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\author{ {\fontspec{Trebuchet MS}Marcin Chrz\k{a}szcz} (Universit\"{a}t Z\"{u}rich)}
\institute{UZH}
\title[Specific \pdf~generation]{Specific \pdf~generation}
\date{\fixme}
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\author{ {\fontspec{Trebuchet MS}Marcin Chrz\k{a}szcz} (Universit\"{a}t Z\"{u}rich)}
\institute{UZH}
\title[Solving linear equation systems with Markov Chain MC]{Solving linear equation systems with Markov Chain MC}
\date{\fixme}
\begin{document}
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\begin{center}
\begin{center}
\begin{columns}
\begin{column}{0.9\textwidth}
\flushright\fontspec{Trebuchet MS}\bfseries \Huge {Solving linear equation systems with Markov Chain MC}
\end{column}
\begin{column}{0.2\textwidth}
%\includegraphics[width=\textwidth]{SHiP-2}
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\quad
\vspace{3em}
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\flushright \vspace{-1.8em} {\fontspec{Trebuchet MS} \Large Marcin Chrząszcz\\\vspace{-0.1em}\small \href{mailto:mchrzasz@cern.ch}{mchrzasz@cern.ch}}
\end{column}
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\vspace{1em}
% \footnotesize\textcolor{gray}{With N. Serra, B. Storaci\\Thanks to the theory support from M. Shaposhnikov, D. Gorbunov}\normalsize\\
\vspace{0.5em}
\textcolor{normal text.fg!50!Comment}{Monte Carlo methods, \\ 21 April, 2016}
\end{center}
\end{frame}
}
\begin{frame}\frametitle{Announcement}
\begin{Large}
There will be no lectures and class on 19$^{th}$ of May
\end{Large}
\end{frame}
\begin{frame}\frametitle{Trivial example}
\begin{minipage}{\textwidth}
\ARROW Lets start with a TRIVIAL example: we want to calculate $S=A+B$. We can rewrite it in:
\begin{align*}
A=p \frac{A}{p}+(1-p) \frac{B}{1-p}
\end{align*}
and one can interpret the sum as expected value of:
\begin{align*}
W=\begin{cases}
\frac{A}{p}~~{ \rm with~propability~} p \\
\frac{A}{1p}~~{ \rm with~propability~} 1-p
\end{cases}
\end{align*}
\ARROW The algorithm:
\begin{itemize}
\item We generate a random variable $W$ and calculate:
\begin{align*}
\hat{S}=\frac{1}{N}\sum_{i=1}^N W_i
\end{align*}
\ARROW The $\hat{S}$ is an unbias estimator of $S$.
\end{itemize}
\end{minipage}
\end{frame}
\begin{frame}\frametitle{Trivial example2 }
\begin{minipage}{\textwidth}
\begin{itemize}
\item Lets say we have a linear equation system:
\end{itemize}
\begin{equation}
\begin{array}{lcl} X & = & pY + (1-p) A \\ Y & = & qX + (1-q)B \end{array} \nonumber
\end{equation}
\begin{itemize}
\item We know $A,B,p,q$; $X$ and $Y$ are meant to be determined.
\item Algorithm:
\begin{enumerate}
\item We choose first element of the first equation with probability $p$ and second with probability $1-p$.
\item We we choose the second one, the outcome of this MCMC is $W=A$.
\item If we choose the first we go to second equation and choose the first element with probability $q$ and the second with $1-q$.
\item We we choose the second one, the outcome of this MCMC is $W=B$.
\item If we choose the first we go to the first equation back again.
\item We repeat the procedure.
\end{enumerate}
\item We can estimate the solution of this system:
\end{itemize}
\begin{equation}
\hat{X} = \dfrac{1}{N}\sum_{i=1} W_i{~}{~}{~}{~}{~} \hat{\sigma_X}=\dfrac{1}{\sqrt{N-1}}\sqrt{\dfrac{1}{N} \sum_{i=1}^N W_i^2-\hat{X}^2} \nonumber
\end{equation}
\end{minipage}
\end{frame}
\begin{frame}\frametitle{Random walk}
\begin{minipage}{\textwidth}
\begin{footnotesize}
\begin{center}
\includegraphics[width=0.8\textwidth]{images/walk.png}
\end{center}
\ARROW We are in the point $x$ and we walk accordingly to the following rules:
\begin{itemize}
\item From point $x$ we walk with probability $p$ to point $y$ or with $1-p$ to $a$.
\item From point $y$ we walk with probability $q$ to point $x$ and with $1-Q$ to $b$.
\item The walks ends when you end up in $a$ or $b$.
