diff --git a/slides/advriskmin/slides-advriskmin-classification-bernoulli.tex b/slides/advriskmin/slides-advriskmin-classification-bernoulli.tex
index e2a1ae20..c608d72a 100644
--- a/slides/advriskmin/slides-advriskmin-classification-bernoulli.tex
+++ b/slides/advriskmin/slides-advriskmin-classification-bernoulli.tex
@@ -231,9 +231,6 @@
 
 
 
-
-
-
 % \vspace*{-0.2cm}
 
 % \begin{eqnarray*}
diff --git a/slides/advriskmin/slides-advriskmin-classification-brier.tex b/slides/advriskmin/slides-advriskmin-classification-brier.tex
index 7bc34838..0a5b54fe 100644
--- a/slides/advriskmin/slides-advriskmin-classification-brier.tex
+++ b/slides/advriskmin/slides-advriskmin-classification-brier.tex
@@ -144,71 +144,6 @@
 
 \end{vbframe}
 
-\begin{vbframe}{Brier score minimization = Gini splitting}
-
-When fitting a tree we minimize the risk within each node $\Np$ by risk minimization and predict the optimal constant. Another approach that is common in literature is to minimize the average node impurity $\text{Imp}(\Np)$. 
-
-\vspace*{0.2cm}
-
-\textbf{Claim:} Gini splitting $\text{Imp}(\Np) = \sum_{k=1}^g \pikN \left(1-\pikN \right)$ is equivalent to the Brier score minimization. 
-
-\begin{footnotesize}
-Note that $\pikN := \frac{1}{n_{\Np}} \sum\limits_{(\xv,y) \in \Np} [y = k]$ 
-\end{footnotesize}
-
-\vspace*{0.2cm}
-
-\begin{footnotesize}
-
-\textbf{Proof: } We show that the risk related to a subset of observations $\Np \subseteq \D$ fulfills 
-
-
-$$
-  \risk(\Np) = n_\Np \text{Imp}(\Np),
-$$
-  
-  where $\text{Imp}$ is the Gini impurity and $\risk(\Np)$ is calculated w.r.t. the (multiclass) Brier score
-
-
-$$
-  L(y, \pix) = \sum_{k = 1}^g \left([y = k] - \pi_k(\xv)\right)^2.
-$$
-
-\framebreak
-
-\vspace*{-0.5cm}
-\begin{eqnarray*}
-\risk(\Np) &=& \sum_{\xy \in \Np}  \sum_{k = 1}^g \left([y = k] - \pi_k(\xv)\right)^2 
-= \sum_{k = 1}^g \sum_{\xy \in \Np} \left([y = k] - \frac{n_{\Np,k}}{n_{\Np }}\right)^2,
-\end{eqnarray*}
-
-by plugging in the optimal constant prediction w.r.t. the Brier score ($n_{\Np,k}$ is defined as the number of class $k$ observations in node $\Np$): 
-$$\hat \pi_k(\xv)= \pikN = \frac{1}{n_{\Np}} \sum\limits_{(\xv,y) \in \Np} [y = k] = \frac{n_{\Np,k}}{n_{\Np }}. $$ 
-
- We split the inner sum and further simplify the expression
-
-\begin{eqnarray*}
-&=& \sum_{k = 1}^{g} \left(\sum_{\xy \in \Np: ~ y = k} \left(1 - \frac{n_{\Np,k}}{n_{\Np }}\right)^2 + \sum_{\xy \in \Np: ~ y \ne k} \left(0 - \frac{n_{\Np,k}}{n_{\Np }}\right)^2\right) \\
-&=& \sum_{k = 1}^g n_{\Np,k}\left(1 - \frac{n_{\Np,k}}{n_{\Np }}\right)^2 + (n_{\Np } - n_{\Np,k})\left(\frac{n_{\Np,k}}{n_{\Np }}\right)^2, 
-\end{eqnarray*}
-
-since for $n_{\Np,k}$ observations the condition $y = k$ is met, and for the remaining $(n_\Np - n_{\Np,k})$ observations it is not. 
-
-
-We further simplify the expression to
-
-% \begin{footnotesize}
-\begin{eqnarray*}
-\risk(\Np) &=&  \sum_{k = 1}^g n_{\Np,k}\left(\frac{n_{\Np } - n_{\Np,k}}{n_{\Np }}\right)^2 + (n_{\Np } - n_{\Np,k})\left(\frac{n_{\Np,k}}{n_{\Np }}\right)^2 \\
-&=& \sum_{k = 1}^g \frac{n_{\Np,k}}{n_{\Np }} \frac{n_{\Np } - n_{\Np,k}}{n_{\Np }} \left(n_{\Np } - n_{\Np,k } + n_{\Np,k}\right) \\
-&=& n_{\Np } \sum_{k = 1}^g \pikN \cdot \left(1 - \pikN \right) = n_\Np \text{Imp}(\Np).
-\end{eqnarray*}
-% \end{footnotesize}
-
-\end{footnotesize}
-
-\end{vbframe}
-
 
 \endlecture