\documentclass[a4paper,11pt]{book}
\usepackage[ngerman]{babel}
%\usepackage[ngerman]{babel}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{amsmath, amsthm, amsfonts, amssymb}
\usepackage{enumerate}
\usepackage[margin=3cm]{geometry}
 
\input{../math.tex}

\newtheorem{thm}{Aufgabe H}
%opening
\title{Übungsblatt 6 W-Theorie}
\author{Max Klinger}
 
\begin{document} 
\section*{Wahrscheinlichkeitstheorie Übungsblatt 6}
\begin{thm}
Seien $\left(X_n\right)_{n\in \NN}$ reelle ZVen, $S_n = \sum^n_{i=1} X_i$. Sind 
\begin{itemize}
 \item $\left\{\sup_{k\in \NN} S_k \leq x\right\} $ mit $x \in \RR$
 \item $\left\{\sup_{k\in \NN} S_k = \infty \right\}$
\item $\left\{\liminf_{n\rightarrow \infty} X_n  \leq x \right\}$ mit $ x\in \RR$
\end{itemize}
terminal?
\end{thm}

\begin{proof}[Beweis]
\begin{enumerate}
 \item $\displaystyle \left\{\sup_{k\in \NN}\int_k \subseteq x \right\} \in \mathcal{T}$ mit $x \in \RR$; nein $(X_i)_{i\geq 1}$ reelle unabh. ZVen. \\ $X_1 \in \{-1, 1\}$ und $X_i \equiv 0, \; i \geq 2; \; \sup_k \int_K = X_1$\\ 
$\Rightarrow \left\{\sup_k \int_k \leq x \right\} \notin \sigma(X_2, X_3, X_4, \dotsc)$\\
$\Rightarrow \{\sup_k \int_k \leq x\}$ ist nicht $\in \mathcal{T}$
 \item $\{\sup_k \int_K = \infty\}$ terminal ?
	Ja, da X \underline{reelle} ZV
\begin{itemize}
 \item $\{\sup_{k\in \NN} \int_K = \infty \} = \{\sup_{k\geq n} \int_K = \infty\};\; \forall n \in \NN$
 \item $\displaystyle \{\sup_{k\geq n} \int_K = \sup_n (X_1 + \dotsc + X_n + \dotsc X_k ) = \infty \} =$\\ 
$\displaystyle \{\sup_k\geq n \sum_{n\leq i \leq k } X_i = \infty \} \in \sigma(X_n, X_{n+1}, \dotsc);\; \forall n \in \NN$ \\$\Rightarrow \{\sup_{k\in \NN} \int_k = \infty\}$ ist terminal
\end{itemize}
\item $\{\liminf_{n\rightarrow \infty} X_n \leq x\}$ ist terminal? Ja.\\
 $\displaystyle \left\{\liminf_{n\rightarrow \infty}\right\} = \bigcap_{k\geq 1} \bigcap_{1\leq j < \infty} \bigcup_{j < i < \infty} \left\{X_i \leq x + \frac{1}{k}\right\}$
Nach Vorlesung (Bsp 3.18) gilt 
$$\bigcap_{1\leq j < \infty} \bigcup_{j \leq i< \infty} \left\{X_i \leq x +\frac{1}{k}\right\} = \bigcap_{N\leq j < \infty} \bigcup_{j\leq i < \infty} \left\{X_i \leq x + \frac{1}{k}\right\} \; \forall k,N \in \NN$$
\begin{align*}
\displaystyle \Rightarrow \{\liminf X_n \leq x\} &= \bigcap_{k\geq 1}\bigcap_{1<j<\infty}\bigcup_{j\leq i <\infty} \left\{X_i \leq x + \frac{1}{2}\right\}\\ &= \underbrace{\bigcap_{k\geq 1} \bigcap_{1<j < \infty} \bigcup_{j<i<\infty} \left\{X_i \leq x + \frac{1}{k}\right\}}_{\sigma(X_n, X_{n+1}, \dotsc}; \; \forall N \in \NN 
\end{align*}
$\displaystyle \Rightarrow \{\liminf X_n \leq 1\} \in \mathcal{T} = \bigcap_{N\geq1} \sigma(X_n, X_{n+1}, \dotsc)$
\end{enumerate}
\end{proof}

\begin{thm}
Sei $n \in \NN, X_n \sim \Poi_n$ und $\left(\frac{X_n}{n}\right)_{n\geq 1} \xrightarrow P c = ?$
\end{thm}

