diff --git a/wtheo9.pdf b/wtheo9.pdf
index dbc30a4..13badf6 100644
Binary files a/wtheo9.pdf and b/wtheo9.pdf differ
diff --git a/wtheo9.tex b/wtheo9.tex
index 60c051b..29cae1c 100644
--- a/wtheo9.tex
+++ b/wtheo9.tex
@@ -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)]