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For, as has been pointed out several times, there is no such thing as a random number ā€” there are only methods to produce random numbers, and a strict arithmetic procedure of course is not such a method.'' - \end{exampleblock} -$\color{PineGreen}\Rrightarrow$ Random number: a given value that is taken by a random variable $ \twoheadrightarrow$ by definition cannot be predicted.\\ -%$\color{PineGreen}\Rrightarrow$ Sequence of random numbers $\twoheadrightarrow$ -$\color{PineGreen}\Rrightarrow$ Sources of truly random numbers: + + +\begin{frame}\frametitle{Markov Chain MC} \begin{itemize} -\item Mechanical -\item Physical +\item Consider a finite possible states: $S_1$, $S_2$, ... +\item And the time steps of time, labelled as $1$, $2$, ... +\item At time $t$ the state is denoted $X_t$. +\item The conditional probability is defined as: \end{itemize} -$\color{PineGreen}\Rrightarrow$ Disadvantages of physical generators: +\begin{equation} +P(X_t=S_j \vert X_{t-1}=S_{j-1},..., X_{1}=S_{1}) \nonumber +\end{equation} \begin{itemize} -\item To slow for typical applications, especially the mechanical ones! -\item Not stable; small changes in boundary conditions might lead to completely different results! +\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. +\end{itemize} +\end{frame} + + + +\begin{frame}\frametitle{Linear Equations} +\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{frame} + +\begin{frame}\frametitle{Neumann-Ulam method} +\begin{itemize} +\item Let's try apply the basic MCMC method to solve a simple linear equation system: +\end{itemize} +\begin{equation} +A \overrightarrow{x} = \overrightarrow{b} \nonumber +\end{equation} +\begin{itemize} +\item The above system can be (always, see linear algebra lecture) translated into system: +\end{itemize} +\begin{equation} +\overrightarrow{x} = \overrightarrow{a} + H \overrightarrow{x} \nonumber +\end{equation} +\begin{itemize} +\item For this method we assume that the norm of the matrix is: +\end{itemize} +\begin{equation} +\Vert H \Vert = \underset{1 \leq i \leq n}{max} \sum_{j=1}^n \vert h_{ij} \vert <1 \nonumber \end{equation} +\begin{itemize} +\item Which we can write in a form: +\end{itemize} +\begin{equation} +(1 -H)\overrightarrow{x}=\overrightarrow{a} \nonumber +\end{equation} +\end{frame} + + + + +\begin{frame}\frametitle{Neumann-Ulam method} +\begin{itemize} +\item The solution would be then: +\end{itemize} +\begin{equation} +\overrightarrow{x}_0=(1 -H)^{-1}\overrightarrow{a} \nonumber +\end{equation} +\begin{itemize} +\item We can Taylor expend this: +\end{itemize} +\begin{equation} +\overrightarrow{x}_0=(1 -H)^{-1}\overrightarrow{a} = \overrightarrow{a} + H \overrightarrow{a} + H^2 \overrightarrow{a} + H^3 \overrightarrow{a} +.... \nonumber +\end{equation} +\begin{itemize} +\item For the $i$-th component of the $\overrightarrow{x}$ vector: +\end{itemize} +\begin{equation} +x_0^i= a_i + \sum_{j=1}^n h_{ij} a_{j_1} + \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_2=1}^n \sum_{j_3=1}^n h_{ij_1} h_{ij_2} h_{ij_3} a_{j_3} + ...