\item You get a ''reward'' $A$ if you end up in point $a$ and $B$ if you end up in $b$.
\item $X$ is expected ''reward'' when you start the walk from $x$, $Y$ when you start from $y$.
\end{itemize}
\ARROW The algorithm above is so-called random walk on the set $\lbrace a,x,y,b \rbrace$\\
\ARROW The described walked can solve the linear equation system that we discussed above.
\end{footnotesize}
\end{minipage}
\end{frame}
\begin{frame}\frametitle{Markov Chain MC}
\begin{footnotesize}
\begin{itemize}
\item Consider a finite (or Countable set) possible states: $S_1$, $S_2$, ...
\item The $X_t$ is the state of the system in the time $t$
\item We are looking at discrete time steps: $1,2,3,...$
\item At time $t$ the state is denoted $X_t$.
\item The conditional probability is defined as:
\end{itemize}
\begin{equation}
P(X_t=S_j \vert X_{t-1}=S_{j-1},..., X_{1}=S_{1}) \nonumber
\end{equation}
\begin{itemize}
\item The Markov chain is then if the probability depends only on previous step.
\end{itemize}
\begin{equation}
P(X_t=S_j \vert X_{t-1}=S_{j-1},..., X_{1}=S_{1}) = P(X_t=S_j \vert X_{t-1}=S_{j-1} )\nonumber
\end{equation}
\begin{itemize}
\item For this reason this reason MCMC is also knows as drunk sailor walk.
\item Very powerful method. Used to solve linear eq. systems, invert matrix, solve differential equations, etc.
\item Also used in physics problems: Brown motions, diffusion, etc.
\end{itemize}
\end{footnotesize}
\end{frame}
\begin{frame}\frametitle{Linear equations system}
\begin{footnotesize}
\ARROW Lets start from a linear equation system:
\begin{align*}
\textbf{A} \overrightarrow{x}=\overrightarrow{b},~~~~~\det \textbf{A} \neq 0,
\end{align*}
where $\textbf{A}=(a_{ij},i,j=1,2,...,n$ -matrix, $\overrightarrow{b}=(b_1,b_2,...,b_n)$-vector, $\overrightarrow{x}=(x_1,x_2,...,x_n)$ - vector of unknowns.\\
\ARROW The solution we mark as $\overrightarrow{x}^0 = (x_1^0, x_2^0,..., x_n^0)$\\
\ARROW The above system can be transformed into the iterative representation:
\begin{align*}
\overrightarrow{x}=\overrightarrow{a} + \textbf{H} \overrightarrow{x}
\end{align*}
where $\textbf{H}$ is a matrix, $\overrightarrow{a}$ is a vector.\\
\ARROW We assume that the matrix norm:
\begin{align*}
\Vert H \Vert= \max_{1 \leq i \leq n} \sum_{j=1}^n \vert h_{h_{ij}} \vert <1
\end{align*}
\pause
\ARROW We can always change transform every system to the iteration form: $\textbf{A}=\textbf{V}-\textbf{W}$.
\begin{align*}
(\textbf{V} - \textbf{W} )\overrightarrow{x} = \overrightarrow{b}~~~~ \mapsto ~~~~ \overrightarrow{x} = \textbf{V}^{-1} \overrightarrow{b} + \textbf{V}^{-1} \textbf{W} \overrightarrow{x}
\end{align*}
\end{footnotesize}
\end{frame}
\begin{frame}\frametitle{Linear equations system}
\begin{footnotesize}
\ARROW Now we further modify the equation system:
\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.
\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}+ ...
\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}
\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}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}\frametitle{Neumann-Ulam method}
\begin{footnotesize}
\begin{itemize}
\item To do so we add to our matrix an additional column of the matrix:
\end{itemize}
\begin{equation}
h_{i,0} = 1-\sum_{j=1}^n h_{ij} > 0 \nonumber
\end{equation}
\begin{itemize}
\item The system has states: $\lbrace 0,1,2...,n\rbrace$
\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 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}
\item The path $\gamma = (i_0, i_1, i_2, ..., i_k, 0)$ is called trajectory.
\item For each trajectory we assign a number:
\begin{align*}
X(\gamma)=X(i_0, i_1, i_2, ..., i_k, 0)=\frac{a_{i_k}}{h_{i_k 0}}
\end{align*}
\end{itemize}
\end{footnotesize}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\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 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)
\end{align*}
where $\gamma_k$ is a trajectory of length $k$, which starts in $i_0=s$ and $P(\gamma)$ is the probability of occurrence of this trajectory.
