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[PS] ch7
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\date{HS 2026}
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\begin{document}
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\subtext{Basiert auf dem Skript von V. Tassion}
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\subtext{Basiert auf dem Skript von V. Tassion, Slides von E.W. Farkas (FS26)}
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\input{parts/00_prereq.tex}
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\newpage
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@@ -19,20 +19,30 @@
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\section{Zufallsvariablen}
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\input{parts/02_variables.tex}
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\newpage
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\section{Erwartungswert}
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\input{parts/03_expectation.tex}
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\newpage
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\section{Gemeinsame Verteilungen}
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\input{parts/04_joint-distributions.tex}
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\newpage
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\section{Erwartungswert}
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\input{parts/03_expectation.tex}
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\newpage
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\section{Ungleichungen}
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\input{parts/05_inequalities.tex}
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\newpage
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\section{Grenzwertsätze}
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\input{parts/05_limits.tex}
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\input{parts/06_limits.tex}
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\newpage
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\section{Statistik}
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\input{parts/06_stats.tex}
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\input{parts/07_stats.tex}
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\newpage
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\section{Tabellen}
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\input{parts/08_tables.tex}
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\end{document}
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@@ -1,3 +1,5 @@
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\subsection{Allgemeiner Erwartungswert}
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\definition \textbf{Erwartungswert} (nicht-negativ)
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$$
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\E[X] = \int_{0}^{\infty} \biggl( 1-F_X(x) \biggr)\ dx
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@@ -11,59 +13,36 @@ $$
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\theorem $\forall \omega \in \Omega: X(\omega) \geq 0 \implies \E[X] \geq 0$\\
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\subtext{Gleichheit: $\E[X] = 0 \iff X=0$, fast sicher}
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\definition \textbf{Erwartungswert}
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$$
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\E[X] = \E[X_+] - \E[X_-] \quad\text{falls}\quad \E[|X|]<\infty
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$$
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{\scriptsize
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\remark $X$ kein konst. Vorzeichen, nicht $\E[|X|] < \infty $: $\E[X]$ undefiniert
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}
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% Sollte nicht relevant sein
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\subsection{Diskreter Erwartungswert}
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% \definition \textbf{Erwartungswert}
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% $$
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% \E[X] = \E[X_+] - \E[X_-] \quad\text{falls}\quad \E[|X|]<\infty
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% $$
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% {\scriptsize
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% \remark $X$ kein konst. Vorzeichen, nicht $\E[|X|] < \infty $: $\E[X]$ undefiniert
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% }
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\theorem \textbf{Diskreter Erwartungswert}
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\theorem \textbf{Diskreter Erwartungswert}\\
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\subtext{$X: \Omega \to \R,\quad W \cleq \N,\quad \phi: \R \to \R$}
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$$
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\E\bigl[ \phi(X) \bigr] = \sum_{w \in W} \phi(x) \cdot \P[X = x]
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$$
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\subtext{$X: \Omega \to \R,\quad W \cleq \N,\quad \phi: \R \to \R$}
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\begin{center}
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\begin{tabular}{l|l|l}
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$\text{Ber}(p)$ & $\E[X] = p$ & $\V[X] = p(1-p)$\\
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$\text{Poisson}(\lambda)$ & $\E[X] = \lambda$ & $\V[X] = \lambda$\\
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$\text{Bin}(n,p)$ & $\E[X] = n\cdot p$ & $\V[X] = np(1-p)$\\
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$\mathbb{I}_A$ & $\E[\mathbb{I}_A] = \P[A]$ \\
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$\exp(\lambda)$ & $\E[X] = \frac{1}{\lambda}$ \\
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$\mathcal{U}([a,b])$ & $\E[X] = \frac{a+b}{2}$
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\end{tabular}
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\end{center}
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\newpage
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\subsection{Stetiger Erwartungswert}
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\definition \textbf{Erwartungswert} (stetig)
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\definition \textbf{Erwartungswert} (stetig)\\
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\subtext{$X: \Omega \to \R,\quad f(x) \text{ Dichtefunktion}$}
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$$
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\E[X] = \int_{-\infty}^{\infty} x \cdot f(x)\ dx
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$$
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\subtext{$X: \Omega \to \R,\quad f(x) \text{ Dichtefunktion}$}
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\theorem \textbf{Linearität}
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\begin{align*}
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\text{(i)} &\quad \E[\lambda X] &=& \lambda \E[X] \\
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\text{(ii)} &\quad \E[X + Y] &=& \E[X] + \E[Y]
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\end{align*}
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\theorem \textbf{Eigenschaften}\\
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\subtext{$X,Y:\Omega\to\R,\quad\lambda\in\R$}
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\theorem \textbf{Monotonie}
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$$
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X \leq Y \implies \E[X] \leq \E[Y]
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$$
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\theorem \textbf{Multiplikation}\\
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\smalltext{$X,Y$ unabhängig}
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$$
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\E[X \cdot Y] = \E[X]\cdot\E[Y]
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$$
