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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