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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}
\newcommand*{\QEDA}{\hfill\ensuremath{\blacksquare}}%
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\begin{document}
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{
\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 {Specific \pdf~generation}
\end{column}
\begin{column}{0.2\textwidth}
%\includegraphics[width=\textwidth]{SHiP-2}
\end{column}
\end{columns}
\end{center}
\quad
\vspace{3em}
\begin{columns}
\begin{column}{0.44\textwidth}
\flushright \vspace{-1.8em} {\fontspec{Trebuchet MS} \Large Marcin Chrząszcz\\\vspace{-0.1em}\small \href{mailto:mchrzasz@cern.ch}{mchrzasz@cern.ch}}
\end{column}
\begin{column}{0.53\textwidth}
\includegraphics[height=1.3cm]{uzh-transp}
\end{column}
\end{columns}
\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, \\ 14 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{Exponential \pdf }
\begin{footnotesize}
% \begin{exampleblock}{~}
\ARROWR The $X(\theta, \lambda)$:\\
\begin{align*}
\rho_{\theta , \lambda}=\frac{1}{\lambda} e^{- \frac{x- \theta}{\lambda}}
\end{align*}
\ARROWR One can transform the variable:
\begin{align*}
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*}
% \end{exampleblock}
\begin{columns}
\column{0.1in}
{~}\\x
\column{3.5in}
\begin{block}{ \begin{footnotesize}Reverting the \cdf \end{footnotesize}}
\begin{align*}
X^{\prime} = - \ln R,~R \in\mathcal{U}(0,1),~~~\Rightarrow X=\lambda X^{\prime} + \theta
\end{align*}
\end{block}
\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}
\end{footnotesize}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}\frametitle{Gaussian \pdf }
\begin{footnotesize}
\only<1>{
% \begin{exampleblock}{~}
\ARROW The \pdf:
\begin{align*}
\phi_{\mu,\sigma}(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}~~ - \infty <x < \infty
\end{align*}
\ARROW Now we can always transform the variables:
\begin{align*}
x \to x'=(x-\mu)/\sigma~~ \Rightarrow~~ N(\mu,\sigma) \to N(0,1)
\end{align*}
\ARROW First method of based on Central limit theorem. See Lecture 2.\\ Bad for the tails.\\
\ARROW Reverting the \cdf
\begin{itemize}
\item In 1 dim the \cdf~is not revertible :( One can use an approximation (Odeh,Evans 1974):
\end{itemize}
\begin{columns}
\column{3in}
\begin{align*}
\Phi^{-1}(u)=\begin{cases}
g(u),~~~~~~~10^{-20}<u<0.5\\
-g(1-u)~~~0.5<u<1-10^{-20}
\end{cases}
\end{align*}
\column{2in}
\begin{align*}
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*}
}
\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 <x,y < \infty
\end{align*}
\begin{itemize}
\item We change the coordinates: $(x,y)=(r\cos \phi,r \sin \phi)$
\item And we factorize: $\widehat{\rho}(\phi,r)=f(\phi)g(r)$, where $f(\phi)=\frac{1}{2 \pi}$,~$g(r)=re^{\frac{-r^2}{2}}$.
\item The angles is generated flat: $\mathcal{U}(0,2\pi)$ and the $r$ with reverting the \cdf.
\end{itemize}
\ARROW If $U_1,U_2 \in \mathcal{U}(0,1)$:
\begin{align*}
x=\sqrt{-2 \ln U_1}\cos (2\pi U_2)\\
y=\sqrt{-2 \ln U_1}\sin (2\pi U_2)
\end{align*}
\ARROW Accurate and simple to use.\\
\ARROW Time consuming calculations of trigonometrical and logarithm function.
}
\only<3>
{
\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 $W>1$ 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{Breit-Wigner \pdf }
\begin{footnotesize}
\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 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 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{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.
}
\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~$b(n,p)$ :
\begin{align*}
\mathcal{P}\lbrace X=m\rbrace ={n \choose m} p^m (1-p)^{n-m},~~~~m=0,1,2,...,n.
