similar to: Poisson Ridge Regression

Dear everyone, I'm student of Masters in Statistics (Actuarial) from Central University of Rajasthan, India. I am doing a major project work as a part of the degree. My major project deals with fitting a glm model for the data of car insurance. I'm facing the problem of multicollinearity for this data which is visible by the plotting of data. But I'm not able to test it. In the case

Dear everyone I'm dealing with a problem related to Poisson Ridge Regression. If anyone can help me in this regard by telling if any changes in the source code of "glm.fit" may help -- Regards Umesh Khatri [[alternative HTML version deleted]]

Hi list, When I type my question in R console using ? sign (example: ?par, when I want to query for par), the following error message popps up: Error in print.help_files_with_topic("C:/PROGRA~1/R/R-24~1.1/library/maps/chm/map") : CHM file could not be displayed I appreciate if any body comes back to me with the solution. Regards, Alireza [[alternative HTML version deleted]]

Dear all, I want to get the likelihood (or AIC or BIC) of a ridge regression model using lm.ridge from the MASS library. Yet, I can't really find it. As lm.ridge does not return a standard fit object, it doesn't work with functions like e.g. BIC (nlme package). Is there a way around it? I would calculate it myself, but I'm not sure how to do that for a ridge regression. Thank you in

Hello, I have questions regarding penalized Cox regression using survival package (functions coxph() and ridge()). I am using R 2.8.0 on Ubuntu Linux and survival package version 2.35-4. Question 1. Consider the following example from help(ridge): > fit1 <- coxph(Surv(futime, fustat) ~ rx + ridge(age, ecog.ps, theta=1), ovarian) As I understand, this builds a model in which `rx' is

Hi, I am working on a project about hospital efficiency. Due to the high multicolinearlity of the data, I want to fit the model using ridge regression. However, I believe that the data from large hospital(indicated by the number of patients they treat a year) is more accurate than from small hosptials, and I want to put more weight on them. How do I do this with lm.ridge? I know I just need

Hello, I have posted this mail a few days ago but I did it wrong, I hope is right now: I have the following doubts related with lm.ridge, from MASS package. To show the problem using the Longley example, I have the following doubts: First: I think coefficients from lm(Employed~.,data=longley) should be equal coefficients from lm.ridge(Employed~.,data=longley, lambda=0) why it does not happen?

Hi all, I have run a ridge regression as follows: reg=lm.ridge(final$l~final$lag1+final$lag2+final$g+final$u, lambda=seq(0,10,0.01)) Then I enter : select(reg) and it returns: modified HKB estimator is 19.3409 modified L-W estimator is 36.18617 smallest value of GCV at 10 I think it means that it is advisable to

Dear R users. I have a doubt about the use of the sequence option on Ridge regression. I'm trying to understand the use of this option when variables are highly linear correlated. I'm running a model where the variables HtShoes and Ht have high VIF values. My program is written below, but I'm not sure about the correct way of using the sequence option: library (faraway) data (seatpos)

Dear all, I need to calculate likelihood ratio test for ridge regression. In February I have reported a bug where coxph returns unpenalized log-likelihood for final beta estimates for ridge coxph regression. In high-dimensional settings ridge regression models usually fail for lower values of lambda. As the result of it, in such settings the ridge regressions have higher values of lambda (e.g.

Unlike L1 (lasso) regression or elastic net (mixture of L1 and L2), L2 norm regression (ridge regression) does not select variables. Selection of variables would not work properly, and it's unclear why you would want to omit "apparently" weak variables anyway. Frank maths123 wrote > I have a .txt file containing a dataset with 500 samples. There are 10 > variables. > >

Hi, I'm not sure if this is the proper way to ask questions, sorry if not. But here's my problem: I'm trying to do a bootstrap estimate of the mean for some survival data. Is there a way to specifically call upon the rmean value, in order to store it in an object? I've used print(...,print.rmean=T) to print the summary of survfit, but I'm not sure how to access only rmean

Hi, What is the optimum range to look for a value of lambda while doing ridge regression. Can/ should lambda be greater than 1 ? I have conflicting (or what appears conflicting to me) sources that use lambda >= 0, without any upper limit, but that makes the search space infinite.. right ?? So, perhaps my question is: is there an upper limit to lambda. Does the value of lambda convey

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.

Dear list, I have a couple of questions concerning ridge regression. I am using the lm.ridge(...) function in order to fit a model to my microarray data. Thus *model=lm.ridge(...)* I retrieve some coefficients and some scales for each gene. First of all, I would like to ask: the real coefficients of the model are not included in the first argument of the output but in the result of coef(model),

Dear all, I considered an ordinary ridge regression problem. I followed three different ways: 1. estimate beta without any standardization 2. estimate standardized beta (standardizing X and y) and then again convert back 3. estimate beta using lm.ridge() function X<-matrix(c(1,2,9,3,2,4,7,2,3,5,9,1),4,3) y<-t(as.matrix(cbind(2,3,4,5))) n<-nrow(X) p<-ncol(X) #Without

hi all a technical question for those bright statisticians. my question involves ridge regression. definition: n=sample size of a data set X is the matrix of data with , say p variables Y is the y matrix i.e the response variable Z(i,j) = ( X(i,j)- xbar(j) / [ (n-1)^0.5* std(x(j))] Y_new(i)=( Y(i)- ybar(j) ) / [ (n-1)^0.5* std(Y(i))] (note that i have scaled the Y matrix as well) k is

Hi I have some data (v. large amount) with a (0,1) response where I want to minimise the errors in the betas rather than SS or deviance. So can anyone point me to a ridge regression function or equivalent for such a logistic regression case? John -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read

Dear all, For an ordinary ridge regression problem, I followed three different approaches: 1. estimate beta without any standardization 2. estimate standardized beta (standardizing X and y) and then again convert back 3. estimate beta using lm.ridge() function X<-matrix(c(1,2,9,3,2,4,7,2,3,5,9,1),4,3) y<-as.matrix(c(2,3,4,5)) n<-nrow(X) p<-ncol(X) #Without standardization

Dear all I'm familiarising myself with Ridge Regressions in R and the following is bugging me: How does one get p-values for the coefficients obtained from MASS::lm.ridge() output (for a given lambda)? Consider the example below (adapted from PRA [1]): > require(MASS) > data(longley) > gr <- lm.ridge(Employed ~ .,longley,lambda = seq(0,0.1,0.001)) > plot(gr) > select(gr)