From 2b64af7c354e9bb74481d787810fe7c6c4a61796 Mon Sep 17 00:00:00 2001 From: Ameyah Date: Mon, 15 Apr 2019 18:35:14 +0200 Subject: [PATCH 1/9] add defeq command --- texmf/tex/latex/mathoperators/mathoperators.sty | 4 ++++ 1 file changed, 4 insertions(+) diff --git a/texmf/tex/latex/mathoperators/mathoperators.sty b/texmf/tex/latex/mathoperators/mathoperators.sty index 31bc77b..ea0dd25 100644 --- a/texmf/tex/latex/mathoperators/mathoperators.sty +++ b/texmf/tex/latex/mathoperators/mathoperators.sty @@ -353,4 +353,8 @@ \newcommand{\und}{\text{ und }} % sprachliches und für align-Umgebungen \newcommand{\oder}{\text{ oder }} % sprachliches oder für align-Umgebungen +% overset text +\newcommand{\defeq}{\overset{\text{Def}}{=}} % Definition over over = + + \endinput \ No newline at end of file From f84e2110c973dae772b36a5a1a2350c3208e16cd Mon Sep 17 00:00:00 2001 From: Ameyah Date: Mon, 15 Apr 2019 18:35:30 +0200 Subject: [PATCH 2/9] lecture --- .../STOCH/TeX_files/Bedingte_Wheiten.tex | 154 +++++++++++++++++- 1 file changed, 152 insertions(+), 2 deletions(-) diff --git a/4. Semester/STOCH/TeX_files/Bedingte_Wheiten.tex b/4. Semester/STOCH/TeX_files/Bedingte_Wheiten.tex index 6b381a1..60148cb 100644 --- a/4. Semester/STOCH/TeX_files/Bedingte_Wheiten.tex +++ b/4. Semester/STOCH/TeX_files/Bedingte_Wheiten.tex @@ -3,6 +3,7 @@ \section{Bedingte Wahrscheinlichkeiten} \begin{example} + \proplbl{3_1_1} Das Würfeln mit zwei fairen, sechsseitigen Würfeln können wir mit \begin{align} \Omega = \set{(i,j,), i,j \in \set{1,\dots,6}}\notag @@ -29,7 +30,7 @@ \frac{1}{6} = \frac{\abs{A \cap B}}{\abs{B}}.\notag \end{align} Das Eintreten von $B$ führt also dazu, dass das Wahrscheinlichkeitsmaß $\probp$ durch ein neues Wahrscheinlichkeitsmaß $\probp_{B}$ ersetzt werden muss. Hierbei sollte gelten: - \begin{align} %TODO add references! + \begin{align} &\text{Renormierung: }\probp_{B} = 1\label{Renorm}\tag{R}\\ &\text{Proportionalität: Für alle} A \subset \sigF \mit A \subseteq B \text{ gilt } \probp_{B}(A) = c_B \probp(A) \text{ mit einer Konstante } c_B.\label{Prop}\tag{P} @@ -43,7 +44,7 @@ \end{align} \end{lemma} -\begin{proof} %TODO surpress ``Gleichung'' here?! +\begin{proof} Offenbar erfüllt $\probp_{B}$ wie definiert \eqref{Renorm} und \eqref{Prop}. Umgekehrt erfüllt $\probp_{B}$ \eqref{Renorm} und \eqref{Prop}. Dann folgt für $A \in \sigF$: \begin{align} \probp_{B}(A) = \probp_{B}(A\cap B) + \underbrace{\probp_{B}(A\setminus B)}_{= 0, \text{ wegen } \eqref{Renorm}} \overset{\eqref{Prop}}{=} c_B \probp(A \cap B).\notag @@ -57,5 +58,154 @@ % % % % % % % % % % % % % % % % % % % % % % % % % % % 5th lecture % % % % % % % % % % % % % % % % % % % % % % % % % % % +\begin{definition} + \proplbl{3_1_3} + Sei $(\Omega, \sigF, \probp)$ Wahrscheinlichkeitsraum und $B \in \sigF$ mit $\probp(B) > 0$. Dann heißt + \begin{align*} + \probp(A\vert B) := \frac{\probp(A\cap B)}{\probp(B)} \mit A\in \sigF + \end{align*} + die \begriff{bedingte Wheit von $A$ gegeben $B$}. + Falls $\probp(B) = 0$, setze + \begin{align*} + \probp(A \vert B) = 0 \mit \forall A \in \sigF + \end{align*} +\end{definition} +\begin{example} %TODO ref + In der Situation \propref{3_1_1} gilt % + \begin{align*} + A \cap B = \set{(4,4)} + \intertext{und damit} + \probp(A \vert B) = \frac{\probp(A\cap B)}{\probp(B)} = \frac{\frac{1}{36}}{\frac{1}{6}} = \frac{1}{6} + \end{align*} +\end{example} +Aus \propref{3_1_3} ergibt sich +\begin{lemma}[Multiplikationsformel] + \proplbl{3_1_4} + Sei $(\Omega, \sigF, \probp)$ Wahrscheinlichkeitsraum und $A_1, \dots, A_n \in \sigF$. Dann + \begin{align*} + \probp(A_1 \cap \cdots \cap A_n) = \probp(A_1)\probp(A_2 \vert A_n) \dots \probp(A_n \vert A_1 \cap \cdots \cap A_{n-1}) + \end{align*} +\end{lemma} + +\begin{proof} + Ist $\probp(A_1 \cap \dots \cap A_n) = 0$, so gilt auch $\probp(A_n \vert \bigcap_{i=1}^{n-1}) = 0$. Andernfalls sind alle Faktoren der rechten Seite ungleich 0 und + \begin{align*} + \probp(A_1)\probp(A_2 \vert A_1) \dots \probp(A_n \vert \bigcap_{i=1}^{n-1} A_i) \\ + = \probp(A_1) \cdot \frac{\probp(A_1 \cap A_2)}{\probp(A_1)} \dots \frac{\probp(\bigcap_{i=1}^{n} A_i)}{\probp(\bigcap_{i=1}^{n-1}A_i)} = \probp(\bigcap_{i=1}^n A_i) + \end{align*} +\end{proof} %TODO add ref. +Stehen die $A_i$ in \propref{3_1_4} in einer (zeitlichen) Abfolge, so liefert Formel einen Hinweis, wie Wahrscheinlichkeitsmaße für \begriff{Stufenexperimente} konstruiert werden können. Ein \emph{Stufenexperiment} aus $n$ nacheinander ausgeführten Teilexperimenten lässt sich als \begriff{Baumdiagramm} darstellen. + +%TODO Baumdiagramm +%\begin{tikzpicture} +% %TODO +%\end{tikzpicture} + +\begin{proposition}[Konstruktion des Wahrscheinlichkeitsmaßes eines Stufenexperiments] + \proplbl{3_1_6} + Gegeben seinen $n$ Ergebnisräume $\Omega_i = \set{\omega_i (1), \dots, \omega_i (k)}, k \in \N \cup \set{\infty}$ und es sei $\Omega = \bigtimes_{i = 1}^n \Omega_i$ der zugehörige Produktraum. Weiter seinen $\sigF_i$ $\sigma$-Algebren auf $\Omega_i$ und $\sigF = \bigotimes_{i=1}^n \sigF_i$ die Produkt-$\sigma$-Algebra auf $\Omega$. Setze $\omega = (\omega_1,\dots,\omega_n)$ und + \begin{align*} + [\omega_1,\dots,\omega_n]:= \set{\omega_1}\times \dots \times \set{\omega_n} \times \Omega_{m-1} \times \Omega_{n},\quad m\le n\\ + \probp(\set{\omega_m}[\omega_1,\dots,\omega_{m-1}]) + \end{align*} %TODO check indices, m-1 instead of m+1? + für die Wheit in der $m$-ten Stufe des Experiments $\omega_m$ zu beobachten, falls in den vorausgehenden Stufen $\omega_1,\dots,\omega_{m-1}$ beobachten wurden. Dann definiert + \begin{align*} + \probp(\set{\omega}) := \probp(\set{\omega_1})\prod_{m=2}^{n}\probp\brackets{\set{\omega_m} \mid [\omega_1, \dots, \omega_{m-1}]} + %TODO maybe wrong here. check + \end{align*} + ein Wahrscheinlichkeitsmaß auf $(\Omega, \sigF, \probp)$. +\end{proposition} + +\begin{example}[\person{Polya}-Urne] + Gegeben sei eine Urne mit $s$ Schwarze und $w$ weiße Kugeln. Bei jedem Zug wird die gezogene Kugel zusammen mit $c\in \N_0\cup \set{-1}$ weiteren Kugeln derselben Farbe zurückgelegt. + \begin{itemize} %TODO seen both in chapter 2.2, but big bracket behind. + \item $c=0$: Urnenmodell mit Zurücklegen + \item $c=-1$: Urnenmodell ohne Zurücklegen + \end{itemize} + Beide schon in Kapitel 2.2 gesehen.\\ + Sei $c\in \N$. (Modell für zwei konkurrierende Populationen) Ziehen wir $n$-mal, so erhalten wir ein $n$-Stufenexperiment mit + \begin{align*} + \Omega = \set{0,1}^n \mit \text{ 0 = ``weiß'', 1 = ``schwarz''}\mit (\Omega_i = \set{0,1}) + \intertext{Zudem gelten im ersten Schritt} + \probp(\set{0}) = \frac{w}{s+w} \und \probp(\set{1}) = \frac{s}{w+s} + \intertext{sowie} + \probp(\set{\omega_m} \mid [\omega_1, \dots \omega_{m-1}]) = + \begin{cases} %TODO fix brackets! + \frac{w+c(m-1 - \sum_{i=1}^{m-1}\omega_i)}{s+w+c(m-1)} & \omega_m = 0\\ + \frac{s + c\sum_{i=1}^{m-1}\omega_i}{s+w+c(m-1)} & \omega_m = 1 + \end{cases} + \end{align*} + Mit \propref{3_1_6} folgt als Wahrscheinlichkeitsmaß auf $(\Omega, \pows(\Omega))$ + \begin{align*} + \probp(\set{(\omega_1, \dots, \omega_n)}) &= \probp(\set{\omega_1})\prod_{m=2}^n \probp(\set{\omega_m}\mid [\omega_1,\dots,\omega_{m-1}])\\ + &=\frac{\prod_{i=0}^{l-1}(s+c_i)\prod_{i=0}^{n-l-1}}{\prod_{i=0}^n (s+w+c_i)} \mit l=\sum_{i=1}^n \omega_i. + \intertext{Definiere wir nun die Zufallsvariable} + S_n:\Omega &\to \N_0 \mit (\omega_1, \dots, \omega_n) \mapsto \sum_{i=1}^n \omega_i + \intertext{welche die Anzahl der gezogenen schwarzen Kugeln modelliert, so folgt,} + \probp(S_n = l) &= \binom{n}{l}\frac{\prod_{i=0}^{l-1}(s+c_i) \prod_{j=0}^{n-l-1}(\omega + c_j)}{\prod_{i=0}^n(s+w+c_i)} + \intertext{Mittels $a:= \sfrac{s}{c},b:= \sfrac{w}{c}$ folgt} + \probp(S_n = l) &= \frac{\prod_{i=0}^{l-1}(-a-i)\prod_{i=0}^{-b-j-1}}{\prod_{i=0}^n (-a-b-i)} = \frac{\binom{-a}{l}\binom{-b}{n-l}}{\binom{-a-b}{n}}\\ &\mit l \in \set{0,\dots,n} + \end{align*} + Dies ist die \begriff{\person{Polya}-Urne} auf $\set{0,\dots,n}, n \in \N$ mit Parametern $a,b > 0$. +\end{example} + +\begin{example} + Ein Student beantwortet eine Multiple-Choice-Frage mit 4 Antwortmöglichkeiten, eine davon ist richtig. Er kennt die richtige Antwort mit Wheit $\sfrac{1}{3}$. Wenn er diese kennt, so wählt er diese aus. Andernfalls wählt er zufällig (gleichverteilt) eine Antwort.\\ + Betrachte + \begin{align*} + W = \set{\text{richtige Antwort gewusst}}\\ + R = \set{\text{Richtige Antwort gewählt}} + \intertext{Dann} + \probp(W) = \frac{2}{3}, \probp(R \vert W) = 1, \probp(R \vert W^C) = \frac{1}{4} + \end{align*} + Angenommen, der Student gibt die richtige Antwort. Mit welcher Wahrscheinlichkeit hat er diese gewusst? + \begin{align*} + \probp(W\vert R) = \text{ ?} + \end{align*} +\end{example} + +\begin{proposition} + \proplbl{3_1_9} + Sei $(\Omega, \sigF, \probp)$ Wahrscheinlichkeitsraum und $\Omega = \bigcup_{i \in I} B_i$ eine höchstens abzählbare Zerlegung in paarweise disjunkte Ereignisse. $B_i \in \sigF$. + \begin{enumerate} %TODO set itemize references. or use enumerate? + \item \emph{Satz von der totalen Wahrscheinlihckeit:} Für alle $A \in \sigF$ + \begin{align*} + \probp(A) = \sum_{i\in I} \probp(A\vert B_i)\probp(B_i) + \end{align*} + \item \emph{Satz von \person{Bayes}:} Für alle $A \in \sigF$ mit $\probp(A) > 0$ und alle $h \in I$ + \begin{align*} + \probp(B_h \vert A) = \frac{\probp(A \vert B_h)\probp(B_h)}{\sum_{i\in I}\probp(A\vert B_i)\probp(B_i)} + \end{align*} + \end{enumerate} +\end{proposition} + +\begin{proof} + \begin{enumerate} + \item Es gilt: + \begin{align*} + \sum_{i\in I} \probp(A\vert B_i)\probp(B_i) \defeq \sum_{i\in I}\frac{\probp(A \cap B_i)}{\probp(B_i)}\probp(B_i) = \sum_{i\in I} \probp(A \cap B_i) \overset{\sigma-add.}{=} \probp(A) + \end{align*} + \item + \begin{align*} + \probp(B_h \vert A) \defeq \frac{\probp(A \cap B_h)}{\probp(A)} \defeq \frac{\probp(A \vert B_h)\probp(B_h)}{\probp(A)} + \end{align*} + also mit a) auch b). %TODO add refs + \end{enumerate} +\end{proof} + +\begin{example} + In der Situation von \propref{3_1_3} folgt mit dem \propref{3_1_9} der totalen Wahrscheinlichkeit + \begin{align*} + \probp(R) &= \probp(R \vert W)\probp(W) + \probp(R\vert W^C)\probp(W^C)\\ + &= 1\cdot \frac{2}{3} + \frac{1}{4}\frac{1}{3} = \frac{3}{4} + \intertext{und mit dem \ref{3_1_9}} %Bayes + \probp(W \vert R) &= \frac{\probp(R \vert W)\probp(W)}{\probp(R)} = \frac{1\frac{2}{3}}{\frac{3}{4}} = \frac{8}{9} \text{ für die gesuchte Wahrscheinlichkeit.} + \end{align*} %TODO +% \begin{tikzpicture} +% +% \end{tikzpicture} +\end{example} +% trees in Latex? +% https://tex.stackexchange.com/questions/5447/how-can-i-draw-simple-trees-in-latex/5451 \section{(Un)-abhängigkeit} \label{sec_unabhangigkeit} \ No newline at end of file From 62d61929548bcec577620228c6ffeda893cfbdab Mon Sep 17 00:00:00 2001 From: Ameyah Date: Mon, 15 Apr 2019 20:58:17 +0200 Subject: [PATCH 3/9] tree wips! - add nodes - multicol with 2 or 3 trees for baum_2 - maybe refine node placement? but how ... @henrydatei --- 4. Semester/STOCH/tikz/baum_1.tex | 20 ++++++++++++++++++++ 4. Semester/STOCH/tikz/baum_2.tex | 19 +++++++++++++++++++ 2 files changed, 39 insertions(+) create mode 100644 4. Semester/STOCH/tikz/baum_1.tex create mode 100644 4. Semester/STOCH/tikz/baum_2.tex diff --git a/4. Semester/STOCH/tikz/baum_1.tex b/4. Semester/STOCH/tikz/baum_1.tex new file mode 100644 index 0000000..1637c3a --- /dev/null +++ b/4. Semester/STOCH/tikz/baum_1.tex @@ -0,0 +1,20 @@ +\documentclass{standalone} + +\usepackage{tikz} +\usepackage{tikz-qtree} +\usepackage{../../../texmf/tex/latex/mathoperators/mathoperators} + +\begin{document} + \begin{tikzpicture} + \tikzset{grow'=right,level distance=32pt} + \tikzset{execute at begin node=\strut} + \tikzset{every tree node/.style={anchor=base west}} %TODO add better node placing and the rest of the nodes! + \Tree [.$\bullet$ + \edge node[midway,left]{$\probp(A_1)$}; + [.$A_1$ [.$A_2$ $A_3$ ] [.$A_2^C$ ] ] + \edge node[auto=right]{$\probp(A_2)$}; + [.$A_1^C$ [.$A_2$ ] + [.