\ARROW Yes you guest it lets do Taylor expansion:
\begin{align*}
E \lbrace X(\gamma) \vert i_0=s \rbrace= \sum_{\gamma_0}X(\gamma)P(\gamma) + \sum_{\gamma_1}X(\gamma)P(\gamma)+...+ \sum_{\gamma_k}X(\gamma)P(\gamma)
\end{align*}
\ARROW Now let's examine the elements of the above series.
\end{footnotesize}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}\frametitle{Neumann-Ulam method}
\begin{footnotesize}
$\lbrace \gamma_0 \rbrace$: One trajectory: $\gamma_0=(i_0=s \vert 0)$, $P(\gamma_0)=h_{s,0}$ and $X(\gamma_0)=a_s/h_{s,0}$. So:
\begin{align*}
\sum_{\gamma_1}X(\gamma)P(\gamma) = \frac{a_s}{h_{s,0}} h_{s,0}=a_s
\end{align*}
$\lbrace \gamma_1 \rbrace$: 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:
\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:
\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_1,0}= \sum_{i_1=1}^n \sum_{i_2=1}^n h_{s,i_1} h_{i_1,i_2} a_{i_2}
\end{align*}
\ARROW etc...
\end{footnotesize}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}\frametitle{Neumann-Ulam method}
\begin{footnotesize}
\ARROW After summing up:
\begin{align*}
E \lbrace X(\gamma) \vert i_0=s \rbrace= a_s+ \sum_{i_1 =1}^n h_{s,i_1} a_{i_1}+ \sum_{i_1 =1}^n \sum_{i_2 =1}^n h_{s,i_1} h_{i_1,i_2} a_{i_2}+....\\ + \sum_{i_1 =1}^n \sum_{i_2 =1}^n ... \sum_{i_k =1}^n h_{s,i_1} h_{i_1,i_2}... h_{i_{k-1},i_k}+...
\end{align*}
\ARROW If you compare this expression with the Neumann series we will they are the same so:
\begin{align*}
x_i^0=E \lbrace X(\gamma) \vert i_0=i \rbrace
\end{align*}
\begin{exampleblock}{To sum up:}
We have proven that solving a linear system can be represented by an expectation value of the random variable $X(\gamma)$. The error is computed using standard deviation equation.
\end{exampleblock}
\end{footnotesize}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}\frametitle{Neumann-Ulam method}
\begin{itemize}
\item For example lets try to solve this equation system:
\end{itemize}
\begin{equation}
\overrightarrow{x} =
\left(\begin{array}{c}
1.5 \\
-1.0\\
0.7 \end{array} \right)
+
\left(\begin{array}{ccc}
0.2 & 0.3 & 0.1 \\
0.4 & 0.3 & 0.2 \\
0.3 & 0.1 & 0.1 \end{array} \right) \overrightarrow{x}
\nonumber
\end{equation}
\begin{itemize}
\item The solution is $\overrightarrow{x}_0 = (2.154303, 0.237389, 1.522255)$.
\end{itemize}
\begin{columns}
\column{0.1in}
\column{2.5in}
\begin{itemize}
\item The propability matrix $h_{ij}$ has the shape:
\end{itemize}
\begin{tabular}{|c|cccc|}
\hline
$i/j$ & 1 & 2 & 3 & 4 \\ \hline
1 & 0.2 & 0.3 & 0.1 & 0.4 \\
2 & 0.4 & 0.3 & 0.2 & 0.1 \\
3 & 0.3 & 0.1 & 0.1 & 0.5 \\ \hline
\end{tabular}
\column{2.5in}
\begin{itemize}
\item An example solution:
\end{itemize}
\includegraphics[width=0.95\textwidth]{images/mark.png}
\end{columns}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\iffalse
\begin{frame}\frametitle{Neumann-Ulam dual method}
\begin{minipage}{\textwidth}
\ARROW The main problem with the Neumann-Ulam method is the fact that one has to calculate each of the $x_0^i$ separately.\\
\ARROW The generalization works as follows:
\begin{enumerate}
\item We set randomly a \pdf~of states: $q_1,q_2,q_3,...,q_n$, such that $q_i >0$ and $\sum_{i=1}^n =1$.
\item We choose the starting point accordingly to $q_i$ probability.
\item If in the $t$ moment the point is in position $i_t$ then the with the probability $p(i_{t+1} \vert i_t)=h_{i_{t+1}, i_t}$ the points moves to the $i_{t+1}$ state.
\item For the state $0$ we assign the probability: $h_{0,i_{t}}=1-\sum_{i=1}^n h_{i,i_t}$
\item WARNING HERE THE MATRIX IS TRANSPOSED compared to method.