\end{align*}
\ARROWR The interpretation: number of success with the probability $p$.\\
\ARROW The algorithm (buffon needle):\\
\includegraphics[width=0.45\textwidth]{images/buff.png}\\
\ARROW It requires many ''trials''
}
\only<2>
{
\ARROW If $n$ is large, we can use a discrete \pdf :
\begin{align*}
p_k=\sum_{i=0}^k \mathcal{P}\lbrace X=i\rbrace
\end{align*}
and use the algorithm:\\
\includegraphics[width=0.45\textwidth]{images/buff2.png}
\begin{exampleblock}{Theory hack:}
If $n$ is big one can write it in a form: $n=kl$, where $l$ is NOT a big number. In this case one can generate $k$ numbers from distribution $b(l,p)$ and calculate $m$ as sum of the generated numbers.
\end{exampleblock}
}
\only<3>
{
\begin{exampleblock}{Theory:}
If $U \in \mathcal{U}(0,1)$ then:
\begin{align*}
Y=\Theta(p-U)~~~~~~~~V=\min \lbrace \frac{U}{p},\frac{1-U}{1-p} \rbrace
\end{align*}
are independent and $V \in \mathcal{U}(0,1)$.
\end{exampleblock}
\ARROW This is super nice! We can treat $Y$ as the indicator of success in the Bernoulli trials. And have a new random variable :)\\
\begin{center}
\includegraphics[width=0.45\textwidth]{images/bern.png}
\end{center}
\ARROW E7.5 Please code the above mention Bernaulli \pdf~generation.
}
\end{footnotesize}
\end{frame}
\begin{frame}\frametitle{Poisson \pdf }
\begin{footnotesize}
\ARROW The \pdf~$P(\lambda)$:
\begin{align*}
\mathcal{P}(X=n)=\frac{\lambda^n}{n!}e^{-\lambda},~~~n=0,1,2,...
\end{align*}
\begin{exampleblock}{Theory:}
If $\epsilon_1$, $\epsilon_2$, $\epsilon_3$,..., are from $E(0,1)$ then the random variable:
\begin{align*}
X=\min \lbrace k:\sum_{i=0}^k \epsilon_i > \lambda \rbrace
\end{align*}
has the distribution of $P(\lambda)$.
\end{exampleblock}
\only<1>{
\ARROW The algorithm:\\
\begin{center}
\includegraphics[width=0.45\textwidth]{images/al1.png}
\end{center}
}
\only<2>{
\ARROW The algorithm 2:\\
\begin{center}
\includegraphics[width=0.45\textwidth]{images/al2.png}
\end{center}
}
%\ARROW E7.6 Code the $P(\lambda)$ generation.
\end{footnotesize}
\end{frame}
\begin{frame}\frametitle{Poisson \pdf }
\begin{footnotesize}
\ARROW Reverting the \cdf :
\begin{center}
\includegraphics[width=0.4\textwidth]{images/al3.png}
\end{center}
\ARROW It has problem with large values of $\lambda$, at you need many generations which causes numerical instabilities.\\
\ARROW E7.6 Implement the abovementioned ways of generating $P(\lambda)$.
\end{footnotesize}
\end{frame}
\begin{frame}\frametitle{Geometric \pdf }
\begin{footnotesize}
\ARROW The \pdf of $G(p)$ :
\begin{align*}
\mathcal{P}(X=n)=(1-p) p^n,~~~~n=0,1,2,3...
\end{align*}
\begin{exampleblock}{Theorem:}
If a random variable has a \pdf~of
\begin{align*}
f_{\alpha}(x)=\alpha e^{-\alpha x}
\end{align*}
then $\lfloor x \rfloor$ has a geometric~\pdf:
\begin{align*}
G(e^{-\alpha})
\end{align*}
\end{exampleblock}
\ARROW Algorithm:
\begin{enumerate}
\item Generate a number $U$ from $\mathcal{U}(0,1)$
\item Calculate $X=\lfloor \ln U / \ln p \rfloor$
\end{enumerate}
\ARROW E7.7 Implement the above algorithm.
