From 5f5bfb6e88e8568c599bf573aa83f1ad57cfc0e9 Mon Sep 17 00:00:00 2001 From: Sheppy Date: Wed, 7 Oct 2015 21:13:49 +0200 Subject: [PATCH] mv --- Public/MatheC4/MaC4Cheatsheet.tex | 621 ----------------------- Public/MatheC4/Makefile | 22 - Public/MatheC4/graph.png | Bin 11992 -> 0 bytes Public/ThProg/Koinduktion_reduktion.tex | 111 ---- Public/ThProg/Konfluenz.tex | 49 -- Public/ThProg/Makefile | 11 - Public/ThProg/Polynomordnung.tex | 71 --- Public/ThProg/SS14_Spickzettel.pdf | Bin 95397 -> 0 bytes Public/ThProg/Strukturelle_Induktion.tex | 51 -- Public/ThProg/SystemF.tex | 100 ---- 10 files changed, 1036 deletions(-) delete mode 100644 Public/MatheC4/MaC4Cheatsheet.tex delete mode 100644 Public/MatheC4/Makefile delete mode 100644 Public/MatheC4/graph.png delete mode 100644 Public/ThProg/Koinduktion_reduktion.tex delete mode 100644 Public/ThProg/Konfluenz.tex delete mode 100644 Public/ThProg/Makefile delete mode 100644 Public/ThProg/Polynomordnung.tex delete mode 100644 Public/ThProg/SS14_Spickzettel.pdf delete mode 100644 Public/ThProg/Strukturelle_Induktion.tex delete mode 100644 Public/ThProg/SystemF.tex diff --git a/Public/MatheC4/MaC4Cheatsheet.tex b/Public/MatheC4/MaC4Cheatsheet.tex deleted file mode 100644 index 495d895..0000000 --- a/Public/MatheC4/MaC4Cheatsheet.tex +++ /dev/null @@ -1,621 +0,0 @@ -\documentclass{article} -\usepackage{amsmath} -\usepackage{amssymb} -% -------- Umlaute korrekt ---------------- -\usepackage{ngerman} -\usepackage[latin1, utf8]{inputenc} -\usepackage[ngerman, english]{babel} -%------------------------------------------- - -% TikZ Library -\usepackage{tikz} -\usetikzlibrary{arrows,backgrounds,positioning,fit,calc,petri} -\usetikzlibrary{shapes, shapes.misc} -\usetikzlibrary{decorations.markings,decorations.pathmorphing} - -% Verlinkung von Url und Kapiteln -\usepackage{hyperref} -% Einrueckung unterbinden nach Absatz -\setlength{\parindent}{0pt} - -\DeclareMathSizes{10}{10}{10}{10} -\title{Mathe C4 Merz - Cheatsheet} -\author{greeny, nudelsalat, Sheppy\\September 2015} -\date{Diesen Zusammenfassung kann Fehler enthalten!} -\begin{document} -\maketitle -\tableofcontents -\newpage -\section{Statistik} -\subsection{empirisches arithmetisches Mittel} -\[x_{arith}=\frac{1}{n}\sum_{i=1}^n x_i\] -\subsection{empirischer Median (Zentralwert)} -\[ - x_{median}= - \begin{cases} - \frac{x_{n+1}}{2} & \text{n ungerade} \\ - \frac{x_{n/2} \;\; + x_{(n+1)/2}}{2} & \text{n gerade} - \end{cases} -\] -Wobei der Index f\"ur die n'te Zahl in einer Angabe in Stile von \{A,B,C,...\} steht. -\subsection{empirische korrigierte Varianz} -\[x_{var}=\frac{1}{n-1}\sum_{i=1}^n (x_i-x_{arith})\] -\subsection{Regressionsgerade} -\textbf{Gauss'sche Normalengleichung} -Die Regressionsgerade wird mit der Gauss'schen Normalengleichung gel\"ost. -\begin{align} - \begin{pmatrix} - \sum x_i^2 & \sum x_i \\ - \sum x_i & n - \end{pmatrix} - \begin{pmatrix} - a \\ - b - \end{pmatrix} - = - \begin{pmatrix} - \sum x_i*y_i \\ - \sum y_i - \end{pmatrix} \text{, mit $i \in n$} -\end{align} -$\rightarrow$ Auflösen nach Parametern $a,b$.\\ -\textbf{Regressionsgerade}: -\begin{align} - y(x) = a*x + b -\end{align} - -\subsection{Maximum-Likelyhood Methode} -\textbf{Problembeschreibung}: Man m\"ochte f\"ur einen unbekannten Parameter $\lambda$ -einer Verteilung, die mindestens einen Parameter besitzt, einen Sch\"atzwert bestimmen -mithilfe einer konkreten Stichprobe $(x_1, \ldots, x_n)$. - -\begin{enumerate} - \item Likelihood-Funktion $L(\lambda)$ bilden f\"ur gegebene Verteilung - \begin{align} - L(\lambda) = L(x_1, \ldots, x_n; \lambda) = = \prod_{i=1}^n - \underbrace{f}_{\text{Dichtefunktion}}(x_i, \lambda)\\ - \end{align} - Im Falle von Exponentialverteilung:\\ - \begin{align} - \prod_{i=1}^n \lambda e^{-\lambda x_i} = \lambda^n * e^{-\lambda * \sum_{i=1}^{n}x_i} - \end{align} - \item Funktion $L(\lambda)$ mit $\ln$ multiplizieren\\ - Rechenregeln f\"ur $\ln$: - \begin{itemize} - \item $\ln a^b = b * \ln a$ - \item $\ln (a*b) = \ln a + \ln b$ - \end{itemize} - \item Ableiten nach $\lambda$: $\frac{\partial \ln * L(\lambda)}{\partial \lambda}$ - \item Funktion gleich $0$ setzen und nach $\lambda$ aufl\"osen. -\end{enumerate} - -\subsection{Konfidenzintervalle} -Standartwerte f\"ur Konfidenz: -\begin{align*} - 90\%:z = 1.65\\ - 95\%:z = 1.96\\ - 99\%:z = 2.58 -\end{align*} - -\begin{align} - P(|\bar{x}-\mu| \geq c) = \alpha \\ - \mu \in [\bar{x} - z_{1-\frac{\alpha}{2}} * \frac{\sigma}{\sqrt{n}}] -\end{align} - -\subsection{Kovarianz} -Sind zwei Zufallsvariablen $X_1$, $X_2$ stochastisch unabh\"angig dann -gilt: - -\begin{align} - cov(X_1,X_2) = 0 -\end{align} - -Ansonsten: -\begin{align} - cov(X_1,X_2) = E(X_1X_2) - E(X_1)E(X_2) -\end{align} -\textbf{Erwartungswert}: -\begin{align} - EX = \sum_{k \in \Omega} k * P(X = k) = \int_{-\infty}^{\infty} x * f(x) dx -\end{align} -\textbf{Beispiel}: -Berechnen der Kovarianz der Zufallsvariablen $Z_1 = X_1 - X_2$ und $Z_2 = X_1$, -wenn der Zufallsvektor $(X_1,X_2)$ auf der Menge -\begin{align} - M = \{(x_1,x_2)| 0 \leq x_2 \leq 2 \text{ und } 0 \leq x_1 \leq x_2\} -\end{align} -\textbf{Gesucht}: $cov(Z_1, Z_2)$ -\begin{enumerate} - \item Kovavarianz umformen - \begin{align} - cov(Z_1, Z_2) = cov(X_1-X_2, X_1) = (E(X^2_1)-E(X_1)^2)-(E(X_2X_1)-E(X_2)E(X_1)) - \end{align} - \item Die \textbf{Fl\"ache} $A_M$ unter Funktion berechnen: $A_M = 2$.\\ - \item Die \textbf{Dichtefunktion} ist der Kehrwert von $A_M$ und damit $\frac{1}{2}$. - \begin{align} - f(x_1,x_2) = - \begin{cases} - \frac{1}{2} & x_1,x_2 \in M \\ - 0 & sonst - \end{cases} - \end{align} - \item Jetzt wieder mittels \textbf{Marginalsdichte} $f(x_1)$ und $f(x_2)$ bestimmen. - \begin{align} - f_1(x_1) = \int_{x_1}^2 f(x_1,x_2) dx_2\\ - f_2(x_2) = \int_{0}^{x_2} f(x_1,x_2) dx_1 - \end{align} - \item Berechnung der ben\"otigten Erwartungswerte $E$: - \begin{align} - E(X_i) = \int_{0}^{2} x_i * f_i(x_i) dx\\ - E(X_i^2) = \int_{0}^{2} x_i^2 * f_i(x_i) dx\\ - E(X_1X_2) = \underbrace{\int_0^{2}\int_{0}^{x_2}}_{\text{Integration \"uber $x_1$ und - $x_2$}} x_1*x_2*f(x_1,x_2) dx_1 dx_2 - \end{align} - \item Einsetzen in umgeformte Kovarianzformel (siehe 1) -\end{enumerate} -\subsection{Markov-Ketten} -\begin{itemize} - \item Bei Übergangsmatrix $P \in (\mathbb{R}_{\geq 0})^{r x r}$ sind alle Zeilensummen gleich $1$. - \item Vektor $\vec{u} \in (\mathbb{R}_{\geq 0})^{r}$ mit $||\vec{u}||_1 = 1$ - der - \begin{align} - \vec{u} = P^T \cdot \vec{u} - \end{align} - erfüllt, heißt \textbf{Gleichgewichtszustand/-verteilung}. - \item \textbf{Berechnung} von $\vec{u}$: $\text{Kern}(P^T - \text{ Id}_r)$.\\$\rightarrow$ - Kern wird berechnet durch klassischen Gauß- Algorithmus. Wenn keine - eindeutige Lsg (z.B. $0 = 0$), dann Variable beliebig wählen. Es gibt - immer einen Kern, da Determinante $0$ garantiert ist durch obige\\ - Summenbedingung. - \item Vektoreinträge müssen positiv sein, sonst Fehler. - \item Vektor $\vec{u}$ durch $||\vec{u}||_1$ (Summennorm) teilen. - \begin{align} - ||\vec{u}||_1 := \sum^n_{i=1}|x_i| - \end{align} -\end{itemize} -\section{Mengen} -\subsection{o-Algebra} -- leere Menge enthalten\\ -- alle Kombinationen der Elemente enthalten, die nicht bereits gemeinsamme Elemente haben also z.B. \textbf{NICHT} \{x,y\} und \{y,z\} zu \{x,y,z\} machen\\ -- alle Komplemente enthalten\\ \\ -\textbf{Beispiel:}\\ -Grundmenge = $\{1,2,3,4\}$\\ -NICHT o-Algebra Menge = $\{\{1,2\},\{3\}\}$\\ -o-Algebra Menge = $\{\emptyset ,\{1,2\},\{3\}, - \underbrace{\{1,2,3\}}_{\substack{\{1,2\}\{3\}}}, - \underbrace{\{3,4\}}_{\substack{\neg \{1,2\}}}, - \underbrace{\{4\}}_{\substack{\neg \{1,2,3\}}}, -\{1,2,3,4\},\{1,2,4\}\}$ - -\section{Wahrscheinlichkeiten} -\subsection{W\"urfeln} -\subsubsection{keine 6} -\[ - p_0 = \left( \frac{5}{6} \right)^n , n = \text{Anzahl der W\"urfe} -\] -\subsubsection{mindestens 'x' 6er (Gegenereignis)} -\[ - p_1 = 1 - \left( \frac{5}{6} \right)^n = 1 - p_0 -\] -\[ - p_2 = 1-\left(1 - \left( \frac{5}{6} \right)^n\right)-\left( \frac{5}{6} \right)^n = 1-p_1 -p_0 -\] -\[ - p_x = 1 - \sum_{i=0}^{x-1} p_i -\] -\subsubsection{6er-Pasch bei 2 W\"urfeln} -$Ereignisraum = 6^2 , \text{Anzahl g\"unstiger Ereignisse = 1 , n\"ahmlich (6,6)}$\\ -dann wieder \"uber Gegenereignis: \\ -\[ p=1-\left(\frac{35}{36}\right)^n \] -\subsubsection{genau eine 6 bei n-W\"urfeln/W\"urfen} -\[ p= \frac{n*5^{(n-1)}}{6^n}\]\\ -- $6^n $ ist wie immer die Anzahl der Gesamtm\"oglichkeiten \\ -- es gibt n-Moglichkeiten an der die 6 sein kann \\ -- es bleiben bei den verbleibenden n-1 W\"urfen 5 M\"oglichkeiten -\subsubsection{genau x-6er bei n-W\"urfeln/W\"urfen} -\[ p= \frac{\begin{pmatrix} - n\\k -\end{pmatrix}5^{(n-k)}}{6^n}\]\\ -\[\begin{pmatrix} - n\\k - \end{pmatrix}= \frac{n!}{k!(n-k)!} -\]\\ -$\textbf{oder noch allgemeiner, mit Anzahl M\"oglichkeiten 'z' (z.B. 6 bei W\"urfel):}$\[ - p= \frac{\begin{pmatrix} - n\\k - \end{pmatrix}(z-1)^{(n-k)}}{z^n} -\] -\subsubsection{X-Mal Werfen, min eine 3 unter der Bedingung min. eine 6} -A = min. eine 3 \\ -B = min. eine 6 \\\\ -\textbf{gesucht:} \[P(A|B) = \frac{P(A\cap B)}{P(B)} \] -\[P(B) = 1-P(keine\;6) = 1-\left(\frac{5}{6}\right)^4 = \frac{625}{1296}\] -\textbf{Idee:} -\begin{align*} - P(A\cap B) &= 1-P(\neg (A\cap B))\\ - &= 1-P(\neg A \cup \neg B)\\ - &= 1-P(\neg A) - P(\neg B) + P(\neg A \cap \neg B)\\ - &= 1-P(keine\;3)-P(keine\;6)+P(weder\;3\;noch\;6)\\ - &= 1-\left( \frac{5}{6}\right)^4-\left( \frac{5}{6}\right)^4-\left( \frac{4}{6}\right)^4 = 1- \frac{994}{1296} -\end{align*} -...und das dann nur noch oben einsetzen und fertig. -\[P(A|B) = \frac{P(A\cap B)}{P(B)} \] -also: -\[P(A|B) = \frac{\frac{994}{1296}}{\frac{625}{1296}}\] - -\subsubsection{Seiten mit verschiedenen Wahrscheinlichkeiten} -z.B. 6 Seiten mit normaler Wahrscheinlichkeit $(w_1)$, 8 Seiten mit 1/4 Wahrscheinlichkeit -$(w_2)$, wir exploiten die Tatsache, dass: \\ \[ \sum -(Teil-)Wahrscheinlichkeiten = 1 \]\\ -also:\\ -\begin{equation} -6w_1 + 8w_2 = 1 \end{equation} -\begin{equation} - \frac{1}{4}w_1 = w_2 -\end{equation}\\ -Zwei Gleichungen, zwei Unbekannte, easy mode. - -\section{Bedingte Wahrscheinlichkeiten} -\subsection{Beispiele} -\subsubsection{Krankheitstest} -0,2\% Krank, 95\% der Kranken werden erkannt, 98\% der Gesunden werden richtig erkannt\\ \\ -Ereignis $A_1$: Person ist krank\\ -Ereignis $A_2$: Person ist gesund\\ -Ereignis $B$: Test identifiziert Person als krank.\\ - -\textbf{Wie viele als Krank erkannte wirklich krank?}\\ -\[ - P(A_1 | B ) = \frac{P(B|A_1)*P(A_1)} - {P(B|A_1)*P(A_1)+P(B| A_2)*P(A_2)} = - \frac{0,95*0,002}{0,95*0,002+0,002*0,998} = 8,7\% -\] - -\vspace*{10pt} - -L\"osung mittels \textbf{Formel von Bayes}: -\begin{align} - P(B_k/A) = \frac{P(A|B_k) * P(B_k)}{\sum_{j \in J} P(A|B_j) * P(B_j)} -\end{align} -Dieser Vorgang wird auch \textbf{R\"uckw\"artsinduktion} genannt. Angenommen man -kennt die Wahrscheinlichkeit eines Ereignis unter einer gewissen Bedingung (hier Test -schl\"agt zu $x\%$ an unter Bedingung Person ist krank $P(B|A_1)$ oder Person ist gesund -$P(B|A_2)$), dann kann man die umgekehrte bedingte Wahrscheinlichkeit -mit dieser Formel berechnen. Hier: Wie wahrscheinlich ist es, dass Person krank ist, unter -Bedingung, dass Test das gemeldet hat $P(A_1|B)$. - -\subsubsection{min. eine 6 unter Bedingung verschiedene Augenzahlen} -\[ - P(min. eine 6|verschiedene Augenzahlen) = \frac{\text{M\"oglichkeiten verschiedene Augenzahlen - UND min. eine 6}}{\text{M\"oglichkeiten verschiedene Augenzahlen}} -\]\\ -\[ - p=\frac{n*(6-1)!