$A_2^C$ + ] ] ] + \end{tikzpicture} +\end{document} \ No newline at end of file diff --git a/4. Semester/STOCH/tikz/baum_2.tex b/4. Semester/STOCH/tikz/baum_2.tex new file mode 100644 index 0000000..6376819 --- /dev/null +++ b/4. Semester/STOCH/tikz/baum_2.tex @@ -0,0 +1,19 @@ +\documentclass{standalone} + +\usepackage{tikz} +\usepackage{tikz-qtree} +\usepackage{../../../texmf/tex/latex/mathoperators/mathoperators} + +%TODO how to do two or three trees next to eatch other +% https://tex.stackexchange.com/questions/258850/putting-a-tree-next-to-another-tree + +\begin{document} + \begin{ + \begin{tikzpicture} + \tikzset{every tree node/.style={align=center,anchor=north}} + \Tree [.S [.NP Det\\the N\\cat ] + [.VP V\\sat + [.PP P\\on + [.NP Det\\the N\\mat ] ] ] ] + \end{tikzpicture} +\end{document} \ No newline at end of file From b4cfdf6cef5a7c5a353ec5bec0cd09f241047251 Mon Sep 17 00:00:00 2001 From: henrydatei <31316025+henrydatei@users.noreply.github.com> Date: Mon, 15 Apr 2019 22:40:23 +0100 Subject: [PATCH 4/9] Update baum_1.tex --- 4. Semester/STOCH/tikz/baum_1.tex | 38 +++++++++++++++++++------------ 1 file changed, 24 insertions(+), 14 deletions(-) diff --git a/4. Semester/STOCH/tikz/baum_1.tex b/4. Semester/STOCH/tikz/baum_1.tex index 1637c3a..10e5e23 100644 --- a/4. Semester/STOCH/tikz/baum_1.tex +++ b/4. Semester/STOCH/tikz/baum_1.tex @@ -1,20 +1,30 @@ \documentclass{standalone} \usepackage{tikz} -\usepackage{tikz-qtree} -\usepackage{../../../texmf/tex/latex/mathoperators/mathoperators} +\usepackage{amssymb} +\usetikzlibrary{trees} \begin{document} - \begin{tikzpicture} - \tikzset{grow'=right,level distance=32pt} - \tikzset{execute at begin node=\strut} - \tikzset{every tree node/.style={anchor=base west}} %TODO add better node placing and the rest of the nodes! - \Tree [.$\bullet$ - \edge node[midway,left]{$\probp(A_1)$}; - [.$A_1$ [.$A_2$ $A_3$ ] [.$A_2^C$ ] ] - \edge node[auto=right]{$\probp(A_2)$}; - [.$A_1^C$ [.$A_2$ ] - [.$A_2^C$ - ] ] ] + \begin{tikzpicture}[level distance=3cm, + level 1/.style={sibling distance=3cm}, + level 2/.style={sibling distance=1.5cm}, + grow=right, sloped] + \node {} + child {node {$A_1^C$} + child {node {$A_2^C$}} + child {node {$A_2$} edge from parent node[above] {$\mathbb{P}(A_2\mid A_1^C)$}} + edge from parent + node[below] {$\mathbb{P}(A_1^C)$} + } + child {node {$A_1$} + child {node {$A_2^C$}} + child {node {$A_2$} + child {node {$A_3^C$}} + child {node {$A_3$} edge from parent node[above] {$\mathbb{P}(A_3\mid A_1\cap A_2)$}} + edge from parent + node[above] {$\mathbb{P}(A_2\mid A_1)$}} + edge from parent + node[above] {$\mathbb{P}(A_1)$} + }; \end{tikzpicture} -\end{document} \ No newline at end of file +\end{document} From 079d44ae80fa206ae23afff272e00b9b29990f8d Mon Sep 17 00:00:00 2001 From: henrydatei <31316025+henrydatei@users.noreply.github.com> Date: Mon, 15 Apr 2019 22:53:50 +0100 Subject: [PATCH 5/9] Update baum_2.tex --- 4. Semester/STOCH/tikz/baum_2.tex | 49 +++++++++++++++++++++++-------- 1 file changed, 36 insertions(+), 13 deletions(-) diff --git a/4. Semester/STOCH/tikz/baum_2.tex b/4. Semester/STOCH/tikz/baum_2.tex index 6376819..6fa003b 100644 --- a/4. Semester/STOCH/tikz/baum_2.tex +++ b/4. Semester/STOCH/tikz/baum_2.tex @@ -1,19 +1,42 @@ \documentclass{standalone} \usepackage{tikz} -\usepackage{tikz-qtree} -\usepackage{../../../texmf/tex/latex/mathoperators/mathoperators} - -%TODO how to do two or three trees next to eatch other -% https://tex.stackexchange.com/questions/258850/putting-a-tree-next-to-another-tree +\usepackage{amssymb} +\usetikzlibrary{trees} +\usetikzlibrary{arrows, decorations.pathmorphing} +\usepackage{xfrac} \begin{document} - \begin{ - \begin{tikzpicture} - \tikzset{every tree node/.style={align=center,anchor=north}} - \Tree [.S [.NP Det\\the N\\cat ] - [.VP V\\sat - [.PP P\\on - [.NP Det\\the N\\mat ] ] ] ] + \begin{tikzpicture}[level distance=1.5cm, + level 1/.style={sibling distance=3cm}, + level 2/.style={sibling distance=1.5cm}, + decoration = {snake, pre length=1pt, post length=1pt}] + \node at (0,0) {} + child {node {$\omega$} + child {node {$R$} edge from parent node[left] {$1$}} + child {node {$R^C$} edge from parent node[right] {$0$}} + edge from parent + node[left] {$\sfrac{2}{3}$} + } + child {node {$\omega^C$} + child {node {$R$} edge from parent node[left] {$\sfrac{1}{4}$}} + child {node {$R^C$} edge from parent node[right] {$\sfrac{3}{4}$}} + edge from parent + node[right] {$\sfrac{1}{3}$} + }; + \draw[->,decorate] (3,-1.5) -- (5,-1.5); + \node at (8,0) {} + child {node {$R$} + child {node {$\omega$} edge from parent node[left] {$\sfrac{8}{9}$}} + child {node {$\omega^C$} edge from parent node[right] {$\sfrac{1}{9}$}} + edge from parent + node[left] {$\sfrac{3}{4}$} + } + child {node {$R^C$} + child {node {$\omega$} edge from parent node[left] {$0$}} + child {node {$\omega^C$} edge from parent node[right] {$1$}} + edge from parent + node[right] {$\sfrac{1}{4}$} + }; \end{tikzpicture} -\end{document} \ No newline at end of file +\end{document} From 64afb4064161e2f5b8ecc915a010bceb68622332 Mon Sep 17 00:00:00 2001 From: Ameyah Date: Wed, 17 Apr 2019 14:01:33 +0200 Subject: [PATCH 6/9] Lecture 17 April 2019 --- .../STOCH/TeX_files/Bedingte_Wheiten.tex | 206 +++++++++++++++++- 4. Semester/STOCH/Vorlesung STOCH.tex | 12 + Material/CAT/symbol_center.pdf | Bin 0 -> 18548 bytes Material/CAT/symbol_center.tex | 23 ++ Material/CAT/symbol_center_diagram.pdf | Bin 0 -> 18470 bytes Material/CAT/symbol_center_diagram.tex | 11 + Material/CAT/symbol_center_diagram_test.pdf | Bin 0 -> 37141 bytes Material/CAT/symbol_center_diagram_test.tex | 21 ++ 8 files changed, 265 insertions(+), 8 deletions(-) create mode 100644 Material/CAT/symbol_center.pdf create mode 100644 Material/CAT/symbol_center.tex create mode 100644 Material/CAT/symbol_center_diagram.pdf create mode 100644 Material/CAT/symbol_center_diagram.tex create mode 100644 Material/CAT/symbol_center_diagram_test.pdf create mode 100644 Material/CAT/symbol_center_diagram_test.tex diff --git a/4. Semester/STOCH/TeX_files/Bedingte_Wheiten.tex b/4. Semester/STOCH/TeX_files/Bedingte_Wheiten.tex index 60148cb..69f8ee2 100644 --- a/4. Semester/STOCH/TeX_files/Bedingte_Wheiten.tex +++ b/4. Semester/STOCH/TeX_files/Bedingte_Wheiten.tex @@ -1,4 +1,4 @@ -\chapter[Bedingte Wahrscheinlichkeiten und (Un)-abbhängigkeit]{Bedingte Wheiten und (Un)-abbhängigkeit} +\chapter[Bedingte Wahrscheinlichkeiten und (Un)-abbhängigkeit]{Bedingte Wkeiten und (Un)-abbhängigkeit} \chaptermark{Bedingte Wheiten und (Un)-abbhängigkeit} \section{Bedingte Wahrscheinlichkeiten} @@ -167,15 +167,15 @@ Stehen die $A_i$ in \propref{3_1_4} in einer (zeitlichen) Abfolge, so liefert Fo \begin{proposition} \proplbl{3_1_9} - Sei $(\Omega, \sigF, \probp)$ Wahrscheinlichkeitsraum und $\Omega = \bigcup_{i \in I} B_i$ eine höchstens abzählbare Zerlegung in paarweise disjunkte Ereignisse. $B_i \in \sigF$. + Sei $(\Omega, \sigF, \probp)$ Wahrscheinlichkeitsraum und $\Omega = \bigcup_{i \in I} B_i$ eine höchstens abzählbare Zerlegung in paarweise disjunkte Ereignisse $B_i \in \sigF$. \begin{enumerate} %TODO set itemize references. or use enumerate? - \item \emph{Satz von der totalen Wahrscheinlihckeit:} Für alle $A \in \sigF$ + \item \emph{Satz von der totalen Wahrscheinlichkeit:} Für alle $A \in \sigF$ \begin{align*} - \probp(A) = \sum_{i\in I} \probp(A\vert B_i)\probp(B_i) + \probp(A) = \sum_{i\in I} \probp(A\vert B_i)\probp(B_i) \label{eq:totWkeit}\tag{totale Wahrscheinlichkeit} \end{align*} \item \emph{Satz von \person{Bayes}:} Für alle $A \in \sigF$ mit $\probp(A) > 0$ und alle $h \in I$ \begin{align*} - \probp(B_h \vert A) = \frac{\probp(A \vert B_h)\probp(B_h)}{\sum_{i\in I}\probp(A\vert B_i)\probp(B_i)} + \probp(B_h \vert A) = \frac{\probp(A \vert B_h)\probp(B_h)}{\sum_{i\in I}\probp(A\vert B_i)\probp(B_i)} \label{eq:bayes}\tag{Bayes} \end{align*} \end{enumerate} \end{proposition} @@ -195,11 +195,11 @@ Stehen die $A_i$ in \propref{3_1_4} in einer (zeitlichen) Abfolge, so liefert Fo \end{proof} \begin{example} - In der Situation von \propref{3_1_3} folgt mit dem \propref{3_1_9} der totalen Wahrscheinlichkeit + In der Situation von \propref{3_1_3} folgt mit dem \propref{3_1_9} \eqref{eq:totWkeit} der totalen Wahrscheinlichkeit \begin{align*} \probp(R) &= \probp(R \vert W)\probp(W) + \probp(R\vert W^C)\probp(W^C)\\ &= 1\cdot \frac{2}{3} + \frac{1}{4}\frac{1}{3} = \frac{3}{4} - \intertext{und mit dem \ref{3_1_9}} %Bayes + \intertext{und mit dem \propref{3_1_9} \eqref{eq:bayes}} %Bayes \probp(W \vert R) &= \frac{\probp(R \vert W)\probp(W)}{\probp(R)} = \frac{1\frac{2}{3}}{\frac{3}{4}} = \frac{8}{9} \text{ für die gesuchte Wahrscheinlichkeit.} \end{align*} %TODO % \begin{tikzpicture} @@ -208,4 +208,194 @@ Stehen die $A_i$ in \propref{3_1_4} in einer (zeitlichen) Abfolge, so liefert Fo \end{example} % trees in Latex? % https://tex.stackexchange.com/questions/5447/how-can-i-draw-simple-trees-in-latex/5451 -\section{(Un)-abhängigkeit} \label{sec_unabhangigkeit} \ No newline at end of file +\section{(Un)-abhängigkeit} \label{sec_unabhangigkeit} +In vielen Fällen besagt die Intuition über verschiedene Zufallsexperimente/ Ereignisse, dass diese sich \emph{nicht} gegenseitig beeinflussen. Für solche $A,B \in \F$ mit $\probp(A) > 0, \probp(B) > 0$ sollte gelten +\begin{align*} + \probp(A\mid B) = \probp(B), \quad \probp(B\mid A) = \probp(B). +\end{align*} +\begin{definition}[(stochastisch)unabhängig] + \proplbl{3_2_11} + $(\Omega, \F, \probp)$ WRaum. Zwei Ereignisse $A,B \in \F$ heißt \begriff{(stochastisch) unabhängig bezüglich $\probp$}, falls + \begin{align*} + \probp(A\cap B) = \probp(A)\probp(B) + \end{align*} + Wir schreiben auch $A B$. +\end{definition} +\begin{example} + Würfeln mit 2 fairen, sechsseitigen Würfel: + \begin{align*} + \Omega = \set{(i,j) \mid i,j \in\set{1,\dots,n}},\quad \F = \pows(\Omega), \quad \probp = \Gleich(\Omega) + \intertext{Betrachte} + A:= \set{(i,j) \in \Omega, i \text{ gerade}}\\ + B:= \set{(i,j) \in \Omega, j > 2} + \end{align*} + In diesem Fall, erwarten wir intuitiv Unabhängigkeit von $A$ und $B$.\\ + In der Tat % start using \P instead of \probp! + \begin{align*} + \P(A) = \frac{1}{2}, \quad \P(B) = \frac{1}{3} \mit \P(A\cap B) = \frac{1}{6} + \intertext{erfüllt} + \P(A\cap B) = \P(A)\P(B). + \end{align*} + Betrachte nun + \begin{align*} + C:= \set{(i,j) \in \Omega, i\neq j = 1}\\ + D:= \set{(i,j) \in \Omega, i = 6} + \intertext{dann gilt} + \P(C) = \frac{1}{6}, \quad \P(D) = \frac{1}{6} + \intertext{und wegen $C \cap D = \set{(6,1)}$ folgt} + \P(C\cap D) = \frac{1}{36} = \frac{1}{6} \frac{1}{6} = \P(C \setminus D) + \end{align*} + $C$ und $D$ sind also \emph{stochastisch} unabhängig, obwohl eine kausale Abhängigkeit vorliegt! +\end{example} +\begin{definition}[unabhängig bezüglich $\P$] + $(\Omega, \F, \P)$ WRaum und $I \neq \emptyset$ endliche Indexmenge. Dann heißt die Familie $(A_i)_{i \in I}$ von Ereignissen in $\F$ \begriff{unabhängig bezüglich $\P$}, falls für alle $J \subseteq I, J \neq \emptyset$ gilt: + \begin{align*} + \P\brackets{\bigcap_{i\in J}A_i} = \prod_{i\in J} \P(A_i) + \end{align*} + Offensichtlich impliziert die Unabhängigkeit einer Familie die paarweise Unabhängigkeit je zweier Familienmitglieder nach \propref{3_2_11}. Umgekehrt gilt dies nicht! +\end{definition} +\begin{example}[Abhängigkeit trotz paarweiser Unabhängigkeit] + Betrachte 2-faches Bernoulliexperiment mit Erfolgswahrscheinlichkeit $\sfrac{1}{2}$, d.h. + \begin{align*} + \Omega = \set{0,1}^2, \quad \F = \pows(\Omega), \quad \P = \Gleich(\Omega) + \intertext{sowie} + A = \set{1}\times \set{0,1} \qquad \text{(Münzwurf: erster Wurf Zahl)}\\ + B = \set{0,1}\times \set{1} \qquad \text{(Münzwurf: zweiter Wurf Zahl)}\\ + C = \set{0,0}\times \set{1,1} \qquad \text{(beide Würfe selbes Ergebnis)} + \end{align*} + Dann gelten + \begin{align*} + \P(A) = \frac{1}{2} = \P(A) = \P(B) + \intertext{und} + \P(A\cap B) = \P(\set{(1,1)}) = \frac{1}{4} = \P(A)\P(B)\\ + \P(A\cap C) = \P(\set{(1,1)}) = \frac{1}{4} = \P(A)\P(C)\\ + \P(B\cap C) = \P(\set{(1,1)}) = \frac{1}{4} = \P(B)\P(C) + \end{align*} + also paarweise Unabhängigkeit.\\ + Aber + \begin{align*} + \P(A\cap B \cap C) = \P(\set{(1,1)}) = \frac{1}{4} + \P(A)\P(B)\P(C) + \intertext{und $A,B,C$ sind \emph{nicht} stochastisch unabhängig.