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\begin{align*}
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\text{(i)} &\quad \E[\lambda X] &=& \lambda \E[X] \\
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\text{(ii)} &\quad \E[X + Y] &=& \E[X] + \E[Y] \\
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\text{(iii)} &\quad X \leq Y &\implies& \E[X] \leq \E[Y] \\
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\text{(iv)} &\quad X \perp Y &\implies& \E[X\cdot Y] = \E[X]\cdot\E[Y]
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\end{align*}
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\theorem \textbf{Dichtefunktion bei Abbildungen}\\
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\smalltext{$\phi:\R\to\R$ stückweise stetig, beschränkt}
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@@ -79,36 +58,6 @@ $$
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$$
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\subtext{Auch verallgemeinert für $X_1,\ldots,X_n$, $\phi_1,\ldots,\phi_n$}
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\newpage
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\subsection{Ungleichungen}
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\theorem \textbf{Markov}\\
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\smalltext{$X \geq 0,\quad g: X(\Omega)\to[0,\infty)$ wachsend}
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$$
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\forall c \in \R \text{ s.d. } g(c)>0:\qquad \P[X \geq c] \leq \frac{\E[g(X)]}{g(c)}
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$$
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\theorem \textbf{Jensen}\\
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\smalltext{$\phi:\R\to\R$ konvex,$\quad \E[\phi(X)],\E[X]$ wohldefiniert}
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$$
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\phi\Bigl(\E[X]\Bigr)\leq\E\Bigl[\phi(X)\Bigr]
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$$
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\lemma \textbf{Dreiecksungleichung}\\
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\subtext{Jensen mit $\phi(X) = |X|$ und $\phi(X) = X^2$}
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$$
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\Bigl\vert\E[X]\Bigr\vert\leq\E\Bigl[\vert X\vert\Bigr] \qquad \E[|X|] \leq \sqrt{\E[X^2]}
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$$
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\theorem \textbf{Chebychev}\\
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\subtext{$Y$ s.d. $\V[Y] < \infty,\quad c>0$}
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$$
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\P\Bigl[ |Y-\E[Y]| \geq c \Bigr] \leq \frac{\V[Y]}{c^2}
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$$
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\theorem \textbf{Chernoff}
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% Slides, ende von ZGS
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\newpage
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\subsection{Varianz}
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@@ -22,8 +22,6 @@ $$
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{\footnotesize
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\remark Nicht umgekehrt: Aus den Randverteilungen lässt sich nichts über die gem. Vert. schliessen.
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}
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\newpage
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\subsection{Gemeinsame Stetige Verteilung}
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\definition \textbf{Gemeinsame Stetige Verteilung}\\
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@@ -33,6 +31,8 @@ $$
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$$
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\subtext{$f$ heisst \textit{gemeinsame Dichte} von $(X_1,\ldots,X_n)$}
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\newpage
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\theorem $\displaystyle\int_{-\infty}^\infty\cdots\int_{-\infty}^\infty f(x_1,\ldots,x_n)\ dx_n\ dx_1 = 1$\\
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\subtext{Umgekehrt existiert für jedes solches $f$ ein Raum $(\Omega, \F, \P)$}
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@@ -0,0 +1,26 @@
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\theorem \textbf{Markov}\\
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\smalltext{$X \geq 0,\quad g: X(\Omega)\to[0,\infty)$ wachsend}
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$$
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\forall c \in \R \text{ s.d. } g(c)>0:\qquad \P[X \geq c] \leq \frac{\E[g(X)]}{g(c)}
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$$
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\theorem \textbf{Jensen}\\
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\smalltext{$\phi:\R\to\R$ konvex,$\quad \E[\phi(X)],\E[X]$ wohldefiniert}
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$$
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\phi\Bigl(\E[X]\Bigr)\leq\E\Bigl[\phi(X)\Bigr]
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$$
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\lemma \textbf{Dreiecksungleichung}\\
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\subtext{Jensen mit $\phi(X) = |X|$ und $\phi(X) = X^2$}
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$$
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\Bigl\vert\E[X]\Bigr\vert\leq\E\Bigl[\vert X\vert\Bigr] \qquad \E[|X|] \leq \sqrt{\E[X^2]}
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$$
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\theorem \textbf{Chebychev}\\
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\subtext{$Y$ s.d. $\V[Y] < \infty,\quad c>0$}
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$$
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\P\Bigl[ |Y-\E[Y]| \geq c \Bigr] \leq \frac{\V[Y]}{c^2}
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$$
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\theorem \textbf{Chernoff}
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% Slides, ende von ZGS
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@@ -1,25 +0,0 @@
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\notation \textbf{Daten} $\{x_i\}_{i=1}^n$\\
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\subtext{Klein, in Abgrenzung zu Zufallsvariablen $X_i$}
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\notation \textbf{Parameter} $\vartheta \in \Theta$\\
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\subtext{Unbekannt, definiert den stoch. Prozess $\P_\vartheta$, dimension undefiniert}
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\definition \textbf{Schätzer}
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$$
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T_l = t_l\Bigl( X_1,\ldots,X_n \Bigr)
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$$
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\smalltext{$t_l$ heisst Schätzfunktion, eine Auswertung $T_l(\omega)$ heisst Schätzwert}
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\definition \textbf{Erwartungstreu}
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$$
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T \text{ Erwartungstreu } \iffdef \forall \vartheta \in \Theta:\quad \E_\vartheta\bigl[ T \bigr] = \vartheta
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$$
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\definition \textbf{Bias} $\quad\E_\vartheta[T] - \vartheta$
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\definition \textbf{MSE} $\quad\text{MSE}_\vartheta[T] = \E_\vartheta\bigl[ (T-\vartheta)^2 \bigr]$
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% def Schätzerfolgen / konsistent
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% bmk 7.7: zerlegung MSE
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@@ -0,0 +1,180 @@
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Wir betrachten Daten $\{x_i\}_{i=1}^n$ als Realisierung von unbekannten Zufallsvariablen $X_1(\omega),\ldots,X_n(\omega)$.