\item The walk ends while you reach $0$ state.
\item For each walk/trajectory $(\gamma=(i_0,i_1,...,i_k,0))$ we assign a vector:
\begin{align*}
\overrightarrow{Y}(\gamma)=\frac{a_{i_0}}{q_{i_0}p(0\vert i_k }) \widehat{e}_{i_k}
\end{align*}
\item The final result is: $\overrightarrow{x}^0=\frac{1}{N}\sum \overrightarrow{Y}$
\end{enumerate}\textbf{
\end{minipage}
\end{frame}
\fi
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}\frametitle{Neumann-Ulam dual method}
\begin{itemize}
\item The problem with Neumann-Ulam method is that you need to repeat it for each of the coordinates of the $\overrightarrow{x}_0$ vector.
\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 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 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:
\end{itemize}
\begin{equation}
\overrightarrow{Y}(\gamma) = \dfrac{a_{i_0}}{ q_{i_{0}} p(0 \vert i_k) } \widehat{e}_{i_{k}} \in \mathcal{R}^n \nonumber
\end{equation}
\item The solution will be : $\overrightarrow{x}^0 = \dfrac{1}{N} \sum \overrightarrow{Y}(\gamma)$
\end{itemize}
\end{frame}
\begin{frame}\frametitle{Neumann-Ulam dual method, prove}
\begin{footnotesize}
\ARROW If $Y_i(\gamma)$ is the i-th component of the $\overrightarrow{Y}(\gamma)$ vector. One needs to show:
\begin{align*}
E\lbrace Y_i(\gamma)\rbrace=x_j^0
\end{align*}
\ARROW From definition:
\begin{align*}
Y_j(i_1,...,i_k,0)=\begin{cases}
\frac{a_{i_k}}{q_{i_0} p(0 \vert i_k)}~~~& i_k=j\\
0 ~~~& i_k \neq j
\end{cases}
\end{align*}
\ARROW The expected value:
\begin{align*}
E \lbrace Y_j (\gamma)=\sum_{ {\rm trajectories}} \frac{a_j}{q_{i_0}p(0 \vert i_k) } P(i_1,i_2,...,i_k,0),
\end{align*}
where $P(i_1,i_2,...,i_k,0)$ is the probability of this trajectory occurring.\\
\ARROW But by our definition the probability:
\begin{align*}
P(i_0,i_1,...,i_{k-1},j,0)=q_{i_k}h_{i_1,i_0}...h_{k,i_{k-1}}p(0 \vert j)
\end{align*}
\ARROW In the end we get:
\begin{align*}
E(Y_j(\gamma))=\sum_{k=0}^{\infty} \sum_{i_{k-1}=1}^n ... \sum_{i_{1}=1}^n \sum_{i_{0}=1}^n h_{j,i_{k-1}} h_{j,i_{k-1}} ... h_{i_2,i_1} h_{i_1,i_0} a_{i_0}
\end{align*}
\end{footnotesize}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}\frametitle{Neumann-Ulam dual method}
\begin{itemize}
\item Let's try to solve the equation system:
\end{itemize}
\begin{equation}
\overrightarrow{x} =
\left(\begin{array}{c}
1.5 \\
-1.0\\
0.7 \end{array} \right)
+
\left(\begin{array}{ccc}
0.2 & 0.3 & 0.1 \\
0.4 & 0.3 & 0.2 \\
0.1 & 0.1 & 0.1 \end{array} \right) \overrightarrow{x}
\nonumber
\end{equation}
\begin{itemize}
\item The solution is: $\overrightarrow{x}_0 = (2.0, 0.0, 1.0)$.
\item Let's put the initial probability as constant:
\end{itemize}
\begin{equation}
q_1=q_2=q_3=\dfrac{1}{3} \nonumber
\end{equation}
\begin{columns}
\column{0.1in}
\column{2.5in}
\begin{itemize}
\item The propability matrix $h_{ij}$ has the shape:
\end{itemize}
\begin{tabular}{|c|cccc|}
\hline
$i/j$ & 1 & 2 & 3 & 4 \\ \hline
1 & 0.2 & 0.4 & 0.1 & 0.3 \\
2 & 0.3 & 0.3 & 0.1 & 0.3 \\
3 & 0.1 & 0.2 & 0.1 & 0.6 \\ \hline
\end{tabular}
\column{2.5in}
\begin{itemize}
\item An example solution:
\end{itemize}
\includegraphics[width=0.95\textwidth]{images/mark2.png}
\end{columns}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}\frametitle{Generalization}
\ARROW Up to now we assumed that each of the matrix elements $h_{i,j} \geq 0$. Now if this is not true:\\
\ARROW We take a probability matrix $P=p_{ij}$ such that:
\begin{align*}
p_{ij} \geq 0~~~p_{ij}=0 \Leftrightarrow h_{ij}=0,~~~~p_{i,0}=1-\sum_j p(i,j) >0.