} + \end{align*} +\end{example} +\begin{definition}[Unabhängige Messräume] + \proplbl{3_2_15} + % started using \O for \Omega and \E for this special generating set E_i + $(\O, \F,\P)$ Wahrscheinlichkeitsraum, $I \neq \emptyset$ Indexmenge und $(E_i, \E_i)$ Messräume + \begin{enumerate} + \item Die Familie $\F_i \subset \F, i \in I$, heißen \emph{unabhängig}, wenn für die $J \subseteq I, I \neq \emptyset, \abs{J} < \infty$ gilt + \begin{align*} + \P\brackets{\bigcap_{i=1} A_i} = \prod_{i\in J} \P(A_i) \qquad \text{ für beliebige } A_i \in \F_i, i \in J + \end{align*} + \item Die Zufallsvariable $X_i: (\O, \F) \to (E_i, \E_i), i \in I$, heißen \emph{unabhängig}, wenn die $\sigma$-Algebren + \begin{align*} + \sigma(X_i) = X^{-1}(\E_i) = \set{\set{X_i \in F}, F \in \E_i}, \quad i \in I + \end{align*} + unabhängig sind. + \end{enumerate} +\end{definition} +\begin{lemma}[Zusammenhang der Definitionen] + \proplbl{3_2_16} + $(\O,\F,\P)$ Wahrscheinlichkeitsraum, $I \neq \emptyset, A \in \F, i \in I$.\\ + Die folgenden Aussagen sind äquivalent: + \begin{enumerate} + \item Die Ereignisse $A_i, i \in I$ sind unabhängig. + \item Die $\sigma$-Algebren $\sigma(A_i), i \in I$ sind unabhängig. + \item Die Zufallsvariablen $\indi_{A_i}, i \in I$ sind unabhängig. + \end{enumerate} +\end{lemma} +\begin{proof} %TODO add ref? + Da die Unabhängigkeit über endliche Teilemengen definiert ist, können wir oBdA $I = \set{1, \dots, n}$ annehmen. + \begin{itemize} + \item Da $\sigma(\indi_{A_i}) = \sigma(A_i)$ folgt die Äquivalenz von 2. und 3. direkt aus \propref{3_2_15}. + \item Zudem ist 2. $\to $ 1. klar! + \item Für 1 $\to$ 2. genügt es zu zeigen, dass + \begin{align*} + A_1, \dots, A_n \text{ unabhängig } &\Rightarrow B_1, \dots, B_n \text{ unabhängig von } B_i \in \set{\emptyset, A_i, A_i^C, \O}. + \intertext{Rekursive folgt dies bereits aus} + A_1,\dots, A_n \text{ unabhängig } &\Rightarrow B_1, A_2, \dots, A_n \text{ unabhängig mit } B_1 \in \set{\emptyset, A_1, A_1^C, \O}. + \end{align*} + Für $B_1 \in \set{\emptyset, A_1, \O}$ ist dies klar.\\ + Sei also $B_1 = A_1^C$ und $J \subseteq I, J \neq \emptyset$. Falls $1 \not \in J$, ist nichts zu zeigen. Sei $1 \in J$, dann gilt mit + \begin{align*} + A &= \bigcap_{i\in J, i \neq 1} A_i + \intertext{sicherlich} + \P\brackets{A_1^C \cap A} &= \P(A \setminus (A_1 \cap A))\\ + &= \P(A) - \P(A_1 \cap A)\\ + &= \prod_{i\in J\setminus \set{1}} \P(A_i) - \prod_{i\in J}(A_i)\\ + &= (1- \P(A_1))\prod_{i\in J\setminus \set{1}} \P(A_i)\\ + &= \P\brackets{A_1^C})\prod_{i\in J\setminus \set{1}} \P(A_i) + \end{align*} + \end{itemize} +\end{proof} +Insbesondere zeigt das \propref{3_2_16}, dass wir in einer Familie unabhängiger Ereignisse beliebig viele Ereignisse durch ihr Komplement, $\emptyset$ oder $\O$ ersetzen können, ohne die Unabhängigkeit zu verlieren. +\begin{proposition} + \proplbl{3_2_17} + $(\O, \F, \P)$ Wahrscheinlichkeitsraum und $\F_i \subseteq \F, i \in I$, seien $\cap$-stabil Familien von Ereignissen. Dann + \begin{align*} + \F_i, i \in I \text{ unabhängig } \Leftrightarrow \sigma(\F_i), i \in I \text{ unabhängig}. + \end{align*} +\end{proposition} + +\begin{proof} + OBdA sei $I = \set{1, \dots, n}$ und das $\O \in \F_i, i \in I$. + \begin{itemize} + \item $\Leftarrow$: trivial, da $\F_i \subseteq \sigma(\F_i)$ und das Weglassen von Mengen erlaubt ist. + \item $\Rightarrow$: zeigen wir rekursive + \begin{enumerate} + \item Wähle $F_i \in \F_i, i = 2, \dots,n$ und defnieren für $F \in \sigma(\F_i)$ die endlichen Maße + \begin{align*} + \mu(F) = \P\brackets{\bigcap_{i=1}^n F_i} \und \nu(F) = \prod_{i=1}^n \P(F_i) + \end{align*} + \item Da die Familien $\F_i$ unabhängig sind, gilt + \begin{align*} + \mu\mid_{\F_1} = \nu\mid_{\F_1} + \end{align*} + Nach Eindeutigkeitssatz für Maße (\proplbl{1_1_19}) folgt $\mu\mid_{\sigma(\F_1)} = \nu\mid_{\sigma(\F_1)}$ also + \begin{align*} + \P(\bigcap_{i=1}^n F_i) = \P(F)\P(F_1)\dots \P(F_n) + \end{align*} + für alle $F \in \sigma(\F_i)$ und $F_i \in \F_i, i = 1, \dots, n$. Da $\O \in \F_i$ für alle $i$ gilt die erhaltene Produktformel auf für alle Teilemenge $J \subseteq I$.\\ + Also sind + \begin{align*} + \sigma(\F_1), \F_2, \dots, \F_n \text{unabhängig} + \end{align*} + \item Wiederholtes Anwenden von $1$ und $2$ liefert den \propref{3_2_17}. + \end{enumerate} + \end{itemize} +\end{proof} + +Mittels \propref{3_2_17} folgen: + +\begin{conclusion} + $(\O,\F,\P)$ Wahrscheinlichkeitsraum und + \begin{align*} + \F_{i,j} \subseteq \F, \quad 1 \le \dots \le n, 1 \le j \le m(i) + \end{align*} + unabhängige, $\cap$-stabile Familien. + Dann sind auch + \begin{align*} + \G_i = \sigma(\F_{i,1},\dots, \F_{i,m(i)}), \quad 1 \le i \le n + \end{align*} + unabhängig. +\end{conclusion} + +\begin{conclusion} + $(\O,\F,\P)$ Wahrscheinlichkeitsraum, und + \begin{align*} + X_{ij}: \O \to E, \quad 1 \le i \le n, 1 \le j \le m(i) + \end{align*} + unabhängige Zufallsvariablen. Zudem seinen $f_i: E^{m(i)} \to \R$ messbar. Dann sind auch die Zufallsvariablen + \begin{align*} + f_i(X_{i1}, \dots, X_{im(i)}), \quad 1 \le i \le n + \end{align*} + unabhängig. +\end{conclusion} + +\begin{example} + $X_1, \dots, X_n$ unabhängige reelle Zufallsvariablen. Dann sind auch + \begin{align*} + Y_1 = X_1, Y_2 = X_2 + \cdots + X_n + \end{align*} + unabhängig. +\end{example} \ No newline at end of file diff --git a/4. Semester/STOCH/Vorlesung STOCH.tex b/4. Semester/STOCH/Vorlesung STOCH.tex index e47e3e1..1f618dd 100644 --- a/4. Semester/STOCH/Vorlesung STOCH.tex +++ b/4. Semester/STOCH/Vorlesung STOCH.tex @@ -4,6 +4,18 @@ \title{\textbf{Stochastik SS 2019}} \author{Dozent: Prof. Dr. \person{Anita Behme}} +% local commands +\undef{\F} +\newcommand{\F}{\mathscr{F}} +\undef{\P} +\newcommand{\P}{\mathbb{P}} +\undef{\O} +\newcommand{\O}{\Omega} +\undef{\G} +\newcommand{\G}{\mathscr{G}} +%\undef{\base} +\newcommand{\E}{\mathcal{E}} + \begin{document} \pagenumbering{roman} \pagestyle{plain} diff --git a/Material/CAT/symbol_center.pdf b/Material/CAT/symbol_center.pdf new file mode 100644 index 0000000000000000000000000000000000000000..ec537353cbdc3427e52c1f627130bddb949daa7a GIT binary patch literal 18548 zcma&OW0Y)Nv#ndUZLV5n+qSXFwr$(CZQHhO+qUbhcfa30d*A!(s!jwHXC#>T^}m8E!PCfNF)-SaHi0Y%8M=4 zB`DZ*RG{|jJG534^;vB6>bkugeBVU)Bs@G4VL>^%d)0N=vL!Cia6oy`fT8q5FdbeF z07IMn@Rn5ZhY&uwjPM39OeFKMJXtpaxaAd^4=-EZh-H#Sx*_7NRIW@7BLrI`&xwk~ zit48gzI&mBk7`~P85bJpd~&Odjl8TBX``n^kG`N*ZTsS;h?R(}CQspg071GKp_u?O$v6 z*CjSz^3cl;A2e_A&!Xi3dy2qo7W4wsjTnVT$xI|)SJJ+fp*{b-rr0ai~RI3`F< 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z=(&~VqCYiew&kfs0=GXwTPz;X(Ca4}P3V5$HCBryZf%}^*}#-^qw=<1A3O$^Y@Gd8sV z^^wr@S_1d6po^JV0OJEyotYtqy#@y6rs#G6#Vmm70Y$HYfrS~mc?JfSz#FO1)md6% zh#8s~qr1<*($Ex`R#EM+G&V;M10!Q2jBqgm1`N7hV*?9d%%PeG?0=e~`@`7C(h``P zQ0z4}HZnvH17lNQhZ0?#nV~s`d1gky0eiT4B}IwAi@%DvY-|+tgEOmALFq+5C_leM z!3dO}fTfCOURu6_5h$yGI4%m&Hby2cj!rJdj;2OVt}ejdt+5->&u(t!Zf*u{&Zb62 hb_#@*#3}#_q2iLnq7raAH8wLbG2v2Gb@g}S0sxQKC7J*L literal 0 HcmV?d00001 diff --git a/Material/CAT/symbol_center_diagram_test.tex b/Material/CAT/symbol_center_diagram_test.tex new file mode 100644 index 0000000..268d178 --- /dev/null +++ b/Material/CAT/symbol_center_diagram_test.tex @@ -0,0 +1,21 @@ +\documentclass{article} +\usepackage{amsmath} +\usepackage{xypic} + +% thats a demo for putting a symbol in the center of a square diagram! + +\newcommand{\refsymbol}{{?}} + +\begin{document} +Here we have a diagram. + \begin{equation} + \xymatrix{ + A \ar[r] \ar[d] + \ar@{}[dr] | {\refsymbol} + & B \ar[d] \\ + C \ar[r] & D + } + \label{eq:diag} + \end{equation} +We want to refer to {\refsymbol} in the square shown in \eqref{eq:diag}. +\end{document} \ No newline at end of file From a7af922c01bc8268442e0ce15306ad6bb5046a6c Mon Sep 17 00:00:00 2001 From: Ameyah Date: Wed, 17 Apr 2019 14:06:55 +0200 Subject: [PATCH 7/9] replaced commands \probp --> \P \vert --> \mid \sigF --> \F \Omega --> \O --- .../STOCH/TeX_files/Bedingte_Wheiten.tex | 126 +++++++++--------- 1 file changed, 63 insertions(+), 63 deletions(-) diff --git a/4. Semester/STOCH/TeX_files/Bedingte_Wheiten.tex b/4. Semester/STOCH/TeX_files/Bedingte_Wheiten.tex index 69f8ee2..6e473bc 100644 --- a/4. Semester/STOCH/TeX_files/Bedingte_Wheiten.tex +++ b/4. Semester/STOCH/TeX_files/Bedingte_Wheiten.tex @@ -6,68 +6,68 @@ \proplbl{3_1_1} Das Würfeln mit zwei fairen, sechsseitigen Würfeln können wir mit \begin{align} - \Omega = \set{(i,j,), i,j \in \set{1,\dots,6}}\notag + \O = \set{(i,j,), i,j \in \set{1,\dots,6}}\notag \end{align} - und $\probp = \Gleich(\Omega)$. Da $\abs{\Omega} = 36$ gilt also + und $\P = \Gleich(\O)$. Da $\abs{\O} = 36$ gilt also \begin{align} - \probp(\set{\omega}) = \frac{1}{36} \quad \forall \omega \in \Omega.\notag + \P(\set{\omega}) = \frac{1}{36} \quad \forall \omega \in \O.\notag \end{align} Betrachte das Ereignis \begin{align} - A = \set{(i,j) \in \Omega : i + j = 8},\notag + A = \set{(i,j) \in \O : i + j = 8},\notag \end{align} dann folgt \begin{align} - \probp(A) = \frac{5}{36}.\notag + \P(A) = \frac{5}{36}.\notag \end{align} Werden die beiden Würfel nach einander ausgeführt, so kann nach dem ersten Wurf eine Neubewertung der Wahrscheinlichkeit von $A$ erfolgen.\\ Ist z.B.: \begin{align} - B = \set{(i,j) \in \Omega, i = 4}\notag + B = \set{(i,j) \in \O, i = 4}\notag \end{align} eingetreten, so kann die Summe 8 nur durch eine weitere 4 realisiert werden, also mit Wahrscheinlichkeit \begin{align} \frac{1}{6} = \frac{\abs{A \cap B}}{\abs{B}}.\notag \end{align} - Das Eintreten von $B$ führt also dazu, dass das Wahrscheinlichkeitsmaß $\probp$ durch ein neues Wahrscheinlichkeitsmaß $\probp_{B}$ ersetzt werden muss. Hierbei sollte gelten: + Das Eintreten von $B$ führt also dazu, dass das Wahrscheinlichkeitsmaß $\P$ durch ein neues Wahrscheinlichkeitsmaß $\P_{B}$ ersetzt werden muss. Hierbei sollte gelten: \begin{align} - &\text{Renormierung: }\probp_{B} = 1\label{Renorm}\tag{R}\\ - &\text{Proportionalität: Für alle} A \subset \sigF \mit A \subseteq B \text{ gilt } - \probp_{B}(A) = c_B \probp(A) \text{ mit einer Konstante } c_B.\label{Prop}\tag{P} + &\text{Renormierung: }\P_{B} = 1\label{Renorm}\tag{R}\\ + &\text{Proportionalität: Für alle} A \subset \F \mit A \subseteq B \text{ gilt } + \P_{B}(A) = c_B \P(A) \text{ mit einer Konstante } c_B.\label{Prop}\tag{P} \end{align} \end{example} \begin{lemma} - Sei $(\Omega, \sigF, \probp)$ Wahrscheinlichkeitsraum und $B \in \sigF$ mit $\probp(B) > 0$. Dann gibt es genau ein Wahrscheinlichkeitsmaß $\probp_B$ auf $(\Omega, \sigF)$ mit den Eigenschaften \eqref{Renorm} und \eqref{Prop}. Dieses ist gegeben durch + Sei $(\O, \F, \P)$ Wahrscheinlichkeitsraum und $B \in \F$ mit $\P(B) > 0$. Dann gibt es genau ein Wahrscheinlichkeitsmaß $\P_B$ auf $(\O, \F)$ mit den Eigenschaften \eqref{Renorm} und \eqref{Prop}. Dieses ist gegeben durch \begin{align} - \probp_{B}(A) = \frac{\probp(A\cap B)}{\probp(B)} \quad \forall A \in \sigF.\notag + \P_{B}(A) = \frac{\P(A\cap B)}{\P(B)} \quad \forall A \in \F.\notag \end{align} \end{lemma} \begin{proof} - Offenbar erfüllt $\probp_{B}$ wie definiert \eqref{Renorm} und \eqref{Prop}. Umgekehrt erfüllt $\probp_{B}$ \eqref{Renorm} und \eqref{Prop}. Dann folgt für $A \in \sigF$: + Offenbar erfüllt $\P_{B}$ wie definiert \eqref{Renorm} und \eqref{Prop}. Umgekehrt erfüllt $\P_{B}$ \eqref{Renorm} und \eqref{Prop}. Dann folgt für $A \in \F$: \begin{align} - \probp_{B}(A) = \probp_{B}(A\cap B) + \underbrace{\probp_{B}(A\setminus B)}_{= 0, \text{ wegen } \eqref{Renorm}} \overset{\eqref{Prop}}{=} c_B \probp(A \cap B).\notag + \P_{B}(A) = \P_{B}(A\cap B) + \underbrace{\P_{B}(A\setminus B)}_{= 0, \text{ wegen } \eqref{Renorm}} \overset{\eqref{Prop}}{=} c_B \P(A \cap B).