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Ziel ist es, die Verteilung von $X_1,\ldots,X_n$ zu finden durch Modellierung.
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Man wählt hierzu eine Famile $(\Omega, \mathcal F, \P_\vartheta)_{\vartheta\in\Theta}$ wobei $(\Omega, \mathcal F)$
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fix ist und $\P_\vartheta$ variabel z.B. $\P_\vartheta = \mathcal{N}(\vartheta_1, \vartheta_2)$ (Modell) und versucht ein optimales $\vartheta$ bzw. $\P_\vartheta$ zu finden.
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\textbf{Statistisches Vorgehen}
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\begin{enumerate}
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\item Datenbeschreibung, intuitives Verständnis von $\{x_i\}_{i=1}^n$
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\item Definition von $\Theta$ und $(\P_\vartheta)_{\vartheta\in\Theta}$
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\item Schätzer nutzen für $(x_1,\ldots,x_n)\mapsto\vartheta\in\Theta$
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\item Überprüfung von $\vartheta$ via statistischem Test
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\item Alternativ: Konfidenzbereich über $\Theta$ finden
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\end{enumerate}
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\subsection{Schätzer}
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\definition \textbf{Schätzer} \subtext{(Estimator)}
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$$
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T_l = t_l\Bigl( X_1,\ldots,X_n \Bigr) \mapsto \vartheta_l \in (\vartheta_1,\ldots,\vartheta_m)
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$$
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\smalltext{$t_l$ heisst \textit{Schätzfunktion}, eine Auswertung $T_l(\omega)$ heisst \textit{Schätzwert}.}\\
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\subtext{(Estimator function, sample estimate)}
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{\footnotesize
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\notation $T = (T_1,\ldots,T_m), \vartheta = (\vartheta_1,\ldots,\vartheta_m)$
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}
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\definition \textbf{Erwartungstreu} \subtext{(unbiased)}
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$$
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T \text{ Erwartungstreu } \iffdef \forall \vartheta \in \Theta:\quad \E_\vartheta\bigl[ T \bigr] = \vartheta
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$$
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\definition \textbf{Erwarteter Schätzfehler} $\quad\E_\vartheta[T] - \vartheta$ \subtext{(Bias)}
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\definition \textbf{MSE} $\quad\text{MSE}_\vartheta[T] = \E_\vartheta\bigl[ (T-\vartheta)^2 \bigr]$ \subtext{(Mean Sq. Error)}
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\definition \textbf{Konsistent}
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$$
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T^{(n)} \text{ konsistent für } \vartheta \iffdef \underset{n\to\infty}{\lim}\P_\vartheta \Bigl[ \vert T^{(n)}-\vartheta \vert > \epsilon \Bigr] = 0
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$$
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\subtext{Intuitiv: Bei beliebig hohem $n$ konvergiert $T^{(n)}$ zu $\vartheta$ unter $\P_\vartheta$}
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\lemma \textbf{Zerlegung des MSE}
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$$
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\text{MSE}_\vartheta[T] = \E_\vartheta\Bigl[ (T-\vartheta)^2 \Bigr] = \V_\vartheta[T] + \underbrace{\Bigl( \E_\vartheta[T]-\vartheta \Bigr)^2}_{=0 \text{ (erwartungstreu)}}
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$$
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{\footnotesize
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\remark $T$ erwartungstreu $\implies \text{MSE}_\vartheta[T]=\V_\vartheta[T]$
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}
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{\scriptsize
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\textbf{Beispiel}: $T = X_n \sim \text{Ber}(\vartheta)$ ist erwartungstreu, da $\E_\vartheta[T]=\vartheta$ ($T \sim \text{Ber}(\vartheta)$).
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Trotzdem ist $T^{(n)}$ nicht konsistent für $\vartheta \in (0,1)$, da $X_n(\omega)\in\{0,1\}$.
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}
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\subsection{Maximum Likelihood Methode (MLE)}
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\textbf{Motivation}: Wie findet man einen guten Schätzer $T$?
