diff --git a/Lectures_my/MC_2016/Lecture7/mchrzasz.log b/Lectures_my/MC_2016/Lecture7/mchrzasz.log
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+++ b/Lectures_my/MC_2016/Lecture7/mchrzasz.log
@@ -1,4 +1,4 @@
-This is XeTeX, Version 3.1415926-2.5-0.9999.3 (TeX Live 2013/Debian) (format=xelatex 2015.4.1)  22 APR 2016 17:42
+This is XeTeX, Version 3.1415926-2.5-0.9999.3 (TeX Live 2013/Debian) (format=xelatex 2015.4.1)  8 MAY 2016 13:16
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diff --git a/Lectures_my/MC_2016/Lecture7/mchrzasz.pdf b/Lectures_my/MC_2016/Lecture7/mchrzasz.pdf
index 5ea4e85..89166fc 100644
--- a/Lectures_my/MC_2016/Lecture7/mchrzasz.pdf
+++ b/Lectures_my/MC_2016/Lecture7/mchrzasz.pdf
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diff --git a/Lectures_my/MC_2016/Lecture7/mchrzasz.tex b/Lectures_my/MC_2016/Lecture7/mchrzasz.tex
index a7a66ae..31c8126 100644
--- a/Lectures_my/MC_2016/Lecture7/mchrzasz.tex
+++ b/Lectures_my/MC_2016/Lecture7/mchrzasz.tex
@@ -428,7 +428,7 @@
 \end{exampleblock}
 \ARROW Prove $(y \leq -1)$:
 \begin{align*}
-\mathcal{P}\lbrace Y\leq y\rbrace =  \frac{1}{2} \mathcal{P}\lbrace X \leq y \rbrace + \frac{1}{2} \mathcal{P}\lbrace \frac{1}{X} \leq \rbrace = 0+\frac{1}{2}\lbrace \frac{1}{y} \leq X <0  \rbrace \\ = \frac{1}{2} \frac{2}{\pi} \int_{1/y}^0 \frac{dr }{1+t^2}=\frac{1}{\pi} \arctan y +\frac{1}{2}~~~{\rm c.d.f~of~}C(0,1) 
+\mathcal{P}\lbrace Y\leq y\rbrace =  \frac{1}{2} \mathcal{P}\lbrace X \leq y \rbrace + \frac{1}{2} \mathcal{P}\lbrace \frac{1}{X} \leq \rbrace = 0+\frac{1}{2}\lbrace \frac{1}{y} \leq X <0  \rbrace \\ = \frac{1}{2} \frac{2}{\pi} \int_{1/y}^0 \frac{dt }{1+t^2}=\frac{1}{\pi} \arctan x +\frac{1}{2}~~~{\rm c.d.f~of~}C(0,1) 
 \end{align*}
 \ARROW The cut-off Breit-Wigner distribution we generate with elimination method using $\mathcal{U}(-1,1)$\\
 \ARROW E7.3 Generate the Brei-Wigner distribution with all described methods.
diff --git a/Lectures_my/MC_2016/Lecture8/mchrzasz.log b/Lectures_my/MC_2016/Lecture8/mchrzasz.log
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+++ b/Lectures_my/MC_2016/Lecture8/mchrzasz.log
@@ -1,4 +1,4 @@
-This is XeTeX, Version 3.1415926-2.5-0.9999.3 (TeX Live 2013/Debian) (format=xelatex 2015.4.1)  21 APR 2016 18:15
+This is XeTeX, Version 3.1415926-2.5-0.9999.3 (TeX Live 2013/Debian) (format=xelatex 2015.4.1)  8 MAY 2016 13:33
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diff --git a/Lectures_my/MC_2016/Lecture8/mchrzasz.tex b/Lectures_my/MC_2016/Lecture8/mchrzasz.tex
index 73ce3ec..3699f76 100644
--- a/Lectures_my/MC_2016/Lecture8/mchrzasz.tex
+++ b/Lectures_my/MC_2016/Lecture8/mchrzasz.tex
@@ -417,15 +417,15 @@
 \begin{align*}
 \overrightarrow{x}=\overrightarrow{a}   + \textbf{H} \overrightarrow{x}   \Rightarrow (\textbf{I} - \textbf{H})\overrightarrow{x}=\overrightarrow{a}
 \end{align*}
-where $\textbf{I}=\delta_{ij}$ - unit matrix, $\delta_{ij}$ is the Kronecker delta.