\notag \end{align} Für $A=B$ folgt zudem aus \eqref{Renorm} \begin{align} - 1 = \probp_{B}(B) = c_B \probp(B)\notag + 1 = \P_{B}(B) = c_B \P(B)\notag \end{align} - also $c_B = \probp(B)^{-1}$. + also $c_B = \P(B)^{-1}$. \end{proof} % % % % % % % % % % % % % % % % % % % % % % % % % % % 5th lecture % % % % % % % % % % % % % % % % % % % % % % % % % % % \begin{definition} \proplbl{3_1_3} - Sei $(\Omega, \sigF, \probp)$ Wahrscheinlichkeitsraum und $B \in \sigF$ mit $\probp(B) > 0$. Dann heißt + Sei $(\O, \F, \P)$ Wahrscheinlichkeitsraum und $B \in \F$ mit $\P(B) > 0$. Dann heißt \begin{align*} - \probp(A\vert B) := \frac{\probp(A\cap B)}{\probp(B)} \mit A\in \sigF + \P(A\mid B) := \frac{\P(A\cap B)}{\P(B)} \mit A\in \F \end{align*} die \begriff{bedingte Wheit von $A$ gegeben $B$}. - Falls $\probp(B) = 0$, setze + Falls $\P(B) = 0$, setze \begin{align*} - \probp(A \vert B) = 0 \mit \forall A \in \sigF + \P(A \mid B) = 0 \mit \forall A \in \F \end{align*} \end{definition} @@ -76,23 +76,23 @@ \begin{align*} A \cap B = \set{(4,4)} \intertext{und damit} - \probp(A \vert B) = \frac{\probp(A\cap B)}{\probp(B)} = \frac{\frac{1}{36}}{\frac{1}{6}} = \frac{1}{6} + \P(A \mid B) = \frac{\P(A\cap B)}{\P(B)} = \frac{\frac{1}{36}}{\frac{1}{6}} = \frac{1}{6} \end{align*} \end{example} Aus \propref{3_1_3} ergibt sich \begin{lemma}[Multiplikationsformel] \proplbl{3_1_4} - Sei $(\Omega, \sigF, \probp)$ Wahrscheinlichkeitsraum und $A_1, \dots, A_n \in \sigF$. Dann + Sei $(\O, \F, \P)$ Wahrscheinlichkeitsraum und $A_1, \dots, A_n \in \F$. Dann \begin{align*} - \probp(A_1 \cap \cdots \cap A_n) = \probp(A_1)\probp(A_2 \vert A_n) \dots \probp(A_n \vert A_1 \cap \cdots \cap A_{n-1}) + \P(A_1 \cap \cdots \cap A_n) = \P(A_1)\P(A_2 \mid A_n) \dots \P(A_n \mid A_1 \cap \cdots \cap A_{n-1}) \end{align*} \end{lemma} \begin{proof} - Ist $\probp(A_1 \cap \dots \cap A_n) = 0$, so gilt auch $\probp(A_n \vert \bigcap_{i=1}^{n-1}) = 0$. Andernfalls sind alle Faktoren der rechten Seite ungleich 0 und + Ist $\P(A_1 \cap \dots \cap A_n) = 0$, so gilt auch $\P(A_n \mid \bigcap_{i=1}^{n-1}) = 0$. Andernfalls sind alle Faktoren der rechten Seite ungleich 0 und \begin{align*} - \probp(A_1)\probp(A_2 \vert A_1) \dots \probp(A_n \vert \bigcap_{i=1}^{n-1} A_i) \\ - = \probp(A_1) \cdot \frac{\probp(A_1 \cap A_2)}{\probp(A_1)} \dots \frac{\probp(\bigcap_{i=1}^{n} A_i)}{\probp(\bigcap_{i=1}^{n-1}A_i)} = \probp(\bigcap_{i=1}^n A_i) + \P(A_1)\P(A_2 \mid A_1) \dots \P(A_n \mid \bigcap_{i=1}^{n-1} A_i) \\ + = \P(A_1) \cdot \frac{\P(A_1 \cap A_2)}{\P(A_1)} \dots \frac{\P(\bigcap_{i=1}^{n} A_i)}{\P(\bigcap_{i=1}^{n-1}A_i)} = \P(\bigcap_{i=1}^n A_i) \end{align*} \end{proof} %TODO add ref. Stehen die $A_i$ in \propref{3_1_4} in einer (zeitlichen) Abfolge, so liefert Formel einen Hinweis, wie Wahrscheinlichkeitsmaße für \begriff{Stufenexperimente} konstruiert werden können. Ein \emph{Stufenexperiment} aus $n$ nacheinander ausgeführten Teilexperimenten lässt sich als \begriff{Baumdiagramm} darstellen. @@ -104,17 +104,17 @@ Stehen die $A_i$ in \propref{3_1_4} in einer (zeitlichen) Abfolge, so liefert Fo \begin{proposition}[Konstruktion des Wahrscheinlichkeitsmaßes eines Stufenexperiments] \proplbl{3_1_6} - Gegeben seinen $n$ Ergebnisräume $\Omega_i = \set{\omega_i (1), \dots, \omega_i (k)}, k \in \N \cup \set{\infty}$ und es sei $\Omega = \bigtimes_{i = 1}^n \Omega_i$ der zugehörige Produktraum. Weiter seinen $\sigF_i$ $\sigma$-Algebren auf $\Omega_i$ und $\sigF = \bigotimes_{i=1}^n \sigF_i$ die Produkt-$\sigma$-Algebra auf $\Omega$. Setze $\omega = (\omega_1,\dots,\omega_n)$ und + Gegeben seinen $n$ Ergebnisräume $\O_i = \set{\omega_i (1), \dots, \omega_i (k)}, k \in \N \cup \set{\infty}$ und es sei $\O = \bigtimes_{i = 1}^n \O_i$ der zugehörige Produktraum. Weiter seinen $\F_i$ $\sigma$-Algebren auf $\O_i$ und $\F = \bigotimes_{i=1}^n \F_i$ die Produkt-$\sigma$-Algebra auf $\O$. Setze $\omega = (\omega_1,\dots,\omega_n)$ und \begin{align*} - [\omega_1,\dots,\omega_n]:= \set{\omega_1}\times \dots \times \set{\omega_n} \times \Omega_{m-1} \times \Omega_{n},\quad m\le n\\ - \probp(\set{\omega_m}[\omega_1,\dots,\omega_{m-1}]) + [\omega_1,\dots,\omega_n]:= \set{\omega_1}\times \dots \times \set{\omega_n} \times \O_{m-1} \times \O_{n},\quad m\le n\\ + \P(\set{\omega_m}[\omega_1,\dots,\omega_{m-1}]) \end{align*} %TODO check indices, m-1 instead of m+1? für die Wheit in der $m$-ten Stufe des Experiments $\omega_m$ zu beobachten, falls in den vorausgehenden Stufen $\omega_1,\dots,\omega_{m-1}$ beobachten wurden. Dann definiert \begin{align*} - \probp(\set{\omega}) := \probp(\set{\omega_1})\prod_{m=2}^{n}\probp\brackets{\set{\omega_m} \mid [\omega_1, \dots, \omega_{m-1}]} + \P(\set{\omega}) := \P(\set{\omega_1})\prod_{m=2}^{n}\P\brackets{\set{\omega_m} \mid [\omega_1, \dots, \omega_{m-1}]} %TODO maybe wrong here. check \end{align*} - ein Wahrscheinlichkeitsmaß auf $(\Omega, \sigF, \probp)$. + ein Wahrscheinlichkeitsmaß auf $(\O, \F, \P)$. \end{proposition} \begin{example}[\person{Polya}-Urne] @@ -126,26 +126,26 @@ Stehen die $A_i$ in \propref{3_1_4} in einer (zeitlichen) Abfolge, so liefert Fo Beide schon in Kapitel 2.2 gesehen.\\ Sei $c\in \N$. (Modell für zwei konkurrierende Populationen) Ziehen wir $n$-mal, so erhalten wir ein $n$-Stufenexperiment mit \begin{align*} - \Omega = \set{0,1}^n \mit \text{ 0 = ``weiß'', 1 = ``schwarz''}\mit (\Omega_i = \set{0,1}) + \O = \set{0,1}^n \mit \text{ 0 = ``weiß'', 1 = ``schwarz''}\mit (\O_i = \set{0,1}) \intertext{Zudem gelten im ersten Schritt} - \probp(\set{0}) = \frac{w}{s+w} \und \probp(\set{1}) = \frac{s}{w+s} + \P(\set{0}) = \frac{w}{s+w} \und \P(\set{1}) = \frac{s}{w+s} \intertext{sowie} - \probp(\set{\omega_m} \mid [\omega_1, \dots \omega_{m-1}]) = + \P(\set{\omega_m} \mid [\omega_1, \dots \omega_{m-1}]) = \begin{cases} %TODO fix brackets! \frac{w+c(m-1 - \sum_{i=1}^{m-1}\omega_i)}{s+w+c(m-1)} & \omega_m = 0\\ \frac{s + c\sum_{i=1}^{m-1}\omega_i}{s+w+c(m-1)} & \omega_m = 1 \end{cases} \end{align*} - Mit \propref{3_1_6} folgt als Wahrscheinlichkeitsmaß auf $(\Omega, \pows(\Omega))$ + Mit \propref{3_1_6} folgt als Wahrscheinlichkeitsmaß auf $(\O, \pows(\O))$ \begin{align*} - \probp(\set{(\omega_1, \dots, \omega_n)}) &= \probp(\set{\omega_1})\prod_{m=2}^n \probp(\set{\omega_m}\mid [\omega_1,\dots,\omega_{m-1}])\\ + \P(\set{(\omega_1, \dots, \omega_n)}) &= \P(\set{\omega_1})\prod_{m=2}^n \P(\set{\omega_m}\mid [\omega_1,\dots,\omega_{m-1}])\\ &=\frac{\prod_{i=0}^{l-1}(s+c_i)\prod_{i=0}^{n-l-1}}{\prod_{i=0}^n (s+w+c_i)} \mit l=\sum_{i=1}^n \omega_i. \intertext{Definiere wir nun die Zufallsvariable} - S_n:\Omega &\to \N_0 \mit (\omega_1, \dots, \omega_n) \mapsto \sum_{i=1}^n \omega_i + S_n:\O &\to \N_0 \mit (\omega_1, \dots, \omega_n) \mapsto \sum_{i=1}^n \omega_i \intertext{welche die Anzahl der gezogenen schwarzen Kugeln modelliert, so folgt,} - \probp(S_n = l) &= \binom{n}{l}\frac{\prod_{i=0}^{l-1}(s+c_i) \prod_{j=0}^{n-l-1}(\omega + c_j)}{\prod_{i=0}^n(s+w+c_i)} + \P(S_n = l) &= \binom{n}{l}\frac{\prod_{i=0}^{l-1}(s+c_i) \prod_{j=0}^{n-l-1}(\omega + c_j)}{\prod_{i=0}^n(s+w+c_i)} \intertext{Mittels $a:= \sfrac{s}{c},b:= \sfrac{w}{c}$ folgt} - \probp(S_n = l) &= \frac{\prod_{i=0}^{l-1}(-a-i)\prod_{i=0}^{-b-j-1}}{\prod_{i=0}^n (-a-b-i)} = \frac{\binom{-a}{l}\binom{-b}{n-l}}{\binom{-a-b}{n}}\\ &\mit l \in \set{0,\dots,n} + \P(S_n = l) &= \frac{\prod_{i=0}^{l-1}(-a-i)\prod_{i=0}^{-b-j-1}}{\prod_{i=0}^n (-a-b-i)} = \frac{\binom{-a}{l}\binom{-b}{n-l}}{\binom{-a-b}{n}}\\ &\mit l \in \set{0,\dots,n} \end{align*} Dies ist die \begriff{\person{Polya}-Urne} auf $\set{0,\dots,n}, n \in \N$ mit Parametern $a,b > 0$. \end{example} @@ -157,25 +157,25 @@ Stehen die $A_i$ in \propref{3_1_4} in einer (zeitlichen) Abfolge, so liefert Fo W = \set{\text{richtige Antwort gewusst}}\\ R = \set{\text{Richtige Antwort gewählt}} \intertext{Dann} - \probp(W) = \frac{2}{3}, \probp(R \vert W) = 1, \probp(R \vert W^C) = \frac{1}{4} + \P(W) = \frac{2}{3}, \P(R \mid W) = 1, \P(R \mid W^C) = \frac{1}{4} \end{align*} Angenommen, der Student gibt die richtige Antwort. Mit welcher Wahrscheinlichkeit hat er diese gewusst? \begin{align*} - \probp(W\vert R) = \text{ ?} + \P(W\mid R) = \text{ ?} \end{align*} \end{example} \begin{proposition} \proplbl{3_1_9} - Sei $(\Omega, \sigF, \probp)$ Wahrscheinlichkeitsraum und $\Omega = \bigcup_{i \in I} B_i$ eine höchstens abzählbare Zerlegung in paarweise disjunkte Ereignisse $B_i \in \sigF$. + Sei $(\O, \F, \P)$ Wahrscheinlichkeitsraum und $\O = \bigcup_{i \in I} B_i$ eine höchstens abzählbare Zerlegung in paarweise disjunkte Ereignisse $B_i \in \F$. \begin{enumerate} %TODO set itemize references. or use enumerate? - \item \emph{Satz von der totalen Wahrscheinlichkeit:} Für alle $A \in \sigF$ + \item \emph{Satz von der totalen Wahrscheinlichkeit:} Für alle $A \in \F$ \begin{align*} - \probp(A) = \sum_{i\in I} \probp(A\vert B_i)\probp(B_i) \label{eq:totWkeit}\tag{totale Wahrscheinlichkeit} + \P(A) = \sum_{i\in I} \P(A\mid B_i)\P(B_i) \label{eq:totWkeit}\tag{totale Wahrscheinlichkeit} \end{align*} - \item \emph{Satz von \person{Bayes}:} Für alle $A \in \sigF$ mit $\probp(A) > 0$ und alle $h \in I$ + \item \emph{Satz von \person{Bayes}:} Für alle $A \in \F$ mit $\P(A) > 0$ und alle $h \in I$ \begin{align*} - \probp(B_h \vert A) = \frac{\probp(A \vert B_h)\probp(B_h)}{\sum_{i\in I}\probp(A\vert B_i)\probp(B_i)} \label{eq:bayes}\tag{Bayes} + \P(B_h \mid A) = \frac{\P(A \mid B_h)\P(B_h)}{\sum_{i\in I}\P(A\mid B_i)\P(B_i)} \label{eq:bayes}\tag{Bayes} \end{align*} \end{enumerate} \end{proposition} @@ -184,11 +184,11 @@ Stehen die $A_i$ in \propref{3_1_4} in einer (zeitlichen) Abfolge, so liefert Fo \begin{enumerate} \item Es gilt: \begin{align*} - \sum_{i\in I} \probp(A\vert B_i)\probp(B_i) \defeq \sum_{i\in I}\frac{\probp(A \cap B_i)}{\probp(B_i)}\probp(B_i) = \sum_{i\in I} \probp(A \cap B_i) \overset{\sigma-add.}{=} \probp(A) + \sum_{i\in I} \P(A\mid B_i)\P(B_i) \defeq \sum_{i\in I}\frac{\P(A \cap B_i)}{\P(B_i)}\P(B_i) = \sum_{i\in I} \P(A \cap B_i) \overset{\sigma-add.}{=} \P(A) \end{align*} \item \begin{align*} - \probp(B_h \vert A) \defeq \frac{\probp(A \cap B_h)}{\probp(A)} \defeq \frac{\probp(A \vert B_h)\probp(B_h)}{\probp(A)} + \P(B_h \mid A) \defeq \frac{\P(A \cap B_h)}{\P(A)} \defeq \frac{\P(A \mid B_h)\P(B_h)}{\P(A)} \end{align*} also mit a) auch b). %TODO add refs \end{enumerate} @@ -197,10 +197,10 @@ Stehen die $A_i$ in \propref{3_1_4} in einer (zeitlichen) Abfolge, so liefert Fo \begin{example} In der Situation von \propref{3_1_3} folgt mit dem \propref{3_1_9} \eqref{eq:totWkeit} der totalen Wahrscheinlichkeit \begin{align*} - \probp(R) &= \probp(R \vert W)\probp(W) + \probp(R\vert W^C)\probp(W^C)\\ + \P(R) &= \P(R \mid W)\P(W) + \P(R\mid W^C)\P(W^C)\\ &= 1\cdot \frac{2}{3} + \frac{1}{4}\frac{1}{3} = \frac{3}{4} \intertext{und mit dem \propref{3_1_9} \eqref{eq:bayes}} %Bayes - \probp(W \vert R) &= \frac{\probp(R \vert W)\probp(W)}{\probp(R)} = \frac{1\frac{2}{3}}{\frac{3}{4}} = \frac{8}{9} \text{ für die gesuchte Wahrscheinlichkeit.} + \P(W \mid R) &= \frac{\P(R \mid W)\P(W)}{\P(R)} = \frac{1\frac{2}{3}}{\frac{3}{4}} = \frac{8}{9} \text{ für die gesuchte Wahrscheinlichkeit.} \end{align*} %TODO % \begin{tikzpicture} % @@ -209,28 +209,28 @@ Stehen die $A_i$ in \propref{3_1_4} in einer (zeitlichen) Abfolge, so liefert Fo % trees in Latex? % https://tex.stackexchange.com/questions/5447/how-can-i-draw-simple-trees-in-latex/5451 \section{(Un)-abhängigkeit} \label{sec_unabhangigkeit} -In vielen Fällen besagt die Intuition über verschiedene Zufallsexperimente/ Ereignisse, dass diese sich \emph{nicht} gegenseitig beeinflussen. Für solche $A,B \in \F$ mit $\probp(A) > 0, \probp(B) > 0$ sollte gelten +In vielen Fällen besagt die Intuition über verschiedene Zufallsexperimente/ Ereignisse, dass diese sich \emph{nicht} gegenseitig beeinflussen. Für solche $A,B \in \F$ mit $\P(A) > 0, \P(B) > 0$ sollte gelten \begin{align*} - \probp(A\mid B) = \probp(B), \quad \probp(B\mid A) = \probp(B). + \P(A\mid B) = \P(B), \quad \P(B\mid A) = \P(B). \end{align*} \begin{definition}[(stochastisch)unabhängig] \proplbl{3_2_11} - $(\Omega, \F, \probp)$ WRaum. Zwei Ereignisse $A,B \in \F$ heißt \begriff{(stochastisch) unabhängig bezüglich $\probp$}, falls + $(\O, \F, \P)$ WRaum. Zwei Ereignisse $A,B \in \F$ heißt \begriff{(stochastisch) unabhängig bezüglich $\P$}, falls \begin{align*} - \probp(A\cap B) = \probp(A)\probp(B) + \P(A\cap B) = \P(A)\P(B) \end{align*} Wir schreiben auch $A B$. \end{definition} \begin{example} Würfeln mit 2 fairen, sechsseitigen Würfel: \begin{align*} - \Omega = \set{(i,j) \mid i,j \in\set{1,\dots,n}},\quad \F = \pows(\Omega), \quad \probp = \Gleich(\Omega) + \O = \set{(i,j) \mid i,j \in\set{1,\dots,n}},\quad \F = \pows(\O), \quad \P = \Gleich(\O) \intertext{Betrachte} - A:= \set{(i,j) \in \Omega, i \text{ gerade}}\\ - B:= \set{(i,j) \in \Omega, j > 2} + A:= \set{(i,j) \in \O, i \text{ gerade}}\\ + B:= \set{(i,j) \in \O, j > 2} \end{align*} In diesem Fall, erwarten wir intuitiv Unabhängigkeit von $A$ und $B$.