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@@ -230,6 +230,8 @@ $$
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{\scriptsize
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{\scriptsize
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\remark $f_X$ (Stetig) ist Intuitiv das Analogon zu $p_x$ (Diskret)
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\remark $f_X$ (Stetig) ist Intuitiv das Analogon zu $p_x$ (Diskret)
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\remark Für stetige $X$ gilt $\forall x:\ \P[X=x] = 0$, da $W_X = \infty$.
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}
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}
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\definition \textbf{Gleichverteilung} $X \sim \mathcal{U}([a,b])$\\
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\definition \textbf{Gleichverteilung} $X \sim \mathcal{U}([a,b])$\\
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@@ -300,5 +302,5 @@ $$
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\lemma \textbf{Standardisierung}\\
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\lemma \textbf{Standardisierung}\\
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\smalltext{$Z \sim \mathcal{N}(m, \rho^2),\quad x\in\R$}
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\smalltext{$Z \sim \mathcal{N}(m, \rho^2),\quad x\in\R$}
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$$
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$$
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\P[x\leq Z] = \Phi\Biggl( \frac{x - m}{\rho} \Biggr)
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\P[x\leq Z] = \Phi\Biggl(\frac{x - \E[Z]}{\sqrt{\V[Z]}}\Biggr) = \Phi\Biggl( \frac{x - m}{\rho} \Biggr)
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$$
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$$
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@@ -87,3 +87,7 @@ $$
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% It's important to him to know \E[X]\E[Y] = \E[X\cdot Y] isn't enough for indep., requires \E[\phi(X)]\E[\psi(X)] instead \forall \phi,\psi
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% It's important to him to know \E[X]\E[Y] = \E[X\cdot Y] isn't enough for indep., requires \E[\phi(X)]\E[\psi(X)] instead \forall \phi,\psi
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\notation \textbf{Bedingte gem. Verteilung}
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$$
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f_{X \vert Y} := \frac{f_{X,Y}(x,y)}{f_Y(y)}
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$$
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@@ -107,18 +107,15 @@ $$
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T = (T_1, T_2) \text{ für } \Bigl(\E_\vartheta[X], \V_\vartheta[X] \Bigr)
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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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$$
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$$
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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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T_1 = \frac{1}{n}\sum_{i=1}^n X_i \qquad T_2 = \frac{1}{n}\sum_{i=1}^n \Bigl( X_i - T_1 \Bigr)^2
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$$
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$$
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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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\definition \textbf{Korr. Empirische Varianz} \subtext{(Unbiased Sample Variance)}
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$$
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$$
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T_1' = T_1 \qquad T_2' = \frac{n}{n-1}T_2
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T_1' = T_1 \qquad T_2' = \frac{1}{n-1}\sum_{i=1}^n \Bigl( X_i - T_1 \Bigr)^2
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$$
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$$
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{\footnotesize
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{\footnotesize
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\remark Erwartungstreu
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\remark $T'$ ist erwartungstreu, $T$ jedoch nicht.
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}
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}
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\subsection{Verteilungen von Schätzern}
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\subsection{Verteilungen von Schätzern}
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@@ -174,7 +171,7 @@ $$
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\bar{X}_n = \sum_{i=1}^{n}X_i,\qquad S^2 = \frac{1}{n-1}\sum_{n}^{i=1}\Bigl( X_i - \bar{X}_n \Bigr)^2
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\bar{X}_n = \sum_{i=1}^{n}X_i,\qquad S^2 = \frac{1}{n-1}\sum_{n}^{i=1}\Bigl( X_i - \bar{X}_n \Bigr)^2
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$$
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$$
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\begin{enumerate}
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\begin{enumerate}
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\item $\bar{X}_n \sim \mathcal{N}\bigl(\mu, \frac{\sigma^2}{n}\bigr)$ also $\frac{\sigma}{\sqrt{n}}(\bar{X}_n - \mu) \sim \mathcal{N}(0,1)$
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\item $\bar{X}_n \sim \mathcal{N}\bigl(\mu, \frac{\sigma^2}{n}\bigr) \implies \displaystyle\frac{\bar{X}_n - \mu}{\frac{\sigma}{\sqrt{n}}} \sim \mathcal{N}(0,1)$
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\item $\frac{n-1}{\sigma^2}S^2 = \frac{1}{\sigma^2}\sum_{i=1}^n\Bigl( X_i - \bar{X}_n \Bigr)^2 \sim \chi^2_{n-1}$
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\item $\frac{n-1}{\sigma^2}S^2 = \frac{1}{\sigma^2}\sum_{i=1}^n\Bigl( X_i - \bar{X}_n \Bigr)^2 \sim \chi^2_{n-1}$
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\item $\bar{X}_n \perp S^2$
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\item $\bar{X}_n \perp S^2$
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\item $\frac{\bar{X}_n-\mu}{\frac{S}{\sqrt{n}}} = \displaystyle\frac{\frac{\bar{X}_n - \mu}{\sigma / \sqrt{n}}}{S / \sigma} = \frac{\frac{\bar{X}_n - \mu}{\sigma / \sqrt{n}}}{\sqrt{\frac{1}{n-1}\frac{n-1}{\sigma^2}S^2}} \sim t_{n-1}$
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\item $\frac{\bar{X}_n-\mu}{\frac{S}{\sqrt{n}}} = \displaystyle\frac{\frac{\bar{X}_n - \mu}{\sigma / \sqrt{n}}}{S / \sigma} = \frac{\frac{\bar{X}_n - \mu}{\sigma / \sqrt{n}}}{\sqrt{\frac{1}{n-1}\frac{n-1}{\sigma^2}S^2}} \sim t_{n-1}$
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@@ -214,9 +211,12 @@ $$
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\subsubsection{Statistisches Vorgehen}
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\subsubsection{Statistisches Vorgehen}
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\definition \textbf{Signifikanzniveau} $\alpha$ s.d. $\underset{\vartheta\in\Theta_0}{\sup}\P_\vartheta[T\in K] \leq \alpha$
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\definition \textbf{Signifikanzniveau} $\alpha$ s.d. $\underset{\vartheta\in\Theta_0}{\sup}\P_\vartheta[T\in K] \leq \alpha$\\
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\subtext{Intuitiv: Chance, dass $\Theta_0$ ($H_0$) in $K$ liegt, ist kleiner als $\alpha$.}
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\definition \textbf{Macht} $\beta: \Theta_A \to [0,1] \quad \vartheta \mapsto \beta(\vartheta) = \P_\vartheta[T \in K]$
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\definition \textbf{Macht} $\beta: \Theta_A \to [0,1] \quad \vartheta \mapsto \beta(\vartheta) = \P_\vartheta[T \in K]$\\
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\subtext{Intuitiv: Chance, dass $\Theta_A$ ($H_A$) im Bereich $K$ liegt.}
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\method \textbf{Asymmetrisches Verfahren}:
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\method \textbf{Asymmetrisches Verfahren}:
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\begin{enumerate}
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\begin{enumerate}
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