add wtheo3, ex. 9,11
This commit is contained in:
BIN
Binary file not shown.
+160
@@ -0,0 +1,160 @@
|
||||
\documentclass[uebung]{../../../lecture}
|
||||
|
||||
\title{Wtheo 0: Übungsblatt 3}
|
||||
\author{Josua Kugler, Christian Merten}
|
||||
\usepackage[]{bbm}
|
||||
\usepackage[]{mathrsfs}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\punkte[9]
|
||||
|
||||
\begin{aufgabe}
|
||||
|
||||
\begin{enumerate}[(a)]
|
||||
\item Beh.: $C_{\alpha, x_m} = \alpha x_m^{\alpha}$.
|
||||
\begin{proof}
|
||||
Es muss gelten
|
||||
\begin{salign*}
|
||||
\int_{-\infty}^{\infty} \mathbbm{f}(x) \d x &= 1
|
||||
\intertext{Damit folgt}
|
||||
\int_{-\infty}^{\infty} C_{\alpha, x_m} x^{-(\alpha +1)} \mathbbm{1}_{\{x \ge x_m\} } \d x
|
||||
&= \int_{x_m}^{\infty} C_{\alpha,x_m} x^{-(\alpha + 1)} \d x \\
|
||||
&= C_{\alpha, x_m} \int_{x_m}^{\infty} x^{-(\alpha +1)} \d x \\
|
||||
&\stackrel{\alpha + 1 > 1}{=} - C_{\alpha, x_m} \frac{1}{\alpha} x^{-\alpha} \Big|_{x_m}^{\infty} \\
|
||||
&= - \frac{C_{\alpha,x_m}}{\alpha} \lim_{z \to \infty} \left[ z^{-\alpha} - x_m^{-\alpha} \right] \\
|
||||
&\stackrel{\alpha > 0}{=} \frac{C_{\alpha, x_m}}{\alpha} x_m^{-\alpha} \\
|
||||
&\stackrel{!}{=} 1
|
||||
\intertext{Damit folgt dann}
|
||||
C_{\alpha,x_m} &= \alpha x_m^{\alpha}
|
||||
.\end{salign*}
|
||||
\end{proof}
|
||||
\item Beh.: $\mathbbm{F}(x) = \left( 1 - \left( \frac{x_m}{x} \right)^{\alpha} \right) \mathbbm{1}_{\{x \ge x_m > 0\} }$
|
||||
\begin{proof}
|
||||
Falls $x < x_m$ ist $\mathbbm{f}(y) = 0$ $\forall y \le x$, also $\mathbb{F}(x) = 0$.
|
||||
Sei also $x \ge x_m$. Dann folgt
|
||||
\begin{salign*}
|
||||
\mathbb{F}(x) &= \int_{x_m}^{x} \alpha x_m^{\alpha} y^{-(\alpha + 1)} \d y \\
|
||||
&= - x_m^{\alpha} y^{-\alpha} \Big|_{x_m}^{x} \\
|
||||
&= - x_m^{\alpha} \left[ x^{-\alpha} - x_m^{-\alpha} \right] \\
|
||||
&= -x_m^{\alpha} x^{-\alpha} + x_m^{\alpha} x_m^{-\alpha} \\
|
||||
&= 1 - \left( \frac{x_m}{x} \right)^{\alpha}
|
||||
\intertext{Insgesamt folgt}
|
||||
\mathbb{F}(x) &= \left( 1 - \left( \frac{x_m}{x} \right)^{\alpha} \right) \mathbbm{1}_{\{x \ge x_m > 0\} }
|
||||
.\end{salign*}
|
||||
\end{proof}
|
||||
\item Beh.: $\mathbb{P}([1,2]) = \frac{1}{2} = \mathbb{P}((2, \infty))$.