\nonumber +\end{equation} +\begin{itemize} +\item One can construct probabilistic behaviour of a system that follows the path of equation above. +\end{itemize} + +\end{frame} + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\begin{frame}\frametitle{Neumann-Ulam method} +\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 $i_t$. +\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 $X(\gamma) = (i_0, i_1, i_2, ..., i_k, 0)$ is called trajectory. +\item It can be proven that $x_i^0 =E \lbrace X (\gamma) \vert i_0=j \rbrace$. \end{itemize} -\end{frame} - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\begin{frame}\frametitle{Random numbers - history remark} -$\color{PineGreen}\Rrightarrow$ In the past there were books with random numbers: -\begin{center} -\includegraphics[width=0.55\textwidth]{images/million-random-digits-open.jpg} - \end{center} -$\color{PineGreen}\Rrightarrow$ It's obvious that they didn't become very popular ;)\\ -$\color{PineGreen}\Rrightarrow$ This methods are comming back!\\ -$\color{PineGreen}\twoheadrightarrow$ Storage device are getting more cheap and bigger (CD, DVD).\\ -$\color{PineGreen}\twoheadrightarrow$ 1995: G. Marsaglia, $650\rm MB$ of random numbers, ''White and Black Noise''. - \end{frame} - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\begin{frame}\frametitle{Pseudorandom numbers} -$\color{PineGreen}\Rrightarrow$ Pseudorandom numbers are numbers that are generated accordingly to strict mathematical formula. \\ -$\color{PineGreen}\looparrowright$ Strictly speaking they are non random numbers, how ever they have all the statistical properties of random numbers.\\ -$\color{PineGreen}\looparrowright$ Discussing those properties is a wide topic so let's just say that without knowing the formula they are generated by one cannot say if those numbers are random or not.\\ -$\color{PineGreen}\Rrightarrow$ Mathematical methods of producing pseudorandom numbers: +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\begin{frame}\frametitle{Neumann-Ulam method, $\rm \color{RubineRed}{Lecture3/Markov}$} \begin{itemize} -\item Good statistical properties of generated numbers. -\item Easy to use and fast! -\item Reproducible! +\item For example lets try to solve this equation system: \end{itemize} -$\color{PineGreen}\Rrightarrow$ Since mathematical pseudorandom genrators are dominantly: pseudorandom $\rightarrowtail$ random. - - -\end{frame} - - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\begin{frame}\frametitle{Middle square generator; von Neumann} -$\color{PineGreen}\Rrightarrow$ The first mathematical generator (middle square) was proposed by von Neumann (1964).\\{~}\\ -$\color{PineGreen}\looparrowright$ Formula: $ -\tcbhighmath[fuzzy halo=0.5mm with PineGreen!50!white,arc=0.1pt, - boxrule=0pt,frame hidden]{ X_n= \lfloor X_{n-1}^2\cdot 10 ^{-m} \rfloor - \lfloor X_{n-1}^2\cdot 10^{-3m} \rfloor} -$ - -$\color{PineGreen}\looparrowright$ where $X_0$ is a constant (seed), $\lfloor\cdot\rfloor$ is the cut-off of a number to integer.\\ -$\color{PineGreen}\Rrightarrow$ Example:\\ -{~}{~}Let's put $m=2$ and $X_0=2045$:\\ +\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{0.1\textwidth} -{~} -\column{0.4\textwidth} -$\color{PineGreen}\looparrowright$ $X_0^2=\underbrace{04}_{\rm rej}1820\underbrace{25}_{\rm rej}$ -\column{0.3\textwidth} - $\Rightarrow X_1=1820$ -\end{columns} -\begin{columns} -\column{0.1\textwidth} -{~} -\column{0.4\textwidth} -$\color{PineGreen}\looparrowright$ $X_1^2=\underbrace{03}_{\rm rej}3124\underbrace{00}_{\rm rej}$ -\column{0.3\textwidth} - $\Rightarrow X_1=3124$ -\end{columns} -$\color{PineGreen}\looparrowright$ Simple generator but unfortunately quite bad generator. Firstly the sequences are very short and strongly dependent on the $X_0$ number. +\column{0.1in} - -\end{frame} -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\begin{frame}\frametitle{Linear generators $\rm \color{RubineRed}{Lecture2/Linear\_gen1}$} -$\color{PineGreen}\Rrightarrow$ This was a first generator written and it's a good example how to not write generators.