-(6-n)!}{6!-n!} -\] -bei 3 W\"urfeln also z.B.:\[ - p=\frac{3*5!-3!}{6!-3!} = \frac{3*5*4}{6*5*4} = 0,5 -\] - -\section{Wahrscheinlichkeitsfunktionen} -\subsection{Eigenschaften von Wahrscheinlichkeitsfunktionen} -\[ \sum_{w \in \Omega} f(w) = 1 \text{ (die Summe aller Wahrscheinlichkeiten ist 1)}\] -und logischerweise: -\[ \forall w\in\Omega . f(w)>=0 \text{ (keine negativen Wahrscheinlichkeiten)} \] -\subsection{Absoluten Momente diskreter Verteilungen} -Ist f\"ur $k \in \{1,2,3,\ldots\}$ die Summe $\sum_{x \in X} |x|^kf(x) < \infty$, -so heisst -\begin{align} - m_k = m_k(P) = \sum_{x \in X} x^kf(x) -\end{align} -das \textbf{k-te absolute Moment} der Verteilung P. - -\subsubsection{Mittelwert, Varianz} -\begin{itemize} - \item Mittelwert: $m_1 = m_1(P) = \sum_{n=0}^\infty n*f(n)$ - \item Varianz: $\widehat{m}_2 = m_2 - m_1^2$ -\end{itemize} -\subsubsection{Momenterzeugende Funktion} -\[ - M(t)=\sum_{n\in\Omega}^{\infty}(e^t)^n * f(n) -\] -- f(n) ist die gegebene Wahrscheinlichkeitsfunktion\\ -- 'n' k\"onnte z.B. definiert sein als $n=\{1,2,3,...\}$ - -\vspace*{15pt} -\textbf{Berechnungsvorschrift} f\"ur das k-te Moment: -\begin{enumerate} - \item Berechne k-te Ableitung $M^k$ von $M(t)$ - \item $m_i = M^{(k)}(0)$ -\end{enumerate} -\subsection{Erzeugende Funktion} -\subsubsection{Wahrscheinlichkeitsfunktion berechnen} -\textbf{Gegegeben:} Eine erzeugende Funktion $\hat{f}(z)$ gegeben. -\begin{align} - \hat{f}(z) = \sum^{\infty}_{k=0} f(k)z^k -\end{align} -\textbf{Gesucht:} Die Funktion $f(k)$ - -M\"oglichkeit 1: Taylorentwicklung -\begin{align} - \hat{f}(z) = \sum_{k=0}^{\infty} \frac{1}{k!}\hat{f}^{(k)}(0)z^k\\ - \Rightarrow f(k) = \frac{1}{k!} \hat{f}^{k}(0) -\end{align} - -M\"oglichkeit 2: Problem auf bekannte diskrete Verteilung zur\"uckf\"uhren (z.B. geometrische -Reihe) -\subsubsection{Mittelwert $m_1$} -\begin{align} - M(t) = \hat{f}(e^t)\\ - m_1 = M'(t)|_{t=0} = \hat{f}'(e^t)e^t|_{t=0} = \hat{f}'(1) -\end{align} -\subsubsection{Varianz $\hat{m}_2$} -\begin{enumerate} - \item Zuerst \textbf{zweites Moment} berechnen: - \begin{align} - m_2 = \hat{f}''(1) + \hat{f}'(1) \rightarrow \text{, falls \textbf{Erzeugende-Funktion} - (hier)}\\ - m_2 = \hat{f}''(0) \rightarrow \text{, falls \textbf{Momenterzeugende-Funktion}} - \end{align} - \item Dann \textbf{Varianz}: - \begin{align} - \hat{m}_2 = m_2 - m_1^2 - \end{align} - Siehe unten f\"ur $m_2$ Berechnungsvorschrift! -\end{enumerate} -\section{Verteilungen und Verteilungsfunktionen} -\subsection{Allgemein} -\subsubsection{Eigenschaften Verteilungsfunktionen} -\begin{itemize} - \item stetig - \item monoton steigend - \item $\lim_{t \to \infty} G(t) = 1, \quad \lim_{t \to -\infty} G(t) = 0$ - \item Dichte $g(t) = G'(t)$ - \item $m_1 = \int_{-\infty}^{\infty}t*g(t)dt$ -\end{itemize} -\subsection{Binominalverteilung} -\subsubsection{Allgemein} -\[ - \mathcal{B}(k | p,n) \enspace \textbf{ oder auch } \enspace B(k;p,n) = -\begin{pmatrix} n \\ k \end{pmatrix} p^k(1-p)^{n-k} \enspace \newline -\text{mit k = 0,1,2,...,n} \] -- wobei diese Funktion die \textbf{kumulierte} Wahrscheinlichkeit angibt, also z.B. -wobei k = 2 die Wahrscheinlichkeit "1 oder 2" -\\ - p ist die Wahrscheinlichkeit f\"ur ein positives Ereignisse -\\ - n ist Anzahl wie oft wir ziehen - -\subsubsection{Beispiel: 500 Druckfehler auf 500 Seiten} -Wie hoch ist die Wahrscheinlichkeit, dass auf einer Seite mindestens 3 Druckfehler -sind? -\[ -1- \sum_{k=0}^{2} \mathcal{B}(k,p,n) \enspace mit \enspace \] \\ -k=0,1,2 (Gegenereignisse)\\ n = 500 -(wir ziehen Fehler "ohne zur\"ucklegen") \\ p=1/500 (die Wahrscheinlichkeit dass -ein Fehler auf einer bestimmten Seite ist)\\ -\begin{equation*} - \begin{split} - 1- \sum_{k=0}^{2} \mathcal{B}(k|1/500,500) - & = 1 - \mathcal{B}(0|1/500,500) - \mathcal{B}(1|1/500,500) - \mathcal{B}(2|1/500,500) \\ - & = 1 - \left( \frac{499}{500} \right) ^{500} - 500\frac{1}{500}\left(\frac{499}{500}\right)^{499} - \frac{500*499}{1*2}\left( \frac{1}{500} \right) ^2 \left( \frac{499}{500} \right) ^{498} \\ & = 0,08 - \end{split} -\end{equation*} -\subsection{Poisson-Verteilung} -\subsubsection{Allgemein} -Ereignisse m\"ussen mit konstanter Rate, unabh\"angig voneinander und in einem festen -Bereich (Modell) stattfinden! -\[ - P_{\lambda}(n) = \frac{\lambda ^n}{n!} e ^{- \lambda} -\] -\subsection{Normal-Verteilung $\mathcal{N}(\mu, \sigma^2)$} -$f(x) = N(\mu, \sigma^2) = \frac{1}{\sqrt{2\pi\sigma^2}}*e^{-\frac{1}{2\sigma^2}(x- -\mu)^2} \quad \quad m_1 = \mu \quad \quad \widehat{m}_2=\sigma^2$ -\subsubsection{$\mathcal{N}(0,1)$-Verteilung} -$f(x) = \frac{1}{\sqrt{2\pi}}*e^{-0.5x^2}$ -\subsection{Exponentiallverteilung} -\textbf{Dichtefunktion}: -\begin{align} f_\lambda(x) = - \begin{cases} - \lambda*e^{-\lambda x} & x \geq 0 \\ - 0 & x < 0 - \end{cases} -\end{align} -\textbf{Verteilungsfunktion}: -\begin{align} F(x) = \int_0^x f_\lambda(t) dt = - \begin{cases} - 1 - e^{-\lambda x} & x \geq 0 \\ - 0 & x < 0 - \end{cases} -\end{align} -\subsection{Laplace-Verteilung} -Zufallsexperimente, bei denen jedes Ergebnis die gleiche Chance hat. \\ -$f(w) = L(\Omega) = \frac{1}{|\Omega|}$ -\subsection{Hypergeometrische Verteilung} -Zufallsexperimente, bei denen man die Ergebnisse als Anzahlen von schwarzen Kugeln unter n gezogenen interpretieren kann. \\ -$f(k) = H(N, K, n) = \frac{\binom{K}{k}*\binom{N-K}{n-k}}{\binom{N}{n}}$ -\subsection{Geometrische Verteilung} -Die geometrische Verteilung beschreibt die Wartezeit für das erstmalige Eintreten -eines Ereignisses unter der Annahme der Ged\"achtnislosigkeit. \\ -$G(p) = f(n) = p*q^{n-1} \quad \quad m_1 = \frac{1}{p}$ -\subsection{Uniform-Verteilung $\mathcal{U}(a,b)$} -\textbf{Dichtefunktion}: -\begin{align} f(x) = - \begin{cases} - \frac{1}{b - a} & a \leq x \leq b \\ - 0 & sonst - \end{cases} -\end{align} -\textbf{Verteilungsfunktion}: -\begin{align} F(x) = - \begin{cases} - 0 & x \leq a\\ - \frac{x - a}{b - a} & a < x < b \\ - 1 & x \geq b\\ - \end{cases} -\end{align} - -\section{Zufallsvariablen} -\subsection{Dichten von Verteilungen von Zufallsvariablen} -\textbf{Problembeschreibung}: Berechnung von Wahrscheinlichkeit des -Ereignisses $(X_1 > a * X_2)$ o.\"a. -Zufallsvariablen $X_1, X_2$ sind dabei stochastisch -unabh\"angig. Die Verteilungen von $X_i$ haben dabei die Dichten -$f_i$.\\ -Somit gilt nach der Marginalsdichte: $f(x_1,x_2) = f_1(x_1)*f_2(x_2)$. - -\begin{align} - P(X_1 > a * X_2) = \int \int_{x_1>a*x_2} f_1(x_1)*f_2(x_2) dx_1dx_2 := I -\end{align} -In Abh\"angigkeit von Reihenfolge, in der die Integration \"uber die Variablen -$x_1$ und $x_2$ durchgef\"uhrt werden, ergeben sich zwei Darstellungen: - -\begin{align} - I = \int_{-\infty}^{\infty} f_1(x_1)F_2(\frac{1}{a}x_1)dx_1\\ - I = \int_{-\infty}^{\infty} f_2(x_2)(1 - F_1(ax_2))dx_2 -\end{align} -Siehe auch L\"osungssammlung Aufgabe $98$ ff. - -\subsubsection{Beispiel} -Die Zufallsvariablen $X_1$ und $X_2$ seien uniform verteilt auf -$[0, 2]$. Berechnen Wahrscheinlichkeit des Ereignisses $(X_1X_2 \leq \frac{1}{2})$. - -\begin{align} - M = \{(x_1, x_2) \in \mathbb{R}^2 | 0 \leq x_1,x_2 \leq 2\}\\ - f(x_1,x_2) = - \begin{cases} - f(x_1)*f(x_2) = \frac{1}{4} & \text{f\"ur } (x_1,x_2) \in