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Abhängig ob $(X_1,\ldots,X_n)$ stetig/diskret:
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\begin{itemize}
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\item Gemeinsame Gewichtsfunktion $p_X(x_1,\ldots,x_n;\vartheta)$
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\item Gemeinsame Dichtefunktion $f_X(x_1,\ldots,x_m;\vartheta)$
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\end{itemize}
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{\footnotesize
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\remark $p(x_1,\ldots,x_n;\vartheta) = \P_\vartheta[X_1=x_1,\ldots,X_n=x_n]$
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\remark Häufig ist $X_k$ i.i.d, d.h. $p_X(\ldots)=\prod_{k=1}^np_X(x_k;\vartheta)$
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}
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\definition \textbf{Likelihood Funktion}
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$$
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L(x_1,\ldots,x_n;\vartheta) = \begin{cases}
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p_X(x_1,\ldots,x_n;\vartheta), & \text{diskret} \\
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f_X(x_1,\ldots,x_n;\vartheta), & \text{stetig}
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\end{cases}
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$$
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\definition \textbf{Maximum Likelihood Schätzer}\\
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\smalltext{Maximierung von $\vartheta \mapsto L(X_1,\ldots,X_n;\vartheta)$}
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$$
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T_\text{ML} = t_\text{ML}(X_1,\ldots,X_n) \in \underset{\vartheta\in\Theta}{\text{arg max}}\Bigl( L(X_1,\ldots,X_n;\vartheta) \Bigr)
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$$
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\remark Äquivalent kann man $\log(L)$ maximieren.\\
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\subtext{Sinnvoll falls $X_k$ i.i.d. weil dann $\log(L) = \sum_{k=1}^{n}\log\Bigl(p_X(x_k;\vartheta)\Bigr)$}
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{\footnotesize
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\textbf{Anwendung}:
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\begin{enumerate}
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\item Gemeinsame Dichte/Verteilung finden
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\item $f(\vartheta) := \ln\Bigl( L(x_1,\ldots,x_n;\vartheta) \Bigr)$
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\item $\vartheta^*$ finden, s.d. $f'(\vartheta^*) = 0$
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\item Argumentieren, dass $\vartheta^*$ das Maximum ist (z.B. via $f''$)
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\end{enumerate}
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}
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\subsection{Momentenschätzer}
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\textbf{Annahme}: $\P_\vartheta$ s.d. $X_i,\ldots,X_n$ i.i.d.
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\definition \textbf{Momentenschätzer}
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$$
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T = (T_1, T_2) \text{ für } \Bigl(\E_\vartheta[X], \V_\vartheta[X] \Bigr)
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$$
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$$
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T_1 = \frac{1}{n}\sum_{i=1}^n X_i \qquad T_2 = \frac{1}{n}\sum_{i=1}^n X_i^2 - \Biggl(\underbrace{\frac{1}{n}\sum_{i=1}^n X_i}_{T_1}\Biggr)^2
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$$