\\ -$\color{PineGreen}\Rrightarrow$ It's highly non stable! -\includegraphics[width=0.8\textwidth]{images/shit.png} - - - -\end{frame} - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\begin{frame}\frametitle{Linear generators} - -$\color{PineGreen}\Rrightarrow$ General equation: -\begin{equation} -\tcbhighmath[fuzzy halo=0.5mm with PineGreen!50!white,arc=0.1pt, - boxrule=0pt,frame hidden]{X_n=(a_1X_{n-1} + a_2X_{n-2}+...+a_kX_{n-k}+c)~{\rm mod}~m ,} \nonumber -\end{equation} -$\color{PineGreen}\looparrowright$ where $a_i,c,m$ are parameters of a generator(integer numbers).\\ -$\color{PineGreen}\looparrowright$ Generator initialization $\color{PineGreen} \rightleftarrows$ setting those parameters.\\ -$\color{PineGreen}\Rrightarrow$ Very old generators. (often used in Pascal, or first C versions): -\begin{columns} -\column{0.1\textwidth} -{~} -\column{0.8\textwidth} -$k=1:~X_n=(aX_{n-1} +c )~{\rm mod} m,$ -$c= -\begin{cases} -=0, {\rm multiplicative geneator}\\ -\neq 0, {\rm mix geneator} -\end{cases} -$ -\column{0.1\textwidth} -{~} - - -\end{columns} -$\color{PineGreen}\Rrightarrow$ The period can be achieved by tuning the seed parameters:\\ -$P_{{\rm max}}= -\begin{cases} -2^{L-2};~{\rm for~} m=2^L\\ -m-1;~{\rm for~} m= {\rm prime~number} -\end{cases} -$ -\end{frame} - - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\begin{frame}\frametitle{Shift register generator} -\begin{small} -$\color{PineGreen}\Rrightarrow$ General equation: -\begin{equation} -\tcbhighmath[fuzzy halo=0.5mm with PineGreen!50!white,arc=0.1pt, - boxrule=0pt,frame hidden]{b_n=(a_1X_{n-1} + a_2X_{n-2}+...+a_kX_{n-k}+c)~{\rm mod}~2 ,} \nonumber -\end{equation} -where $a_i \subset(\lbrace0,1\rbrace )$\\ -$\color{PineGreen}\Rrightarrow$ Super fast and easy to implement due to: $(a+b) ~{\rm mod}~2 = a~{\rm xor}~b$ -\begin{center} -\begin{tabular}{||c|c|c||} -\hline \hline a & b & a xor b\\ \hline -0 & 0 & 0\\ -1 & 0 & 1\\ -0 & 1 & 1\\ -1 & 1 & 0\\ \hline \hline -\end{tabular} -\end{center} -$\color{PineGreen}\Rrightarrow$ Maximal period is $2^k-1$.\\ -$\color{PineGreen}\Rrightarrow$ Example (Tausworths generator):\\ -$a_p=a_q=1$, other $a_i=0$ and $p>q$. Then: $b_n=b_{n-p}~{\rm xor}~b_{n-q}$ - -$\color{PineGreen}\Rrightarrow$ How to get numbers from bits (for example):\\ -$U_i = \sum_{j=1}^L 2^{-j} b_{is+j},~s{ -\includegraphics[width=0.95\textwidth]{images/gen1.png} -} -\only<2>{ -\includegraphics[width=0.95\textwidth]{images/gen2.png} -} -\only<3>{ -\includegraphics[width=0.95\textwidth]{images/gen3.png} -} -\only<4>{ -\includegraphics[width=0.95\textwidth]{images/gen4.png} -} -\only<5>{ -\includegraphics[width=0.95\textwidth]{images/gen5.png} -} -\only<6>{ -\includegraphics[width=0.95\textwidth]{images/gen6.png} -} -\end{small} -\end{frame} -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\begin{frame}\frametitle{Detector simulation} - -\begin{small} -$\color{PineGreen}\Rrightarrow$ Things do not get simpler on the detector side simulation.\\ -$\color{PineGreen}\Rrightarrow$ Lots of effects need to be taken into account: -\begin{columns} -\column{0.2in} -{~} -\column{2in} -$\color{PineGreen}\rightarrowtail$ Bremsstrahlung\\ -$\color{PineGreen}\rightarrowtail$ Interactions with different detector materials\\ -$\color{PineGreen}\rightarrowtail$ Particle identification\\ -$\color{PineGreen}\rightarrowtail$ Showers\\ -\column{3in} -\includegraphics[width=0.95\textwidth]{{images/lhcb2_h-640x408}.jpg} -\end{columns} -$\color{PineGreen}\Rrightarrow$ Example of generators:\\ -$\color{PineGreen}\rightarrowtail$ FLUKA\\ -$\color{PineGreen}\rightarrowtail$ Geant - -\end{small} -\end{frame} - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\begin{frame}\frametitle{Method of Moments} - -\begin{small} -$\color{PineGreen}\Rrightarrow$ Now real cool things!