diff --git a/Lectures_my/MC_2016/Lecture9/mchrzasz.log b/Lectures_my/MC_2016/Lecture9/mchrzasz.log index ac45318..5d05a7d 100644 --- a/Lectures_my/MC_2016/Lecture9/mchrzasz.log +++ b/Lectures_my/MC_2016/Lecture9/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) 25 APR 2016 19:00 +This is XeTeX, Version 3.1415926-2.5-0.9999.3 (TeX Live 2013/Debian) (format=xelatex 2015.4.1) 30 APR 2016 16:15 entering extended mode restricted \write18 enabled. %&-line parsing enabled. diff --git a/Lectures_my/MC_2016/Lecture9/mchrzasz.pdf b/Lectures_my/MC_2016/Lecture9/mchrzasz.pdf index a4305ed..e8f7f4a 100644 --- a/Lectures_my/MC_2016/Lecture9/mchrzasz.pdf +++ b/Lectures_my/MC_2016/Lecture9/mchrzasz.pdf Binary files differ diff --git a/Lectures_my/MC_2016/Lecture9/mchrzasz.synctex.gz b/Lectures_my/MC_2016/Lecture9/mchrzasz.synctex.gz index d83e4ae..be3db69 100644 --- a/Lectures_my/MC_2016/Lecture9/mchrzasz.synctex.gz +++ b/Lectures_my/MC_2016/Lecture9/mchrzasz.synctex.gz Binary files differ diff --git a/Lectures_my/MC_2016/Lecture9/mchrzasz.tex b/Lectures_my/MC_2016/Lecture9/mchrzasz.tex index bb480dc..383569f 100644 --- a/Lectures_my/MC_2016/Lecture9/mchrzasz.tex +++ b/Lectures_my/MC_2016/Lecture9/mchrzasz.tex @@ -334,7 +334,7 @@ \begin{footnotesize} \ARROW For the observed trajectory $\gamma_k=(i,i_1,..,j_k,0)$ we assign the value of: \begin{align*} -X(\gamma_k)=\frac{ h_{ii_1} h_{i_1 i_2}... h_{i_{k-1} i_k} h_{i_k 0}~~ \delta_{i_k j }}{ p_{ii_1} p_{i_1 i_2}... p_{i_{k-1} i_k} ~~p_{i_k 0} p_{i_k 0} } +X(\gamma_k)=\frac{ h_{ii_1} h_{i_1 i_2}... h_{i_{k-1} i_k} ~~ \delta_{i_k j }}{ p_{ii_1} p_{i_1 i_2}... p_{i_{k-1} i_k} ~~p_{i_k 0} } \end{align*} \ARROW The mean is the of all observed $X(\gamma_k)$ is an unbiased estimator of the $(MA)^{-1}_{ij}$.\\ \begin{exampleblock}{Prove:} @@ -371,7 +371,7 @@ \begin{align*} q_1,q_2,...,q_n~{ \rm such~that~}q_i>0,~i=1,2,3...n{\rm ~and~} \sum_{i=1}^n q_i =1. \end{align*} -\ARROW The choose arbitrary the probability matrix $P$ (usual restrictions apply): +\ARROW Then choose arbitrary the probability matrix $P$ (usual restrictions apply): \begin{itemize} \item The initial point we choose with the probability ${q_i}$. \item If in the moment $t$ the point is in the $i_t$ state, then in the time $t+1$ he will be in state $i_{t+1}$ with the probability $p_{i_t,t_{t+1}}$. @@ -381,7 +381,7 @@ \begin{align*} Y(\gamma_k)=\frac{ h_{i_1 i} h_{i_2 i_1}... h_{i_k i_{k-1}} }{ p_{i_1 i} p_{i_2 i_1}... p_{i_k i_{k-1}} } \frac{1}{q_{i_0}p_{i_k 0} } e_{i_k i_0} \in \mathbb{R}^n \times\mathbb{R}^n \end{align*} -\ARROW The mean of $Y(\gamma)$ is an unbias estimator of the $(MA)^{-1}$ matrix.\\ +\ARROW The mean of $Y(\gamma)$ is an unbiased estimator of the $(MA)^{-1}$ matrix.\\ \ARROW The Wasow estimator reads: \begin{align*} Y^{\ast}=\sum_{m=0}^k \frac{ h_{i_1 i} h_{i_2 i_1}... h_{i_m i_{m-1}} }{ p_{i_1 i} p_{i_2 i_1}... p_{i_m i_{m-1}} } e_{i_m i_0} \in \mathbb{R}^n \times\mathbb{R}^n @@ -398,7 +398,7 @@ \begin{frame}\frametitle{Partial differential equations, intro} \begin{minipage}{\textwidth} \begin{footnotesize} -\ARROW Let say we are want to describe a point that walks on the $\mathbb{R}$ axis: +\ARROW Let's say we are want to describe a point that walks on the $\mathbb{R}$ axis: \begin{itemize} \item At the beginning $(t=0)$ the particle is at $x=0$ \item If in the $t$ the particle is in the $x$ then in the time $t+1$ it walks to $x+1$ with the known probability $p$ and to the point $x-1$ with the probability $q=1-p$. @@ -427,7 +427,7 @@ \begin{tiny} \ARROW Now in differential equation language we would say that the particle walks in steps of $\Delta x$ in times: $k\Delta t$, $k=1,2,3....