\\ - In der Tat % start using \P instead of \probp! + In der Tat % start using \P instead of \P! \begin{align*} \P(A) = \frac{1}{2}, \quad \P(B) = \frac{1}{3} \mit \P(A\cap B) = \frac{1}{6} \intertext{erfüllt} @@ -238,8 +238,8 @@ In vielen Fällen besagt die Intuition über verschiedene Zufallsexperimente/ Er \end{align*} Betrachte nun \begin{align*} - C:= \set{(i,j) \in \Omega, i\neq j = 1}\\ - D:= \set{(i,j) \in \Omega, i = 6} + C:= \set{(i,j) \in \O, i\neq j = 1}\\ + D:= \set{(i,j) \in \O, i = 6} \intertext{dann gilt} \P(C) = \frac{1}{6}, \quad \P(D) = \frac{1}{6} \intertext{und wegen $C \cap D = \set{(6,1)}$ folgt} @@ -248,7 +248,7 @@ In vielen Fällen besagt die Intuition über verschiedene Zufallsexperimente/ Er $C$ und $D$ sind also \emph{stochastisch} unabhängig, obwohl eine kausale Abhängigkeit vorliegt! \end{example} \begin{definition}[unabhängig bezüglich $\P$] - $(\Omega, \F, \P)$ WRaum und $I \neq \emptyset$ endliche Indexmenge. Dann heißt die Familie $(A_i)_{i \in I}$ von Ereignissen in $\F$ \begriff{unabhängig bezüglich $\P$}, falls für alle $J \subseteq I, J \neq \emptyset$ gilt: + $(\O, \F, \P)$ WRaum und $I \neq \emptyset$ endliche Indexmenge. Dann heißt die Familie $(A_i)_{i \in I}$ von Ereignissen in $\F$ \begriff{unabhängig bezüglich $\P$}, falls für alle $J \subseteq I, J \neq \emptyset$ gilt: \begin{align*} \P\brackets{\bigcap_{i\in J}A_i} = \prod_{i\in J} \P(A_i) \end{align*} @@ -257,7 +257,7 @@ In vielen Fällen besagt die Intuition über verschiedene Zufallsexperimente/ Er \begin{example}[Abhängigkeit trotz paarweiser Unabhängigkeit] Betrachte 2-faches Bernoulliexperiment mit Erfolgswahrscheinlichkeit $\sfrac{1}{2}$, d.h. \begin{align*} - \Omega = \set{0,1}^2, \quad \F = \pows(\Omega), \quad \P = \Gleich(\Omega) + \O = \set{0,1}^2, \quad \F = \pows(\O), \quad \P = \Gleich(\O) \intertext{sowie} A = \set{1}\times \set{0,1} \qquad \text{(Münzwurf: erster Wurf Zahl)}\\ B = \set{0,1}\times \set{1} \qquad \text{(Münzwurf: zweiter Wurf Zahl)}\\ @@ -280,7 +280,7 @@ In vielen Fällen besagt die Intuition über verschiedene Zufallsexperimente/ Er \end{example} \begin{definition}[Unabhängige Messräume] \proplbl{3_2_15} - % started using \O for \Omega and \E for this special generating set E_i + % started using \O for \O and \E for this special generating set E_i $(\O, \F,\P)$ Wahrscheinlichkeitsraum, $I \neq \emptyset$ Indexmenge und $(E_i, \E_i)$ Messräume \begin{enumerate} \item Die Familie $\F_i \subset \F, i \in I$, heißen \emph{unabhängig}, wenn für die $J \subseteq I, I \neq \emptyset, \abs{J} < \infty$ gilt From a24822f472b26b7035633bde2069e94c4e206aba Mon Sep 17 00:00:00 2001 From: Ameyah Date: Wed, 17 Apr 2019 14:49:56 +0200 Subject: [PATCH 8/9] new structure and figures in. --- .../Bedingte_Wahrscheinlichkeiten.tex | 218 +++++++++++++++--- .../STOCH/TeX_files/Bedingte_Wheiten.tex | 62 ++--- .../STOCH/TeX_files/Erste_Standardmodelle.tex | 2 +- .../unbedingte_Wahrscheinlichkeiten.tex | 191 +++++++++++++++ 4. Semester/STOCH/Vorlesung STOCH.tex | 21 +- 4. Semester/STOCH/tikz/baum_1.tex | 50 ++-- texmf/tex/latex/mathscript/mathscript.cls | 2 +- 7 files changed, 443 insertions(+), 103 deletions(-) create mode 100644 4. Semester/STOCH/TeX_files/unbedingte_Wahrscheinlichkeiten.tex diff --git a/4. Semester/STOCH/TeX_files/Bedingte_Wahrscheinlichkeiten.tex b/4. Semester/STOCH/TeX_files/Bedingte_Wahrscheinlichkeiten.tex index 2549407..9943803 100644 --- a/4. Semester/STOCH/TeX_files/Bedingte_Wahrscheinlichkeiten.tex +++ b/4. Semester/STOCH/TeX_files/Bedingte_Wahrscheinlichkeiten.tex @@ -1,59 +1,211 @@ -\chapter{Bedingte Wahrscheinlichkeiten und (Un)-abbhängigkeit} - \section{Bedingte Wahrscheinlichkeiten} - \begin{example} - Das Würfeln mit $2$ fairen, sechseitigen Würfeln können wir mit + \proplbl{3_1_1} + Das Würfeln mit zwei fairen, sechsseitigen Würfeln können wir mit \begin{align} - \Omega = \set{(i,j, \colon, i,j \in \set{1,\dots,6}} \quad \und \quad \probp = \Gleich(\Omega) \notag - \end{align} - modellieren. Da $\abs{\Omega} = 36$ gilt also - \begin{align} - \probp(\set{\omega}) = \frac{1}{36} \quad \forall \omega \in \Omega.\notag + \O = \set{(i,j,), i,j \in \set{1,\dots,6}}\notag + \end{align} + und $\P = \Gleich(\O)$. Da $\abs{\O} = 36$ gilt also + \begin{align} + \P(\set{\omega}) = \frac{1}{36} \quad \forall \omega \in \O.\notag \end{align} Betrachte das Ereignis \begin{align} - A = \set{(i,j) \in \Omega : i + j = 8},\notag + A = \set{(i,j) \in \O : i + j = 8},\notag \end{align} dann folgt \begin{align} - \probp(A) = \frac{5}{36}.\notag + \P(A) = \frac{5}{36}.\notag \end{align} Werden die beiden Würfel nach einander ausgeführt, so kann nach dem ersten Wurf eine Neubewertung der Wahrscheinlichkeit von $A$ erfolgen.\\ Ist z.B.: \begin{align} - B = \set{(i,j) \in \Omega, i = 4}\notag + B = \set{(i,j) \in \O, i = 4}\notag \end{align} - eingetreten, so kann die Summe $8$ nur durch eine weitere $4$ realisiert werden, also mit Wahrscheinlichkeit + eingetreten, so kann die Summe 8 nur durch eine weitere 4 realisiert werden, also mit Wahrscheinlichkeit \begin{align} - \frac{1}{6} = \frac{\abs{A \cap B}}{\abs{B}}\notag + \frac{1}{6} = \frac{\abs{A \cap B}}{\abs{B}}.\notag \end{align} - Diese Eintreten von $B$ führt also dazu, dass das Wahrscheinlichkeitsmaß $\probp$ durch ein neues Wahrscheinlichkeitsmaß $\probp_{B}$ ersetzt werden muss. Hierbei sollte gelten: - \begin{itemize} - \item[(R)] Renormierung: $\probp_{B} = 1$ - \item[(P)] Proportionalität: Für alle $A \in \sigF$ mit $A \subseteq B$ gilt - \begin{align} - \probp_{B}(A) = c_B \probp(A)\notag - \end{align} - eine Konstante $c_B$. - \end{itemize} + Das Eintreten von $B$ führt also dazu, dass das Wahrscheinlichkeitsmaß $\P$ durch ein neues Wahrscheinlichkeitsmaß $\P_{B}$ ersetzt werden muss. Hierbei sollte gelten: + \begin{align} + &\text{Renormierung: }\P_{B} = 1\label{Renorm}\tag{R}\\ + &\text{Proportionalität: Für alle} A \subset \F \mit A \subseteq B \text{ gilt } + \P_{B}(A) = c_B \P(A) \text{ mit einer Konstante } c_B.\label{Prop}\tag{P} + \end{align} \end{example} \begin{lemma} - Sei $(\Omega, \sigF, \probp)$ WRaum und $B \in \sigF$ mit $\probp(B) > 0$. Dann gibt es genau ein WMaß $\probp_B$ auf $(\Omega, \sigF)$ mit den Eigenschaften (R) und (P). Dieses ist gegeben durch + Sei $(\O, \F, \P)$ Wahrscheinlichkeitsraum und $B \in \F$ mit $\P(B) > 0$. Dann gibt es genau ein Wahrscheinlichkeitsmaß $\P_B$ auf $(\O, \F)$ mit den Eigenschaften \eqref{Renorm} und \eqref{Prop}. Dieses ist gegeben durch \begin{align} - \probp_{B}(A) = \frac{\probp(A\cap B)}{\probp(B)} \quad \forall A \in \sigF.\notag + \P_{B}(A) = \frac{\P(A\cap B)}{\P(B)} \quad \forall A \in \F.\notag \end{align} \end{lemma} \begin{proof} - Offenbar erfüllt $\probp_{B}$ wie definiert (R) und (P). Umgekehrt erfülle $\probp_{B}$ (R) und (P). Dann folgt für $A \in \sigF$: + Offenbar erfüllt $\P_{B}$ wie definiert \eqref{Renorm} und \eqref{Prop}. Umgekehrt erfüllt $\P_{B}$ \eqref{Renorm} und \eqref{Prop}. Dann folgt für $A \in \F$: \begin{align} - \probp_{B}(A) = \probp_{B}(A\cap B) + \underbrace{\probp_{B}(A\setminus B)}_{= 0, \text{ wegen } (R)} \overset{(P)}{=} c_B \probp(A \cap B) \notag + \P_{B}(A) = \P_{B}(A\cap B) + \underbrace{\P_{B}(A\setminus B)}_{= 0, \text{ wegen } \eqref{Renorm}} \overset{\eqref{Prop}}{=} c_B \P(A \cap B).\notag \end{align} - Für $A=B$ folgt zudem aus (R) - \begin{align} - 1 = \probp_{B}(B) = c_B \probp(B) \notag - \end{align} - also $c_B = \probp(B)^{-1}$. -\end{proof} \ No newline at end of file + Für $A=B$ folgt zudem aus \eqref{Renorm} + \begin{align} + 1 = \P_{B}(B) = c_B \P(B)\notag + \end{align} + also $c_B = \P(B)^{-1}$. +\end{proof} + +% % % % % % % % % % % % % % % % % % % % % % % % % % % 5th lecture % % % % % % % % % % % % % % % % % % % % % % % % % % % + +\begin{definition} + \proplbl{3_1_3} + Sei $(\O, \F, \P)$ Wahrscheinlichkeitsraum und $B \in \F$ mit $\P(B) > 0$. Dann heißt + \begin{align*} + \P(A\mid B) := \frac{\P(A\cap B)}{\P(B)} \mit A\in \F + \end{align*} + die \begriff{bedingte Wahrscheinlichkeit von $A$ gegeben $B$}. + Falls $\P(B) = 0$, setze + \begin{align*} + \P(A \mid B) = 0 \mit \forall A \in \F + \end{align*} +\end{definition} + +\begin{example} %TODO ref + In der Situation \propref{3_1_1} gilt % + \begin{align*} + A \cap B &= \set{(4,4)} + \intertext{und damit} + \P(A \mid B) &= \frac{\P(A\cap B)}{\P(B)} = \frac{\frac{1}{36}}{\frac{1}{6}} = \frac{1}{6} + \end{align*} +\end{example} +Aus \propref{3_1_3} ergibt sich +\begin{lemma}[Multiplikationsformel] + \proplbl{3_1_4} + Sei $(\O, \F, \P)$ Wahrscheinlichkeitsraum und $A_1, \dots, A_n \in \F$. Dann + \begin{align*} + \P(A_1 \cap \cdots \cap A_n) = \P(A_1)\P(A_2 \mid A_n) \dots \P(A_n \mid A_1 \cap \cdots \cap A_{n-1}) + \end{align*} +\end{lemma} + +\begin{proof} + Ist $\P(A_1 \cap \dots \cap A_n) = 0$, so gilt auch $\P(A_n \mid \bigcap_{i=1}^{n-1}) = 0$. Andernfalls sind alle Faktoren der rechten Seite ungleich 0 und + \begin{align*} + \P(A_1)\P(A_2 \mid A_1) \dots \P(A_n \mid \bigcap_{i=1}^{n-1} A_i) \\ + = \P(A_1) \cdot \frac{\P(A_1 \cap A_2)}{\P(A_1)} \dots \frac{\P(\bigcap_{i=1}^{n} A_i)}{\P(\bigcap_{i=1}^{n-1}A_i)} = \P(\bigcap_{i=1}^n A_i) + \end{align*} +\end{proof} %TODO add ref. +Stehen die $A_i$ in \propref{3_1_4} in einer (zeitlichen) Abfolge, so liefert Formel einen Hinweis, wie Wahrscheinlichkeitsmaße für \begriff{Stufenexperimente} konstruiert werden können. Ein \emph{Stufenexperiment} aus $n$ nacheinander ausgeführten Teilexperimenten lässt sich als \begriff{Baumdiagramm} darstellen. + +%TODO Baumdiagramm +\begin{center} + \begin{tikzpicture} + \input{./tikz/baum_1} + % \caption{ \propref{3_1_4}} + \end{tikzpicture} +\end{center} + +\begin{proposition}[Konstruktion des Wahrscheinlichkeitsmaßes eines Stufenexperiments] + \proplbl{3_1_6} + Gegeben seinen $n$ Ergebnisräume $\O_i = \set{\omega_i (1), \dots, \omega_i (k)}, k \in \N \cup \set{\infty}$ und es sei $\O = \bigtimes_{i = 1}^n \O_i$ der zugehörige Produktraum. Weiter seinen $\F_i$ $\sigma$-Algebren auf $\O_i$ und $\F = \bigotimes_{i=1}^n \F_i$ die Produkt-$\sigma$-Algebra auf $\O$. Setze $\omega = (\omega_1,\dots,\omega_n)$ und + \begin{align*} + [\omega_1,\dots,\omega_n]:= \set{\omega_1}\times \dots \times \set{\omega_n} \times \O_{m-1} \times \O_{n},\quad m\le n\\ + \P(\set{\omega_m}[\omega_1,\dots,\omega_{m-1}]) + \end{align*} %TODO check indices, m-1 instead of m+1? + für die Wahrscheinlichkeit in der $m$-ten Stufe des Experiments $\omega_m$ zu beobachten, falls in den vorausgehenden Stufen $\omega_1,\dots,\omega_{m-1}$ beobachten wurden. Dann definiert + \begin{align*} + \P(\set{\omega}) := \P(\set{\omega_1})\prod_{m=2}^{n}\P\brackets{\set{\omega_m} \mid [\omega_1, \dots, \omega_{m-1}]} + %TODO maybe wrong here. check + \end{align*} + ein Wahrscheinlichkeitsmaß auf $(\O, \F, \P)$. +\end{proposition} + +\begin{example}[\person{Polya}-Urne] + Gegeben sei eine Urne mit $s$ Schwarze und $w$ weiße Kugeln. Bei jedem Zug wird die gezogene Kugel zusammen mit $c\in \N_0\cup \set{-1}$ weiteren Kugeln derselben Farbe zurückgelegt. + \begin{itemize} %TODO seen both in chapter 2.2, but big bracket behind. + \item $c=0$: Urnenmodell mit Zurücklegen + \item $c=-1$: Urnenmodell ohne Zurücklegen + \end{itemize} + Beide schon in Kapitel 2.2 gesehen.