\end{footnotesize}
\end{frame}
\begin{frame}\frametitle{Equal division of interval}
\begin{footnotesize}
\ARROW The method of equal division of an $(0,1)$ interval (the \pdf ):
\begin{align*}
\mathcal{P}(X=k)=p_k,~~~~k=1,2,3...,K
\end{align*}
\ARROW Some times the inverting the \cdf~might be slow. This happens for large values of $K$. \\
\ARROW A more efficient method: \\
\begin{itemize}
\item The interval $(0,1)$ we divide in $K+1$ bins: $(\frac{i-1}{K+1}, \frac{i}{K+1})$, which are equal size and we number them: $1,2,...,K+1$.
.
\item The random variable $U \in \mathcal{U}(0,1)$ falls into bin $\lfloor (K+1)U \rfloor$.
\item We create a sequence: $ q_j =\sum_{k=0}^j p_k$, $j=0,1,....,K$.
\item And a companioning one: $g_j = \max \lbrace j:q_j < \frac{i}{K+1} \rbrace$, $i=0,1,2,...$
\end{itemize}
\begin{center}
\includegraphics[width=0.45\textwidth]{images/al4.png}
\end{center}
\end{footnotesize}
\end{frame}
\begin{frame}\frametitle{Multidimensional generation}
\begin{footnotesize}
\ARROW Let $\overrightarrow{X}$ be a $m$ dimensional variable with a \pdf~of $f(x_1,x_2,x_3,...,x_m)$.\\
\ARROW To generate a \pdf like that we use the elimination method. \\
\ARROW The problem with this is that for large dimensions we can have problems :(\\
\ARROW Example:\\
\begin{itemize}
\item Generate a flat \pdf~on the hyper circle $K_m(0,1)$ with the accept reject method.
\item The probability of accepting event:
\begin{align*}
p_m=\pi^{m/2} / \left[ 2^m \Gamma(m/2+1) \right]
\end{align*}
\end{itemize}
\begin{center}
\begin{tabular}{||c|c|c||}
\hline \hline
$m$ & $p_m$ & $N_m=1/p_m$ \\ \hline
$2$ & $7.854 \cdot 10^{-1}$ & $1.27$\\
$5$ & $1.645 \cdot 10^{-1}$ & $6.08$\\
$10$ & $2.490 \cdot 10^{-3}$ & $4.015 \cdot 10^2$ \\
$20$ & $2.461 \cdot 10^{-8}$ & $4.063 \cdot 10^7$ \\
$50$ & $1.537 \cdot 10^{-28}$ & $6.507 \cdot 10^{28}$ \\ \hline \hline
\end{tabular}\\{~}\\
\end{center}
\ARROW Good luck simulating $10^{28}$ points ;)
\end{footnotesize}
\end{frame}
\begin{frame}\frametitle{Multidimensional generation}
\begin{footnotesize}
\only<1>{
\ARROW Uniform distribution on a simplex:
\begin{exampleblock}{Theorem:}
If $U_1,U_2,...,U_m~\in~\mathcal{U}(0,1)$ and $U_{1:m},U_{2:m},...,U{m:m}$. The a random variable:
\begin{align*}
X_1=U_{1:m},~X_2=U_{2:m}-U_{1:m},...,X_m=U_{m:m}-U_{m-1:m}
\end{align*}
has a uniform distribution on a simplex:
\begin{align*}
W_m=\lbrace (x_1,x_2,...,x_m): \sum_{j=1}^m x_j \leq 1,~x_j\geq 0,~j=1,2,...,m
\end{align*}
\end{exampleblock}
}
\only<2>{
\ARROW Uniform distribution on a simplex surface:
\begin{exampleblock}{Theorem:}
If $U_1,U_2,...,U_{m-1}~\in~\mathcal{U}(0,1)$ and $U_{1:m-1},U_{2:m-1},...,U{m-1:m-1}$. The a random variable:
\begin{align*}
X_1=U_{1:m-1},~X_{m-1}=U_{m-1:m-1}-U_{m-2:m-1},~X_m=1-U_{m-1:m-1}
\end{align*}
has a uniform distribution on a simplex surface:
\begin{align*}
W_m=\lbrace (x_1,x_2,...,x_m): \sum_{j=1}^m x_j = 1,~x_j \geq 0,~j=1,2,...,m
\end{align*}
\end{exampleblock}
}
\end{footnotesize}
\end{frame}
\backupbegin
\begin{frame}\frametitle{Backup}
\end{frame}
\backupend
\end{document}
mchrzasz