add ana, update whteo
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\documentclass[uebung]{../../../lecture}
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\title{Einführung in die Wahrscheinlichkeitstheorie und Statistik: Übungsblatt 1}
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\author{Christian Merten}
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\title{Wtheo 0: Übungsblatt 1}
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\author{Josua Kugler, Christian Merten}
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\newcommand{\IP}{\mathbb{P}}
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\usepackage[]{mathrsfs}
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\begin{document}
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@@ -134,31 +136,50 @@
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\begin{aufgabe}
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Sei $(\Omega, \mathcal{A}, \mathbb{P})$ ein Wahrscheinlichkeitsraum.
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\begin{enumerate}[(a)]
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\item Sei $n \in \N$ und $A_1, \ldots, A_n \in \mathcal{A}$.
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Beh.:
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\item Der Induktionsanfang ist offensichtlich wahr, $\IP(A_1) = (-1)^0 \cdot \IP(A_1)$. Gelte die Behauptung also für ein $n\in \N$. Dann folgern wir
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\begin{align*}
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\IP\left(\bigcup_{j=1}^{n+1} A_j\right) =& \IP\left(\bigcup_{j=1}^{n} A_j\right) + \IP(A_{n+1}) - \IP\left(\bigcup_{j=1}^{n} A_j \cap A_{n+1}\right)\\
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=& \sum_{j = 1}^{n} \left((-1)^{j-1} \cdot \sum_{\{k_1, \dots, k_n\} \subset \{1,\dots, n\}} \IP(A_{k_1} \cap \dots \cap A_{k_j})\right) + \IP(A_{n+1})\\
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&- \IP\left(\bigcup_{j=1}^{n} (A_j \cap A_{n+1})\right)\\
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=& \sum_{j = 1}^{n} \left((-1)^{j-1} \cdot \sum_{\{k_1, \dots, k_n\} \subset \{1,\dots, n\}} \IP(A_{k_1} \cap \dots \cap A_{k_j})\right) + \IP(A_{n+1})\\
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&- \sum_{j = 1}^{n} \left((-1)^{j-1} \cdot \sum_{\{k_1,\dots, k_j\} \subset \{1,\dots, n\}} \IP((A_{k_1} \cap A_{n+1}) \cap \dots \cap (A_{k_j} \cap A_{n+1}))\right)\\
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=& \sum_{j = 1}^{n} \left((-1)^{j-1} \cdot \sum_{\{k_1, \dots, k_n\} \subset \{1,\dots, n\}} \IP(A_{k_1} \cap \dots \cap A_{k_j})\right) + \IP(A_{n+1})\\
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&- \sum_{j = 1}^{n} \left((-1)^{j-1} \cdot \sum_{\{k_1, \dots, k_j\} \subset \{1,\dots, n\}} \IP(A_{k_1} \cap \dots \cap A_{k_j} \cap A_{n+1})\right)\\
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=& \sum_{j = 1}^{n} \left((-1)^{j-1} \cdot \sum_{\substack{\{k_1, \dots, k_n\} \subset \{1,\dots, n+1\}\\\forall i\colon k_i \neq n+1}} \IP(A_{k_1} \cap \dots \cap A_{k_j})\right) + \IP(A_{n+1})\\
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&+ \sum_{j = 2}^{n+1} \left((-1)^{j-1} \cdot \sum_{\substack{\{k_1, \dots, k_j\} \subset \{1,\dots, n+1\}\\\exists i\colon k_i = n+1}} \IP(A_{k_1} \cap \dots \cap A_{k_j})\right)\\
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=& \sum_{j = 1}^{n} \left((-1)^{j-1} \cdot \sum_{\substack{\{k_1, \dots, k_n\} \subset \{1,\dots, n+1\}\\\forall i\colon k_i \neq n+1}} \IP(A_{k_1} \cap \dots \cap A_{k_j})\right)\\
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&+ \sum_{j = 1}^{n+1} \left((-1)^{j-1} \cdot \sum_{\substack{\{k_1, \dots, k_j\} \subset \{1,\dots, n+1\}\\\exists i\colon k_i = n+1}} \IP(A_{k_1} \cap \dots \cap A_{k_j})\right)\\
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\end{align*}
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Für $j = n+1$ gilt $\{k_1,\dots, k_j\} = \{1,\dots, n+1\}$. Daher können wir die beiden Summen im letzten Schritt einfach zusammenfassen und erhalten
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\[
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\mathbb{P}\left( \bigcup_{j=1}^{n} A_n \right)
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= \sum_{j=1}^{n} \left( (-1)^{j-1} \cdot \sum_{\{k_1, \ldots, k_j\} \subseteq \{1, \ldots, n\} }
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\mathbb{P}(A_{k_1} \cap \ldots \cap A_{k_j}) \right)
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.\]
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\begin{proof}
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Per Induktion über $n$. Sei $n=1$: Dann ist $\mathbb{P}(\bigcup_{j=1}^{1} A_j) = \mathbb{P}(A_1)$.
