diff --git a/mr/ub8/mr8.pdf b/mr/ub8/mr8.pdf
index 091b875..547420f 100644
--- a/mr/ub8/mr8.pdf
+++ b/mr/ub8/mr8.pdf
Binary files differ
diff --git a/mr/ub8/mr8.tex b/mr/ub8/mr8.tex
index 0b37338..cc02fd0 100644
--- a/mr/ub8/mr8.tex
+++ b/mr/ub8/mr8.tex
@@ -75,8 +75,7 @@
 \textbf{Exercise \arabic{n}: #1} (#2 Punkte)\\}
 
 \newcommand{\Normal}[3]{\mathcal{N}\left(#1,#2,#3\right)}
-\newcommand{\Normalf}[3]{\frac{1}{#3 \cdot\sqrt{2\pi}}e^{-\frac{\left(#1 - #2 \right)^2}{2 #3 ^2}}}
-\newcommand{\Normalfs}[3]{\frac{1}{#3}e^{-\frac{\left(#1 - #2 \right)^2}{2 #3 ^2}}}
+\newcommand{\Normalf}[3]{\frac{\left|#3\right|^{-\frac{1}{2}}}{\sqrt{2\pi}}e^{-\frac{\left(#1 - #2 \right)^2\cdot \left(#3\right)^{-1}}{2}}}
 
 
 
