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\institute{UZH} -\title[Arbitrary \pdf~generation]{Arbitrary \pdf~generation} +\title[Specific \pdf~generation]{Specific \pdf~generation} \date{\fixme} \newcommand*{\QEDA}{\hfill\ensuremath{\blacksquare}}% \newcommand*{\QEDB}{\hfill\ensuremath{\square}}% @@ -216,14 +217,14 @@ \begin{document} \tikzstyle{every picture}+=[remember picture] -{ +{x \setbeamertemplate{sidebar right}{\llap{\includegraphics[width=\paperwidth,height=\paperheight]{bubble2}}} \begin{frame}[c]%{\phantom{title page}} \begin{center} \begin{center} \begin{columns} \begin{column}{0.9\textwidth} - \flushright\fontspec{Trebuchet MS}\bfseries \Huge {Arbitrary \pdf~generation} + \flushright\fontspec{Trebuchet MS}\bfseries \Huge {Specific \pdf~generation} \end{column} \begin{column}{0.2\textwidth} %\includegraphics[width=\textwidth]{SHiP-2} @@ -245,500 +246,264 @@ \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, \\ 7 April, 2016} + \textcolor{normal text.fg!50!Comment}{Monte Carlo methods, \\ 14 April, 2016} \end{center} \end{frame} } -\begin{frame}\frametitle{Reverting the \cdf} +\begin{frame}\frametitle{Exponential \pdf } \begin{footnotesize} - \begin{alertblock}{~} -\ARROW Let $U$ be a random variable from $\mathcal{U}(0,1)$\\ -\ARROW Now let $F$ be a non decreasing function such that: +% \begin{exampleblock}{~} +\ARROWR The $X(\theta, \lambda)$:\\ \begin{align*} -F(- \infty) =0~~~~F(\infty)=1 +\rho_{\theta , \lambda}=\frac{1}{\lambda} e^{- \frac{x- \theta}{\lambda}} \end{align*} -then: +\ARROWR One can transform the variable: \begin{align*} -X=F^{-1}(U) +x \to x^{\prime} =\frac{x-\theta}{\lambda}~~ \Rightarrow~~ E(\theta, \lambda) \to E(0,1): \rho_{0,1}=e^{-x^{\prime},} x^{\prime} \geq 0 \end{align*} -has a \pdf~ distribution with a \cdf~function of F. - \end{alertblock} -\ARROW Prove: -\begin{align*} -F(x)=\mathcal{P}(U\leq F(x))=\mathcal{P}(F^{-1}(U) \leq x) = \mathcal{P}(X \leq x) ~~~~ \QEDB -\end{align*} -\ARROW So it looks very simple if $x_1,X_2,...,X_n$ are random variables from $\mathcal{U}(0,1)$ then: $\lbrace X_i=F^{-1}(x_i)\rbrace,i=1,...,n$ is the sequence that has a \cdf~distirbution of $F$. - - - \end{footnotesize} - -\end{frame} - -\begin{frame}\frametitle{Reverting the \cdf, examples} - \begin{footnotesize} -\ARROW The exponential distribution: $E(0,1)$. \\ -\ARROW The \pdf : $\rho(X)=e^{-X},~X \geq 0$.\\ -\ARROW The \cdf : $F(x) = \int_0^x e^{-X} dX= 1- e^{-x}.$\\ -\ARROW Now let $R \in \mathcal{U}(0,1):~R=F(X)=1-e^{-X} \longrightarrow X = -\ln(1-R)$\\ -\ARROW Now we can play a trick: if $R \in \mathcal{U}(0,1)$ then $1-R$ also in $\mathcal{U}(0,1)$. \\ -\ARROW In the end we get: $X=-\ln(R)$ \\ - {~}\\ -\ARROW Use the reverting to generate the following distributions: -\begin{itemize} -\item E 6.1 $\rho(X)= \dfrac{c}{X}$ on the interval $[a,b]$, where $0 p(x): & \text{miss}. -\end{cases} -\end{align*} +\end{block} -\column{2.5in} -\includegraphics[width=0.75\textwidth]{images/result.png} +\begin{block}{ \begin{footnotesize}Monolitic series method\end{footnotesize}} +\begin{enumerate} +\item Generate a sequence: $U_1,U_2,... \in \mathbb{U}(0,1)$ +\item We look at series: $U_1 \geq U_2 \geq U_3 ...\geq U_n < U_{n+1}$, which we then order with numbers: $0,1,2,3,...