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{\footnotesize
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\remark Allgemein nicht erwartungstreu
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}
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\definition \textbf{Empirische Varianz} \subtext{(Unbiased Sample Variance)}
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$$
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T_1' = T_1 \qquad T_2' = \frac{n}{n-1}T_2
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$$
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{\footnotesize
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\remark Erwartungstreu
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}
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\subsection{Verteilungen von Schätzern}
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\textbf{Motivation}: Die Verteilung eines Schätzers $T$.\\
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\subtext{Es gibt wenig exakte Aussagen, daher approximativ via ZGS}
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{\scriptsize
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\textbf{Beispiel}: $T = \frac{1}{n}\sum_{i=1}^n X_i$\\
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ZGS $\implies T \sim \mathcal{N}\Bigl(n\E_\vartheta[X_i],n\V_\vartheta[X_i]\Bigr)$ (für grosse $n$)
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}
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\definition $\chi^2$\textbf{-Verteilung}\\
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\smalltext{Parameter $m$: \textit{Freiheitsgrade}}
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$$
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X \sim \chi^2_m \iffdef f_X(x) = \frac{1}{2^{\frac{m}{2}}\Gamma(\frac{m}{2})}x^{\frac{m}{2}-1}e^{-\frac{x}{2}} \quad (x \geq 0)
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$$
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\definition \textbf{Euler'sche Gammafunktion}
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$$
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\Gamma(x) = \int_0^\infty t^{x-1}e^{-t}\text{d}t \qquad \forall n \in \N_0: \Gamma(n+1) = n!
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$$
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\remark \textbf{Entstehung der } $\chi^2$\textbf{-Verteilung}\\
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\smalltext{$X_1,\ldots,X_m$ i.i.d. s.d. $X_i \sim \mathcal{N}(0,1)$}
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$$
|
||||
\sum_{i=1}^{m} X_i^2 \sim \chi_m^2
|
||||
$$
|
||||
|
||||
\newpage
|
||||
|
||||
\definition \textbf{Student'sche} $t$-\textbf{Verteilung}
|
||||
$$
|
||||
X \sim t_m \iffdef f_X(x) = \frac{\Gamma(\frac{m+1}{2})}{\sqrt{m\pi}\Gamma(\frac{m}{2})}\Biggl( 1+\frac{x^2}{m} \Biggr)^{-\frac{m+1}{2}}
|
||||
$$
|
||||
|
||||
\remark \textbf{Entstehung der } $t$-\textbf{Verteilung}\\
|
||||
\smalltext{$X\sim\mathcal{N}(0,1)$ und $Y \sim \chi^2_m$}
|
||||
$$
|
||||
\frac{X}{\sqrt{\frac{1}{m}Y}} \sim t_m
|
||||
$$
|
||||
|
||||
{\footnotesize
|
||||
\remark $m=1 \implies t_1$ ist Cauchy Verteilung
|
||||
|
||||
\remark $\underset{m\to\infty}{\lim} t_m = \mathcal{N}(0,1)$
|
||||
|
||||
\remark $t_m$ ist symmetrisch um $0$ (wie $\mathcal{N}$)
|
||||
}
|
||||
|
||||
\theorem \textbf{Aussagen für Normalverteilungen}\\
|
||||
\smalltext{$X_1,\ldots,X_n \sim \mathcal{N}(\mu, \sigma^2)$}
|
||||
$$
|
||||
\bar{X}_n = \sum_{i=1}^{n}X_i,\qquad S^2 = \frac{1}{n-1}\sum_{n}^{i=1}\Bigl( X_i - \bar{X}_n \Bigr)^2
|
||||
$$
|
||||
\begin{enumerate}
|
||||
\item $\bar{X}_n \sim \mathcal{N}\bigl(\mu, \frac{\sigma^2}{n}\bigr)$ also $\frac{\sigma}{\sqrt{n}}(\bar{X}_n - \mu) \sim \mathcal{N}(0,1)$
|
||||