\\ -$\color{PineGreen}\Rrightarrow$ Let's consider we want to study a rare decay: $\PB^{\pm} \to \PK^{\pm} \Pmu \Pmu$. The decay is described by the following PDF: -\begin{equation} -\dfrac{1}{\Gamma}\dfrac{d^2\Gamma}{dq^2 d\cos \theta_l} =\dfrac{3}{4}(1-F_H)(1-\cos^2 \theta_l)+F_H/2 + A_{FB}\cos \theta_l \nonumber -\end{equation} -$\color{PineGreen}\Rrightarrow$ PDF by construction is normalized: $\int_{-1}^{1} \dfrac{1}{\Gamma}\dfrac{d^2\Gamma}{dq^2 d\cos \theta_l} =1$ -\begin{columns} -\column{0.1in} -{~} -\column{2.2in} -\begin{itemize} -\item Normally we do a likelihood fit and we are done. -\item There is a second way! -\end{itemize} -\column{2.8in} -\includegraphics[width=0.95\textwidth]{images/Kmumu_LL.png} -\end{columns} - -\end{small} -\end{frame} - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\begin{frame}\frametitle{Method of Moments} -\begin{small} -$\color{PineGreen}\Rrightarrow$ Let's calculate the integrals: -\begin{equation} -\int_{-1}^{1} \dfrac{1}{\Gamma}\dfrac{d^2\Gamma}{dq^2 d\cos \theta_l} \cdot \cos \theta_l = \dfrac{2}{3}A_{FB} \nonumber -\end{equation} -\begin{equation} -\int_{-1}^{1} \dfrac{1}{\Gamma}\dfrac{d^2\Gamma}{dq^2 d\cos \theta_l} \cdot \cos^2 \theta_l = \dfrac{1}{5} + \dfrac{2 F_H}{15} \nonumber -\end{equation} -$\color{PineGreen}\Rrightarrow$ So we can get our parameters that we searched for by doing a integration. So now what?\pause \\ -$\color{PineGreen}\Rrightarrow$ Well nature is the best random number generator so let's take the data and treat and calculate the integral estimates: -\begin{equation} -\int_{-1}^{1} \dfrac{1}{\Gamma}\dfrac{d^2\Gamma}{dq^2 d\cos \theta_l} \cdot \cos \theta_l = \dfrac{2}{3}A_{FB} = \dfrac{1}{N} \sum_{i=1}^N \cos \theta_{l,i} \nonumber -\end{equation} -\begin{equation} -\int_{-1}^{1} \dfrac{1}{\Gamma}\dfrac{d^2\Gamma}{dq^2 d\cos \theta_l} \cdot \cos^2 \theta_l = \dfrac{1}{5} + \dfrac{2 F_H}{15}= \dfrac{1}{N} \sum_{i=1}^N \cos^2 \theta_{l,i} \nonumber -\end{equation} - -\end{small} -\end{frame} - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\begin{frame}\frametitle{Method of Moments} -\begin{small} -$\color{PineGreen}\Rrightarrow$ So what did we do? -\begin{itemize} -\item We have just estimated a parameters of interests without using any fit!! -\end{itemize} -$\color{PineGreen}\Rrightarrow$ Pros and cones of method of moments: -\begin{itemize} -\item {\color{PineGreen}{Are very immune to bias.}} -\item {\color{PineGreen}{Do not suffer from boundary problems.}} -\item {\color{PineGreen}{Require less statistic to work then likelihood fit.}} -\item {\color{PineGreen}{They always have a Gaussian error}}. -\item {\color{Red}{Estimator has a larger uncertainty.