search for: rnegbin

Dear R users, I'm wanting to sample from the negative binomial distribution using the rnegbin function from the MASS library to create artificial samples for the purpose of doing some power calculations. However, I would like to work with samples that come from a negative binomial distribution that includes only values greater than or equal to 1 (a truncated negative binomial), and I c...

I want to generate 1,000 random samples of sample size=1,000 from ZINB. I know there is a rnegbin() to generate random samples from NB, and I know I can use the following process: do i=1 to 1000 n=0 do i=1 to 1000 if runi(1)>0.1 then x(i) = 0; else x(i)=rnegbin(); n=n+1; if n>1000 then stop; end; output; end; Anybody can help me out with the R code? Thanks very much ahead of tim...

...### mean function logit model p0 <- exp(alpha0 + alpha1*trt)/(1 + exp(alpha0 + alpha1*trt)) ### 0 / 1 based on p0 y1 <- as.numeric(runif(n)>p0) ### mean function count portion mu <- exp(beta0 + beta1*trt) ### estimate counts using NB dist require(MASS, quietly = TRUE) y2 <- rnegbin(n, mu = mu, theta = theta) ### if y2 = 0, draw new value while(sum(y2==0)>0){ y2[which(y2==0)] <- rnegbin(length(which(y2==0)), mu=mu[which(y2==0)], theta = theta) } y<-y1*y2 data.frame(trt=trt,y=y) } #alpha0, alpha1 is the parameter for zero part #beta0,beta1 is the parameter for...

...tionship between the two possible parametrizations of the negative binomial and found that for this fuction I must use: prob = theta/(theta+mu) and size = theta Theta, however, is not an integer. So how can size (which is the number of successes) equal theta? I know there is a function rnegbin which does what I want. I would still like to raise the above question (probably a more statistical one than R). Thanks a lot for any comments. Hadassa [[alternative HTML version deleted]]

Hi, Is there a Zipf-like distribution RNG in R? Thanks, Weiguang

Dear All, I am trying to fit some data both as a negative binomial and a zero inflated binomial. For the first case, I have no particular problems, see the small snippet below library(MASS) #a basic R library set.seed(123) #to have reproducible results x4 <- rnegbin(500, mu = 5, theta = 4) #Now fit and check that we get the right parameters fd <- fitdistr(x4, "Negative Binomial") summary(fd) #and mu and theta are as expected #now I add artificially some zeros x5 <- sample(c(x4,rep(0,100))) However, when I artificially add some zeros (case...

...alysis. Is there any tutorial available? anova.negbin Likelihood Ratio Tests for Negative Binomial GLMs glm.convert Change a Negative Binomial fit to a GLM fit glm.nb Fit a Negative Binomial Generalized Linear Model negative.binomial Family function for Negative Binomial GLMs rnegbin Simulate Negative Binomial Variates theta.md Estimate theta of the Negative Binomial gam.neg.bin GAMs with the negative binomial distribution dnb2 Density for negative binomial, used in mmlcr theta.mmmod Estimate theta of the Negative Binomial by Moments NegBinomial The...

...the F-test was more appropriate when dispersion is estimated)? Secondly, after a bit of muddling with the negative binomial pdf, I concluded that, like for the Poisson, phi is actually 1. This result is borne out by simulations. Is this correct? # an example in R 1.81 with library(MASS) y<-rnegbin(n=100,mu=1,theta=1) x<-1:length(y) model<-glm(y~x,family=neg.bin(1)) summary(model)$dispersion [1] 1.288926 anova(model,test='Chisq") #... Df Deviance Resid. Df Resid. Dev P(>|Chi|) NULL 99 102.038 x 1 0.185 98 101.853...

What does 'with ties in' mean? with some identical elements (par ex., au moins une paire ex-equo) HTH ____________________ Ken Knoblauch Inserm U371, Cerveau et Vision Department of Cognitive Neurosciences 18 avenue du Doyen Lepine 69675 Bron cedex France tel: +33 (0)4 72 91 34 77 fax: +33 (0)4 72 91 34 61 portable: 06 84 10 64 10 http://www.lyon.inserm.fr/371/

...<- with(mydata1, tapply(value, treatments, mean)) now I need the confidence intervals for the means And then again for a negative binomial GLM require(MASS) mydata2 <- data.frame('treatments'=gl(4, 4, 64, LETTERS[1:4]), 'species'=gl(4, 1, 64, letters[1:4]), 'value'=rnegbin(64,1,1)) model2 <- glm.nb(value ~ treatments*species, data=mydata2) means2 <- with(cbind('x'=gl(16,4),mydata2), tapply(value, x, mean)) Can I get the confidence intervals for each mean in means1 and means2, according to the family used? Thanks in advance, all the best! Rafael....

I need to estimate the parameters for negative binomial distribution (pdf) using maximun likelihood, I also need to estimate the parameter for the Poisson by ML, which can be done by hand, but later I need to conduct a likelihood ratio test between these two distributions and I don't know how to start! I'm not an expert programmer in R. Please help -- View this message in context:

...gt; > > anova.negbin > Likelihood Ratio Tests for Negative Binomial GLMs > glm.convert > Change a Negative Binomial fit to a GLM fit > glm.nb > Fit a Negative Binomial Generalized Linear Model > negative.binomial > Family function for Negative Binomial GLMs > rnegbin > Simulate Negative Binomial Variates > theta.md > Estimate theta of the Negative Binomial > gam.neg.bin > GAMs with the negative binomial distribution > dnb2 > Density for negative binomial, used in mmlcr > theta.mmmod > Estimate theta of the Negative Binomial b...

...quot;Alevel1",4),rep("Alevel2",4),rep("Alevel3",4),rep("Alevel4",4)),4),                       "B"=c(rep("Blevel1",16),rep("Blevel2",16),rep("Blevel3",16),rep("Blevel4",16)),                       "value"=rnegbin(64, 10, 10)) # fitting the model with interaction (a + b + a:b) model <- glm.nb(value ~ A*B, data=my.data) anova(model, test="F")      Df Deviance Resid. Df Resid. Dev      F Pr(>F) NULL                    63     80.639 A     3    1.374        60     79.265 0.4581 0.7115 B     3 ...