\item $\frac{n-1}{\sigma^2}S^2 = \frac{1}{\sigma^2}\sum_{i=1}^n\Bigl( X_i - \bar{X}_n \Bigr)^2 \sim \chi^2_{n-1}$
|
||||
\item $\bar{X}_n \perp S^2$
|
||||
\item $\frac{\bar{X}_n-\mu}{\frac{S}{\sqrt{n}}} = \displaystyle\frac{\frac{\bar{X}_n - \mu}{\sigma / \sqrt{n}}}{S / \sigma} = \frac{\frac{\bar{X}_n - \mu}{\sigma / \sqrt{n}}}{\sqrt{\frac{1}{n-1}\frac{n-1}{\sigma^2}S^2}} \sim t_{n-1}$
|
||||
\end{enumerate}
|
||||
@@ -0,0 +1,179 @@
|
||||
\subsection{Integrale \& Ableitungen}
|
||||
|
||||
{\small
|
||||
\begin{center}
|
||||
\begin{tabular}{c||c||c}
|
||||
$F(x)$ & $f(x)$ & $f'(x)$ \\
|
||||
\hline
|
||||
\hline
|
||||
$x$ & $c$ & $0$\\
|
||||
$\frac{1}{a+1}\cdot x^{a+1}$ & $x^a$ & $a\cdot x^{a-1}$\\
|
||||
$\frac{1}{a\cdot(n+1)}\cdot(ax+b)^{n+1}$ & $(ax+b)^n$ & $n \cdot(ax+b)^{n-1}\cdot a$\\
|
||||
$\ln|x|$ & $\frac{1}{x}=x^{-1}$ & $-\frac{1}{x^2} = -x^{-2}$\\
|
||||
$\frac{2}{3}\cdot x^{\frac{3}{2}}$ & $\sqrt{x} = x^{\frac{1}{2}}$ & $\frac{1}{2\sqrt{x}} = \frac{1}{2}\cdot x^{-\frac{1}{2}}$\\
|
||||
$\frac{n}{n+1}\cdot x^{\frac{1}{n}+1}$ & $\sqrt[\leftroot{0} n]{x} = x^\frac{1}{n}$ & $\frac{1}{n}\cdot x^{\frac{1}{n}-1}$\\
|
||||
\hline
|
||||
$e^x$ & $e^x$ & $e^x$\\
|
||||
$\frac{1}{\ln(a)}\cdot a^x$ & $a^x$ & $a^x\cdot \ln(a)$\\
|
||||
$x\cdot (\ln|x|-1)$ & $\ln|x|$ & $\frac{1}{x} = x^{-1}$\\
|
||||
$\frac{x}{\ln(a)}\cdot(\ln|x|-1)$ & $\log_a(x)$ & $\frac{1}{x\cdot\ln(a)}$\\
|
||||
\hline
|
||||
$-\cos(x)$ & $\sin(x)$ & $\cos(x)$\\
|
||||
$\sin(x)$ & $\cos(x)$ & $-\sin(x)$\\
|
||||
$-\ln|\cos(x)|$ & $\tan(x)$ & $\frac{1}{\cos(x)^2}= 1 + \tan(x)^2$\\
|
||||
$\ln|\sin(x)|$ & $\cot(x)$ & $-\frac{1}{\sin(x)^2}$\\
|
||||
$x \cdot\arcsin(x) + \sqrt{1 - x^2}$ & $\arcsin(x)$ & $\frac{1}{\sqrt{1-x^2}}$\\
|
||||
$x \cdot \arccos(x)-\sqrt{1-x^2}$ & $\arccos(x)$ & $-\frac{1}{\sqrt{1-x^2}}$\\
|
||||
$x \cdot \arctan(x)-\frac{\ln(x^2+1)}{2}$ & $\arctan(x)$ & $\frac{1}{x^2+1}$\\
|
||||
\hline
|
||||
$\sinh(x)$ & $\cosh(x)$ & $\sinh(x)$\\
|
||||
$\cosh(x)$ & $\sinh(x)$ & $\cosh(x)$\\
|
||||
$\ln|\cosh(x)|$ & $\tanh(x)$ & $\frac{1}{\cosh(x)^2} = 1-\tanh(x)^2$\\
|
||||
& $\text{arcsinh}(x)$ & $\frac{1}{\sqrt{x^2+1}}$\\
|
||||
& $\text{arccosh}(x)$ & $\frac{1}{\sqrt{x^2}-1}$\\
|
||||
& $\text{arctanh}()$ & $\frac{1}{1-x^2}$\\
|
||||
\hline
|
||||
\end{tabular}
|
||||
\end{center}
|
||||
}
|
||||
|
||||
\newpage
|
||||
|
||||
\subsection{Weitere Integrale \& Ableitungen}
|
||||
\begin{center}
|
||||
\begin{tabular}{c||c}
|
||||
|
||||
$F(x)$ & $f(x)$\\
|
||||
\hline
|
||||
\hline
|
||||
$\frac{1}{a}\ln|ax+b|$ & $\frac{1}{ax+b}$ \\
|
||||
$\frac{ax}{c}-\frac{ad-bc}{c^2}\ln|cx+d|$ & $\frac{a(cx+d) - c(ax+b)}{(cx+d)^2}$\\
|
||||
$\frac{x}{2}f(x)+\frac{a^2}{2}\ln|x+f(x)|$ & $\sqrt{a^2+x^2}$ \\
|
||||
$\frac{x}{2}f(x)-\frac{a^2}{2}\arcsin(\frac{x}{|a|})$ & $\sqrt{a^2-x^2}$ \\
|
||||
$\frac{x}{2}f(x) - \frac{a^2}{2}\ln|x + f(x)|$ & $\sqrt{x^2-a^2}$\\
|
||||
$\ln(x + \sqrt{x^2 \pm a^2})$ & $\frac{1}{\sqrt{x^2 \pm a^2}}$\\
|
||||
$\arcsin(\frac{x}{|a|})$ & $\frac{1}{\sqrt{a^2-x^2}}$ \\
|
||||
$\frac{1}{a} \cdot\arctan(\frac{x}{a})$ & $\frac{1}{x^2+a^2}$\\
|
||||
\hline
|
||||
$-\frac{1}{a}\cos(ax+b)$ & $\sin(ax+b)$ \\
|
||||
$\frac{1}{a}\sin(ax+b)$ & $\cos(ax+b)$ \\
|
||||
\hline
|
||||
$x^x$ & $x^x \cdot (1 + \ln|x|)$\\
|
||||
$(x^x)^x$ & $(x^x)^x(x+2x\ln|x|)$\\
|
||||
$x^{(x^x)}$ & $x^{(x^x)}(x^{x-1}+\ln|x|\cdot x^x(1+\ln|x|)$\\
|
||||
\hline
|
||||
$\frac{1}{2}(x-\frac{1}{2}\sin(2x))$ & $\sin(x)^2$\\
|
||||
$\frac{1}{2}(x+\frac{1}{2}\sin(2x))$ & $\cos(x)^2$\\
|
||||
\end{tabular}
|
||||
\end{center}
|
||||
|
||||
%\subsection{Tyler, The Creator}
|
||||
% \begin{center}
|
||||
% \includegraphics[width=0.3\linewidth]{tyler2.png}
|
||||
% \end{center}
|
||||
|
||||
\newpage
|
||||
\subsection{Grenzwerte: Folgen}
|
||||
\subtext{Credits: D. Camenisch}
|
||||
\begin{center}
|
||||
\begin{tabular}{ l || l }
|
||||
