Add slide transitions to exercise solutions
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@@ -421,6 +421,8 @@
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\begin{frame}
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\frametitle{Aufgabe 1: Punktschätzer}
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\vspace*{-15mm}
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Die Anzahl der Studierenden, die zur Mittagszeit in der KIT-Mensa
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essen gehen, sei näherungsweise Poissonverteilt mit unbekanntem
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Parameter $\lambda > 0$, wobei $\lambda$ die mittlere Ankunftsrate an
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@@ -440,21 +442,31 @@
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\item Bestimmen Sie die Log-Likelihoodfunktion für $N$
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Messwerte und damit den ML-Schätzer für die
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Ankunftsrate $\lambda$.
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\vspace*{5mm}
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\pause
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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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\prod_{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!} e^{-\lambda} \right)
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\prod_{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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\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 \right) \right]
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\lambda \right) - \sum_{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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\pause
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\begin{gather*}
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\left.
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\begin{array}{l}
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@@ -462,7 +474,8 @@
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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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\displaystyle\frac{\partial^2 l_{\bm{x}}(\lambda)}{\partial
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\displaystyle\frac{\partial^2
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l_{\bm{x}}(\lambda)}{\partial
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\lambda^2} = - \frac{1}{\lambda^2} \sum_{i=1}^{N} x_i < 0
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\end{array}
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% tex-fmt: off
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@@ -481,13 +494,17 @@
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% TODO: Erwartungswert Rechenregeln in Zusammenfassung
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% TODO: Tschebyscheff Ungleichung in Theorie und Zusammenfassung
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\begin{frame}
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\frametitle{}
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\frametitle{Aufgabe 1: Punktschätzer}
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\vspace*{-10mm}
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\begin{enumerate}%
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% tex-fmt: off
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[a{)}]
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% tex-fmt: on
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\setcounter{enumi}{1}
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\item Zeigen Sie, dass der Schätzer erwartungstreu ist.
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\pause
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\begin{gather*}
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E(\hat{\lambda}_\text{ML}) = E \left(\frac{1}{N}
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\sum_{i=1}^{N} X_i \right)
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@@ -496,21 +513,40 @@
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\hspace{7mm}\Rightarrow\hspace{7mm} \text{Schätzer
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ist erwartungstreu}
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\end{gather*}
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\item Ist der ML-Schätzer konsistent?
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\pause
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\vspace*{-5mm}
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\item Ist der ML-Schätzer konsistent? \\[-5mm]
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\pause
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\begin{minipage}{0.24\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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\begin{gather*}
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E\left( \lvert \hat{\lambda}_\text{ML} - \lambda
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\rvert > \varepsilon
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\right)
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\end{gather*}
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\end{minipage}%
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\pause %
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\begin{minipage}{0.22\textwidth}
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\begin{gather*}
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= E\left( \lvert \hat{\lambda}_\text{ML} -
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E\left(\hat{\lambda}_\text{ML}\right) \rvert > \varepsilon
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E\left(\hat{\lambda}_\text{ML}\right) \rvert
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> \varepsilon
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\right)
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\le
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\frac{V\left(\hat{\lambda}_\text{ML}\right)}{\varepsilon^2} \\
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\frac{V\left(\hat{\lambda}_\text{ML}\right)}{\varepsilon^2}
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\end{gather*}
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\end{minipage} \\[2mm]
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\pause
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\begin{gather*}
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V\left(\hat{\lambda}_\text{ML}\right) = V \left(
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\frac{1}{N} \sum_{i=1}^{N} X_i \right) =
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\frac{1}{N^2} \sum_{i=1}^{N} V(X_i) =
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\frac{N\lambda}{N^2} = \frac{\lambda}{N}
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\end{gather*}
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\pause
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\begin{gather*}
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E\left( \lvert \hat{\lambda}_\text{ML} - \lambda
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\rvert > \varepsilon
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@@ -521,7 +557,10 @@
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\hspace{7mm} \Rightarrow \hspace{7mm} \text{Schätzer
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ist konsistent}
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\end{gather*}
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\pause
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\vspace*{-5mm}
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\item Ist der ML-Schätzer effizient?