M \\ - 0 & \text{sonst} - \end{cases} -\end{align} -Borelsche Menge: -\begin{align} - B = \{(x_1, x_2) \in \mathbb{R}^2 | x_1*x_2 \leq \frac{1}{2}\ = x_2 \leq \frac{1}{2 x_1}\} \\ - P(x_1x_2 \leq \frac{1}{2}) = \int_B f(x_1, x_2) d(x_1, x_2) = \int 1_B*1_M*\frac{1}{4} - d(x_1,x_2)\\ - \int 1_{B \cap M} * \frac{1}{4} d(x_1,x_2) -\end{align} -Schnittmenge aus $B$ und $M$: -\begin{center} - \includegraphics[scale=0.3]{graph.png} -\end{center} -\begin{align} - B \cap M = \{(x_1, x_2) \in \mathbb{R}^2\ | (0 \leq x_1 \leq \frac{1}{4} \wedge 0 \leq x_2 \leq - 2) \vee (\frac{1}{4} \leq x_1 \leq 2 \wedge 0 \leq x_2 \leq \frac{1}{2x_1})\}\\ - \longrightarrow - P(x_1x_2 \leq \frac{1}{2}) = \int^{\frac{1}{4}}_{0} \int^2_0 \frac{1}{4} dx_1dx_2 - + \int^2_{\frac{1}{4}} \int^{\frac{1}{2x_1}}_0 \frac{1}{4} dx_2 dx_1 -\end{align} -\subsection{Erwartungswert $\varepsilon$ diskreter Zufallsvariablen} -Falls der Erwartungswert einer diskreten Zufallsvariablen $X$ auf -einem diskreten Wahrscheinlichkeitsraum $(\Omega, \mathcal{A}, P)$ existiert, -ist -\begin{align} - \varepsilon_P X = \sum_{\omega \in \Omega} X(\omega)P\{\omega\} -\end{align} - -\section{Marginaldichte - Beispielrechnung} -\[ - f(x_1,x_2)= - \begin{cases} - ce^{-(2x_1+3x_2)} & x_1 > 0 \: und \: 0 < x_2 =stealth',semithick}, - post/.style={->,shorten >=1pt,>=stealth',semithick}, - mid/.style={-,shorten >=1pt,>=stealth',semithick}, - place/.style={circle,draw=red!50,fill=red!20,thick}] - - \node[place] (A) at ( 0,0)[label=above:Before] {$(\Omega, A, P) $}; - \node[place] (B) at ( 2,0) {$(R^n, B_n, P^X)$} - edge [pre] node [auto] {X} (A); - \node[place, align=center] (C) at ( 2,-3) {$(R^m, B_m, P^G$} - edge [pre] node [auto] {$Y = G \circ X$} (A) - edge [pre] node [auto] {$G$} (B); - \end{tikzpicture} -\end{center} -\vspace*{7pt} -\textbf{Gegeben:} Man hat stochastisch unabhängige Zufallsvariablen $X_1, \ldots , X_n$ gegeben mit -Art der Verteilung. - -\textbf{Gesucht:} Verteilung von Zufallsvariable $Y$, die sich aus $X_i$ berechnen lässt. - -\textbf{Beispiel:}\\ -Welche Verteilung besitzt -\begin{align} - Y = \frac{X_1}{X_1 + X_2} -\end{align} -falls $X_1$ und $X_2$ exponentiell verteilt mit Paramter $\lambda$ und stochastisch -unabhängig sind. - -\begin{enumerate} - \item Wegen Unabhängigkeit der Variablen $X_1$ und $X_2$ besitzt $P^X$ - die Dichte $f(x_1,x_2) = f_1(x_1)f_2(x_2)$. - \item $M = {(x_1, x_2); x_1 > 0 \text{ und } x_2 > 0}$\\ - $\longrightarrow$ Wertebereich von $x_n$ anhand von Verteilung ermitteln. - \item Gleichungen $G(x)$ definieren: - \begin{align} - y_1 &= \frac{x_1}{x_1 + x_2}\\ - y_2 &= x_2 - \end{align} - \item Funktionaldeterminante ($J_{G}(x)$) der Abbildung $G$ berechnen - \begin{align} - J_{G}(x) = - \text{det} \begin{pmatrix} - \frac{\partial G_1}{\partial x_1} (x) & \cdots & \frac{\partial G_1}{\partial x_n} (x) \\ - \vdots & \ddots & \vdots \\ - \frac{\partial G_n}{\partial x_1} (x) & \cdots & \frac{\partial G_n}{\partial x_n} (x) \\ - \end{pmatrix}\\ - J_{G}(x_1,x_2) = - \text{det} \begin{pmatrix} - \frac{x_2}{(x_1 + x_2)^2} & * \\ - 0 & 1 \\ - \end{pmatrix} = \frac{x_2}{(x_1 + x_2)^2} - \end{align} - \item Umkehrabbildung $G^*$ berechnen. Alle Zufallsvariablen werden - werden mittels Funktionen verändert: z.B: $y_1 = x_1/x_2$. - Jede i-te Funktion nach $x_i$ auflösen. - \begin{align} - x1 = \frac{y_1y_2}{1 - y_1}\\ - x_2 = y_2 - \end{align} - \item Gesuchte Funktion: $g(y) = f(G^*(y))\frac{1}{|J_G(G^*(y))|}$\\ - $\longrightarrow$ Setze für alle $x_i$ dementsprechend $y_i$ ein und multipliziere - mit Kehrwehrt von Funktionaldeterminante. - \begin{align} - g(y_1,y_2) = \lambda^2e^{-\frac{\lambda}{1 - y_1}}\frac{y_2}{(1-y_1)^2} - \end{align} - \item Mit Marginaldichte $g_1(y_1)$ berechnen:\\ - \begin{align} - g_1(y_1) = \frac{\lambda}{1 - y_1} \int^\infty_0 y_2\frac{\lambda}{1 - y_1} - e^{-\frac{\lambda}{1 - y_1}} dy_2\\ - = \frac{\lambda}{1 - y_1} m_1 (\varepsilon(\frac{\lambda}{1 - y_1}))\\ - = 1 - \end{align} - $\longrightarrow$ Da Mittelwert der $\varepsilon$-Verteilung gerade Kehrwert des - Paramters ist. - \item Folgerung: Dichte $g_1$ ist also die der Uniform-Verteilung ($U(0,1)$). -\end{enumerate} - -\end{document} 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\title{Ko-Rekursion/-Induktion} - \date{ } - \begin{document} - \maketitle - \section{Korekursion} - \textbf{Gegeben:} - \begin{fleqn} - \begin{align*} - co&data Signal\:where \\ - & \enspace currentSample : Signal \rightarrow Int \\ - & \enspace discardSample : Signal \rightarrow Signal - \end{align*} - \end{fleqn} - \textbf{sowie:} - \begin{fleqn} - \begin{align*} - ¤tSample(flat\:x) = x & - &discardSample(flat\:x) = flat\:x \\ - ¤tSample(square\:x\:y) & - &discardSample(square\:x\:y) - \end{align*} - \end{fleqn} - \textbf{Es soll gelten:}\\ - sampler t s\enspace = s , wenn $t>0$ und 0 sonst\\ - sowie insbesondere: - \begin{fleqn} - \begin{align*} - &sampler:\:Signal\rightarrow Signal\rightarrow Signal \\ - &sampler(square\:0\:1)(square\:x\:0) = flat\:0 \\ - &sampler(square\:1\:0)(flat\:x) = square\:x\:0 \\ - \end{align*} - \end{fleqn} - \textbf{Schritt 1:}\\ - $\rightarrow$ sampler Funktion in codata-Funktionen einsetzen - \begin{fleqn} - \begin{align*} - ¤tSample(sampler\:t\:s)\\ - &discardSample(sampler\:t\:s) - \end{align*} - \end{fleqn} - \textbf{Schritt 2:}\\ - $\rightarrow$ Bedingung von oben bei erster Funktion anwenden also: - \begin{fleqn} - \begin{align*} - currentSample(sampler\:t\:s) = \:&if\:(currentSample\:\:t>0)\\ - &then\:(currentSample\:\:s)\\ - &else\:(flat\:\:0) - \end{align*} - \end{fleqn} - \textbf{sowie zweite Formel zum Aufteilen verwenden:} - \begin{fleqn} - \begin{align*} - discardSample(sampler\:t\:s) = sampler(discardSample\:\:t)(discardSample\:\:s) - \end{align*} - \end{fleqn} - \newpage - \section{Ko-Induktion:} - \textbf{Induktionsanfang:}\\ - $\rightarrow$ anhand erster Formel 'R' aufstellen - \[ - R = \{ - \underbrace{ - sampler(square\:0\:1)(square\:x\:0)}_{linke\:Seite},\underbrace{flat\:0}_{rechte\:Seite} \:|\: \underbrace{x\in Int}_{von\:oben}\} - \] - $\rightarrow$ "R ist Bisimulation" hinschreiben, linke Seite aufloesen\\ - \[ sampler(square\:0\:1)(square\:x\:0) = 0 \] - $\rightarrow$ wenn hier am Endde kein Term rauskommt, dann koennen wir einfach die \textbf{rechte Seite bei der normalen Funktion} einsetzen und selbige aufloesen: - \[ currentSample(flat\;0) = 0 \] - $\rightarrow$ sehr gut, die rechten Seiten sind gleich wir sind hier also fertig\\ \\ - \textbf{zweite Bedingung:} - \[discardSample(sampler(square\:1\:0)(square\:0\:x)) = sampler(square\:1\:0)(square\:0\:x)\] - \textit{das ist doof denn:} - \[discardSample(flat 0) = 0\] - \textbf{Schritt 3:}\\ - \[R' = R \;\cup\;\{\underbrace{sampler(square\:1\:0)(square\:0\:x)}_{aufgeloeste\;2.\;Bedingung},flat\;0\:|\:x\in Int\}\] - wir muessen jetzt zeigen: - \[ - currentSample(\underbrace{sampler(square\:1\:0)(square\:0\:x)}_{hier\;=\;0}) = \underbrace{currentSample(flat\;0)}_{hier\;=\;0} - \] - passt also, jetzt wie oben auch zweite Funktion: - \[ - discardSample(sampler(square\:1\:0)(square\:0\:x)) = \underbrace{sampler(square\:1\:0)(square\:0\:x)}_{wieder\;linke\;Seite\;in\;R'}\]\[ - discardSample(flat\;0) = \underbrace{flat\;0}_{wieder\;rechte\;Seite} - \] - und fertig!