[PS] More fixes

semester4/ps/ps-jh/parts/03_expected-value/05_variance.tex

@@ -1,4 +1,6 @@
 \subsection{Varianz}
+$\E[\cX^2] = \sum_{x \in W} x^2 \cdot p_\cX(x)$ ($\cX$ diskret)
+
 \shortdefinition $\cX$ mit $\E[\cX^2] < \8$, $\V[\cX] = \E[(\cX - \E[\cX])^2]$
 
 \shortdefinition[Standardabweichung] $\sigma(\cX) = \sqrt{\V[\cX]}$
@@ -11,7 +13,8 @@ $\V[\cX] = \E[\cX^2] - \E[\cX]^2 = a^2\E[1_\Omega] - a^2 = 0$
 % Task 4.42 (needs proof?)
 \shortremark $\E[\cX] < \8$, dann $\V[\cX] \geq 0$ mit $=$ g.d.w. $\cX$ konst; zudem $\V[a \cX] = a^2 \V[\cX]$ und $\V[\cX + a] = \V[\cX]$
 
-\shortproposition $\cX_k$ paarw. unabh. $\V\left[ \sum_{k = 1}^{n} \cX_k \right] = \sum_{k = 1}^{n} \V[\cX_k]$
+\shortproposition $\cX_k$ paarw. unabh. $\V\left[ \sum_{k = 1}^{n} \cX_k \right] = \sum_{k = 1}^{n} \V[\cX_k]$.
+Falls $\cX_k$ abhängig, dann gilt $\neq$
 
 \shortexample Varianz von bekannten Verteilungen
 \begin{itemize}
@@ -21,5 +24,3 @@ $\V[\cX] = \E[\cX^2] - \E[\cX]^2 = a^2\E[1_\Omega] - a^2 = 0$
     \item $\cX \sim \cU([a, b])$, $\V[\cX] = \frac{(b - a)^2}{12}$
     \item $\cX \sim \cN(\mu, \sigma^2)$, $\V[\cX] = \sigma^2$
 \end{itemize}
-
-\shortcorollary[Cheb.] $\V[\cY]$ end. $\forall c > 0$ gilt: $\P[|\cY - \E[\cY]| \geq c] \leq \frac{\V[\cY]}{c^2}$