$\lim_{x\to\infty} \frac{1}{x} = 0$ & $\lim_{x\to\infty} 1 + \frac{1}{x} = 1$ \\
|
||||
\hline
|
||||
$\lim_{x \to \infty} e^x = \infty$ & $\lim_{x \to - \infty} e^x = 0$ \\
|
||||
\hline
|
||||
$\lim_{x \to \infty} e^{-x} = 0$ & $\lim_{x \to - \infty} e^{-x} = \infty$ \\
|
||||
\hline
|
||||
$\lim_{x \to \infty} \frac{e^x}{x^m} = \infty$ & $\lim_{x \to - \infty} xe^x = 0$ \\
|
||||
\hline
|
||||
$\lim_{x \to \infty} \ln(x) = \infty$ & $\lim_{x \to 0} \ln(x) = - \infty$ \\
|
||||
\hline
|
||||
$\lim_{x \to \infty} (1 + x)^{\frac{1}{x}} = 1$ & $\lim_{x \to 0} (1 + x)^{\frac{1}{x}} = e$ \\
|
||||
\hline
|
||||
$\lim_{x\to\infty} \left(1 + \frac{1}{x}\right)^b = 1$ & $\lim_{x\to\infty} \left(1 + \frac{1}{x}\right)^b = 1$ \\
|
||||
\hline
|
||||
$\lim_{x\to\infty} x^aq^x = 0, \; \forall 0 \leq q < 1$ & $\lim_{x\to\infty} n^\frac{1}{n} = 1$ \\
|
||||
\hline
|
||||
$\lim_{x\to \pm \infty} \left(1+\frac{1}{x}\right)^x = \mathrm{e}$ & $\lim_{x\to \infty} \left(1-\frac{1}{x}\right)^x = \frac{1}{\mathrm{e}}$ \\
|
||||
\hline
|
||||
$\lim_{x \to \pm \infty} \left(1+\frac{k}{x}\right)^{mx} = e^{km}$ & $\lim_{x \to 0} \frac{\sin{x}}{x} = 1$ \\
|
||||
\hline
|
||||
$\lim_{x\to 0} \frac{1}{\cos(x)} = 1$ & $\lim_{x\to 0} \frac{\cos{x}-1}{x} = 0$ \\
|
||||
\hline
|
||||
$\lim_{x\to 0} \frac{\log{1-x}}{x} = -1$ & $\lim_{x\to 0} x\log{x} = 0$ \\
|
||||
\hline
|
||||
$\lim_{x\to 0} \frac{1-\cos{x}}{x^2} = \frac{1}{2}$ & $\lim_{x\to 0} \frac{\mathrm{e}^x-1}{x} = 1$ \\
|
||||
\hline
|
||||
$\lim_{x\to 0} \frac{x}{\arctan{x}} = 1$ & $\lim_{x\to\infty} \arctan{x} = \frac{\pi}{2}$ \\
|
||||
\hline
|
||||
$\lim_{x \to \infty} \left(\frac{x}{x + k}\right)^x = e^{-k}$ & $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$ \\
|
||||
\hline
|
||||
$\lim_{x \to 0} \frac{a^x - 1}{x} = \ln(a) \; \forall a > 0$ & $\lim_{x \to 0} \frac{e^{ax} - 1}{x} = a$ \\
|
||||
\hline
|
||||
$\lim_{x \to 0} \frac{\ln(x + 1)}{x} = 1$ & $\lim_{x \to 1} \frac{\ln(x)}{x - 1} = 1$ \\
|
||||
\hline
|
||||
$\lim_{x \to \infty} \frac{\ln(x)}{x} = 0$ & $\lim_{x \to \infty} \frac{\log(x)}{x^a} = 0$ \\
|
||||
\hline
|
||||
$\lim_{x \to \infty} \sqrt[x]{x} = 1$ & $\lim_{x \to \infty} \frac{2x}{2^x} = 0$ \\
|
||||
\hline
|
||||
$\lim_{x \to \frac{\pi^-}{2}} \tan{x} = +\infty$ & $\lim_{x \to \frac{\pi^+}{2}} \tan{x} = -\infty$ \\
|
||||
\hline
|
||||
$\lim_{x \to \infty} \frac{\sin{x}}{x} = 0$ & $\lim_{x \to 0^+} x\ln{x} = 0$\\
|
||||
\hline
|
||||
\end{tabular}
|
||||
\end{center}
|
||||
\subsection{Grenzwerte: Reihen}
|
||||
\subtext{Credits: D. Camenisch}
|
||||
\begin{center}
|
||||
\begin{tabular}{ l || l }
|
||||
$\sum_{i = 1}^{n} i = \frac{n(n+1)}{2}$ & $\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$ \\
|
||||
\hline
|
||||
$\sum_{i=1}^{n} i^3 = \frac{n^2(n+1)^2}{4}$ & $\sum_{i = 1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ \\
|
||||
\hline
|
||||
$\sum_{i = 1}^{\infty} \frac{1}{n(n+1)} = 1$ & $\sum_{i = 1}^\infty z^i = \frac{1 - z^{i + 1}}{1 - z}$ \\
|
||||
\hline
|
||||
\end{tabular}
|
||||
\end{center}
|
||||
|
||||
\newpage
|
||||
\subsection{Verteilungen}
|
||||
\subtext{Credits: N. Wehrli \& S. Metzker}
|
||||
|
||||
\renewcommand*{\arraystretch}{2}
|
||||
\begin{center}
|
||||
|
||||
\begin{tabularx}{\textwidth}{|l|l|l|X|X|X|X|}
|
||||
\hline
|
||||
Verteilung & Notation & Parameter & \( \E[X] \) & \( \Var(X) \) & \( p_X(t)/f_X(t) \) & \( F_X(t) \) \\
|
||||
\hline
|
||||
\hline
|
||||
Gleichverteilung & unbekannt & \makecell[l]{\( n \): Anzahl Ereignisse \\ (\( x_i \): Ereignisse)} & \( \frac{1}{n} \sum_{i=1}^{n} x_i \) & \( \frac{1}{n} \sum_{i=1}^{n} x_i^2 - \frac{1}{n^2} \left(\sum_{i=1}^{n} x_i \right)^2 \) & \( \frac{1}{n} \) & \( \frac{|\{k:x_k \leq t\}|}{n} \) \\
|
||||
\hline
|
||||
Bernoulli & $\text{Ber}(p)$ & \( p: \) ErfolgsW'keit & \( p \) & \( p \cdot (1-p) \) & \( p^t(1-p)^{1-t} \) & \( 1-p \) für \( 0 \leq t < 1 \) \\
|
||||
\hline
