\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] $\sigma(\cX) = \sqrt{\V[\cX]}$

\shortremark $\V[\cX] = \E[\cX^2] - \E[\cX]^2$

\shortexample $\cX$ determ. Z.V (= konst) mit Wert $a$, also $\cX = a1_\Omega$. Dann: $\E[\cX] = a \E[1_\Omega] = a \P[\Omega] = a$ und\\
$\V[\cX] = \E[\cX^2] - \E[\cX]^2 = a^2\E[1_\Omega] - a^2 = 0$

% Task 4.42 (needs proof?)
\shortremark $\E[\cX] < \8$, dann $\V[\cX] \geq 0$ mit $=$ g.d.w. $\cX$ konst; \hl{zudem} $\V[a \cX] = a^2 \V[\cX]$ und $\V[\cX + a] = \V[\cX]$

\shortproposition $\cX_k$ paarw. unabh. $\V\left[ \sum_{k = 1}^{n} \cX_k \right] = \sum_{k = 1}^{n} \V[\cX_k]$.
Falls $\cX_k$ abhängig, dann gilt $\neq$

\newpage
\shortexample Varianz von bekannten Verteilungen
\begin{itemize}
    \item $\cX \sim \text{Ber}(p)$, $\V[\cX] = p (1 - p)$
    \item $\cX \sim \text{Bin}(n, p)$, $\V[\cX] = n p (1 - p)$
    \item $\cX \sim \text{NBin}(r, p)$, $\V[\cX] = \frac{r(1 - p)}{p^2}$
    \item $\cX \sim \text{Geom}(p)$, $\V[\cX] = \frac{1 - p}{p^2}$
    \item $\cX \sim \text{H}(n, r, m)$, $\V[\cX] = m\frac{rn - r^2}{n^2} \frac{n - m}{n - 1}$
    \item $\cX \sim \text{Poisson}(\lambda)$, $\V[\cX] = \lambda = \E[\cX]$
    \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}