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\pause
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\begin{gather*}
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J\left( \lambda \right) = - E
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\left(
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@@ -529,7 +568,10 @@
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(\lambda) \right)
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= - E \left( \frac{1}{\lambda^2} \sum_{i=1}^{N} X_i \right)
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= \frac{1}{\lambda^2} \sum_{i=1}^{N} E\left( X_i
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\right) = \frac{N}{\lambda} \\[5mm]
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\right) = \frac{N}{\lambda}
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\end{gather*}
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\pause
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\begin{gather*}
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V\left( \hat{\lambda}_\text{ML} \right)
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% tex-fmt: off
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\overset{\text{c)}}{=}
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@@ -642,6 +684,7 @@
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\vspace*{5mm}
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\pause
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\begin{minipage}{0.25\textwidth}
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\phantom{a}
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\end{minipage}
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@@ -653,7 +696,9 @@
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\end{itemize}
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\end{minipage}
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\vspace*{5mm}
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\pause
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\vspace*{15mm}
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\begin{figure}[H]
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\centering
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\includegraphics[scale=1.4]{res/boxplot.pdf}
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@@ -671,6 +716,7 @@
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\vspace*{-10mm}
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\pause
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\begin{align*}
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\text{Median: } \hspace{5mm}&5 \\
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\text{Untere Quartilsgrenze: } \hspace{5mm}&3 \\
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@@ -720,6 +766,7 @@
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\end{enumerate}
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%
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\vspace*{-3mm}
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\pause
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\begin{minipage}{0.5\textwidth}
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\begin{gather*}
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z_1 =
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@@ -729,13 +776,22 @@
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\rightarrow
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\begin{pmatrix}
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22 & 24 & 25 & 25 & 26 & 28
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\end{pmatrix} \\[5mm]
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\overline{z} = \frac{1}{N} \sum_{i=1}^{N} z_{1,i} = 25 \\
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s^2 = \frac{1}{N-1} \sum_{i=1}^{N} \left( z_{1,i} -
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\overline{z} \right)^2 = 4 \\[5mm]
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\end{pmatrix}
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\end{gather*}
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\vspace*{-10mm}
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\pause
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\begin{gather*}
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z_{3/4} - z_{1/4} = 26 - 24 = 2
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\end{gather*}
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\vspace*{-8mm}
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\pause
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\begin{gather*}
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\overline{z} = \frac{1}{N} \sum_{i=1}^{N} z_{1,i} = 25 \\
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s^2 = \frac{1}{N-1} \sum_{i=1}^{N} \left( z_{1,i} -
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\overline{z} \right)^2 = 4
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\end{gather*}
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\end{minipage}%
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\pause
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\begin{minipage}{0.5\textwidth}
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\begin{gather*}
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z_1 =
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@@ -745,11 +801,19 @@
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\rightarrow
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\begin{pmatrix}
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0 & 22 & 24 & 25 & 25 & 26 & 28 & 129
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\end{pmatrix} \\[5mm]
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\end{pmatrix}
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\end{gather*}
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\vspace*{-10mm}
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\pause
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\begin{gather*}
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z_{3/4} - z_{1/4} = \frac{26 + 28}{2} - \frac{22 + 24}{2} = 4
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\end{gather*}
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\vspace*{-8mm}
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\pause
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\begin{gather*}
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\overline{z} = \frac{1}{N} \sum_{i=1}^{N} z_{1,i} = 34{,}875 \\
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s^2 = \frac{1}{N-1} \sum_{i=1}^{N} \left( z_{1,i} -
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\overline{z} \right)^2 = 1525{,}84 \\[5mm]
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z_{3/4} - z_{1/4} = \frac{26 + 28}{2} - \frac{22 + 24}{2} = 4
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\overline{z} \right)^2 \approx 1525{,}84
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\end{gather*}
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\end{minipage}
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\end{frame}
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