\\ + Sei $c\in \N$. (Modell für zwei konkurrierende Populationen) Ziehen wir $n$-mal, so erhalten wir ein $n$-Stufenexperiment mit + \begin{align*} + \O = \set{0,1}^n \mit \text{ 0 = ``weiß'', 1 = ``schwarz''}\mit (\O_i = \set{0,1}) + \intertext{Zudem gelten im ersten Schritt} + \P(\set{0}) = \frac{w}{s+w} \und \P(\set{1}) = \frac{s}{w+s} + \intertext{sowie} + \P(\set{\omega_m} \mid [\omega_1, \dots \omega_{m-1}]) = + \begin{cases} %TODO fix brackets! + \frac{w+c(m-1 - \sum_{i=1}^{m-1}\omega_i)}{s+w+c(m-1)} & \omega_m = 0\\ + \frac{s + c\sum_{i=1}^{m-1}\omega_i}{s+w+c(m-1)} & \omega_m = 1 + \end{cases} + \end{align*} + Mit \propref{3_1_6} folgt als Wahrscheinlichkeitsmaß auf $(\O, \pows(\O))$ + \begin{align*} + \P(\set{(\omega_1, \dots, \omega_n)}) &= \P(\set{\omega_1})\prod_{m=2}^n \P(\set{\omega_m}\mid [\omega_1,\dots,\omega_{m-1}])\\ + &=\frac{\prod_{i=0}^{l-1}(s+c_i)\prod_{i=0}^{n-l-1}}{\prod_{i=0}^n (s+w+c_i)} \mit l=\sum_{i=1}^n \omega_i. + \intertext{Definiere wir nun die Zufallsvariable} + S_n:\O &\to \N_0 \mit (\omega_1, \dots, \omega_n) \mapsto \sum_{i=1}^n \omega_i + \intertext{welche die Anzahl der gezogenen schwarzen Kugeln modelliert, so folgt,} + \P(S_n = l) &= \binom{n}{l}\frac{\prod_{i=0}^{l-1}(s+c_i) \prod_{j=0}^{n-l-1}(\omega + c_j)}{\prod_{i=0}^n(s+w+c_i)} + \intertext{Mittels $a:= \sfrac{s}{c},b:= \sfrac{w}{c}$ folgt} + \P(S_n = l) &= \frac{\prod_{i=0}^{l-1}(-a-i)\prod_{i=0}^{-b-j-1}}{\prod_{i=0}^n (-a-b-i)} = \frac{\binom{-a}{l}\binom{-b}{n-l}}{\binom{-a-b}{n}}\\ &\mit l \in \set{0,\dots,n} + \end{align*} + Dies ist die \begriff{\person{Polya}-Urne} auf $\set{0,\dots,n}, n \in \N$ mit Parametern $a,b > 0$. +\end{example} + +\begin{example} + Ein Student beantwortet eine Multiple-Choice-Frage mit 4 Antwortmöglichkeiten, eine davon ist richtig. Er kennt die richtige Antwort mit Wahrscheinlichkeit $\sfrac{1}{3}$. Wenn er diese kennt, so wählt er diese aus. Andernfalls wählt er zufällig (gleichverteilt) eine Antwort.\\ + Betrachte + \begin{align*} + W = \set{\text{richtige Antwort gewusst}}\\ + R = \set{\text{Richtige Antwort gewählt}} + \intertext{Dann} + \P(W) = \frac{2}{3}, \P(R \mid W) = 1, \P(R \mid W^C) = \frac{1}{4} + \end{align*} + Angenommen, der Student gibt die richtige Antwort. Mit welcher Wahrscheinlichkeit hat er diese gewusst? + \begin{align*} + \P(W\mid R) = \text{ ?} + \end{align*} +\end{example} + +\begin{proposition} + \proplbl{3_1_9} + Sei $(\O, \F, \P)$ Wahrscheinlichkeitsraum und $\O = \bigcup_{i \in I} B_i$ eine höchstens abzählbare Zerlegung in paarweise disjunkte Ereignisse $B_i \in \F$. + \begin{enumerate} %TODO set itemize references. or use enumerate? + \item \emph{Satz von der totalen Wahrscheinlichkeit:} Für alle $A \in \F$ + \begin{align*} + \P(A) = \sum_{i\in I} \P(A\mid B_i)\P(B_i) \label{eq:totWkeit}\tag{totale Wahrscheinlichkeit} + \end{align*} + \item \emph{Satz von \person{Bayes}:} Für alle $A \in \F$ mit $\P(A) > 0$ und alle $h \in I$ + \begin{align*} + \P(B_h \mid A) = \frac{\P(A \mid B_h)\P(B_h)}{\sum_{i\in I}\P(A\mid B_i)\P(B_i)} \label{eq:bayes}\tag{Bayes} + \end{align*} + \end{enumerate} +\end{proposition} + +\begin{proof} + \begin{enumerate} + \item Es gilt: + \begin{align*} + \sum_{i\in I} \P(A\mid B_i)\P(B_i) \defeq \sum_{i\in I}\frac{\P(A \cap B_i)}{\P(B_i)}\P(B_i) = \sum_{i\in I} \P(A \cap B_i) \overset{\sigma-add.}{=} \P(A) + \end{align*} + \item + \begin{align*} + \P(B_h \mid A) \defeq \frac{\P(A \cap B_h)}{\P(A)} \defeq \frac{\P(A \mid B_h)\P(B_h)}{\P(A)} + \end{align*} + also mit a) auch b). %TODO add refs + \end{enumerate} +\end{proof} + +\begin{example} + In der Situation von \propref{3_1_3} folgt mit dem \propref{3_1_9} \eqref{eq:totWkeit} + \begin{align*} + \P(R) &= \P(R \mid W)\P(W) + \P(R\mid W^C)\P(W^C)\\ + &= 1\cdot \frac{2}{3} + \frac{1}{4}\frac{1}{3} = \frac{3}{4} + \intertext{und mit dem \propref{3_1_9} \eqref{eq:bayes}} %Bayes + \P(W \mid R) &= \frac{\P(R \mid W)\P(W)}{\P(R)} = \frac{1\frac{2}{3}}{\frac{3}{4}} = \frac{8}{9} \text{ für die gesuchte Wahrscheinlichkeit.} + \end{align*} %TODO compile as pdf and include it. not working. +% \begin{center} +% \begin{tikzpicture} +% \input{./tikz/baum_2} +% \caption{ \propref{3_1_3}} +% \end{tikzpicture} +% \end{center} +\end{example} \ No newline at end of file diff --git a/4. Semester/STOCH/TeX_files/Bedingte_Wheiten.tex b/4. Semester/STOCH/TeX_files/Bedingte_Wheiten.tex index 6e473bc..38408ff 100644 --- a/4. Semester/STOCH/TeX_files/Bedingte_Wheiten.tex +++ b/4. Semester/STOCH/TeX_files/Bedingte_Wheiten.tex @@ -1,5 +1,5 @@ \chapter[Bedingte Wahrscheinlichkeiten und (Un)-abbhängigkeit]{Bedingte Wkeiten und (Un)-abbhängigkeit} -\chaptermark{Bedingte Wheiten und (Un)-abbhängigkeit} +\chaptermark{Bedingte Wahrscheinlichkeiten und (Un)-abbhängigkeit} \section{Bedingte Wahrscheinlichkeiten} \begin{example} @@ -64,7 +64,7 @@ \begin{align*} \P(A\mid B) := \frac{\P(A\cap B)}{\P(B)} \mit A\in \F \end{align*} - die \begriff{bedingte Wheit von $A$ gegeben $B$}. + die \begriff{bedingte Wahrscheinlichkeit von $A$ gegeben $B$}. Falls $\P(B) = 0$, setze \begin{align*} \P(A \mid B) = 0 \mit \forall A \in \F @@ -74,9 +74,9 @@ \begin{example} %TODO ref In der Situation \propref{3_1_1} gilt % \begin{align*} - A \cap B = \set{(4,4)} + A \cap B &= \set{(4,4)} \intertext{und damit} - \P(A \mid B) = \frac{\P(A\cap B)}{\P(B)} = \frac{\frac{1}{36}}{\frac{1}{6}} = \frac{1}{6} + \P(A \mid B) &= \frac{\P(A\cap B)}{\P(B)} = \frac{\frac{1}{36}}{\frac{1}{6}} = \frac{1}{6} \end{align*} \end{example} Aus \propref{3_1_3} ergibt sich @@ -98,9 +98,12 @@ Aus \propref{3_1_3} ergibt sich Stehen die $A_i$ in \propref{3_1_4} in einer (zeitlichen) Abfolge, so liefert Formel einen Hinweis, wie Wahrscheinlichkeitsmaße für \begriff{Stufenexperimente} konstruiert werden können. Ein \emph{Stufenexperiment} aus $n$ nacheinander ausgeführten Teilexperimenten lässt sich als \begriff{Baumdiagramm} darstellen. %TODO Baumdiagramm -%\begin{tikzpicture} -% %TODO -%\end{tikzpicture} +\begin{center} + \begin{tikzpicture} + \input{./tikz/baum_1} + % \caption{ \propref{3_1_4}} + \end{tikzpicture} +\end{center} \begin{proposition}[Konstruktion des Wahrscheinlichkeitsmaßes eines Stufenexperiments] \proplbl{3_1_6} @@ -109,7 +112,7 @@ Stehen die $A_i$ in \propref{3_1_4} in einer (zeitlichen) Abfolge, so liefert Fo [\omega_1,\dots,\omega_n]:= \set{\omega_1}\times \dots \times \set{\omega_n} \times \O_{m-1} \times \O_{n},\quad m\le n\\ \P(\set{\omega_m}[\omega_1,\dots,\omega_{m-1}]) \end{align*} %TODO check indices, m-1 instead of m+1? - für die Wheit in der $m$-ten Stufe des Experiments $\omega_m$ zu beobachten, falls in den vorausgehenden Stufen $\omega_1,\dots,\omega_{m-1}$ beobachten wurden. Dann definiert + für die Wahrscheinlichkeit in der $m$-ten Stufe des Experiments $\omega_m$ zu beobachten, falls in den vorausgehenden Stufen $\omega_1,\dots,\omega_{m-1}$ beobachten wurden. Dann definiert \begin{align*} \P(\set{\omega}) := \P(\set{\omega_1})\prod_{m=2}^{n}\P\brackets{\set{\omega_m} \mid [\omega_1, \dots, \omega_{m-1}]} %TODO maybe wrong here. check @@ -151,7 +154,7 @@ Stehen die $A_i$ in \propref{3_1_4} in einer (zeitlichen) Abfolge, so liefert Fo \end{example} \begin{example} - Ein Student beantwortet eine Multiple-Choice-Frage mit 4 Antwortmöglichkeiten, eine davon ist richtig. Er kennt die richtige Antwort mit Wheit $\sfrac{1}{3}$. Wenn er diese kennt, so wählt er diese aus. Andernfalls wählt er zufällig (gleichverteilt) eine Antwort.\\ + Ein Student beantwortet eine Multiple-Choice-Frage mit 4 Antwortmöglichkeiten, eine davon ist richtig. Er kennt die richtige Antwort mit Wahrscheinlichkeit $\sfrac{1}{3}$. Wenn er diese kennt, so wählt er diese aus. Andernfalls wählt er zufällig (gleichverteilt) eine Antwort.\\ Betrachte \begin{align*} W = \set{\text{richtige Antwort gewusst}}\\ @@ -195,19 +198,20 @@ Stehen die $A_i$ in \propref{3_1_4} in einer (zeitlichen) Abfolge, so liefert Fo \end{proof} \begin{example} - In der Situation von \propref{3_1_3} folgt mit dem \propref{3_1_9} \eqref{eq:totWkeit} der totalen Wahrscheinlichkeit + In der Situation von \propref{3_1_3} folgt mit dem \propref{3_1_9} \eqref{eq:totWkeit} \begin{align*} \P(R) &= \P(R \mid W)\P(W) + \P(R\mid W^C)\P(W^C)\\ &= 1\cdot \frac{2}{3} + \frac{1}{4}\frac{1}{3} = \frac{3}{4} \intertext{und mit dem \propref{3_1_9} \eqref{eq:bayes}} %Bayes \P(W \mid R) &= \frac{\P(R \mid W)\P(W)}{\P(R)} = \frac{1\frac{2}{3}}{\frac{3}{4}} = \frac{8}{9} \text{ für die gesuchte Wahrscheinlichkeit.} - \end{align*} %TODO -% \begin{tikzpicture} -% -% \end{tikzpicture} + \end{align*} + \begin{center} + \begin{tikzpicture} + \input{./tikz/baum_2} +% \caption{ \propref{3_1_3}} + \end{tikzpicture} + \end{center} \end{example} -% trees in Latex? -% https://tex.stackexchange.com/questions/5447/how-can-i-draw-simple-trees-in-latex/5451 \section{(Un)-abhängigkeit} \label{sec_unabhangigkeit} In vielen Fällen besagt die Intuition über verschiedene Zufallsexperimente/ Ereignisse, dass diese sich \emph{nicht} gegenseitig beeinflussen. Für solche $A,B \in \F$ mit $\P(A) > 0, \P(B) > 0$ sollte gelten \begin{align*} @@ -215,7 +219,7 @@ In vielen Fällen besagt die Intuition über verschiedene Zufallsexperimente/ Er \end{align*} \begin{definition}[(stochastisch)unabhängig] \proplbl{3_2_11} - $(\O, \F, \P)$ WRaum. Zwei Ereignisse $A,B \in \F$ heißt \begriff{(stochastisch) unabhängig bezüglich $\P$}, falls + $(\O, \F, \P)$ Wahrscheinlichkeitsraum. Zwei Ereignisse $A,B \in \F$ heißt \begriff{(stochastisch) unabhängig bezüglich $\P$}, falls \begin{align*} \P(A\cap B) = \P(A)\P(B) \end{align*} @@ -224,31 +228,31 @@ In vielen Fällen besagt die Intuition über verschiedene Zufallsexperimente/ Er \begin{example} Würfeln mit 2 fairen, sechsseitigen Würfel: \begin{align*} - \O = \set{(i,j) \mid i,j \in\set{1,\dots,n}},\quad \F = \pows(\O), \quad \P = \Gleich(\O) + \O &= \set{(i,j) \mid i,j \in\set{1,\dots,n}},\quad \F = \pows(\O), \quad \P = \Gleich(\O) \intertext{Betrachte} - A:= \set{(i,j) \in \O, i \text{ gerade}}\\ - B:= \set{(i,j) \in \O, j > 2} + A&:= \set{(i,j) \in \O, i \text{ gerade}}\\ + B&:= \set{(i,j) \in \O, j > 2} \end{align*} In diesem Fall, erwarten wir intuitiv Unabhängigkeit von $A$ und $B$.\\ In der Tat % start using \P instead of \P! \begin{align*} - \P(A) = \frac{1}{2}, \quad \P(B) = \frac{1}{3} \mit \P(A\cap B) = \frac{1}{6} + \P(A) &= \frac{1}{2}, \quad \P(B) = \frac{1}{3} \mit \P(A\cap B) = \frac{1}{6} \intertext{erfüllt} - \P(A\cap B) = \P(A)\P(B). + \P(A\cap B) &= \P(A)\P(B). \end{align*} Betrachte nun \begin{align*} - C:= \set{(i,j) \in \O, i\neq j = 1}\\ - D:= \set{(i,j) \in \O, i = 6} + C&:= \set{(i,j) \in \O, i\neq j = 1}\\ + D&:= \set{(i,j) \in \O, i = 6} \intertext{dann gilt} \P(C) = \frac{1}{6}, \quad \P(D) = \frac{1}{6} \intertext{und wegen $C \cap D = \set{(6,1)}$ folgt} - \P(C\cap D) = \frac{1}{36} = \frac{1}{6} \frac{1}{6} = \P(C \setminus D) + \P(C\cap D) &= \frac{1}{36} = \frac{1}{6} \frac{1}{6} = \P(C \setminus D) \end{align*} $C$ und $D$ sind also \emph{stochastisch} unabhängig, obwohl eine kausale Abhängigkeit vorliegt! \end{example} \begin{definition}[unabhängig bezüglich $\P$] - $(\O, \F, \P)$ WRaum und $I \neq \emptyset$ endliche Indexmenge. Dann heißt die Familie $(A_i)_{i \in I}$ von Ereignissen in $\F$ \begriff{unabhängig bezüglich $\P$}, falls für alle $J \subseteq I, J \neq \emptyset$ gilt: + $(\O, \F, \P)$ Wahrscheinlichkeitsraum und $I \neq \emptyset$ endliche Indexmenge. Dann heißt die Familie $(A_i)_{i \in I}$ von Ereignissen in $\F$ \begriff{unabhängig bezüglich $\P$}, falls für alle $J \subseteq I, J \neq \emptyset$ gilt: \begin{align*} \P\brackets{\bigcap_{i\in J}A_i} = \prod_{i\in J} \P(A_i) \end{align*} @@ -259,9 +263,9 @@ In vielen Fällen besagt die Intuition über verschiedene Zufallsexperimente/ Er \begin{align*} \O = \set{0,1}^2, \quad \F = \pows(\O), \quad \P = \Gleich(\O) \intertext{sowie} - A = \set{1}\times \set{0,1} \qquad \text{(Münzwurf: erster Wurf Zahl)}\\ - B = \set{0,1}\times \set{1} \qquad \text{(Münzwurf: zweiter Wurf Zahl)}\\ - C = \set{0,0}\times \set{1,1} \qquad \text{(beide Würfe selbes Ergebnis)} + A &= \set{1}\times \set{0,1} \qquad \text{(Münzwurf: erster Wurf Zahl)}\\ + B &= \set{0,1}\times \set{1} \qquad \text{(Münzwurf: zweiter Wurf Zahl)}\\ + C &= \set{0,0}\times \set{1,1} \qquad \text{(beide Würfe selbes Ergebnis)} \end{align*} Dann gelten \begin{align*} diff --git a/4. Semester/STOCH/TeX_files/Erste_Standardmodelle.tex b/4. Semester/STOCH/TeX_files/Erste_Standardmodelle.tex index edc03a0..a3784e3 100644 --- a/4. Semester/STOCH/TeX_files/Erste_Standardmodelle.tex +++ b/4. Semester/STOCH/TeX_files/Erste_Standardmodelle.tex @@ -154,7 +154,7 @@ Gegeben: Urne mit $N$ Kugeln verschiedener Farben aus $E$, $\abs{E} \ge 2$. Es w die \begriff{Hypergeometrische Verteilung} mit Parametern $N,W,n$. Wir schreiben $\Hyper(N,W,n)$. \end{definition} -\section{\person{Poisson}-Approximation und \person{Poisson}-Verteilung} +\section{Poisson-Approximation und \person{Poisson}-Verteilung} $\Bin(n,p)$ ist zwar explizit und elementar definiert, jedoch für große $n$ mühsam auszuwerten. Für seltene Ereignisse ($n$ groß, $p$ klein) verwende daher: \begin{proposition}[Poisson-Approximation] diff --git a/4. Semester/STOCH/TeX_files/unbedingte_Wahrscheinlichkeiten.tex b/4. Semester/STOCH/TeX_files/unbedingte_Wahrscheinlichkeiten.tex new file mode 100644 index 0000000..74f1473 --- /dev/null +++ b/4. Semester/STOCH/TeX_files/unbedingte_Wahrscheinlichkeiten.tex @@ -0,0 +1,191 @@ +\section{(Un)-abhängigkeit} \label{sec_unabhangigkeit} +In vielen Fällen besagt die Intuition über verschiedene Zufallsexperimente/ Ereignisse, dass diese sich \emph{nicht} gegenseitig beeinflussen. Für solche $A,B \in \F$ mit $\P(A) > 0, \P(B) > 0$ sollte gelten +\begin{align*} +\P(A\mid B) = \P(B), \quad \P(B\mid A) = \P(B). +\end{align*} +\begin{definition}[(stochastisch)unabhängig] + \proplbl{3_2_11} + $(\O, \F, \P)$ Wahrscheinlichkeitsraum. Zwei Ereignisse $A,B \in \F$ heißt \begriff{(stochastisch) unabhängig bezüglich $\P$}, falls + \begin{align*} + \P(A\cap B) = \P(A)\P(B) + \end{align*} + Wir schreiben auch $A B$. +\end{definition} +\begin{example} + Würfeln mit 2 fairen, sechsseitigen Würfel: + \begin{align*} + \O &= \set{(i,j) \mid i,j \in\set{1,\dots,n}},\quad \F = \pows(\O), \quad \P = \Gleich(\O) + \intertext{Betrachte} + A&:= \set{(i,j) \in \O, i \text{ gerade}}\\ + B&:= \set{(i,j) \in \O, j > 2} + \end{align*} + In diesem Fall, erwarten wir intuitiv Unabhängigkeit von $A$ und $B$.