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[PS] Many small fixes
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@@ -13,7 +13,7 @@ $\V[\cX] = \E[\cX^2] - \E[\cX]^2 = a^2\E[1_\Omega] - a^2 = 0$
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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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Falls \bi{mermals} selbe Z.V. (e.g. $\cX - \cY - \cY$, ist $\cX - 2\cY$, dann $\V = 2^2 \V[\cY]$)
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Falls \bi{mermals} selbe Z.V. (e.g. $\cX - \cY - \cY$, ist $\cX - 2\cY$, dann $\V = \V[\cX] + 2^2 \V[\cY]$)
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\newpage
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\shortexample Varianz von bekannten Verteilungen
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