|
||||
\begin{proof}
|
||||
Mit $\alpha = x_m = 1$ folgt $\mathbb{F}(x) = \left( 1 - \frac{1}{x} \right) \mathbbm{1}_{\{x \le 1\} }$. Damit folgt
|
||||
\begin{salign*}
|
||||
\mathbb{P}([1,2]) &= \mathbb{F}(2) - \mathbb{F}(1) = 1 - \frac{1}{2} - 1 + 1 = \frac{1}{2} \\
|
||||
\mathbb{P}((2, \infty)) &= 1 - \mathbb{P}((-\infty, 2]) = 1 - \mathbb{F}(2) = 1 - 1 +\frac{1}{2} = \frac{1}{2}
|
||||
.\end{salign*}
|
||||
\end{proof}
|
||||
\end{enumerate}
|
||||
\end{aufgabe}
|
||||
|
||||
\stepcounter{aufgabe}
|
||||
|
||||
\begin{aufgabe}
|
||||
Zunächst ist zu bemerken, dass mit $\mathscr{E} := \{ (a, \infty] \mid a \in \R\} $ nach VL
|
||||
gilt $\sigma(\mathscr{E}) = \overline{\mathscr{B}}$. Damit ist
|
||||
$f\colon \Omega \to \overline{\R}$ genau dann $(\mathscr{A}, \overline{\mathscr{B}})$ messbar,
|
||||
wenn $f^{-1}(\mathscr{E}) \subseteq \mathscr{A}$.
|
||||
\begin{enumerate}[(a)]
|
||||
\item
|
||||
\begin{enumerate}[(1)]
|
||||
\item Sei $m \in \N$. Beh.: Folgende Abbildungen sind $(\mathscr{A}, \overline{\mathscr{B}})$
|
||||
messbar:
|
||||
\begin{enumerate}[(i)]
|
||||
\item $\sup_{n \ge m} X_n \colon \Omega \to \overline{\R}$
|
||||
\item $\inf_{n \ge m} X_n \colon \Omega \to \overline{\R}$
|
||||
\end{enumerate}
|
||||
\begin{proof}
|
||||
\begin{enumerate}[(i)]
|
||||
\item Sei $a \in \R$ bel. Für $x \in \R$ gilt dann
|
||||
\[
|
||||
\sup_{n \ge m} X^{n}(x) > a \iff \exists n \ge m\colon X^{n}(x) > a
|
||||
.\]
|
||||
Damit folgt da $X^{n}$ $(\mathscr{A}, \overline{\mathscr{B}})$-messbar
|
||||
und $\mathscr{A}$ $\sigma$-Algebra:
|
||||
\begin{salign*}
|
||||
(\sup_{n \ge m} X_n)^{-1}((a, \infty]))
|
||||
&= \{ x \in \Omega \mid \sup_{n \ge m} X^{n}(x) > a\} \\
|
||||
&= \{ x \in \Omega \mid \exists n \ge m\colon X^{n} > a\} \\
|
||||
&= \bigcup_{n \ge m} \{ x \in \Omega \mid X^{n}(x) > a\} \\
|
||||
&= \bigcup_{n \ge m} \underbrace{(X^{n})^{-1}((a, \infty])}_{\in \mathscr{A}}
|
||||
\in \mathscr{A}
|
||||
.\end{salign*}
|
||||
Also $\sup_{n \ge m} X^{n}$ $(\mathscr{A}, \overline{\mathscr{B}})$-messbar.
|
||||
\item Sei $a \in \R$. Für $x \in \R$ gilt dann
|
||||
\[
|
||||
\inf_{n \ge m} X^{n}(x) < a \iff \exists n \ge m\colon X^{n}(x) < a
|
||||
.\]
|
||||
Damit folgt da $X^{n}$ $(\mathscr{A}, \overline{\mathscr{B}})$-messbar
|
||||
und $\mathscr{A}$ $\sigma$-Algebra:
|
||||
\begin{salign*}
|
||||
(\inf_{n \ge m} X_n)^{-1}([-\infty, a))
|
||||
&= \{ x \in \Omega \mid \inf_{n \ge m} X^{n}(x) < a\} \\
|
||||
&= \{ x \in \Omega \mid \exists n \ge m\colon X^{n} < a\} \\
|
||||
&= \bigcup_{n \ge m} \{ x \in \Omega \mid X^{n}(x) < a\} \\
|
||||
&= \bigcup_{n \ge m} \underbrace{(X^{n})^{-1}([-\infty, a))}_{\in \mathscr{A}}
|
||||
\in \mathscr{A}
|
||||
.\end{salign*}
|
||||
Da auch $\sigma(\{ [-\infty, a) \mid a \in \R \}) = \overline{\mathscr{B}}$
|
||||
folgt
|
||||
also $\inf_{n \ge m} X^{n}$ $(\mathscr{A}, \overline{\mathscr{B}})$-messbar.
