similar to: log likelihood

What objective function is maximized by survreg with the default Weibull model? I'm getting finite parameters in a case that has the likelihood maximzed at Infinite, so it can't be a simple maximum likelihood. Consider the following: ############################# > set.seed(3) > Stress <- rep(1:3, each=3) > ch.life <- exp(9-3*Stress) > simLife <- rexp(9,

Hello! I have a matrix with data and a column indicating whether it is censored or not. Is there a way to apply weibull and exponential maximum likelihood estimation directly on the censored data, like in the paper: Backtesting Value-at-Risk: A Duration-Based Approach, P Chrisoffersen and D Pelletier (October 2003) page 8? The problem is that if I type out the code as below the likelihood

Hi, Is it normal to get intercept in the list of covariates in the output of survreg function with standard error, z, p.value etc? Does it mean that intercept was fitted with the covariates? Does Value column represent coefficients or some thing else? Regards, ------------------------------------------------- tmp = survreg(Surv(futime, fustat) ~ ecog.ps + rx, ovarian,

Hi all, I'm using the function "fitdistr" (library MASS) to fit a distribution to given data. What I have to do further, is getting the log-Likelihood-Value from this estimation. Is there any simple possibility to realize it? Regards, Carsten

I was wondering if someone familiar with survival analysis can help me with the following. I would like to fit a Weibull curve, that may be dependent on a covariate, my dataframe "labdata" that has the fields "cov", "time", and "censor". Do I do the following? wieb<-survreg(Surv(labdata$time, labadata$censor)~labdata$cov,

Dear R-users, I need to use the aftreg function in package 'eha' to estimate failure times for left truncated survival data. Apparently, survreg still cannot fit such models. Both functions should be fitting the accelerated failure time (Weibull) model. However, as G?ran Brostr?m points out in the help file for aftreg, the parameterisation is different giving rise to different

Hi All, I have conducted the following survival analysis which appears to be OK (thanks BRipley for solving my earlier problem). > surv.mod1 <- survreg( Surv(timep1, relall6)~randgrpc, data=Dataset, dist="weibull", scale = 1) > summary(surv.mod1) Call: survreg(formula = Surv(timep1, relall6) ~ randgrpc, data = Dataset, dist = "weibull", scale = 1)

Dear R-users I have some datasets, all left-censoring, and I would like to fit distributions to (weibull,exponential, etc..). I read one solution using the function survreg in the survival package. i.e survreg(Surv(...)~1, dist="weibull") but it returns only the scale parameter. Does anyone know how to successfully fit the exponential, weibull etc... distributions to left-censoring

Dear R-experts: I am using survreg() to estimate the parameters of a Weibull density having right-censored observations. Some observations are weighted. To do that I regress the weighed observations against a column of ones. When I enter the data as 37 weighted observations, the parameter estimates are exactly the same as when I enter the data as the corresponding 70 unweighted observations.

Hi, I'm dealing with wind data and I'd like to model their distribution in order to simulate data to fill-in missing values. Wind direction are typically following a vonmises distribution and wind speeds follow a weibull distribution. I'd like to build a joint distribution of directions and speeds as a VonMises-Weibull bivariate distribution. First is this a stupid question? I'm

Dear all, I'd like to discuss about a possible bug in function StructTS of stats package. It seems that the function returns wrong value of the log-likelihood, as the added constant to the relevant part of the log-likelihood is misspecified. Here is an simple example: > data(Nile) > fit <- StructTS(Nile, type = "level") > fit$loglik [1] -367.5194 When computing the

Dear list, I would like to simulate individual survival times from a model that has been fitted using the survreg procedure (library survival). Output shown below. My plan is to extract the shape and scale arguments for use with rweibull() since my error terms are assumed to be Weibull, but it does not make any sense. The mean survival time is easy to predict, but I would like to simulate

hi Suppose I have a mixture of 2 distributions generated by rtwonormals <- function(npnt,m1,s1,m2,s2,p2){ rv<-vector(npnt,mode="numeric") for( i in seq(1:npnt)){ if(runif(1,0,1)<=p2){ rv[i]<-rnorm(1,m2,s2) } else{ rv[i]<-rnorm(1,m1,s1) } } return(rv) } x <- rtwonormals(50000,0,100,500,500,0.05) #and I try to fit these with (based on thread: [R]

Hello, is there a quick way of estimating Weibull parameters for some data points that are assumed to be Weibull-distributed? I guess I'm just too lazy to set up a Maximum-Likelihood estimation... ...but maybe there is a simpler way? Thanks for any hint (and yes, I've read help(Weibull) ;) Kaspar Pflugshaupt -- Kaspar Pflugshaupt Geobotanical Institute ETH Zurich, Switzerland

Hi! Recently I try to find the method maximum likelihood for gamma,weibull,Pearson type III,Kappa Distribution, mixed exponential distribution, skew distribution. I have tried function ms() for gamma two parameters and weibull two parameters.It works but not for Pearson type III. I have problem to find the likelihood function for mixed exponential distribution and kappa distribution. So can

Two related questions. First, I am fitting a model with a single predictor, and then a null model with only the intercept. In theory, the fitted model should have a higher log-likelihood than the null model, but that does not happen. See the output below. My first question is, how can this happen? > m Call: glm(formula = school ~ sv_conform, family = binomial, data = dat, weights =

Can some kind person point out my error here? I'm trying to set up a grid for a countour plot of a likelihood function. > u <- rnorm(20,9.5,2.5) > # sample of size 20 from N(9.5,2.5^2) > loglik <- function(th1,th2) { + n <- length(u) + -(n/2)*log(2*pi*th2^2)-0.5*sum((u-th1)^2/th2^2) + } > x <- seq(4.5,14.5,len=50) > y <- seq(0.5,6,len=50) > f <-

Hello, I'm trying to calculate the Maximum likelihood estimators for a dataset which contains censored data. I started by using the function "nlm", but isn't there a separate method for doing this for e.g. the "weibull" and the "log-normal" distribution? Thanks, Olivia [[alternative HTML version deleted]]

It seems to me that R returns the unpenalized log-likelihood for the ratio likelihood test when ridge regression Cox proportional model is implemented. Is this as expected? In the example below, if I am not mistaken, fit$loglik[2] is unpenalized log-likelihood for the final estimates of coefficients. I would expect to get the penalized log-likelihood. I would like to check if this is as expected.

Hi there, I'm working with clustered data sets and trying to calculate log-likelihood (and/or AIC, AICc) for my models. In using the gee and geese packages one gets Wald test output; but apparently there is no no applicable method for "logLik" (log-likelihood)calculation. Is anyone aware of a way to calculate log-likelihood for GEE models? Thanks for the help, Bruce