}} - - -\end{itemize} - -\end{small} -\end{frame} -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\begin{frame}\frametitle{Method of Moments, uncertainty estimator} -\begin{small} -\begin{columns} -\column{0.1in} -{~} \column{2.5in} -$\color{PineGreen}\Rrightarrow$ It can be proven that Method of Moments estimator converges slower then the maximum likelihood fit. -\column{2.5in} -\includegraphics[width=0.95\textwidth]{images/S7.png} +\begin{itemize} +\item An example solution: +\end{itemize} +\includegraphics[width=0.95\textwidth]{images/mark.png} + \end{columns} + +\end{frame} + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\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, $\rm \color{RubineRed}{Lecture3/Markov2}$} +\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} -\includegraphics[width=0.95\textwidth]{images/S8F_950.png} +\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} -\includegraphics[width=0.95\textwidth]{images/S8.png} +\begin{itemize} +\item An example solution: +\end{itemize} +\includegraphics[width=0.95\textwidth]{images/mark2.png} + \end{columns} - -\end{small} \end{frame} -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\begin{frame}\frametitle{Other application of MC - testing your analysis} -\begin{small} -$\color{PineGreen}\Rrightarrow$ Probably the biggest application of MC methods in HEP are validations of your experimental methodology. The procedure is as follows: + +\begin{frame}\frametitle{Look elsewhere effect, $\rm \color{RubineRed}{Lecture3/LEE}$} \begin{itemize} -\item Define your analysis methodology: selection, efficiency corrections, parameters you want to measure. -\item Simulate an assembly of simulation events for different values of parameters you want to measure. -\item Do the analysis on this pseudo data. -\item See if you are getting back what you have simulated. +\item Look elsewhere effect addresses the following problem: +\begin{itemize} +\item Imagine you observed a $3\sigma$ deviation in one of the observable that you measured. +\item Before you get excited one needs to understand if given the fact that you had so many measurements this might happen! +\end{itemize} +\item Example: Let's say we have measured 50 observables. What is the probability to observed 1 that is $3\sigma$ away from theory prediction? +\item Let's simulate 50 Gaussian distribution centred at 0 and width of 1. We count how simulations where at least one of the 50 numbers have the absolute value $>3$. +\item More complicated example: what if you observed 3 in a row $2\sigma$ fluctuations among 50 measurements? +\item This kind of studies are the best solvable by MC simulations. \end{itemize} -\end{small} + \end{frame} -\begin{frame}\frametitle{Testing your analysis, $\rm \color{RubineRed}{Lecture2/Test\_met}$} -\begin{small} -$\color{PineGreen}\Rrightarrow$ Probably the biggest application of MC methods in HEP are validations of your experimental methodology. The procedure is as follows: + + + + + +\begin{frame}\frametitle{Travelling Salesman Problem} \begin{itemize} -\item Define your analysis methodology: selection, efficiency corrections, parameters you want to measure. -\item Simulate an assembly of simulation events for different values of parameters you want to measure. -\item Do the analysis on this pseudo data. -\item See if you are getting back what you have simulated. +\item Salesman starting from his base has to visit $n-1$ other locations and return to base headquarters. The problem is to find the shortest way. +\item For large $n$ the problem can't be solver by brutal force as the complexity of the problem is $(n-1)!