\\ + In der Tat % start using \P instead of \P! + \begin{align*} + \P(A) &= \frac{1}{2}, \quad \P(B) = \frac{1}{3} \mit \P(A\cap B) = \frac{1}{6} + \intertext{erfüllt} + \P(A\cap B) &= \P(A)\P(B). + \end{align*} + Betrachte nun + \begin{align*} + C&:= \set{(i,j) \in \O, i\neq j = 1}\\ + D&:= \set{(i,j) \in \O, i = 6} + \intertext{dann gilt} + \P(C) = \frac{1}{6}, \quad \P(D) = \frac{1}{6} + \intertext{und wegen $C \cap D = \set{(6,1)}$ folgt} + \P(C\cap D) &= \frac{1}{36} = \frac{1}{6} \frac{1}{6} = \P(C \setminus D) + \end{align*} + $C$ und $D$ sind also \emph{stochastisch} unabhängig, obwohl eine kausale Abhängigkeit vorliegt! +\end{example} +\begin{definition}[unabhängig bezüglich $\P$] + $(\O, \F, \P)$ Wahrscheinlichkeitsraum und $I \neq \emptyset$ endliche Indexmenge. Dann heißt die Familie $(A_i)_{i \in I}$ von Ereignissen in $\F$ \begriff{unabhängig bezüglich $\P$}, falls für alle $J \subseteq I, J \neq \emptyset$ gilt: + \begin{align*} + \P\brackets{\bigcap_{i\in J}A_i} = \prod_{i\in J} \P(A_i) + \end{align*} + Offensichtlich impliziert die Unabhängigkeit einer Familie die paarweise Unabhängigkeit je zweier Familienmitglieder nach \propref{3_2_11}. Umgekehrt gilt dies nicht! +\end{definition} +\begin{example}[Abhängigkeit trotz paarweiser Unabhängigkeit] + Betrachte 2-faches Bernoulliexperiment mit Erfolgswahrscheinlichkeit $\sfrac{1}{2}$, d.h. + \begin{align*} + \O = \set{0,1}^2, \quad \F = \pows(\O), \quad \P = \Gleich(\O) + \intertext{sowie} + A &= \set{1}\times \set{0,1} \qquad \text{(Münzwurf: erster Wurf Zahl)}\\ + B &= \set{0,1}\times \set{1} \qquad \text{(Münzwurf: zweiter Wurf Zahl)}\\ + C &= \set{0,0}\times \set{1,1} \qquad \text{(beide Würfe selbes Ergebnis)} + \end{align*} + Dann gelten + \begin{align*} + \P(A) = \frac{1}{2} = \P(A) = \P(B) + \intertext{und} + \P(A\cap B) = \P(\set{(1,1)}) = \frac{1}{4} = \P(A)\P(B)\\ + \P(A\cap C) = \P(\set{(1,1)}) = \frac{1}{4} = \P(A)\P(C)\\ + \P(B\cap C) = \P(\set{(1,1)}) = \frac{1}{4} = \P(B)\P(C) + \end{align*} + also paarweise Unabhängigkeit.\\ + Aber + \begin{align*} + \P(A\cap B \cap C) = \P(\set{(1,1)}) = \frac{1}{4} + \P(A)\P(B)\P(C) + \intertext{und $A,B,C$ sind \emph{nicht} stochastisch unabhängig.} + \end{align*} +\end{example} +\begin{definition}[Unabhängige Messräume] + \proplbl{3_2_15} + % started using \O for \O and \E for this special generating set E_i + $(\O, \F,\P)$ Wahrscheinlichkeitsraum, $I \neq \emptyset$ Indexmenge und $(E_i, \E_i)$ Messräume + \begin{enumerate} + \item Die Familie $\F_i \subset \F, i \in I$, heißen \emph{unabhängig}, wenn für die $J \subseteq I, I \neq \emptyset, \abs{J} < \infty$ gilt + \begin{align*} + \P\brackets{\bigcap_{i=1} A_i} = \prod_{i\in J} \P(A_i) \qquad \text{ für beliebige } A_i \in \F_i, i \in J + \end{align*} + \item Die Zufallsvariable $X_i: (\O, \F) \to (E_i, \E_i), i \in I$, heißen \emph{unabhängig}, wenn die $\sigma$-Algebren + \begin{align*} + \sigma(X_i) = X^{-1}(\E_i) = \set{\set{X_i \in F}, F \in \E_i}, \quad i \in I + \end{align*} + unabhängig sind. + \end{enumerate} +\end{definition} +\begin{lemma}[Zusammenhang der Definitionen] + \proplbl{3_2_16} + $(\O,\F,\P)$ Wahrscheinlichkeitsraum, $I \neq \emptyset, A \in \F, i \in I$.\\ + Die folgenden Aussagen sind äquivalent: + \begin{enumerate} + \item Die Ereignisse $A_i, i \in I$ sind unabhängig. + \item Die $\sigma$-Algebren $\sigma(A_i), i \in I$ sind unabhängig. + \item Die Zufallsvariablen $\indi_{A_i}, i \in I$ sind unabhängig. + \end{enumerate} +\end{lemma} +\begin{proof} %TODO add ref? + Da die Unabhängigkeit über endliche Teilemengen definiert ist, können wir oBdA $I = \set{1, \dots, n}$ annehmen. + \begin{itemize} + \item Da $\sigma(\indi_{A_i}) = \sigma(A_i)$ folgt die Äquivalenz von 2. und 3. direkt aus \propref{3_2_15}. + \item Zudem ist 2. $\to $ 1. klar! + \item Für 1 $\to$ 2. genügt es zu zeigen, dass + \begin{align*} + A_1, \dots, A_n \text{ unabhängig } &\Rightarrow B_1, \dots, B_n \text{ unabhängig von } B_i \in \set{\emptyset, A_i, A_i^C, \O}. + \intertext{Rekursive folgt dies bereits aus} + A_1,\dots, A_n \text{ unabhängig } &\Rightarrow B_1, A_2, \dots, A_n \text{ unabhängig mit } B_1 \in \set{\emptyset, A_1, A_1^C, \O}. + \end{align*} + Für $B_1 \in \set{\emptyset, A_1, \O}$ ist dies klar.\\ + Sei also $B_1 = A_1^C$ und $J \subseteq I, J \neq \emptyset$. Falls $1 \not \in J$, ist nichts zu zeigen. Sei $1 \in J$, dann gilt mit + \begin{align*} + A &= \bigcap_{i\in J, i \neq 1} A_i + \intertext{sicherlich} + \P\brackets{A_1^C \cap A} &= \P(A \setminus (A_1 \cap A))\\ + &= \P(A) - \P(A_1 \cap A)\\ + &= \prod_{i\in J\setminus \set{1}} \P(A_i) - \prod_{i\in J}(A_i)\\ + &= (1- \P(A_1))\prod_{i\in J\setminus \set{1}} \P(A_i)\\ + &= \P\brackets{A_1^C})\prod_{i\in J\setminus \set{1}} \P(A_i) + \end{align*} + \end{itemize} +\end{proof} +Insbesondere zeigt das \propref{3_2_16}, dass wir in einer Familie unabhängiger Ereignisse beliebig viele Ereignisse durch ihr Komplement, $\emptyset$ oder $\O$ ersetzen können, ohne die Unabhängigkeit zu verlieren. +\begin{proposition} + \proplbl{3_2_17} + $(\O, \F, \P)$ Wahrscheinlichkeitsraum und $\F_i \subseteq \F, i \in I$, seien $\cap$-stabil Familien von Ereignissen. Dann + \begin{align*} + \F_i, i \in I \text{ unabhängig } \Leftrightarrow \sigma(\F_i), i \in I \text{ unabhängig}. + \end{align*} +\end{proposition} + +\begin{proof} + OBdA sei $I = \set{1, \dots, n}$ und das $\O \in \F_i, i \in I$. + \begin{itemize} + \item $\Leftarrow$: trivial, da $\F_i \subseteq \sigma(\F_i)$ und das Weglassen von Mengen erlaubt ist. + \item $\Rightarrow$: zeigen wir rekursive + \begin{enumerate} + \item Wähle $F_i \in \F_i, i = 2, \dots,n$ und defnieren für $F \in \sigma(\F_i)$ die endlichen Maße + \begin{align*} + \mu(F) = \P\brackets{\bigcap_{i=1}^n F_i} \und \nu(F) = \prod_{i=1}^n \P(F_i) + \end{align*} + \item Da die Familien $\F_i$ unabhängig sind, gilt + \begin{align*} + \mu\mid_{\F_1} = \nu\mid_{\F_1} + \end{align*} + Nach Eindeutigkeitssatz für Maße (\proplbl{1_1_19}) folgt $\mu\mid_{\sigma(\F_1)} = \nu\mid_{\sigma(\F_1)}$ also + \begin{align*} + \P(\bigcap_{i=1}^n F_i) = \P(F)\P(F_1)\dots \P(F_n) + \end{align*} + für alle $F \in \sigma(\F_i)$ und $F_i \in \F_i, i = 1, \dots, n$. Da $\O \in \F_i$ für alle $i$ gilt die erhaltene Produktformel auf für alle Teilemenge $J \subseteq I$.\\ + Also sind + \begin{align*} + \sigma(\F_1), \F_2, \dots, \F_n \text{unabhängig} + \end{align*} + \item Wiederholtes Anwenden von $1$ und $2$ liefert den \propref{3_2_17}. + \end{enumerate} + \end{itemize} +\end{proof} + +Mittels \propref{3_2_17} folgen: + +\begin{conclusion} + $(\O,\F,\P)$ Wahrscheinlichkeitsraum und + \begin{align*} + \F_{i,j} \subseteq \F, \quad 1 \le \dots \le n, 1 \le j \le m(i) + \end{align*} + unabhängige, $\cap$-stabile Familien. + Dann sind auch + \begin{align*} + \G_i = \sigma(\F_{i,1},\dots, \F_{i,m(i)}), \quad 1 \le i \le n + \end{align*} + unabhängig. +\end{conclusion} + +\begin{conclusion} + $(\O,\F,\P)$ Wahrscheinlichkeitsraum, und + \begin{align*} + X_{ij}: \O \to E, \quad 1 \le i \le n, 1 \le j \le m(i) + \end{align*} + unabhängige Zufallsvariablen. Zudem seinen $f_i: E^{m(i)} \to \R$ messbar. Dann sind auch die Zufallsvariablen + \begin{align*} + f_i(X_{i1}, \dots, X_{im(i)}), \quad 1 \le i \le n + \end{align*} + unabhängig. +\end{conclusion} + +\begin{example} + $X_1, \dots, X_n$ unabhängige reelle Zufallsvariablen. Dann sind auch + \begin{align*} + Y_1 = X_1, Y_2 = X_2 + \cdots + X_n + \end{align*} + unabhängig. +\end{example} \ No newline at end of file diff --git a/4. Semester/STOCH/Vorlesung STOCH.tex b/4. Semester/STOCH/Vorlesung STOCH.tex index 1f618dd..7e85e87 100644 --- a/4. Semester/STOCH/Vorlesung STOCH.tex +++ b/4. Semester/STOCH/Vorlesung STOCH.tex @@ -5,14 +5,13 @@ \author{Dozent: Prof. Dr. \person{Anita Behme}} % local commands -\undef{\F} -\newcommand{\F}{\mathscr{F}} -\undef{\P} -\newcommand{\P}{\mathbb{P}} -\undef{\O} -\newcommand{\O}{\Omega} -\undef{\G} -\newcommand{\G}{\mathscr{G}} +\renewcommand{\F}{\mathscr F} +%\undef{\P} +\renewcommand{\P}{\mathbb{P}} +%\undef{\O} +\renewcommand{\O}{\Omega} +%\undef{\G} +\renewcommand{\G}{\mathscr{G}} %\undef{\base} \newcommand{\E}{\mathcal{E}} @@ -38,8 +37,10 @@ \input{./TeX_files/Grundbegriffe_Wtheorie} % Erste Standardmodelle der Wahrscheinlichkeitstheorie \input{./TeX_files/Erste_Standardmodelle} -% Bedingte Wahrscheinlichkeiten und (Un)-abbhängigkeit} -\input{./TeX_files/Bedingte_Wheiten} +\chapter[Bedingte Wahrscheinlichkeiten und (Un)-abbhängigkeit]{Bedingte Wkeiten und (Un)-abbhängigkeit} +\chaptermark{Bedingte Wahrscheinlichkeiten und (Un)-abbhängigkeit} +\input{./TeX_files/Bedingte_Wahrscheinlichkeiten} +\input{./TeX_files/unbedingte_Wahrscheinlichkeiten} \chapter{Test} diff --git a/4. Semester/STOCH/tikz/baum_1.tex b/4. Semester/STOCH/tikz/baum_1.tex index 10e5e23..14fda79 100644 --- a/4. Semester/STOCH/tikz/baum_1.tex +++ b/4. Semester/STOCH/tikz/baum_1.tex @@ -1,30 +1,22 @@ -\documentclass{standalone} - -\usepackage{tikz} -\usepackage{amssymb} -\usetikzlibrary{trees} - -\begin{document} - \begin{tikzpicture}[level distance=3cm, - level 1/.style={sibling distance=3cm}, - level 2/.style={sibling distance=1.5cm}, - grow=right, sloped] - \node {} - child {node {$A_1^C$} - child {node {$A_2^C$}} - child {node {$A_2$} edge from parent node[above] {$\mathbb{P}(A_2\mid A_1^C)$}} +\begin{tikzpicture}[level distance=3cm, +level 1/.style={sibling distance=3cm}, +level 2/.style={sibling distance=1.5cm}, +grow=right, sloped] +\node {} +child {node {$A_1^C$} + child {node {$A_2^C$}} + child {node {$A_2$} edge from parent node[above] {$\mathbb{P}(A_2\mid A_1^C)$}} + edge from parent + node[below] {$\mathbb{P}(A_1^C)$} +} +child {node {$A_1$} + child {node {$A_2^C$}} + child {node {$A_2$} + child {node {$A_3^C$}} + child {node {$A_3$} edge from parent node[above] {$\mathbb{P}(A_3\mid A_1\cap A_2)$}} edge from parent - node[below] {$\mathbb{P}(A_1^C)$} - } - child {node {$A_1$} - child {node {$A_2^C$}} - child {node {$A_2$} - child {node {$A_3^C$}} - child {node {$A_3$} edge from parent node[above] {$\mathbb{P}(A_3\mid A_1\cap A_2)$}} - edge from parent - node[above] {$\mathbb{P}(A_2\mid A_1)$}} - edge from parent - node[above] {$\mathbb{P}(A_1)$} - }; - \end{tikzpicture} -\end{document} + node[above] {$\mathbb{P}(A_2\mid A_1)$}} + edge from parent + node[above] {$\mathbb{P}(A_1)$} +}; +\end{tikzpicture} \ No newline at end of file diff --git a/texmf/tex/latex/mathscript/mathscript.cls b/texmf/tex/latex/mathscript/mathscript.cls index a5bc8ed..d91d04c 100644 --- a/texmf/tex/latex/mathscript/mathscript.cls +++ b/texmf/tex/latex/mathscript/mathscript.cls @@ -112,7 +112,7 @@ \usepgfplotslibrary{fillbetween} \RequirePackage{pgf} \RequirePackage{tikz} -\usetikzlibrary{patterns,arrows,calc,decorations.pathmorphing,backgrounds, positioning,fit,petri,decorations.fractals} +\usetikzlibrary{patterns,arrows,calc,decorations.pathmorphing,backgrounds, positioning,fit,petri,decorations.fractals,trees} \usetikzlibrary{matrix} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% From