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[PS] More fixes
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@@ -3,11 +3,11 @@
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\shortremark $\E[\cX]$ immer definiert und endlich oder unendlich
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\shorttheorem $\cX$ n.-neg. Dann: $\E[\cX] \geq 0$. $=$, wenn $\cX = 0$ fast sicher
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\shorttheorem $\cX$ n.neg. Dann: $\E[\cX] \geq 0$. $=$, wenn $\cX = 0$ fast sicher
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\shortdefinition $\E[\cX] = \E[\cX_+] - \E[\cX_-]$ mit $\cX_-$ auch n.-neg.
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\shortremark $|\cX| = \cX_+ + \cX_-$. Für $\cX \geq 0$ ist $\E[\cX]$ immer definiert.
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\shortremark $|\cX| = \cX_+ + \cX_-$; $\cX = \cX_+ - \cX_-$. Für $\cX \geq 0$ ist $\E[\cX]$ immer definiert.
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Falls $\cX$ kein konst. Vorzeichen, $\E[\cX]$ undef.
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\shortremark $\E[\cX] = \int_{0}^{\8} (1 - F_\cX(x)) \dx x - \int_{-\8}^{0} F_\cX(x)$
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@@ -8,7 +8,7 @@
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\subsubsection{Beispiele}
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\begin{itemize}
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\item $\cX \sim \text{Ber}(p)$: $\E[\cX] = p$
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\item $\cX \sim \text{Ber}(p)$: $\E[\cX] = p$ ($\E[1_A] = \P[A]$)
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\item $\cX \sim \text{Bin}(n, p)$: $\E[\cX] = np$
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\item $\cX \sim \text{Poisson}(\lambda)$: $\E[\cX] = \lambda$
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\end{itemize}
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@@ -1,12 +1,13 @@
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\subsection{Stetige Zufallsvariablen}
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\shortdefinition $\cX$ stetig $\E[\cX] = \int_{-\8}^{\8} x f_\cX(x) \dx x$
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\shorttheorem $\E[\varphi(\cX)] = \int_{-\8}^{\8} \varphi(x) f_\cX(x) \dx x$, falls int. wohldefiniert
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\shorttheorem $\E[\varphi(\cX)] = \int_{-\8}^{\8} \varphi(x) f_\cX(x) \dx x$, falls int. wohldef.
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\shortlemma[Int über gauss. Glockenk.] $\int_{-\8}^{\8} e^{\frac{-x^2}{2\sigma^2}} \dx x = \sqrt{2 \pi \sigma^2}$
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\subsubsection{Beispiele}
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% TODO: Consider if need derivation of them here and prev section as well
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% TODO: Also add the ones proven in exercises
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\shortlemma[Int über gauss. Glockenk.] $\int_{-\8}^{\8} e^{\frac{-x^2}{2\sigma^2}} \dx x = \sqrt{2 \pi \sigma^2}$
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\begin{itemize}
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\item $\cX \sim \cU([a, b])$, $a < b$: $\E[\cX] = \frac{a + b}{2}$
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\item $\cX \sim \text{Exp}(\lambda)$, $\lambda > 0$: $\E[\cX] = \frac{1}{\lambda}$
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@@ -14,7 +14,5 @@ $\E\left[ \prod_{k = 1}^n \cX_k \right] = \prod_{k = 1}^n \E[\cX_k]$
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\shorttheorem $f : \R \rightarrow \R_+$ mit $\int_{-\8}^{\8} f(x) \dx x = 1$. Dann ist äquivalent:
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\bi{(1)} $\cX$ stetig mit Dichte $f$ und \bi{(2)} für jede stückweise stetige Abb. $\varphi : \R \rightarrow \R$ gilt $\E[\varphi(\cX)] = \int_{-\8}^{\8} \varphi(x) f(x) \dx x$
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\shorttheorem äquivalent: \bi{1} $\cX, \cY$ unabhängig, für alle $\varphi, \psi : \R \rightarrow \R$: $\E[\varphi(\cX) \psi(\cX)] = \E[\varphi(\cX)] \E[\psi]$
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\shorttheorem äquivalent: \bi{(1)} $\cX_i$ unabhängig,\\
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\bi{(2)} $\forall \varphi_i$: $\E[\varphi_1(\cX_1) \cdots \varphi_n(\cX_n)] = \E[\varphi_1(\cX_1)] \cdots \E[\varphi_n(\cX_n)]$
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@@ -4,3 +4,5 @@
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\shorttheorem[Jensensche] $\varphi: \R \rightarrow \R$ konvex, und falls $\E[\varphi(\cX)]$ und $\E[\cX]$ wohldefiniert: $\varphi(\E[\cX]) \leq \E[\varphi(\cX)]$
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\shorttheorem[Dreieck] $\varphi(x) = |x|$, dann $|\E[\cX]| \leq \E[|\cX|]$. $\varphi(x) = x^2$, dann $\E[|\cX|] \leq \sqrt{\E[\cX^2]}$
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\shortcorollary[Cheb.] $\V[\cY]$ end. $\forall c > 0$ gilt: $\P[|\cY - \E[\cY]| \geq c] \leq \frac{\V[\cY]}{c^2}$
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@@ -1,4 +1,6 @@
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\subsection{Varianz}
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$\E[\cX^2] = \sum_{x \in W} x^2 \cdot p_\cX(x)$ ($\cX$ diskret)
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\shortdefinition $\cX$ mit $\E[\cX^2] < \8$, $\V[\cX] = \E[(\cX - \E[\cX])^2]$
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\shortdefinition[Standardabweichung] $\sigma(\cX) = \sqrt{\V[\cX]}$
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@@ -11,7 +13,8 @@ $\V[\cX] = \E[\cX^2] - \E[\cX]^2 = a^2\E[1_\Omega] - a^2 = 0$
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% Task 4.42 (needs proof?)
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\shortremark $\E[\cX] < \8$, dann $\V[\cX] \geq 0$ mit $=$ g.d.w. $\cX$ konst; zudem $\V[a \cX] = a^2 \V[\cX]$ und $\V[\cX + a] = \V[\cX]$
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\shortproposition $\cX_k$ paarw. unabh. $\V\left[ \sum_{k = 1}^{n} \cX_k \right] = \sum_{k = 1}^{n} \V[\cX_k]$
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\shortproposition $\cX_k$ paarw. unabh. $\V\left[ \sum_{k = 1}^{n} \cX_k \right] = \sum_{k = 1}^{n} \V[\cX_k]$.
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Falls $\cX_k$ abhängig, dann gilt $\neq$
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\shortexample Varianz von bekannten Verteilungen
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\begin{itemize}
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@@ -21,5 +24,3 @@ $\V[\cX] = \E[\cX^2] - \E[\cX]^2 = a^2\E[1_\Omega] - a^2 = 0$
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\item $\cX \sim \cU([a, b])$, $\V[\cX] = \frac{(b - a)^2}{12}$
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\item $\cX \sim \cN(\mu, \sigma^2)$, $\V[\cX] = \sigma^2$
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\end{itemize}
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\shortcorollary[Cheb.] $\V[\cY]$ end. $\forall c > 0$ gilt: $\P[|\cY - \E[\cY]| \geq c] \leq \frac{\V[\cY]}{c^2}$
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