda^2} \sum_{i=1}^{N} E\left( X_i
|
||||
= E \left( \frac{1}{\lambda^2} \nsum_{i=1}^{N} X_i \right)
|
||||
= \frac{1}{\lambda^2} \nsum_{i=1}^{N} E\left( X_i
|
||||
\right) = \frac{N}{\lambda}
|
||||
\end{gather*}
|
||||
\pause
|
||||
@@ -786,8 +1083,8 @@
|
||||
\vspace*{-8mm}
|
||||
\pause
|
||||
\begin{gather*}
|
||||
\overline{z} = \frac{1}{N} \sum_{i=1}^{N} z_{1,i} = 25 \\
|
||||
s^2 = \frac{1}{N-1} \sum_{i=1}^{N} \left( z_{1,i} -
|
||||
\overline{z} = \frac{1}{N} \nsum_{i=1}^{N} z_{1,i} = 25 \\
|
||||
s^2 = \frac{1}{N-1} \nsum_{i=1}^{N} \left( z_{1,i} -
|
||||
\overline{z} \right)^2 = 4
|
||||
\end{gather*}
|
||||
\end{minipage}%
|
||||
@@ -811,8 +1108,8 @@
|
||||
\vspace*{-8mm}
|
||||
\pause
|
||||
\begin{gather*}
|
||||
\overline{z} = \frac{1}{N} \sum_{i=1}^{N} z_{1,i} = 34{,}875 \\
|
||||
s^2 = \frac{1}{N-1} \sum_{i=1}^{N} \left( z_{1,i} -
|
||||
\overline{z} = \frac{1}{N} \nsum_{i=1}^{N} z_{1,i} = 34{,}875 \\
|
||||
s^2 = \frac{1}{N-1} \nsum_{i=1}^{N} \left( z_{1,i} -
|
||||
\overline{z} \right)^2 \approx 1525{,}84
|
||||
\end{gather*}
|
||||
\end{minipage}
|
||||
|
||||
Reference in New Issue
Block a user