$. +\item First series which length $n$ is odd we take as integral part of a number. The decimal part is taken as $R_1$. +\end{enumerate} +\end{block} +\column{0.1in} +{~}\\ +\column{1.3in} + +\ARROW E7.1 Write the two above generators of $E(0,1)$. Compare \cdf~and \pdf \end{columns} -\ARROW In the end we managed to generate a \pdf : $f(x)=\cos(x)$. - \end{footnotesize} - -\end{frame} - -\begin{frame}\frametitle{Elimination method, example for the Buffon needle} - \begin{footnotesize} -\begin{center} -\includegraphics[width=0.65\textwidth]{images/result.png} -\end{center} -\ARROW We generate the points from $(U_1,U_2)$ from $\mathcal{U}(0,\dfrac{\pi}{2}) \times \mathcal{U}(0,1)$.\\ -\ARROW Points that are below the function we keep.\\ -\ARROW The ones above we reject.\\ -\ARROW The new variable will have the $f(x)$ \pdf. - - - \end{footnotesize} \end{frame} -\begin{frame}\frametitle{Elimination method, multi dimension case} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + + +\begin{frame}\frametitle{Gaussian \pdf } \begin{footnotesize} -\ARROW Let $X=(X_1,X_2,...,X_m)$ be a $m$ dimensional random variable with a \pdf of $f(x_1,x_2,...,x_m)$ of the domain of $\Omega \in \mathcal{R}^m$ and $f(x_1,x_2,..x_m){ +% \begin{exampleblock}{~} +\ARROW The \pdf: \begin{align*} -\psi(x,y)=c\sqrt{1-(x^2+y^2)},~~x^2+y^2\leq 1 +\phi_{\mu,\sigma}(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}~~ - \infty 0,~n>1.$\\ -We can write down the functions: $g_Y(x)=ye^{-xy}$ and $h(y)=ny^{-(n+1)}$. -\begin{enumerate} -\item The numbers $Y$ should be generated from the \pdf $h(y)$. One can use the Reverting \cdf~for example: $Y=(1-U)^{1/n}$, where $U \in \mathcal{U}(0,1)$. -\item The numbers $X$ of a \pdf $g_Y(x)$ are generated from the exponential \pdf~$E(0,\frac{1}{y}): -X=- \frac{1}{Y}\ln V$, where $V \in \mathcal{U}(0,1)$. -\end{enumerate} - - - - \end{footnotesize} - -\end{frame} - - -\begin{frame}\frametitle{Discrete superposition of distribution} - \begin{footnotesize} -\ARROW Let's: -\begin{align*} -f(x) = \sum_{i=1}^{\infty} p_i g_i(x) -\end{align*} -where $p_i$ discrete \pdf . $p_i\geq 0$ and $\sum_{i=1}^{\infty} p_1 =1$\\ -\ARROW The algorithm: -\begin{enumerate} -\item Generate the $i$ number accordingly to the $p_i$ \pdf . Using Reverting \cdf~method. -\item For a given $i$ generate the number $X$ from the $g_i(x)$. -\end{enumerate} -\ARROW Example: $f(x)=\dfrac{5}{12}\left[ 1+ (x-1)^4\right],~0\leq x \leq 2.$\\ -$\rightarrow~f(x)=\dfrac{5}{6}g_1(x)+1/6 g_2(x),$\\ -where $g_1(x)=\dfrac{1}{2}$ and $g_2(x)=\dfrac{5}{2}(x-1)^4$, so $p_1=\dfrac{5}{6}$ and $p_2=\dfrac{1}{6}$. -\begin{align*} -X= \begin{cases} -2 U_2~~~~{ \rm if~ } U_1 < \dfrac{5}{6}\\ -1+(2U_2-1)^{\frac{1}{5}}~~~~{ \rm if~ } U_1 < \dfrac{5}{6} +\Phi^{-1}(u)=\begin{cases} +g(u),~~~~~~~10^{-20}0$ such as: -\begin{align*} -f_i(x) \leq c_i g_i(x) \forall_x . -\end{align*} -\ARROW The algorithm: -\begin{enumerate} -\item Generate $i$ accordingly to the distribution $p_i$ -\item For a given $i$ generate $X$ from the \pdf~$g_i(x)$. -\item Generate a $U\in \mathcal{U}(0,1)$. -\item If $c_i U g_i(X) \leq f_i(X)$ then we accept the $X$ if not then we start over. -\end{enumerate} -\ARROW Alternative: -\begin{enumerate} -\item One can also calculate the weights: $w(X)=\dfrac{f_i(X)}{g_i(X)}$. -\item Find the maximal weight: $w_i^{max}$. -\item If $w_i(X) \geq U w_i^{max}$ we accept $X$. If not then start over -\end{enumerate} - - - - \end{footnotesize} - -\end{frame} - -\begin{frame}\frametitle{Combination, warnings, example} - \begin{footnotesize} -\ARROW We need a different maximal weight for each ''branch''. This can be done either numerically or analytically.