\end{align*}
\ARROW To solve the system we construct a Markov Chain with the $P$ matrix as probabilities of transitions.\\
\ARROW The probability of a trajectory is equal ($i_0=i$):
\begin{align*}
P(\gamma_k)=p_{i,i_1}p_{i_1,i_2}...p_{i_k,0}
\end{align*}
\ARROW The trajectory we assign the number:
\begin{align*}
X(\gamma_k)=\nu_{i,i_1} \nu_{i_1,i_2} ..., \nu_{i_{k-1},i_k} \frac{a_{i_k}}{p_{i_k,0}}
\end{align*}
where
\begin{align*}
\nu_{i,j}=\begin{cases}
h_{ij}/p_{ij},~~~& p_{ij} \neq 0\\
1 ~~~& p_{ij} = 0
\end{cases}
\end{align*}
\end{frame}
\begin{frame}\frametitle{Generalization, prove}
\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
\end{align*}
\ARROW The probability is given by the formula:
\begin{align*}
P \lbrace X(\gamma_k) \rbrace = P \lbrace X(\gamma_k)= \nu_{i,i_1} \nu_{i_1,i_2} ..., \nu_{i_{k-1},i_k} \frac{a_{i_k}}{p_{i_k,0}} \rbrace \\ = p_{i,i_1}...,p_{i_{k-1},i_k}p_{i_k,0}
\end{align*}
\ARROW However:
\begin{align*}
X(\gamma_k) P \lbrace X(\gamma_k) \rbrace=h_{i,i_1} h_{i_1,i_2} ... h_{i_{k-1},i_k} a_{i_k}
\end{align*}
so:
\begin{align*}
\sum_{\gamma_k} X(\gamma_k)P \lbrace X(\gamma_k) \rbrace = \sum_{i_1=1} ... \sum_{i_k=1} h_{i,i_1} h_{i_1,i_2} ... h_{i_{k-1}, i_k} a_{i_k}
\end{align*}
\end{frame}
\begin{frame}\frametitle{Generalization, the algorithm}
\ARROW We set the $P$ matrix in a arbitrary way.\\
\ARROW If in the $t$ moment the point is in the $i_t$ state the with the probability $p_{i_t, i_{t+1}}$ he can go to $i_{t+1}$ state. \\
\ARROW We stop the walk once we reach $0$.\\
\ARROW For the given trajectory we assign the value: $X(\gamma_k)$\\
\ARROW We repeat the procedure $N$ times and take the mean and RMS.\\
\ARROW We repeat this also for every of the $\overrightarrow{x}^0$ vector components.
\end{frame}
\begin{frame}\frametitle{Wasow method}
\begin{footnotesize}
\ARROW The main problem with the Neumann-Ulam methods is the fact that each time we estimate only one of the part of the taylor expansion.\\
\ARROW W.Wasow (1956) was smarter:
\begin{itemize}
\item For the trajectory: $\gamma(i_0,i_1,...,i_k,0)$ we look at begging trajectories:
\begin{align*}
(i_0),~(i_0,i_1),~(i_0,i_1,...,i_k)
\end{align*}
and for each we associate a number:
\begin{align*}
(i_0,i_1,i_2,...,i_m),~0 \leq m \leq k
\end{align*}
we assign a number:
\begin{align*}
\nu_{i_0,i_1} \nu_{i_1,i_2}...\nu_{i_{m-1},i_m}a_{i_m}
\end{align*}
\end{itemize}
\ARROW For the trajectory we define:
\begin{align*}
X^{\ast}(\gamma)=\sum_{m=0}^k nu_{i_1,i_2}...\nu_{i_{m-1},i_m}a_{i_m}
\end{align*}
\ARROW One can prove that:
\begin{align*}
E \lbrace X^{\ast}(\gamma) \vert i_0=i \rbrace =x_i^0
\end{align*}
\end{footnotesize}
\end{frame}
\begin{frame}\frametitle{Homework}
\ARROW E8.(1,2,3) Please solve the linear equation system from slide 16 using Wasow, Neumann-Ulam, dual Neumann-Ulam methods.
\end{frame}
\backupbegin
\begin{frame}\frametitle{Backup}
\end{frame}
\backupend
\end{document}
mchrzasz