\begin{proof}[Beweis]
Sei $Y_1, \dotsc , Y_n$ unabhängig und $Y_j \sim \Poi_1 ;\;\forall i$ $\Rightarrow \sum^n_{i=1} Y_i \sim \Poi_n $.\\
Wir können das Gesetz der großen Zahlen anwenden (wegen $\Poi_1$ und unabh\"angig)
$$ \frac{1}{n} \sum^N_{1 \leq i \leq N} Y_i \xrightarrow P \EE[Y_1] = 1; \Rightarrow \frac{X_n}{\omega} \xrightarrow P 1 $$
\end{proof}

\begin{thm}$\mu_n, \mu$ W'Maße $\mu_n \left(]-\infty, x]\right) = F_n (x),\; \mu \left(]-\infty, x]\right) = F(x)$
\begin{itemize}
 \item F stetig
\item $\mu_n\xrightarrow w \mu $
\end{itemize}
Z.Z.: $\displaystyle \sup_{x \in \RR} \abs{F_n(x) - F(x)} \rightarrow 0$
\end{thm}

\begin{proof}
\begin{itemize}
\item $F_n(x) \rightarrow F(x)$ f\"ur $x \in \RR$\\
Widerspruchsannahme $\displaystyle \exists \epsilon > 0; \; \exists (n_k)_{k\in \NN} \; \exists (x_{n_k})_{k\in \NN} \; \abs{F_{n_k} (x_{n_k}) -F (x_{n_k}) } \geq \epsilon$
\begin{description}
 \item [1.Fall:] $\left(x_{n_k}\right)_{k\in\NN}$ sei unbeschränkt\\
 \underline{1. Unterfall} $x_{n_k}\nearrow \infty$ monoton. 
$F(n) \xrightarrow{n\rightarrow \infty} 1 $, d.h. 
$\exists N \in \NN \; 1 - F(N) < \frac{\epsilon}{3}$ \\
\underline{wegen} $F_{n_k} (N) \rightarrow F(N) \; \exists N^{'} \in \NN :\; \abs{F_{n_k} - F(N)} < \frac{\epsilon}{3} \; \forall k \geq \NN$. \\
\begin{align*}
\Rightarrow \abs{F(x)-F_{n_k} (x)} &\leq \abs{1-F(x)}+\abs{1-F_{n_k}(x)} \leq \underbrace{1-F(x)}_{\frac{\epsilon}{3}} + 1 - F_{n_k} (x) \\ 
&\leq 1-F(N) + 1 - F_{n_k}(N) \\
&< \frac{\epsilon}{3} + 1 - F(N) + F(N ) - F_{n_k} (N) < \epsilon 
\end{align*}
\underline{2. Unterfall} $ x_{n_k} \searrow -\infty$ monoton analoger Widerspruch
\item[2.Fall:] ($x_{n_k}$) sei beschränkt o. E. $x_{n_k} \xrightarrow {k\rightarrow \infty}x$\\
Da F stetig $\exists \delta > 0 \abs{x-y} \leq \delta \Rightarrow \abs{F(x) - F(y)} \leq \frac{\epsilon}{5}$\\
Da $F_n \rightarrow F$ punktweise $\exists N \in \NN: \abs{F_{n_k} (x-\delta) - F(x-\delta)} < \frac{\epsilon}{5}; \; \forall k \geq N$ $\abs{F_{n_k}(x+\delta) - F(x+\delta }< \frac{\epsilon}{5}; \; \forall k \geq N $\\
$y\in [x-\frac{\delta}{2}, x+\frac{\delta}{2}]\Rightarrow$
\begin{align*}
\abs{F(y) - F_{n_k}(y)} &\leq \abs{F(y) - F(x- \delta} + \abs{F(x-\delta) - F_{n_k}(x - \delta)} \\
&\hspace{12 pt} + \abs{F_{n_k}(x-\delta) - F_{n_k}(y)} \\
& < \frac{\epsilon}{5} + \frac{\epsilon}{5} + \underbrace{F_{n_k}(y)}_{F(x-\delta) + \epsilon/5} -  \underbrace{F_{n_k}(x-\delta) }_{F(x-\delta) + \epsilon/5} \vspace{10pt} \\
&< \frac{\epsilon}{5} + \frac{\epsilon}{5} + \frac{\epsilon}{5} _ F(x+\delta) - F(x- \delta) + \frac{\epsilon}{5} \\
&= 4 \frac{\epsilon}{5} + F(x+\delta) -F(x-\delta) \\
&< \epsilon 
\end{align*}
\end{description}

\end{itemize}
\end{proof}
\end{document}