|
||||
\end{enumerate}
|
||||
\end{proof}
|
||||
\item Beh.: $\limsup_{n \to \infty} X^{n}$ und $\liminf_{n \to \infty} X^{n}$ sind
|
||||
$(\mathscr{A}, \overline{\mathscr{B}})$-messbar.
|
||||
|
||||
\begin{proof}
|
||||
Definiere $f_m \coloneqq \sup_{n \ge m} X_n$. Dann ist $f_m$ messbar nach (1)(i)
|
||||
und
|
||||
\[
|
||||
\limsup_{n \to \infty} X_n = \inf_{m \ge 1} \sup_{n \ge m} X_n = \inf_{m \ge 1} f_m
|
||||
\] $(\mathscr{A}, \overline{\mathscr{B}})$-messbar nach (1)(ii).
|
||||
|
||||
Definiere nun $h_m \coloneqq \inf_{n \ge m} X_n$. $h_m$ messbar nach (1) (ii)
|
||||
$\forall m \in \N$. Dann ist auch
|
||||
\[
|
||||
\liminf_{n \to \infty} X_n = \sup_{m \ge 1} \inf_{n \ge m} X_n = \sup_{m \ge 1} h_m
|
||||
\] $(\mathscr{A}, \overline{\mathscr{B}})$-messbar nach (1)(i).
|
||||
\end{proof}
|
||||
\item Beh.: $\lim_{n \to \infty} X_n$ $(\mathscr{A}, \overline{\mathscr{B}})$-messbar.
|
||||
\begin{proof}
|
||||
Sei $X = \lim_{n \to \infty} X_n$. Dann ist
|
||||
$X = \liminf_{n \to \infty} X_n = \limsup_{n \to \infty} X_n$, also
|
||||
$X$ $(\mathscr{A}, \overline{\mathscr{B}})$-messbar nach (2).
|
||||
\end{proof}
|
||||
\end{enumerate}
|
||||
\item Beh.: $Y$ $(\mathscr{A}, \mathscr{B})$-messbar.
|
||||
\begin{proof}
|
||||
Beachte, dass es für $\mathscr{B}$ genügt, die offenen Intervalle
|
||||
$(a, \infty)$ für $a \in \R$ zu betrachten.
|
||||
|
||||
Sei $a \in \R$ bel.
|
||||
\begin{itemize}
|
||||
\item Falls $a \le 0$, dann ist $0, 1 \in (a, \infty)$, also
|
||||
$Y^{-1}((a, \infty)) = \Omega \in \mathscr{A}$.
|
||||
\item Falls $0 < a < 1$: Dann ist $1 \in (a, \infty)$ und $0 \not\in (a, \infty)$. Damit
|
||||
folgt
|
||||
\begin{salign*}
|
||||
Y^{-1}((a, \infty)) &= \{ \omega \in \Omega \mid X_1(\omega) > X_2(\omega) \} \\
|
||||
&= \{ \omega \in \Omega \mid X_1(\omega) - X_2(\omega) > 0\} \\
|
||||
&= \{ \omega \in \Omega \mid (X_1 - X_2)(\omega) \in (0, \infty]\} \\
|
||||
&= (X_1 - X_2)^{-1}((0, \infty])
|
||||
.\end{salign*}
|
||||
Da $X_1, X_2$ $(\mathscr{A}, \overline{\mathscr{B}})$-messbar, ist
|
||||
nach VL auch $X_1 - X_2$ $(\mathscr{A}, \overline{\mathscr{B}})$ messbar.
|
||||
Da weiter $(0, \infty] \in \overline{\mathscr{B}}$, folgt also
|
||||
$(X_1 - X_2)^{-1}((0, \infty]) \in \mathscr{A}$.
|
||||
\item Falls $a \ge 1$, dann ist $0, 1 \not\in (a, \infty)$, also
|
||||
$Y^{-1}((a, \infty)) = \infty \in \mathscr{A}$.
|
||||
\end{itemize}
|
||||
\end{proof}
|
||||
\end{enumerate}
|
||||
\end{aufgabe}
|
||||
|
||||
\end{document}
|
||||
Reference in New Issue
Block a user