poisson

How to calculate a Poisson-Matrix in Matlab analogous to Excel or R and sum up certain parts?

由 邮差的信 提交于 2019-12-11 17:39:36
问题 I've already posted a similar question under this link but it hasn't worked maybe due to the fact that I've expressed myself not clearly enough or I was not able to comprehend the answer. However this question is more precise since if I've realised my project using R (R Studio). Unfortunately I've recognised that R Studio calculates my Poisson-Matrix as slow as Excel that's why I am forced to realise my project using Matlab. Code in R: a <- Table$ValueA b <- Table1$ValueB lapply(a,function(a)

Calculating the expected number of zeros under Poisson and Negative Binomial distributions in R

由 假装没事ソ 提交于 2019-12-11 10:39:30
问题 This question was migrated from Cross Validated because it can be answered on Stack Overflow. Migrated 5 years ago . I know the maths, but how to calculate the expected number of zeros under Poisson and Negative Binomial distributions in R? Is there any function to do so? 回答1: You say you know the mathematics, in which case you already know you're calculating n p(0) , right? You already know n . So isn't your question "How do I evaluate p(0) = P(X=0) for a Poisson or negative binomial in R?"

Poisson density curve, histogram and shaded area with ggplot2?

由 青春壹個敷衍的年華 提交于 2019-12-11 07:36:20
问题 How to make: a density curve and histogram displaying a poisson distribution with lambda = 2.5; and a density curve with shaded area showing P(X >= 4 with lambda = 2.5) the x axis should be 0 to 10 回答1: Poisson distribution is a discrete probability distribution (function is defined only at integer values). So instead of a line it is better represented with points at integer values. To color a specific range under a function one can use geom = "area" and xlim = c(min(range), max(range) :

zeroinfl “system is computationally singular” whereas no correlation in predictors

由 蹲街弑〆低调 提交于 2019-12-11 03:17:32
问题 I am trying to model count data on the number of absence days by worker in a year (dependant variable). I have a set of predictors, including information about workers, about their job, etc..., and most of them are categorical variables. Consequently, there is a large number of coefficient to estimate (83), but as I have more than 600 000 rows, I think it should not be problematic. In addition, I have no missing values in my dataset. My dependant variable contains lot of zero values, so I

Add simulated poisson distributions to a ggplot

由 限于喜欢 提交于 2019-12-10 15:53:50
问题 I have made a poisson regression and then visualised the model: library(ggplot2) year <- 1990:2010 count <- c(29, 8, 13, 3, 20, 14, 18, 15, 10, 19, 17, 18, 24, 47, 52, 24, 25, 24, 31, 56, 48) df <- data.frame(year, count) my_glm <- glm(count ~ year, family = "poisson", data = df) my_glm$model$fitted <- predict(my_glm, type = "response") ggplot(my_glm$model) + geom_point(aes(year, count)) + geom_line(aes(year, fitted)) Now I want to add these simulated Poisson distributions to the plot:

How do I generate discrete random events with a Poisson distribution?

由 浪子不回头ぞ 提交于 2019-12-09 10:11:46
问题 I'm aware of Knuth's algorithm for generating random Poisson distributed numbers (below in Java) but how do I translate that into calling a method, generateEvent() , randomly over time? int poissonRandomNumber(int lambda) { double L = Math.exp(-lambda); int k = 0; double p = 1; do { k = k + 1; double u = Math.random(); p = p * u; } while (p > L); return k - 1; } 回答1: If you are looking to simulate the inter-event arrival time, you want the exponential distribution. Take a look at Pseudorandom

Poisson equation using FFTW with rectanguar domain

由 天大地大妈咪最大 提交于 2019-12-08 19:15:40
I am trying to solve Poisson equation by using FFTW library with rectangular domain(-4<=x<=4 and -2<=y<=2). I have correct result if domain is square and it is wrong if domain is rectangular. Please give me some suggestion. Thank you so much. Here is my code. #include <stdio.h> #include <math.h> #include <cmath> #include <fftw3.h> #include <iostream> #include <vector> using namespace std; int main() { int N1=64; int N2=32; double pi = 3.141592653589793; double L1 = 8.0; double dx = L1/(double)(N1-1); double L2= 4.0; double dy=L2/(double)(N2-1); std::vector<double> in1(N1*N2,0.0); std::vector

Fit poisson distribution to data

由 旧巷老猫 提交于 2019-12-08 07:42:34
问题 I have plotted a histogram and would like to fit a poisson distribution to the histogram. To do this, I have passed the x and y histogram coordinate vector to the poissfit() function to estimate lambda. For example, here is what I've done: expecteddata = cat(2,x,y) [l,lci] = poissfit(expecteddata) My output looks like so: l = 44.3766 0.0130 lci = 42.8887 0.0003 45.8645 0.0724 I'm assuming the the lambda I'm interested in for plotting would be 0.013 (I think my lambda is so small because my

Poisson equation using FFTW with rectanguar domain

由 北城余情 提交于 2019-12-08 06:54:53
问题 I am trying to solve Poisson equation by using FFTW library with rectangular domain(-4<=x<=4 and -2<=y<=2). I have correct result if domain is square and it is wrong if domain is rectangular. Please give me some suggestion. Thank you so much. Here is my code. #include <stdio.h> #include <math.h> #include <cmath> #include <fftw3.h> #include <iostream> #include <vector> using namespace std; int main() { int N1=64; int N2=32; double pi = 3.141592653589793; double L1 = 8.0; double dx = L1/(double

hand-rolled R code for Poisson MLE

由 我怕爱的太早我们不能终老 提交于 2019-12-08 04:46:24
问题 I'm attempting to write my own function to understand how the Poisson distribution behaves within a Maximum Likelihood Estimation framework (as it applies to GLM). I'm familiar with R's handy glm function, but wanted to try and hand-roll some code to understand what's going on: n <- 10000 # sample size b0 <- 1.0 # intercept b1 <- 0.2 # coefficient x <- runif(n=n, min=0, max=1.5) # generate covariate values lp <- b0+b1*x # linear predictor lambda <- exp(lp) # compute lamda y <- rpois(n=n,
订阅 poisson