|
||||
Binomial & $\text{Bin}(n,p)$ & \makecell[l] {\( n \): Anzahl Versuche \\ \( p: \) ErfolgsW'keit } & \( np \) & \( np(1-p) \) & \( \binom{n}{t}p^t(1-p)^{n-t} \) & \( \sum_{k=0}^{t} \binom{n}{k} p^k(1-p)^{n-k} \) \\
|
||||
\hline
|
||||
Geometrisch & $\text{Geo}(p)$ & \makecell[l] { \( p \): ErfolgsW'keit \\ (\( t: \) Anzahl Versuche)} & \( \frac{1}{p} \) & \( \frac{1-p}{p^2} \) & \( p(1-p)^{t-1} \) & \( 1-(1-p)^t\) \\
|
||||
\hline
|
||||
Poisson& $\text{Poisson}(\lambda)$ & \makecell[l]{ \( \lambda \): Erwartungswert \\ und Varianz} & \( \lambda \) & \( \lambda \) & \( \frac{\lambda^t}{t!}e^{-\lambda} \) & \( e^{-\lambda} \sum_{k=0}^{t} \frac{\lambda^{k}}{k!} \) \\
|
||||
\hline
|
||||
\makecell[l]{Gleichverteilung\\(im Intervall)} & $U \sim \mathcal{U}([0,1])$ & \( [a,b] \): Intervall & \( \frac{a+b}{2} \) & \( \frac{1}{12}(b-a)^2 \) & \(\begin{cases} \frac{1}{b-a} &a \le x \le b \\ 0 & \text{sonst}\end{cases}\) & \(\begin{cases} 0 & x\le a \\ \frac{t-a}{b-a} & a < x < b \\ 1 & x \ge b \end{cases}\) \\
|
||||
\hline
|
||||
Exponentialv. & $ \text{Exp}(\lambda)$ & \( \lambda: \frac{1}{\E[X]} \) & \( \frac{1}{\lambda} \) & \( \frac{1}{\lambda^2} \) & \( \begin{cases} \lambda e^{-\lambda t} & t \geq 0 \\ 0 & t < 0 \end{cases} \) & \( \begin{cases} 1-e^{-\lambda t} & t>0 \\ 0 & t \leq 0\end{cases}\) \\
|
||||
\hline
|
||||
Normalverteilung & $\mathcal{N}\left(\mu, \sigma^2\right)$ & \makecell[l]{\( \mu: \E[X] \) \\ \( \sigma^2 \): Varianz} & \( \mu \) & \( \sigma ^2 \) & \( \frac{1}{\sqrt{2\pi \sigma^2} }e^{-{\frac{(t-\mu)^2}{2\sigma^2} }} \) & \( \frac{1}{\sigma {\sqrt{2\pi}}} \int_{-\infty}^t e^{-\frac{1}{2}\left( \frac{y-\mu}{\sigma} \right) ^2} \mathrm{d} y \) \\
|
||||
\hline
|
||||
\( \chi ^2 \)-Verteilung & $\chi_{m}^{2}$ & \( n \): Freiheitsgrad & \( n \) & \( 2n \) & \( \frac{1}{2^{\frac{n}{2}}\Gamma (\frac{n}{2})} t^{\frac{n}{2}-1} e^{-\frac{t}{2}} \text{ für } t>0\) & \(P\left( \frac{n}{2}, \frac{t}{2}\right) \) \\
|
||||
\hline
|
||||
t-Verteilung & $t_{m}$ & \( n \): Freiheitsgrad & \( \begin{cases} 0 & n>1 \\ \text{undef.} & \text{sonst} \end{cases} \) & \( \begin{cases} \frac{n}{n-2} & n> 2 \\ \infty & 1<n \leq 2 \\ \text{undef.} & \text{sonst} \end{cases} \) & \( \frac{\Gamma \left( \frac{n+1}{2} \right) }{\sqrt{n\pi } \cdot \Gamma (\frac{n}{2})} \left( 1+ \frac{t^2}{n} \right) ^{- \frac{n+1}{2}} \) & zu kompliziert \\
|
||||
\hline
|
||||
Negativbinomial & \(\operatorname{NBin}(r, p)\) & \(r \in \mathbb{N}\), \(p \in [0,1]\) & $\frac{r}{p}$ & $\frac{r(1-p)}{p^2}$ & $\binom{k-1}{r-1} p^r (1-p)^{k-r}$ & zu kompliziert \\
|
||||
\hline
|
||||
|
||||
Cauchy-Verteilung & $\operatorname{Cauchy}\left(x_0, \gamma\right)$ & \(x_0 \in \mathbb{R}\), \(\gamma > 0\) & Existiert nicht & Existiert nicht & \(\frac{1}{\pi} \frac{\gamma}{\gamma^2 + (x-x_0)^2}\) & \(\frac{1}{2} + \frac{1}{\pi} \arctan\left(\frac{x - x_0}{\gamma}\right)\) \\
|
||||
\hline
|
||||
|
||||
Hypergeometrisch & \(\mathrm{H}(n, r, m)\) & \(n \in \mathbb{N}\), \(m, r \in \{1, \ldots, n\}\) & $m\frac{r}{n}$ & $m \frac{r}{n}\left(1-\frac{r}{n}\right) \frac{n-m}{n-1}$ & $\frac{\binom{r}{k} \binom{n-r}{m-k}}{\binom{n}{m}}$ & $\sum_{y=0}^k \frac{\binom{r}{y} \binom{n-r}{m-y}}{\binom{n}{m}}$ \\
|
||||
\hline
|
||||
\end{tabularx}
|
||||
|
||||
\end{center}
|
||||
@@ -62,6 +62,7 @@
|
||||
\def \E{\mathbb{E}}
|
||||
\def \I{\mathbb{I}}
|
||||
\def \V{\mathbb{V}}
|
||||
\def \Var{\mathbb{V}}
|
||||
|
||||
% Titles
|
||||
\def \definition{\colorbox{lightgray}{Def} }
|
||||
|
||||
@@ -38,4 +38,7 @@
|
||||
|
||||
% Flexible graphs / visualisations inside latex
|
||||
\usepackage{tikz}
|
||||
\usetikzlibrary{positioning, arrows.meta, calc, matrix}
|
||||
\usetikzlibrary{positioning, arrows.meta, calc, matrix}
|
||||
|
||||
\usepackage{makecell}
|
||||
\usepackage{tabularx}
|
||||
Reference in New Issue
Block a user