$: \begin{align*} -\nu(x,t+\Delta)=p\nu(x-\Delta x,t)+q\nu(x+\Delta x,t). +\nu(x,t+\Delta t)=p\nu(x-\Delta x,t)+q\nu(x+\Delta x,t). \end{align*} \ARROW To solve this equation we need to expand the $\nu(x,t)$ funciton in the Taylor series: \begin{align*} @@ -441,7 +441,7 @@ \begin{align*} (p-q) \frac{\Delta x }{\Delta t} \to 2 c,~~~~~~\frac{ (\Delta x)^2}{\Delta t } \to 2D, \end{align*} -\ARROW We get the Fokker-Planck equation for the difusion with current: +\ARROW We get the Fokker-Planck equation for the diffusion with current: \begin{align*} \frac{\partial \nu(x,t)}{\partial t } = -2c \frac{\partial \nu(x,t) }{\partial x} + D \frac{\partial^2 \nu(x,t)}{\partial x^2} \end{align*} @@ -461,8 +461,8 @@ \begin{frame}\frametitle{Laplace equation, Dirichlet boundary conditions} \begin{minipage}{\textwidth} \begin{footnotesize} -\ARROW The aforementioned example show the way to solve the partial differential equation using Markov Chain MC. \\ -\ARROW We will see how different classes of partial differential equations can be approximated with a Markov Chain MC, which expectation value is the solution of the equation. +\ARROW The aforementioned example shows the way to solve the partial differential equation using Markov Chain MC. \\ +\ARROW We will see how different classes of partial differential equations can be approximated with a Markov Chain MC, whose expectation value is the solution of the equation. \ARROW The Laplace equation: \begin{align*} \frac{\partial^2 u }{\partial x_1^2 } +\frac{\partial^2 u }{\partial x_2^2 }+...+\frac{\partial^2 u }{\partial x_k^2 }=0 @@ -474,7 +474,7 @@ u(x_1,x_2,...,x_k)=f(x_1,x_2,...,x_k) \in \Gamma(D) \end{align*} \end{exampleblock} -\ARROW Now I am lazzy so I put $k=2$ but it's the same for all k! +\ARROW Now I am lazy so I put $k=2$ but it's the same for all k! \end{footnotesize} @@ -507,7 +507,7 @@ \end{columns} \ARROW We express the second derivatives with the discrete form: \begin{align*} -\frac{ \frac{u(x+h)-u(x)}{h} -\frac{u(x)-u(x-h) }{h} }{h} = \frac{u(x+h)=2u(x)+u(x-h)}{h^2} +\frac{ \frac{u(x+h)-u(x)}{h} -\frac{u(x)-u(x-h) }{h} }{h} = \frac{u(x+h)-2u(x)+u(x-h)}{h^2} \end{align*} \ARROW Now we choose the units so $h=1$. @@ -530,9 +530,9 @@ \end{align*} where $f^{\ast}(x,y)$ is the discrete equivalent of $f(x,y)$ function. \end{exampleblock} -\ARROW We consider a random walk over the lattice $D^{\ast} \cup \Gamma(D^{\ast}$. +\ARROW We consider a random walk over the lattice $D^{\ast} \cup \Gamma(D^{\ast})$. \begin{itemize} -\item In the $t=0$ we are in some point $(\xi,\eta) \in D^{\ast}$ +\item In the $t=0$ we are in some point $(\xi,\eta) \in D^{\ast})$ \item If at the $t$ the particle is in $(x,y)$ then at $t+1$ it can go with equal probability to any of the four neighbour lattices: $(x-1,y)$, $(x+1,y)$, $(x,y-1)$, $(x,y+1)$. \item If the particle at some moment gets to the edge $\Gamma(D^{\ast}$ then the walk is terminated. \item For the particle trajectory we assign the value of: $\nu(\xi,\eta)=f^{\ast}(x,y)$, where $(x,y)\in \Gamma(D^{\ast})$. @@ -554,7 +554,7 @@ \begin{align} p_{\xi,\eta}(x,y)=\begin{cases} 1,~~(x,y)=\xi,\eta)\\ -0,~~(x,y)\neq =\xi,\eta) +0,~~(x,y)\neq \xi,\eta) \end{cases}\label{eq:trivial} \end{align} \item The point $(\xi,\eta) \in D^{\ast}$: @@ -690,7 +690,7 @@ \end{align*} \ARROW The differential equation: \begin{align*} -\frac{u(x+h,t-l) -2u(x,t-l)+u(x-h,t-l}{h^2}=c \frac{u(x,t)-u(x,t-l)}{l} +\frac{u(x+h,t-l) -2u(x,t-l)+u(x-h,t-l}{h^2})=c \frac{u(x,t)-u(x,t-l)}{l} \end{align*} @@ -717,7 +717,7 @@ \item If in a given time step $t$ particle is at $x$ then with $50:50$ chances it can go to $x-h$ or $x+h$ and time $t-l$. \item The particle ends the walk in two situations: \begin{itemize} -\item If it reaches the $x=0$ or $x=a$. In this case we assign to a given trajectory a value of $f_(t)$ or $f_2(t)$, where $t$ is the actuall ''weight''. +\item If it reaches the $x=0$ or $x=a$. In this case we assign to a given trajectory a value of $f_1(t)$ or $f_2(t)$, where $t$ is the actuall ''weight''. \item If the ''weight'' of the particle is equal zero. in this case we assign as a value of the trajectory the $g(x)$, where $x$ is the actual position of the particle. \end{itemize} \end{itemize} @@ -734,7 +734,7 @@ \begin{footnotesize} \ARROW Repeat the above procedure $N$ times. The expected value of a function $u$ in $(\xi,\tau)$ point is the mean of observed values. -\begin{exampleblock}{Digresion:} +\begin{exampleblock}{Digression:} The 1-dim calse can be treated as a 2-dim $(x,t)$, where the area is unbounded in the $t$ dimension. The walk is terminated after maximum $\tau/l$ steps. \end{exampleblock}