$ +\item There exist simplified numerical solutions assuming factorizations. Unfortunately even those require anonymous computing power. +\item Can MC help? YES :) +\item The minimum distance $l$ has to depend on 2 factors: $P$ the area of the city the Salesman is travelling and the density of places he wants to visit: $\dfrac{n}{P}$ +\item Form this we can assume: \end{itemize} -\includegraphics[width=0.4\textwidth]{{images/mean_estimator}.png} -\end{small} +\begin{equation} +l \sim P^a (\dfrac{n}{P})^b=P^{a-b}n^b. \nonumber +\end{equation} + \end{frame} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\begin{frame}\frametitle{Wrap up} -\begin{small} -$\color{PineGreen}\Rrightarrow$ Things to remember: + + + + + + +\begin{frame}\frametitle{Traveling Salesman Problem} \begin{itemize} -\item Computer cannot produce random numbers, only pseudorandom numbers. -\item We use pseudorandon numbers as random numbers if they are statistically acting the same as random numbers. -\item Linear generators are not commonly used nowadays. -\item State of the art generators are the ones based on Kolomogorows theorem. -\item MC methods used to simulate physics process, detector response and validating the estimators. +\item From dimension analysis: \end{itemize} -\end{small} +\begin{equation} +a-b=\dfrac{1}{2}. \nonumber +\end{equation} +\begin{itemize} +\item To get $l$ we need square root of area. +\item From this it's obvious: +\end{itemize} +\begin{equation} +l \sim P^a (\dfrac{n}{P})^b=P^{0.5}n^{a-0.5}. \nonumber +\end{equation} +\begin{itemize} +\item Now we can multiply the area by alpha factor that keeps the density constant then: +\end{itemize} +\begin{equation} +l \sim \alpha^0.5 \alpha6{a-0.5} = \alpha^a \nonumber +\end{equation} +\begin{itemize} +\item In this case the distance between the clients will not change, but the number of clients will increase by $\alpha$ so: +\end{itemize} +\begin{equation} +l \sim \alpha \nonumber +\end{equation} +\begin{itemize} +\item In the end we get: $a=1$ +\end{itemize} \end{frame} +\begin{frame}\frametitle{Traveling Salesman Problem} +\begin{itemize} +\item In total: +\end{itemize} +\begin{equation} +l \sim k (nP)^{0.5}\nonumber +\end{equation} +\begin{itemize} +\item Of course the k depends on the shape of the area and locations of client. However for large $n$ the k starts loosing the dependency. It's an asymptotically free estimator. +\item To use the above formula we need to somehow calculate k. +\item How to estimate this? Well make a TOY MC: take a square put uniformly $n$ points. Then we can calculate $l$. Then it's trivial: +\end{itemize} +\begin{equation} +k= l(nP)^{-0.5} \nonumber +\end{equation} +\end{frame} + + +\begin{frame}\frametitle{Traveling Salesman Problem} +\begin{itemize} +\item This kind of MC experiment might require large CPU power and time. The adventage is that once we solve the problem we can use the obtained k for other cases (it's universal constant!). +\item It turns out that: +\end{itemize} +\begin{equation} +k \sim \dfrac{3}{4} \nonumber +\end{equation} +\begin{itemize} +\item Ok, but in this case we can calculate $l$ but not the actual shortest way! Why the hell we did this exercise?! +\item Turns out that for most of the problems we are looking for the solution that is close to smallest $l$ not the exact minimum. +\end{itemize} +\end{frame} + +\begin{frame}\frametitle{War Games} + +\begin{itemize} +\item S. Andersoon 1966 simulated for Swedish government how would a tank battle look like. +\item Each of the sides has 15 tanks. that they allocate on the battle field. +\item The battle is done in time steps. +\item Each tank has 5 states: +\begin{itemize} +\item OK +\item Tank can only shoot +\item Tank can only move +\item Tank is destroyed +\item Temporary states +\end{itemize} +\item This models made possible to +\end{itemize} +\end{frame} + \backupbegin \begin{frame}\frametitle{Backup}