\\\\ - \begin{tiny} - \copyright\ Joint-Troll-Expert-Group (JTEG) 2015 - \end{tiny} - \newpage - - - - - - - - - - - - - - - - -\end{document} \ No newline at end of file diff --git a/Public/ThProg/Konfluenz.tex b/Public/ThProg/Konfluenz.tex deleted file mode 100644 index 22cd5f0..0000000 --- a/Public/ThProg/Konfluenz.tex +++ /dev/null @@ -1,49 +0,0 @@ - \documentclass{article} - \usepackage{amsmath} - \usepackage{nccmath} - \DeclareMathSizes{10}{10}{10}{10} - \setlength{\parindent}{0pt} - \title{Konfluenz} - \date{ } - \begin{document} - \maketitle - \section{Matching Table:} - Formeln: - \begin{align} - x \Uparrow ( y \Uparrow z) & \rightarrow_{0} - \; x \Uparrow (y \Downarrow y) - \\ - x \Downarrow ( x \Downarrow y ) & \rightarrow_0 - \; x \Downarrow y - \end{align} - - alle Regeln m\"ussen gegen alle anderen gematched werden - daher f\"ur \"ubersichtlichkeit:\\ \\ - \begin{tabular}{l c r} - & 1 & 2 \\ - 1 & ? & ? \\ - 2 & ? & ? \\ - \end{tabular} - - - - - - - - - - - - - - - - - - - - - \begin{tiny} - \copyright\ Joint-Troll-Expert-Group (JTEG) 2015 - \end{tiny} -\end{document} \ No newline at end of file diff --git a/Public/ThProg/Makefile b/Public/ThProg/Makefile deleted file mode 100644 index 89fe154..0000000 --- a/Public/ThProg/Makefile +++ /dev/null @@ -1,11 +0,0 @@ -all: Konfluenz.pdf Koinduktion_reduktion.pdf Struckturelle_Induktion.pdf Polynomordnung.pdf SystemF.pdf - -continuous: $(PDF).continuous - -%.continuous: %.pdf - latexmk -jobname=$(@:%.continuous=%) -pvc -pdf $(@:%.continuous=%).tex - -%.pdf: %.tex - latexmk -jobname=$(@:%.pdf=%) -pdf $< - -.PHONY: all continuous diff --git a/Public/ThProg/Polynomordnung.tex b/Public/ThProg/Polynomordnung.tex deleted file mode 100644 index de82172..0000000 --- a/Public/ThProg/Polynomordnung.tex +++ /dev/null @@ -1,71 +0,0 @@ - \documentclass{article} - \usepackage{amsmath} - \usepackage{nccmath} - \usepackage{ulem} - \DeclareMathSizes{10}{10}{10}{10} - \setlength{\parindent}{0pt} - \title{Ko-Rekursion/-Induktion} - \date{Oktober 2015} - \begin{document} - \maketitle - \section{In Funktionsschreibweise bringen} - Infixnotation: - \[x \uparrow ( y \uparrow z) \rightarrow_{0}\; x \uparrow (y \downarrow y)\]\\ - Funktionsschreibweise: - \[ P_{\uparrow}(x,P_{\uparrow}(y,z)) \rightarrow_{0} \; P_{\uparrow}(x,P_{\downarrow}(y,y))\] - \section{Kontext ggf. kuerzen} - \[ P_{\uparrow}(x,P_{\uparrow}(y,z)) \rightarrow_{0} \; P_{\uparrow}(x,P_{\downarrow}(y,y))\] - \[\xout{P(x},P_{\uparrow}(y,z)) \rightarrow_{0} \; \xout{P(x},P_{\downarrow}(y,y)))\] - \[P_{\uparrow}(y,z) \rightarrow_{0} \;P_{\downarrow}(y,y)\] - \section{Polynom finden} - \[P_{\uparrow}(y,z) \rightarrow_{0} \;P_{\downarrow}(y,y)\] - Ein "+" zwischen die Parameter setzen und Multiplikator vor beide sodass gilt: - \[P_{\uparrow}(y,z) > \;P_{\downarrow}(y,y)\;\;\textbf{\underline{bzw:}}\;\;ay+bx>cy+dy\] - \textbf{Hinweise:}\\ - - linke Seite ist syntaktisch echt gr\"osser als die Rechte ist aussreichendes Kriterium also z.B.: - \[ P_{\downarrow}( P_{\downarrow}(x,y) ) > P_{\downarrow}(x,y) \] - - niemals Minuswerte \\ - - niemals '0' als Multiplikator\\\\ - \textbf{hier:} - \[ay+bx>cy+dy\] - \textbf{Wir nehmen an dass wir 'y' hoch genug setzen damit bx irrelevant wird:}\\ - \[ay>cy+dy = a>c+d \] - \\ \textbf{das ist nun trivial, wir raten:} - \begin{fleqn} - \begin{align*} - &c = 1 \\ - &d = 1 \\ - &a = c+d+1 = 3 - \end{align*} - \end{fleqn} \\ - \textbf{und damit die Polynome:}\\ - \[P\downarrow = x_1+x_2 \;\;\; und \;\;\; P\uparrow = 3x_1+x_2 \] \\ - \section{Dom\"anen und Grenzf\"alle} - Unsere Polynome gelten potentiell f\"ur kleine Werte nicht, hier, nur f\"ur die 0 nicht, daher geben wir als Dom\"ane an: - \[ A = N\setminus\{0\} \] -und Funktionsdom\"ane:\\ - \[\mathcal{A} = \{P(*),P(*)\}\] - - - - - - - - - - - - - - - - - - - - - - - - \end{document} \ No newline at end of file diff --git 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