@@ -105,30 +104,39 @@
         That is we assume no variance and fully trust the measurement.\pagebreak\\
 \Aufgabe{Excercise2}{8}
 \begin{enumerate}[(a)]
-    \item %TODO Herleitung
-        \begin{itemize}
-            \item $p(x_{t-1})=\mathcal{N}\left(x_{t-1},\hat{x}_{t-1},\sigma_{t-1}\right)$
-            \item $p(z_t)=\mathcal{N}\left(z_t,cx_t,r_t\right)$
-            \item $p(x_t)=\mathcal{N}\left(x_t,a\mu_{t-1},q_t\right)$
-        \end{itemize}
+    \item $\Normal{x_t}{\mu_t}{\Sigma_t}=\eta\Normal{z_t}{Cx_t}{Q_t}\Normal{x_t}{\bar{\mu}_t}{\bar{\Sigma}_t}$
     \item \begin{align*}
-            \Normal{x_t}{ax_{t-1}}{q_t}&=\eta\Normal{z_t}{cx_t}{r_t}\Normal{x_{t-1}}{\hat{x}_{t-1}}{\sigma_{t-1}}\\
-            \Normalf{x_t}{ax_{t-1}}{q_t}&=\eta\Normalf{z_t}{cx_t}{r_t}\Normalf{x_{t-1}}{\hat{x}_{t-1}}{\sigma_{t-1}}\\
-            \Normalfs{x_t}{ax_{t-1}}{q_t}&=\eta\Normalfs{z_t}{cx_t}{r_t}\Normalfs{x_{t-1}}{\hat{x}_{t-1}}{\sigma_{t-1}}
-        \end{align*}
-        %TODO
-	% \item $p(x_t|z_{t,\dots1},u_{t,\dots1})\sim N(x_t,\mu_{x,t},\sigma_{x,t})\\
-	% \mu_{x,t}= a\mu_{x,t-1}+bu_t+\epsilon_t\\
-	% \sigma_{x,t}= |a|\sigma_{x,t-1}\\
-	% \\
-	% p(x_{t}|z_{t,\dots1},u_{t,\dots1})\sim N(x_{t-1},\mu_{x,t-1},\sigma_{x,t-1})\\
-	% \mu_{x,t-1}= a\mu_{x,t-2}+bu_{t-1}+\epsilon_{t-1}\\
-	% \sigma_{x,t-1}= |a|\sigma_{x,t-2}\\
-	% \\
-	% p(z_t|x-t)\sim N(z_t,\mu_z,\sigma_z)\\
-	% \mu_z=c*\mu_{x,t-1}+\delta_t \\
-	% \sigma_z=|c|\sigma_{x,t-1}
-	% $
+        &\Normalf{x_t}{\mu_t}{\Sigma_t}\\&=\eta\Normalf{z_t}{Cx_t}{Q_t}\Normalf{x_t}{\bar{\mu}_t}{\bar{\Sigma}_t}\\
+        &=\eta\frac{1}{2\pi}\left|Q_t\right|^{-\frac{1}{2}}\cdot\left|\bar{\Sigma}_t\right|^{-\frac{1}{2}}\cdot e^{-\frac{1}{2}\left(z_t^2Q_t^{-1}-2z_tCx_tQ_t^{-1}+\left(Cx\right)^2Q_t^{-1}+x_t^2\bar\Sigma_t^{-1}-2x_t\mu_t\bar\Sigma_t^{-1}+\mu_t^2\bar\Sigma_t^{-1}\right)}\\
+        &\eta_1e^{-\frac{1}{2}\Sigma_t^{-1}x_t^2+\Sigma^{-1}_tx_t\mu_t-\frac{1}{2}\Sigma_t^{-1}\mu_t^2}\\&=\eta_2 e^{-\frac{1}{2}\left(z_t^2Q_t^{-1}-2z_tCx_tQ_t^{-1}+\left(Cx\right)^2Q_t^{-1}+x_t^2\bar\Sigma_t^{-1}-2x_t\mu_t\bar\Sigma_t^{-1}+\mu_t^2\bar\Sigma_t^{-1}\right)}\\
+    \end{align*}
+    \item $\Sigma_t = \left(C^2Q_t^{-1}+\bar\Sigma_t^{-1}\right)^{-1}$\\
+    $\Sigma_t = \left(z_tCQ_t^{-1}+\bar\mu_t\bar\Sigma^{-1}\right)^{-1}\mu_t$
+    \item From the lecture and from above we get:\\
+    \begin{math}
+    \begin{array}{lllr}
+        \Sigma_t&=\left(I-K_tC\right)\bar\Sigma_t\\
+        &=\bar\Sigma_t-\frac{\bar\Sigma_t^2C^2}{C^2\bar\Sigma_t+Q_t}\\
+        \Sigma_t&=\left(C^2Q_t^{-1}+\bar\Sigma_t^{-1}\right)^{-1}\\
+        &=\frac{1}{\frac{C^2}{Q_t}+\frac{1}{\bar\Sigma_t}}\\
+        &=\frac{Q_t\bar\Sigma_t}{C^2\bar\Sigma_t+Q_t}\\
+        \Rightarrow & \bar\Sigma_t-\frac{\bar\Sigma_t^2C^2}{C^2\bar\Sigma_t+Q_t} &= \frac{Q_t\bar\Sigma_t}{C^2\bar\Sigma_t+Q_t}\\
+        \Leftrightarrow & I-\frac{\bar\Sigma_tC^2}{C^2\bar\Sigma_t+Q_t} &= \frac{Q_t}{C^2\bar\Sigma_t+Q_t}\\
+        \Leftrightarrow & Q_t &= C^2\bar\Sigma_t+Q_t-\bar\Sigma_tC^2\\
+        \Leftrightarrow & Q_t &= Q_t &\qed
+    \end{array}\end{math}
+    \item From the lecture and from above we get:\\
+    \begin{math}
+    \begin{array}{llr}
+    \bar\mu+K(z-C\bar\mu) &= \Sigma\left(\frac{zC}{Q}+\bar\mu\bar\Sigma^{-1}\right)\\
+    \bar\mu+K(z-C\bar\mu) &= (I-KC)\bar\Sigma\left(\frac{zC}{Q}+\bar\mu\bar\Sigma^{-1}\right)\\
+    \bar\mu+K(z-C\bar\mu) &= \bar\Sigma\left(\frac{zC}{Q}+\bar\mu\bar\Sigma^{-1}\right)-KC\bar\Sigma\left(\frac{zC}{Q}+\bar\mu\bar\Sigma^{-1}\right)\\
+    K(z-C\bar\mu) &= z\bar\Sigma\frac{C}{Q}-zKC\bar\Sigma\frac{C}{Q}-KC\bar\mu\\
+    Kz &= z\bar\Sigma\frac{C}{Q}-zKC\bar\Sigma\frac{C}{Q}\\
+    QK &= \bar\Sigma C - K\bar\Sigma C^2\\
+    K(Q+C^2\bar\Sigma) &= \bar\Sigma C\\
+    K&=K&\qed
+    \end{array}\end{math}
 \end{enumerate}
 
 \end{document}