\\ -\ARROW Example: -\begin{itemize} -\item Let our \pdf~be in a form of: -\end{itemize} -\begin{align*} -f(x)=\sum_{i=1}^n c_ix^i,~~0 \leq x \leq 1 -\end{align*} -but $\exists_{j \in \lbrace 1,...,n \rbrace}: c_j <0$. -We can do a transmigration: -\begin{align*} -cj=c^+_j -c^-_j: c^+_j, c^-_j>0 -\end{align*} -\ARROW So: -\begin{align*} -f(x)=\sum_{i=1}^n c_i x^i \leq \sum_{i=1}^n c_i^+ x^i = g(x) -\end{align*} - \ARROW Now we use the weight function: - \begin{align*} - \bar{g}(x)=\frac{g(x)}{\sum_{i=1}^n c_i^+} - \end{align*} - - - \end{footnotesize} - -\end{frame} - -\begin{frame}\frametitle{Combination, example} - \begin{footnotesize} -\ARROW The algorithm: -\begin{itemize} -\item Generate $X$ accordingly to $\bar{g}(x)$ \pdf. The same we did in the discrete superposition method. -\item Generate $U \in \mathcal{U}(0,1)$ -\item If $Ug(X) \leq f(X)$ then we accept the $X$. If not start over. -\end{itemize} -\ARROW Example $N(0,1)$: - \begin{align*} - f(x) = \sqrt{\dfrac{2}{\pi}} e^{-\frac{x^2}{2}},~x \geq 0. - \end{align*} -\ARROW We write the function as: -\begin{align*} -f(x) \alpha_1 g_1(x)h_1(x)+\alpha_2 g_2(x)h_2(x) -\end{align*} -where: -\begin{columns} -\column{2.5in} -\begin{align*} -\alpha_1=\sqrt{\frac{2}{\pi}} -\end{align*} -\begin{align*} -g_1=\begin{cases} -1,~0\leq x \leq 1\\ -0,~x \geq 1\\ -\end{cases} -\end{align*} - -\column{2.5in} -\begin{align*} -\alpha_2=\sqrt{\frac{1}{2 \pi}} -\end{align*} -\begin{align*} -g_2=\begin{cases} -2e^{-2(x-1)},~x \geq 1\\ -0,~0\leq x \leq 1\\ -\end{cases} -\end{align*} - - -\end{columns} - - \end{footnotesize} - -\end{frame} -\begin{frame}\frametitle{Combination, example} - \begin{footnotesize} - -\begin{columns} -\column{2.5in} - -\begin{align*} -h_1(x)=e^{\frac{-x^2}{2}} -\end{align*} - -\column{2.5in} - -\begin{align*} -h_2(x)=e^{-\frac{(x-2)^2}{2}} +g(u)=t-\dfrac{L(t)}{M(t)},\\ t=\sqrt{-2 \ln u} \end{align*} \end{columns} -{~}\\{~}\\ +\begin{align*} +L(t)=0.322232431088 + t + 0.342242088547 t^2\\ + 0.0204231210245 t^3 + 0.0000453642210148 t^4 +\end{align*} +\begin{align*} +M(t)=0.099348462606 + 0.588581570495 t + 0.531103462366 t^2\\ + 0.10353775285 t^3 + 0.0038560700634 t^4 +\end{align*} +} -\ARROW The variable $X$ is chosen: +\only<2>{ +\ARROW Reverting the \cdf~in 2 dim: +\begin{align*} +\phi(x,y)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2+y^2}{2}},~~~ - \infty +{ +\ARROW The Marsaglia \& Bray method (1964): +\begin{itemize} +\item If $U_1,U_2 \in \mathcal{U}(-1,1)$ are independent random variables, and $U_1^2+U_2^2 \leq 1$ then: +\begin{align*} +X_1=U_1\sqrt{\frac{-2 \ln (U_1^2+U_2^2)}{U_1^2+U_2^2}},~~Y_1=X_1\frac{U_2}{U_1} +\end{align*} +have the distribution of $N(0,1)$. +\end{itemize} +\ARROW The algorithm: +\begin{itemize} +\item Generate $R_1,R_2 \in \mathcal{U}(0,1)$ and calculate the $U_1=2R_1-1,~U_2=2R_2-1$ +\item Calculate $W=U_1^2+U_2^2$. +\item If $W>0$ start over. +\item Calculate the $X=U_1 Z$ and $Y=U_2 Z$, where $Z=\sqrt{\frac{-2 \ln W}{W}}$ +\end{itemize} +\ARROW E7.2 Generate $N(0,1)$ using \cdf~reverting and Marsaglia \& Bray method. + + +} + + \end{footnotesize} \end{frame} -\begin{frame}\frametitle{Acceptance - complement, Kornal \& Peterson 1981} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + +\begin{frame}\frametitle{Breit-Wigner \pdf } \begin{footnotesize} -\ARROW If $f=f_1+f_2$ is a \pdf~ of a random variable $X$ and $p=\int f_1(x) dx.