search for: possion

Make a plot in R where you compare the probability distributions for the Binomial distributions with N=1000 trials and succes probability 0.005 and the Possion Distribution with rate lambda=5. My answer is b<-binom(x, 1000, 0.005, log = FALSE) plot(b) p<-dpois(x, lambda, log = FALSE) plot(p) *Is this the right way to solve the problem?* -- View this message in context: http://r.789695.n4.nabble.com/Bionomial-and-possion-tp4641531.html Sent f...

...ad a book from WOOD, there's an example which is talking about the pollutant. library(gamair) library(mgcv) y<-gam(death~s(time,bs="cr",k=200)+s(pm10median,bs="cr")+s(so2median,bs="cr")+s(o3median,bs="cr")+s(tmpd,bs="cr"),data=chicago,family=Possion) lag.sum<-function(a,10,11) {n<-length(a) b<-rep(0,n-11) for(i in 0:(11-10)) b<-b+a[(i+1):(n-11+i)] b} death<-chicago$death[4:5114] time<-chicago$time[4:5114] o3<-lag.sum(chicago$o3median,0,3) tmp<-lag.sum(chicago$tmpd,0,3) pm10<-lag.sum(log(chicago$pm10median+40),0,3)...

Dear all, I would like to model the relationship between y and x. y is binary variable, and x is a count variable which may be possion-distribution. I think it is better to divide x into intervals and change it to a factor before calling glm(y~x,data=dat,family=binomail). I try to use rpart. As y is binary, I use "class" method and get the following result. > rpart(y~x,data=dat,method="class") n=778 (22 ob...

Hi - I have a likelihood function that involves sums of two possions: L = a*dpois(Xi,theta1)*dpois(Yi,theta2)+b*(1-c)*a*dpois(Xi,theta1+theta3)*dpois(Yi,theta2) where a,b,c,theta1,theta2,theta3 are parameters to be estimated. (Xi,Yi) are observations. However, Xi and Yi are usually big (> 20000). This causes dpois to returns 0 depending on values of theta1, th...

A quick question! The number of episodes per year of otitis media follows a Possion distribution with lambda = 1.6 episodes per year. Wouldn't the probability of getting 3 or more episodes of otitis media in the first 2 years of life be: > ppois(q=3, lambda=1.6*2, lower.tail = TRUE, log.p = FALSE) [1] 0.6025197 I am confused with the lambda and 3 or more.. thx much

...dvance. -Sean #data simulation (modified from code at http://markmail.org/message/j3zmgrklihe73p4p) set.seed(100) m <- 5 n <- 100 N <- n*m #X <- cbind(1,runif(N)) X <- cbind(1,rnorm(N)) X <- cbind(runif(N),rnorm(N)) id <- rep(1:n,each=m) # Z <- kronecker(diag(n),rep(1,m)) #Possion with group level heterogeneity z <- rpois(N, exp(X%*%matrix(c(1,2)) + Z%*%matrix(rnorm(n)))) #2*rnorm(n*m) is added to each observation to create overdispersion z.overdis <- rpois(N, exp(X%*%matrix(c(1,2)) + Z%*%matrix(rnorm(n)) + 2*rnorm(n*m))) #without observation-level random shock i.e.,...

...dvance. -Sean #data simulation (modified from code at http://markmail.org/message/j3zmgrklihe73p4p) set.seed(100) m <- 5 n <- 100 N <- n*m #X <- cbind(1,runif(N)) X <- cbind(1,rnorm(N)) X <- cbind(runif(N),rnorm(N)) id <- rep(1:n,each=m) # Z <- kronecker(diag(n),rep(1,m)) #Possion with group level heterogeneity z <- rpois(N, exp(X%*%matrix(c(1,2)) + Z%*%matrix(rnorm(n)))) #2*rnorm(n*m) is added to each observation to create overdispersion z.overdis <- rpois(N, exp(X%*%matrix(c(1,2)) + Z%*%matrix(rnorm(n)) + 2*rnorm(n*m))) #without observation-level random shock i.e.,...

Hi I'm doing a glm analysis and I have two doubts (at least :) 1) When I run the function it gives a lot of warnings (see below) what they mean ? (may be I'm ignorant about this analysis ...) glm.poisson<-glm(log(Jkij+1)~fac.ano+fac.tri+fac.icesr+fac.mat+fac.ano:fac.icesr+fac.ano:fac.tri,family=poisson()) warnings() 40: non-integer x = 1.252763 41: non-integer x = 1.864785 42:

Hello, from a search of the archives and functions, I am looking for information on creating random correlated counts from a multivariate Poisson distribution.  I can not seem to find a function that does this. Perhaps, it has not yet  been created. Has anyone created an R package that does this.   thanks,   Jourdan Gold     [[alternative HTML version deleted]]