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Sei nun $n \in \N$ und Behauptung gezeigt für $k \le n$. Dann gilt
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\begin{salign*}
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\mathbb{P}\left( \bigcup_{j=1}^{n+1} A_j \right)
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=& \mathbb{P}\left(\bigcup_{j=1}^{n} A_j \cup A_{n+1}\right) \\
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\stackrel{(*)}{=}& \mathbb{P}\left( \bigcup_{j=1}^{n} A_j \right)
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+ \mathbb{P}(A_{n+1}) - \mathbb{P}\left( \bigcup_{j=1}^{n} A_j \cap A_{n+1} \right) \\
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\stackrel{\text{I.V.}}{=}&
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\sum_{j=1}^{n} \left( (-1)^{j-1} \sum_{\{k_1, \ldots, k_j\} \subseteq \{1, \ldots, n\}}
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\mathbb{P}(A_{k_1} \cap \ldots \cap A_{k_j})\right)
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+ \mathbb{P}(A_{n+1}) \\
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&- \sum_{j=1}^{n} \left( (-1)^{j-1} \sum_{\{k_1, \ldots, k_j\} \subseteq \{1, \ldots, n\} }
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\mathbb{P}(A_{k_1} \cap A_{n+1} \cap \ldots \cap A_{k_j} \cap A_{n+1}) \right) \\
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=& \sum_{j=1}^{n+1} \left( (-1)^{j-1} \sum_{\{k_1, \ldots, k_j\}\subseteq \{1, \ldots, n\} }
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\mathbb{P}(A_{k_1} \cap \ldots \cap A_{k_j})\right)
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.\end{salign*}
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\end{proof}
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\IP\left(\bigcup_{j=1}^{n+1} A_j\right) = \sum_{j = 1}^{n+1} \left((-1)^{j-1} \cdot \sum_{\{k_1, \dots, k_n\} \subset \{1,\dots, n\}} \IP(A_{k_1} \cap \dots \cap A_{k_j})\right),
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\]
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was zu zeigen war.
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% \item Sei $n \in \N$ und $A_1, \ldots, A_n \in \mathcal{A}$.
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% Beh.:
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% \[
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% \mathbb{P}\left( \bigcup_{j=1}^{n} A_n \right)
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% = \sum_{j=1}^{n} \left( (-1)^{j-1} \cdot \sum_{\{k_1, \ldots, k_j\} \subseteq \{1, \ldots, n\} }
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% \mathbb{P}(A_{k_1} \cap \ldots \cap A_{k_j}) \right)
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% .\]
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% \begin{proof}
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% Per Induktion über $n$. Sei $n=1$: Dann ist $\mathbb{P}(\bigcup_{j=1}^{1} A_j) = \mathbb{P}(A_1)$.
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% Sei nun $n \in \N$ und Behauptung gezeigt für $k \le n$. Dann gilt
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% \begin{salign*}
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% \mathbb{P}\left( \bigcup_{j=1}^{n+1} A_j \right)
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% =& \mathbb{P}\left(\bigcup_{j=1}^{n} A_j \cup A_{n+1}\right) \\
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% \stackrel{(*)}{=}& \mathbb{P}\left( \bigcup_{j=1}^{n} A_j \right)
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% + \mathbb{P}(A_{n+1}) - \mathbb{P}\left( \bigcup_{j=1}^{n} A_j \cap A_{n+1} \right) \\
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% \stackrel{\text{I.V.}}{=}&
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% \sum_{j=1}^{n} \left( (-1)^{j-1} \sum_{\{k_1, \ldots, k_j\} \subseteq \{1, \ldots, n\}}
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% \mathbb{P}(A_{k_1} \cap \ldots \cap A_{k_j})\right)
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% + \mathbb{P}(A_{n+1}) \\
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% &- \sum_{j=1}^{n} \left( (-1)^{j-1} \sum_{\{k_1, \ldots, k_j\} \subseteq \{1, \ldots, n\} }
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% \mathbb{P}(A_{k_1} \cap A_{n+1} \cap \ldots \cap A_{k_j} \cap A_{n+1}) \right) \\
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% =& \sum_{j=1}^{n+1} \left( (-1)^{j-1} \sum_{\{k_1, \ldots, k_j\}\subseteq \{1, \ldots, n\} }
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% \mathbb{P}(A_{k_1} \cap \ldots \cap A_{k_j})\right)
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% .\end{salign*}
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% \end{proof}
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\item Beh.: Die Wahrscheinlichkeit für $n \to \infty$ ist $1 - \frac{1}{e}$.
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\begin{proof}
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Setze $\Omega \coloneqq \{ (g_1, \ldots, g_n) \mid g_1, \ldots, g_n \in \{1, \ldots, n\},
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