ad36b433c8adc7b154540c49dcbdaf93bbfd7c5e Mon Sep 17 00:00:00 2001 From: Ameyah Date: Wed, 17 Apr 2019 19:57:08 +0200 Subject: [PATCH 9/9] more typos and formating --- .../Korper/Alg_Korpererweiterungen.tex | 46 +++++++++--------- .../TeX_files/Korper/Korpererweiterungen.tex | 16 +++--- 4. Semester/ALGZTH/Vorlesung ALGZTH.pdf | Bin 397791 -> 409163 bytes .../Bedingte_Wahrscheinlichkeiten.tex | 18 +++---- 4. Semester/STOCH/Vorlesung STOCH.tex | 2 +- 4. Semester/STOCH/tikz/baum_2.tex | 12 +---- 6 files changed, 41 insertions(+), 53 deletions(-) diff --git a/4. Semester/ALGZTH/TeX_files/Korper/Alg_Korpererweiterungen.tex b/4. Semester/ALGZTH/TeX_files/Korper/Alg_Korpererweiterungen.tex index 6c7fb49..bfd4af8 100644 --- a/4. Semester/ALGZTH/TeX_files/Korper/Alg_Korpererweiterungen.tex +++ b/4. Semester/ALGZTH/TeX_files/Korper/Alg_Korpererweiterungen.tex @@ -1,6 +1,6 @@ \section{Algebraische Körpererweiterungen} -Sei $L \vert K$ eine Körpererweiterung. +Sei $L \mid K$ eine Körpererweiterung. \begin{definition}[algebraisch, transzendent] Sei $\alpha \in L$. Gibt es ein $0 \neq f \in K$ mit $f(\alpha) = 0$, so heißt $\alpha$ \begriff{algebraisch} über $K$, andernfalls \begriff{transzendent} über $K$. @@ -49,8 +49,8 @@ Sei $L \vert K$ eine Körpererweiterung. \begin{definition}[Monimalpolynom, Grad] Sei $\alpha \in L$ algebraisch über $K$, $\Ker(\phi_\alpha) = (f_\alpha)$ mit $f_\alpha \in K$ normiert und irreduzibel. \begin{enumerate} - \item $\MinPol(\alpha\vert K) f_\alpha$, das \begriff{Minimalpolynom} von $\alpha$ über $K$. - \item $\deg(\alpha\vert K) :\Leftrightarrow \deg(f_\alpha)$, der \begriff{Grad} von $\alpha$ über $K$. + \item $\MinPol(\alpha\mid K) := f_\alpha$, das \begriff{Minimalpolynom} von $\alpha$ über $K$. + \item $\deg(\alpha\mid K) :\Leftrightarrow \deg(f_\alpha)$, der \begriff{Grad} von $\alpha$ über $K$. \end{enumerate} \end{definition} @@ -61,8 +61,8 @@ Sei $L \vert K$ eine Körpererweiterung. \item $\alpha$ transzendent über $K$ \\ $\Rightarrow K[\alpha] \cong K$, $K(\alpha) \cong_K K(X)$, $[K(\alpha) : K] = \infty$. \item $\alpha$ algebraisch über $K$ \\ - $\Rightarrow K[\alpha] = K(\alpha) \cong \lnkset{K}{\MinPol{\alpha}{K}}$ , $[ K(\alpha) \colon K)] = \deg(\alpha \vert K) < \infty$ und \\ - $1, \alpha, \dots , \alpha^{\deg(\alpha \vert K) -1}$ ist $K$-Basis von $K(\alpha)$. + $\Rightarrow K[\alpha] = K(\alpha) \cong \lnkset{K}{\MinPol(\alpha\mid K)}$ , $[ K(\alpha) \colon K)] = \deg(\alpha \mid K) < \infty$ und \\ + $1, \alpha, \dots , \alpha^{\deg(\alpha \mid K) -1}$ ist $K$-Basis von $K(\alpha)$. \end{enumerate} \end{proposition} @@ -71,7 +71,7 @@ Sei $L \vert K$ eine Körpererweiterung. \item $\Ker(\phi_\alpha) = (0) \Rightarrow \phi_\alpha$ ist Isomorphismus (da zusätzlich injektiv) \\ $\Rightarrow K(\alpha) \cong_K \Quot(K[\alpha]) \cong_K \Quot(K) = K(X)$ \\ $\Rightarrow [K(\alpha) \colon K] = [K(x) \colon K] = \infty$ - \item Sei $f = f_\alpha = \MinPol(\alpha \vert K)$, $n = \deg(\alpha \vert K) = \deg(f)$. + \item Sei $f = f_\alpha = \MinPol(\alpha \mid K)$, $n = \deg(\alpha \mid K) = \deg(f)$. \begin{itemize} \item $f$ irreduzibel $\Rightarrow (f) \neq (0)$ prim ${\xRightarrow{\text{GEO II.4.7}}} (f)$ ist maximal \\ $\Rightarrow K[\alpha] \cong \lnkset{K}{(f)}$ ist Körper $\Rightarrow K[\alpha] = K(\alpha)$ @@ -96,7 +96,7 @@ Sei $L \vert K$ eine Körpererweiterung. \begin{enumerate}[label=(\alph*)] \item $p \in \Z$ prim $\Rightarrow$ $\sqrt{p} \in \C$ ist algebraisch über $\Q$. \\ Da $f(X) = X^2 - p$ irreduzibel in $\Q$ ist (GEO II.7.3), ist $\MinPol(\sqrt{p}:\Q) = X^2 - p$, $[\Q(\sqrt{p}) : \Q] = 2$. - \item Sei $\zeta_p = e^{\frac{2\pi i}{p}} \in \C$ ($p \in \N$ prim). Da $\Phi_p = \frac{X^p-1}{X-1} = X^{p-1} + X^{p-2} + \cdots + X + 1 \in \Q$ irreduzibel in $\Q$ ist (GEO II.7.9), ist $\MinPol(\zeta_p \vert \Q) = \Phi_p$, $[\Q(\zeta_p) : \Q] = p-1$. Daraus folgt schließlich $[\C : \Q \ge [\Q(\zeta_p) : \Q] = p-1 \enskip \forall p \Rightarrow [\C : \Q] = \infty \Rightarrow [R : \Q] = \infty$. + \item Sei $\zeta_p = e^{\frac{2\pi i}{p}} \in \C$ ($p \in \N$ prim). Da $\Phi_p = \frac{X^p-1}{X-1} = X^{p-1} + X^{p-2} + \cdots + X + 1 \in \Q$ irreduzibel in $\Q$ ist (GEO II.7.9), ist $\MinPol(\zeta_p \mid \Q) = \Phi_p$, $[\Q(\zeta_p) : \Q] = p-1$. Daraus folgt schließlich $[\C : \Q \ge [\Q(\zeta_p) : \Q] = p-1 \enskip \forall p \Rightarrow [\C : \Q] = \infty \Rightarrow [R : \Q] = \infty$. \item $e \in \R$ ist transzendent über $\Q$ (\person{Hermite} 1873), $\pi \in \R$ ist transendent über $\Q$ (\person{Lindemann} 1882). \\ Daraus folgt: $[R : \Q] \ge [\Q(\pi): \Q] = \infty$. Jedoch ist unbekannt, ob z.B. $\pi + e$ transzendent ist. @@ -104,12 +104,12 @@ Sei $L \vert K$ eine Körpererweiterung. \end{example} \begin{definition} - $L \vert K$ ist \begriff{algebraisch} $:\Leftrightarrow$ jedes $\alpha \in L$ ist algebraisch über $K$. + $L \mid K$ ist \begriff{algebraisch} $:\Leftrightarrow$ jedes $\alpha \in L$ ist algebraisch über $K$. \end{definition} \begin{proposition} \proplbl{1_2_10} - $L \vert K$ endlich $\Rightarrow$ $L \vert K$ algebraisch. + $L \mid K$ endlich $\Rightarrow$ $L \mid K$ algebraisch. \end{proposition} \begin{proof} @@ -118,7 +118,7 @@ Sei $L \vert K$ eine Körpererweiterung. \begin{conclusion} \proplbl{1_2_11} - Ist $L = K(\alpha_1, \dots, \alpha_n)$ mit $\alpha_1, \dots, \alpha_n$ algebraisch über $K$, so ist $L \vert K$ endlich, insbesondere algebraisch. + Ist $L = K(\alpha_1, \dots, \alpha_n)$ mit $\alpha_1, \dots, \alpha_n$ algebraisch über $K$, so ist $L \mid K$ endlich, insbesondere algebraisch. \end{conclusion} \begin{proof} @@ -139,8 +139,8 @@ Sei $L \vert K$ eine Körpererweiterung. \begin{conclusion} Es sind äquivalent: \begin{enumerate} - \item $L \vert K$ ist endlich. - \item $L \vert K$ ist endlich erzeugt und algebraisch. + \item $L \mid K$ ist endlich. + \item $L \mid K$ ist endlich erzeugt und algebraisch. \item $L = K(\alpha_1, \dots , \alpha_n)$ mit $\alpha_1, \dots, \alpha_n$ algebraisch über $K$. \end{enumerate} \end{conclusion} @@ -169,19 +169,19 @@ Sei $L \vert K$ eine Körpererweiterung. \proplbl{1_2_14} Seien $K \subseteq L \subseteq M$ Körper. Dann gilt: \begin{align*} - M\vert K \text{ algebraisch } \Leftrightarrow M\vert L \text{ algebraisch und } L \vert K \text{ algebraisch } + M\mid K \text{ algebraisch } \Leftrightarrow M\mid L \text{ algebraisch und } L \mid K \text{ algebraisch } \end{align*} \end{proposition} \begin{proof} \begin{itemize} \item[($\Rightarrow$)] klar, siehe \propref{1_2_3}. - \item[($\Leftarrow$)] Sei $\alpha \in M$. Schreibe $f=\MinPol(\alpha \vert L) = \sum_{i=0}^{n} a_i x^i, \quad L_0 := K(a_0,\dots,a_n)$\\ + \item[($\Leftarrow$)] Sei $\alpha \in M$. Schreibe $f=\MinPol(\alpha \mid L) = \sum_{i=0}^{n} a_i x^i, \quad L_0 := K(a_0,\dots,a_n)$\\ $\Rightarrow f \in L_0[x]$\\ $\Rightarrow [L_0(\alpha): L_0] \le \deg(f) \le \infty$\\ $\Rightarrow [K(\alpha: K)] \le [K(a_0,\dots,a_n,\alpha):K] = \underbrace{[L_0(\alpha):L_0]}_{< \infty}\underbrace{[L_0 :K]}_{< \text{ nach } \propref{1_2_7}}$\\ $\Rightarrow \alpha$ abgebraisch über $K$\\ - $\overset{\alpha \text{ bel.}}{\Rightarrow} M \vert K$ algebraisch. + $\overset{\alpha \text{ bel.}}{\Rightarrow} M \mid K$ algebraisch. \end{itemize} \end{proof} @@ -190,13 +190,15 @@ Sei $L \vert K$ eine Körpererweiterung. \end{conclusion} \begin{proof} %TODO find a good way to format the RIGHTARROWS? - \item $\alpha, \beta \in \tilde{K}:\\ - \Rightarrow K(\alpha, \beta)\vert K$ endlich, insbesondere algebraisch\\ - $\Rightarrow \alpha + \beta, \alpha - \beta, \alpha \cdot \beta, \alpha^{-1} \in K(\alpha,\beta)$ alle algebraisch über $K$, also $K(\alpha, \beta) \subseteq \tilde{K}$. - \item $\alpha \in L$ algebraisch über $\tilde{K}$:\\ - $\Rightarrow \tilde{K}(\alpha)\vert \tilde{K}$ algebraisch\\ - $\Rightarrow \tilde{K}\vert K$ algebraisch - $\overset{\propref{1_2_14}}{\Rightarrow} \tilde{K}(\alpha\vert K)$ algebraisch, insbesondere $\alpha \in \tilde{K}$. + \begin{itemize} + \item $\alpha, \beta \in \tilde{K}:\\ + \Rightarrow K(\alpha, \beta)\mid K$ endlich, insbesondere algebraisch\\ + $\Rightarrow \alpha + \beta, \alpha - \beta, \alpha \cdot \beta, \alpha^{-1} \in K(\alpha,\beta)$ alle algebraisch über $K$, also $K(\alpha, \beta) \subseteq \tilde{K}$. + \item $\alpha \in L$ algebraisch über $\tilde{K}$:\\ + $\Rightarrow \tilde{K}(\alpha)\mid \tilde{K}$ algebraisch\\ + $\Rightarrow \tilde{K}\mid K$ algebraisch + $\overset{\propref{1_2_14}}{\Rightarrow} \tilde{K}(\alpha\mid K)$ algebraisch, insbesondere $\alpha \in \tilde{K}$. + \end{itemize} \end{proof} \begin{definition}[relative algebraische Abschluss] diff --git a/4. Semester/ALGZTH/TeX_files/Korper/Korpererweiterungen.tex b/4. Semester/ALGZTH/TeX_files/Korper/Korpererweiterungen.tex index d29c8ea..269efa0 100644 --- a/4. Semester/ALGZTH/TeX_files/Korper/Korpererweiterungen.tex +++ b/4. Semester/ALGZTH/TeX_files/Korper/Korpererweiterungen.tex @@ -59,11 +59,11 @@ Sei $K,L,M$ Körper. \end{proof} \begin{definition}[Körpererweiterung] - Ist $K$ ein Teilkörper von $L$, so nennt man $L$ eine \begriff{Köpererweiterung} von $K$, auch geschrieben $L\vert K$. + Ist $K$ ein Teilkörper von $L$, so nennt man $L$ eine \begriff{Köpererweiterung} von $K$, auch geschrieben $L \mid K$. \end{definition} \begin{definition}[$K$-Homomorphismus] - Seien $L_1\vert K$ und $L_2 \vert K$ Körpererweiterungen. + Seien $L_1 \mid K$ und $L_2 \mid K$ Körpererweiterungen. \begin{enumerate} \item Ein Ringhomomorphismus $\phi\colon L_1 \to L_2$ ist ein $K$-Homomorphismus, wenn $\phi\vert_K = \id_K$ (i.Z. $\phi: L_1 \to L_2$) \item $\Hom_K(L_1,L_2) = \set{\phi \mid \phi: L_1 \to L_2 \text{ ist $K$-Homomorphismus}}$ @@ -72,11 +72,11 @@ Sei $K,L,M$ Körper. \end{definition} \begin{remark} - $L\vert K$ eine Körpererweiterung, so wird $L$ durch Einschränkung der Multiplikation zu einem $K$-Vektorraum. + $L\mid K$ eine Körpererweiterung, so wird $L$ durch Einschränkung der Multiplikation zu einem $K$-Vektorraum. \end{remark} \begin{definition}[Körpergrad] - $[L:K]:= \dim_k(L) \in \N \cup \{\infty\}$, der \begriff{Körpergrad} der Körpererweiterungen $L\vert K$. + $[L:K]:= \dim_k(L) \in \N \cup \{\infty\}$, der \begriff{Körpergrad} der Körpererweiterungen $L\mid K$. \end{definition} \begin{example} @@ -93,7 +93,7 @@ Sei $K,L,M$ Körper. (``Körpergrad ist multiplikativ'') \end{proposition} -\begin{proof} %TODO maybe make unnumbered lemma here for the claim? +\begin{proof} %TODO maybe make unnumbered lemma here for the claim? add align! Behauptung: Sei $x_1, \dots, x_n \in L$ $K$-linear unabhängig und $y_1, \dots, y_m \in M$ $L$-linear unabhängig\\ $\Rightarrow x_i y_j, i \in \set{1,\dots,n}, j \in \set{1, \dots, m}$ $K$-linear unabhängig.\\ Beweis: $\sum_{i,j} \lambda_{ij}x_i y_j = 0$ mit $\lambda_{ij} \in K$\\ @@ -111,13 +111,13 @@ Sei $K,L,M$ Körper. \end{proof} \begin{definition}[Körpergrad endlich] - $L\vert K$ endlich $:\Leftrightarrow [L:K] < \infty$. + $L\mid K$ endlich $:\Leftrightarrow [L:K] < \infty$. \end{definition} %%%%%%%%%%%%%%%%%%%%%%%%%%%% 2nd lecture %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{definition}[Unterring, Teilkörper] - Sei $L\vert K$ eine Körpererweiterung $a_1, a_2, \dots, a_n \in L$. + Sei $L\mid K$ eine Körpererweiterung $a_1, a_2, \dots, a_n \in L$. \begin{enumerate} \item $K[a_1, \dots, a_n]$ ist kleinster \begriff{Unterring} von $L$, der $K \cup \set{a_1, \dots, a_n}$ enthält (``$a_1, \dots, a_n$ über $K$ erzeugt'') \item $K[a_1, \dots, a_n]$ ist kleinster \begriff{Teilkörper} von $L$, der $K \cup \set{a_1, \dots, a_n}$ enthält (``von ``$a_1, \dots, a_n$ über $K$ erzeugte'', ``$a_1, \dots, a_n$'' zu $K$ adjungieren) @@ -129,7 +129,7 @@ Sei $K,L,M$ Körper. \begin{remark} \proplbl{1_1_15} \begin{enumerate}[label=(\alph*)] - \item $L\vert K$ endlich $\Rightarrow L\vert K$ endlich erzeugt. + \item $L\mid K$ endlich $\Rightarrow L\mid K$ endlich erzeugt. \item $K[a_1, \dots, a_n]$ ist das Bild des Homomorphismus \begin{align} \begin{cases} diff --git a/4. Semester/ALGZTH/Vorlesung ALGZTH.pdf b/4. 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