similar to: gamlss for censored and truncated distributions

Dear R helpers, I have following loss data and I need to fit LEFT truncated Log Normal distribution to this data which is Truncated at 1000000. dat = c(1333834,5710254,9987567,7809469,6940935,3473671,1270209,1102523,1124002, 5830159,4302300,3925242,2638409,2324421,7238436,9088709,7439250,4976551,4864319, 8741334,1863770,7098310,4942288,4971829,4986372) library(gamlss.tr) gen.trun(5, LOGNO)

Hi, I'm a bit puzzled by the gamlss fitting of exponential and lognormal data. Gamlss seems to think that exponentially distributed data fits better with a lognormal distribution, and vice versa. For example, X <- rexp(1000) Gexp <- gamlss(X~1,family=EXP) # X~1 is X tilde 1 GAMLSS-RS iteration 1: Global Deviance = 2037.825 GAMLSS-RS iteration 2: Global Deviance = 2037.825 Glno

Hi, I'm playing around with gamlss and don't entirely understand the sigma result from an attempted lognormal fit. In the example below, I've created lognormal data with mu=10 and sigma=2. When I try a gamlss fit, I get an estimated mu=9.947 and sigma=0.69 The mu estimate seems in the ballpark, but sigma is very low. I get similar results on repeated trials and with Normal and

Hi everyone, I''m trying to use function vcov to extract the covariance matrix from a GAMLSS object. But I''m getting some strange errors and I was hoping someone could help me out? Vcov works with the same model for lm and glm objects, but not gamlss objects. I''ve searched various help sites to no avail. Its very possible the reason is that vcov failed though,

Hi, I am running a negative binomial model using Gamlss and when I try to include random effect, I get the following message: Warning messages: 1: In vcov.gamlss(object, "all") : addive terms exists in the mu formula standard errors for the linear terms maybe are not appropriate 2: In vcov.gamlss(object, "all") : addive terms exists in the sigma formula standard

Dear all, I am using gamlss (Package gamlss version 4.0-0, R version 2.10.1, Windows XP Service Pack 3 on a HP EliteBook) to relate bird counts to habit variables. However, most models fail because “the global deviance is increasing” and I am not sure what causes this behaviour. The dataset consists of counts of birds (duck) and 5 habit variables measured in the field (n= 182). The dependent

Hello: I'd like to know how and if the GLM convergence problems are addressed in gamlss. For simplicity, let's focus on Normal and Negative Binomial with log link. The convergence issues of the glm() function were alleviated in 2011 when glm2 package was released. Package gamlss was released in 2012, so it might still use the glm-like solution or call glm() directly. Is that the case or

Hola a tod en s, tengo una duda que la comunidad R me puede ayudar. Estoy trabajando con gamlss, porque tengo una variable respuesta con valores entre 0 y 1 e incluidos estos. La distribución que utilizo com gamlss para este caso es "beta inflated" (Stasinopulos and Rigby 2007. Journal of Statistical Software 23(7)). El modelo que intento correr es: m1<-gamlss(Teleosteos ~

If the code you are running in parallel is complicated, maybe foreach is not sophisticated enough to find all the variables you refer to. Maybe use parallel::clusterExport yourself? But be a aware that passing parameters is much safer than directly accessing globals in parallel processing, so this might just be your warning to not do that anyway. -- Sent from my phone. Please excuse my brevity.

Dear All: I have succeeded in fitting a GAMLSS.dist model to growth data I am working with it. My aim is to create a matrix of predicted percentiles and the corresponding the fitted model's sigma mu nu by agebins. Q: How do it generate these parameters as in L M S per Cole and Green 1992? Here are my working codes. Name of fitted model is gamlssfit > Agebin<-seq(6,36,6)

Hello all: Please help me with this "can't find object" issue. I'm trying to get leave-one-out predicted values for Beta-binomial regression. It may be the gamlss issue because the code seems to work when %do% is used. I have searched for similar issues, but haven't managed to figure it out. This is on Windows 10 platform. Thanks in advance, Nik #

Dear all and Mikis I have the opportunity to compare fits with the 'classical' glm and gamlss and no smoother of any kind just the same model formula (both with the binomial family). I get exactly the same coefficients but very different residuals, gamlss giving residuals which are extremely close to 'normal' and glm very far... How can this be ? Thanks in advance for

Hi, I'm not sure how use curve(dexGAUS(… None of the following four works: rt<- rexGAUS(100, mu=300, nu=100, sigma=35) m1<-gamlss(rt~1, family=exGAUS) curve(dexGAUS(rt=x, mu=300 ,sigma=35,nu=100), 100, 600, main = "The ex- GAUS density mu=300 ,sigma=35,nu=100") curve(dexGAUS(x=rt, mu=300 ,sigma=35,nu=100), 100, 600, main = "The ex- GAUS density mu=300

Dear All, I'd like to set up a loop whereby successive parameter values are used as start values in gamlss (yes I know this isn't usually necessary ! - unfortunately for my truncated data it is), to return the estimated parameters etc. giving the lowest AIC value. I notice that mu.start can take a vector of values - but I can't find any information as to what the function actually

Hi all. In a GLM in which g(mu) = b0 + b1X1 + b2X2 + b3X3 + b4X4 + b5X5 + b6X6 + b7X7, if I want to test if b1 + b5 = b2 + b6, I can use the contrast package or multicomp package. How can I do a similar test if I am fitting a GAMLSS using the gamlss package? Thank you for your help. Gustavo

Hello Mikis: Thanks a lot, it worked. Could you tell me what the problem was? Regards, Nik ----- Original Message ----- From: r-help-request at r-project.org To: "r-help" Sent: Sunday, March 11, 2018 6:00:02 AM Subject: R-help Digest, Vol 181, Issue 11 Send R-help mailing list submissions to r-help at r-project.org To subscribe or unsubscribe via the World Wide Web, visit

Hi Folks! Just started using the gamlss package and I tried a simple code example (see below). Why the negative sigma? John > y <- rt(100, df=1)> m1<-fitDist(y, type="realline")Warning messages:1: In MLE(ll3, start = list(eta.mu = eta.mu, eta.sigma = eta.sigma, : possible convergence problem: optim gave code=1 false convergence (8)2: In MLE(ll4, start = list(eta.mu =

Hello! I've just build one-inflated beta regression model using package GAMLSS. It all worked very nicely but now I want to make prediction using it. I use typical function predict() give all necessary arguments (my new data is in data frame and all relevant columns have same names as before). Unfortunately it ends with following error: Error in nrow(x) : (subscript) logical subscript too

Dear Nik Try the following code loo_predict.mu <- function(model.obj, input.data) { yhat <- foreach(i = 1 : nrow(input.data), .packages="gamlss", .combine = rbind) %dopar% { updated.model.obj <- update(model.obj, data = input.data[-i, ]) predict(updated.model.obj, what = "mu", data = input.data[-i, ], newdata = input.data[i,], type =

Hi List, Is there anyway to estimate the parameters with integral forms? Here is the function. The idea is to calculate the real mean from the censored mean given normal distribution. I have the function as trun.mean<- function(mu,sigma,t) # t is the threshold { f <- function(x) (1/(sigma*sqrt(2*pi)))*exp(-(x-mu)^2/(2*sigma^2)); pdf.fun <- function(x) x*f(x); mut <-