+where $\textbf{I}=\delta_{ij}$ - unit matrix, $\delta_{ij}$ is the Kronecker delta.\\
 \ARROW What one can do is to represent the solution in terns of Neumann series:
 \begin{align*}
-\overrightarrow{x}^0=(\textbf{I}-\textbf{H})^{-1}\overrightarrow{a}= \overrightarrow{a} \textbf{H}  \overrightarrow{a} + \textbf{H}^2  \overrightarrow{a}+ \textbf{H}^3 \overrightarrow{a}+ ...
+\overrightarrow{x}^0=(\textbf{I}-\textbf{H})^{-1}\overrightarrow{a}= \overrightarrow{a} + \textbf{H}  \overrightarrow{a} + \textbf{H}^2  \overrightarrow{a}+ \textbf{H}^3 \overrightarrow{a}+ ...
 \end{align*}
 \ARROW So for the $i^{th}$ component we have:
 \begin{align*}
-x_i^0=a_i+\sum_{j=1}^nh_{ij} a_j + \sum_{j_1 =1}^n \sum_{j_2 =1}^n h_{ij_1} h_{ij_2} a_{j_2} \\
-+...+\sum_{j_1 =1}^n ...\sum_{j_n =1}^n  h_{ij_1}... h_{ij_n} a_{j_n}
+x_i^0=a_i+\sum_{j=1}^nh_{ij} a_j + \sum_{j_1 =1}^n \sum_{j_2 =1}^n h_{ij_1} h_{j_1 j_2} a_{j_2} \\
++...+\sum_{j_1 =1}^n ...\sum_{j_n =1}^n  h_{ij_1}... h_{j_{n-1} j_n} a_{j_n}
 \end{align*}
 \ARROW We will construct a probabilistic interpretation using MCMC and then we show that the expected value is equal to the above formula.
 \end{footnotesize}
@@ -451,7 +451,7 @@
 \item State at $t$ time is denoted as $i_t(i_t=0,1,2,...,n;t=0,1,....)$
 \item We make a random walk accordingly to to the following rules:
 \begin{itemize}
-\item At the begging of the walk ($t=0$) we are at $i_0$. 
+\item At the beginning of the walk ($t=0$) we are at $i_0$. 
 \item In the $t$ moment we are in the $i_t$ position then in $t+1$ time stamp we move to state $i_{t+1}$ with the probability $h_{i_t i_{t+1}}$. 
 \item We stop walking if we are in state $0$. 
 \end{itemize}
@@ -472,7 +472,7 @@
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \begin{frame}\frametitle{Neumann-Ulam method}
 \begin{footnotesize}
-\ARROW The $X(\gamma)$ variable is a random variable from: $\lbrace a_1/h_{1,0},a_2/h_{2,0},...,a_n/h_{n,0} \rbrace$. The probability that $X(\gamma) a_j/h_{j,0}$ is equal to the probability that the last non zero state of the $\gamma$ trajectory is $j$.\\
+\ARROW The $X(\gamma)$ variable is a random variable from: $\lbrace a_1/h_{1,0},a_2/h_{2,0},...,a_n/h_{n,0} \rbrace$. The probability that $X(\gamma)= a_j/h_{j,0}$ is equal to the probability that the last non zero state of the $\gamma$ trajectory is $j$.\\
 \ARROW The expected value of the $X(\gamma)$ trajectory if the trajectory begins from $i_0=s$ is:
 \begin{align*}
 E \lbrace X(\gamma) \vert i_0=s \rbrace=\sum_{k=0}^{\infty} \sum_{ \lbrace \gamma_k \rbrace} X(\gamma) P(\gamma)
@@ -499,13 +499,13 @@
 
 
 
-$\lbrace \gamma_1 \rbrace$: Trajectories: $\gamma_1=(i_0,i_1=s \vert 0),~i_1 \neq 0$,  $P(\gamma_1)= P(s,i_1,0)=h_{s,i_1}h_{i_1,0} $ and $X(\gamma_1)=a_{i_1}/h_{i_1,0}$. So:
+$\lbrace \gamma_1 \rbrace$: Trajectories: $\gamma_1=(i_0=s,i_1 \vert 0),~i_1 \neq 0$,  $P(\gamma_1)= P(s,i_1,0)=h_{s,i_1}h_{i_1,0} $ and $X(\gamma_1)=a_{i_1}/h_{i_1,0}$. So:
 \begin{align*}
 \sum_{\gamma_1}X(\gamma)P(\gamma) = \sum_{i_1=1}^n \frac{a_{i_1}}{h_{i_1,0}} h_{s,i_1} h_{i_1,0}= \sum_{i=1}^n h_{s,i_1}a_{i_1}
 \end{align*}
 
 
-$\lbrace \gamma_2 \rbrace$: Trajectories: $\gamma_2=(i_0,i_1,i_2=s \vert 0),~i_1,i_2 \neq 0$,  $P(\gamma_2)= P(s,i_1, i_2,0)=h_{s,i_1}h_{i_1,i_2}h_{i_1,0} $ and $X(\gamma_2)=a_{i_2}/h_{i_2,0}$. So:
+$\lbrace \gamma_2 \rbrace$: Trajectories: $\gamma_2=(i_0=s,i_1,i_2 \vert 0),~i_1,i_2 \neq 0$,  $P(\gamma_2)= P(s,i_1, i_2,0)=h_{s,i_1}h_{i_1,i_2}h_{i_1,0} $ and $X(\gamma_2)=a_{i_2}/h_{i_2,0}$. So:
 \begin{align*}
 \sum_{\gamma_2}X(\gamma)P(\gamma) = \sum_{i_1=1}^n  \sum_{i_2=1}^n  \frac{a_{i_2}}{h_{i_2,0}} h_{s,i_1} h_{i_1,i_2} h_{i_2,0}= \sum_{i_1=1}^n  \sum_{i_2=1}^n  h_{s,i_1} h_{i_1,i_2}   a_{i_2}
 \end{align*}
@@ -624,9 +624,9 @@
 \item The dual method calculates the whole $\overrightarrow{x}_0$ vector.
 \item The algorithm:
 \begin{itemize}
-\item On the indexes: $\lbrace0,1,...,n\rbrace$ we set a probability distribution:\\ $q_1, q_2,..., q_n$, $q_i>0$ and $\sum_i=1^n q_i=1$. 
+\item On the indexes: $\lbrace0,1,...,n\rbrace$ we set a probability distribution:\\ $q_1, q_2,..., q_n$, $q_i>0$ and $\sum_{i=1^n} q_i=1$. 
 \item The starting point we select from $q_i$ distribution.
-\item If in $t$ time we are in $i_t$ state then with probability $p(i_{t+1} \vert i_t) = h_{i_{t+1},i_{t}}$ in $t+1$ we will be in state $i_1$. For  $i_{t+1}=0$ we define the probability: $h_{0,i_{t}}=1-\sum_{j=1}^n h_{j,i_{t}}$. Here we also assume that $h_{j,i_{t}} > 0$.
+\item If in $t$ time we are in $i_t$ state then with probability $p(i_{t+1} \vert i_t) = h_{i_{t+1},i_{t}}$ in $t+1$ we will be in state $i_{t+1}$. For  $i_{t+1}=0$ we define the probability: $h_{0,i_{t}}=1-\sum_{j=1}^n h_{j,i_{t}}$. Here we also assume that $h_{j,i_{t}} > 0$.
 \item NOTE: there the matrix is transposed compared to previous method: $H^{T}$.
 \item Again we end our walk when we are at state $0$. 
 \item For the trajectory: $\gamma = (i_0, i_1,...,i_k, 0)$, we assign the vector:
@@ -751,7 +751,7 @@
 \begin{frame}\frametitle{Generalization, proof}
 \ARROW For a $X(\gamma)$ trajectory the expected value is:
 \begin{align*}
-E \lbrace X(\gamma_k) = \sum_{k=0}^{\infty}\sum_{\gamma_k} X(\gamma_k)P \lbrace X(\gamma_k) \rbrace 
+E \lbrace X(\gamma_k) \rbrace = \sum_{k=0}^{\infty}\sum_{\gamma_k} X(\gamma_k)P \lbrace X(\gamma_k) \rbrace 
 \end{align*}
 \ARROW The probability is given by the formula:
 \begin{align*}
@@ -792,7 +792,7 @@
 \ARROW W.Wasow (1956) was smarter:
 \begin{itemize}
 
-\item For the trajectory: $\gamma(i_0,i_1,...,i_k,0)$ we look trajectories begging with:
+\item For the trajectory: $\gamma(i_0,i_1,...,i_k,0)$ we look trajectories:
 \begin{align*}
 (i_0),~(i_0,i_1),~(i_0,i_1,...,i_k)
 \end{align*}