$\\ -\ARROW $f_1/p_1$ and $f_2(1-p)$ are the \pdf of a discrete function.\\ -\ARROW Now if the functions $f_1$ and $f_2$ are easy to generate from then: + \only<1>{ +\ARROW The \pdf : +\begin{align*} +f_{\theta,\lambda}(x)=\frac{\lambda}{\pi} \frac{1}{\lambda^2+(x-\theta)^2},~~~~- \infty < x< \infty +\end{align*} +\ARROW The variable transformation: +\begin{align*} +x \to x^{\prime}~~~ \Rightarrow~~~ C(\theta,\lambda) \to C(0,1) +\end{align*} +\ARROW The reverting \cdf : \begin{itemize} -\item We choose the numbers with the probability distribution $f_1/p_1$ from the \pdf~$f_1$. -\item We choose the numbers with the probability distribution $f_2/(1-p_1)$ from the \pdf~$f_2$. +\item The \cdf +\begin{align*} +F(x)=\frac{1}{\pi} \arctan x + \frac{1}{2}\\ +\Rightarrow X= \tan \left( \pi \left[U-\frac{1}{2}\right]\right),~~~U \in \mathcal{U}(0,1) +\end{align*} \end{itemize} -\ARROW The method is super powerful if $f_2= { \rm const}$. - +\ARROW A statistical digression: There is no expected value of the Cauchy function. The variance is infinite. +} + +\only<2>{ +\ARROW One can use a cut-off Cauchy method $C_u(0,1)$: +\begin{align*} +f_u(x)=\begin{cases} +\frac{2}{\pi} \frac{1}{1+x^2},~~~\vert x \vert \leq 1,\\ +0,~~~~~~~~~~~\vert x \vert> 1,\\ +\end{cases} +\end{align*} +\begin{exampleblock}{Theorem:} +If a random variable $X$ has a cuf-off Cauchy distribution $C_u(0,1)$, then the new random variable $Y$, which is with $50~\%$ equal $X$ and with $50\%$ equal $1/X$ has a ''normal'' Cauchy distribution. +\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) +\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. + + + + + +} + \end{footnotesize} \end{frame} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + +\begin{frame}\frametitle{$x^n$ \pdf } + \begin{footnotesize} + \only<1>{ +\ARROW The \pdf : +\begin{align*} +f_{1}(x) &=n x^{n-1}\\ +f_{2}(x) &=n(1-x)^{n-1} +\end{align*} +where $0\leq x \leq 1,~n\in \mathbb{N}$ +\ARROW Revert the \cdf: +\begin{align*} +X=U^{1/n}~~~\longrightarrow &f_1,~~~& U\in \mathcal{U}(0,1)\\ +Y=1-U^{1/n}~~~\longrightarrow &f_2,~~~& U\in \mathcal{U}(0,1) +\end{align*} +\ARROWR Disadvantage: The operation $U^{1/n}$ is time consuming. +\ARROW Second method: +\begin{itemize} +\item Generate $U_1,U_2,...U_n \in \mathcal{U}(0,1)$. +\item $X=\max\lbrace U_1,U_2,...U_n \rbrace$ has \pdf~of $f_1$. +\item $Y=\min\lbrace U_1,U_2,...U_n \rbrace$ has \pdf~of $f_2$. +\end{itemize} + +\ARROW E7.4 Generate the $f_1$ and $f_2$ \pdf~ with two methods. +} + + \end{footnotesize} + +\end{frame} + + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + +\begin{frame}\frametitle{Bernoulli \pdf } + \begin{footnotesize} + \only<1>{ +\ARROW The \pdf : +\begin{align*} +f_{1}(x) &=n x^{n-1}\\ +f_{2}(x) &=n(1-x)^{n-1} +\end{align*} +where $0\leq x \leq 1,~n\in \mathbb{N}$ +\ARROW Revert the \cdf: +\begin{align*} +X=U^{1/n}~~~\longrightarrow &f_1,~~~& U\in \mathcal{U}(0,1)\\ +Y=1-U^{1/n}~~~\longrightarrow &f_2,~~~& U\in \mathcal{U}(0,1) +\end{align*} +\ARROWR Disadvantage: The operation $U^{1/n}$ is time consuming. +\ARROW Second method: +\begin{itemize} +\item Generate $U_1,U_2,...U_n \in \mathcal{U}(0,1)$. +\item $X=\max\lbrace U_1,U_2,...U_n \rbrace$ has \pdf~of $f_1$. +\item $Y=\min\lbrace U_1,U_2,...U_n \rbrace$ has \pdf~of $f_2$. +\end{itemize} + +\ARROW E7.4 Generate the $f_1$ and $f_2$ \pdf~ with two methods. +} + + \end{footnotesize} + +\end{frame} + + + + + \backupbegin