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JosuaKugler
2021-01-27 12:24:25 +01:00
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@@ -3,6 +3,7 @@
\title{Wtheo 0: Übungsblatt 9}
\author{Josua Kugler, Christian Merten}
\newcommand{\E}{\mathbb{E}}
\renewcommand{\P}{\mathbb{P}}
\usepackage[]{mathrsfs}
\newcommand{\cov}{\mathbb{C}\text{ov}}
\newcommand{\var}{\mathbb{V}\text{ar}}
@@ -81,7 +82,66 @@
\end{enumerate}
\end{aufgabe}
\stepcounter{aufgabe}
\begin{aufgabe}
\begin{enumerate}[(a)]
\item Es gilt
\begin{salign*}
\mathbb{F}^{Z_p}(x) &= \P^{Z_p}((-\infty, x])\\
&= \P^{(-1)^{V_p}\cdot Y}((-\infty, x] \cap \{V_p = 0\}) + \P^{(-1)^{V_p}\cdot Y}((-\infty, x] \cap \{V_p = 1\})\\
&= \P^Y((-\infty, x] \cap \{V_p = 0\}) + \P^{-Y}((-\infty, x] \cap \{V_p = 1\})\\
&\stackrel{Y \indep V_p}{=} \P(\{Y \leq x\}) \cdot \P(\{V_p = 0\}) + \P(\{-Y \leq x\}) \cdot \P(\{V_p = 1\})\\
&\stackrel{\text{Symmetrie } N_{(0,1)}}{=} \P(\{Y \leq x\}) \cdot \P(\{V_p = 0\}) + \P(\{Y \leq x\}) \cdot \P(\{V_p = 1\})\\
&= \P(\{Y \leq x\}) \cdot \P(\{V_p = 0\}\cup \{V_p = 1\})\\
&= \P(\{Y \leq x\})\\
&= \mathbb{F}^{Y}(x)\\
\end{salign*}
Daher gilt $Z_p \sim Y \sim N_{(0,1)}$.
\item Es gilt
\begin{salign*}
\P(\{Y < -1, Z_p < -1\}) &= \P(\{Y < -1\} \cap \{V_p = 0\})
&\stackrel{Y \indep V_p}{=} \P(\{Y < -1\}) \cdot \P(\{V_p = 0\})\\
&= (1-p) \cdot \P(\{Y < -1\})
\end{salign*} und völlig analog
\begin{salign*}
\P(\{Y < -1, Z_p > 1\}) &= \P(\{Y < -1\} \cap \{V_p = 1\})
&\stackrel{Y \indep V_p}{=} \P(\{Y < -1\}) \cdot \P(\{V_p = 1\})\\
&= p \cdot \P(\{Y < -1\})
\end{salign*}
Angenommen, $Y \indep Z_p$. Dann gilt
\begin{salign*}
\P(\{Y < -1, Z_p < -1\}) &= \P(\{Y< -1\})\P(\{Z_p < -1\})\\
(1-p) \cdot \P(\{Y < -1\}) &\stackrel{\text{(a)}}{=} \P(\{Y < -1\})^2\\
(1-p) &= \P(\{Y < -1\})
\end{salign*} und völlig analog
\begin{salign*}
\P(\{Y < -1, Z_p > 1\}) &= \P(\{Y< -1\})\P(\{Z_p > 1\})\\
p \cdot \P(\{Y < -1\}) &\stackrel{\text{(a), Symmetrie } N_{(0,1)}}{=} \P(\{Y < -1\})^2\\
p &= \P(\{Y < -1\})
\end{salign*}
Nun führen wir eine Fallunterscheidung durch.
Für $p = \frac{1}{2}$ folgt $\P(\{Y < -1\}) < \P(\{Y < 0\}) \leq \frac{1}{2}$, Widerspruch zu $p = \P(\{Y < -1\})$.
Für $p \neq \frac{1}{2}$ erhalten wir aus $p = \P(\{Y < -1\}) = (1-p)$ ebenfalls einen Widerspruch.
Daher ist $Y \not \indep Z_p$.
\item Es gilt
\begin{salign*}
\E(YZ_p) &= \int_\R \int_\R yz \mathbbm{f}^{Y, Z_p}(y, z) \d{y}\d{z}\\
&\stackrel{Z_p = (-1)^{V_p}Y}{=} \int_{0,1} \int_\R y^2 (-1)^v \mathbbm{f}^{Y, V_p}(y, v) \d{y}\d{v}\\
&\stackrel{Y \indep V_p}{=} (1-p) \cdot \int_\R y^2 \mathbbm{f}^Y(y) \d{y} + p\cdot \int_\R -y^2 \mathbbm{f}^Y(y) \d{y}\\
&= (1 - 2p) \int_\R y^2 \mathbbm{f}^Y(y) \d{y}
\end{salign*}
Außerdem gilt
\begin{salign*}
\E(y)\E(Z_p) &= \int_\R y\mathbbm{f}^Y(y) \d{y} \int_\R z\mathbbm{f}^{Z_p}(z) \d{z}\\
&\stackrel{Z_p = (-1)^{V_p}Y}{=} \int_\R y\mathbbm{f}^Y(y) \d{y} \cdot \int_{0,1} \int_\R y (-1)^v \mathbbm{f}^{Y, V_p}(y, v) \d{y}\d{v}\\
&\stackrel{Y \indep V_p}{=} (1-p) \cdot \left(\int_\R y \mathbbm{f}^Y(y) \d{y}\right)^2 - p\cdot \left(\int_\R y^2 \mathbbm{f}^Y(y) \d{y}\right)^2\\
&= (1 - 2p) \left(\int_\R y \mathbbm{f}^Y(y) \d{y}\right)^2
\end{salign*}
Für $p = 0$ erhalten wir daher
\begin{align*}
\cov(Y, Z_p) = \E(YZ_p) - \E(Y)\E(Z_p) = 0 - 0 = 0.
\end{align*}
\end{enumerate}
\end{aufgabe}
\begin{aufgabe}
\begin{enumerate}[(a)]