similar to: GCV in lm.ridge (MASS) (PR#10755)

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 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)

Hi, I'm a beginner in the field, I have to perform the ridge regression with lm.ridge for many datasets, and I wanted to do it in an automatic way. In which way I can automatically choose lambda ? As said, right now I'm using lm.ridge MASS function, which I found quite simple and fast, and I've seen that among the returned values there are HKB estimate of the ridge constant and L-W

Hi all, I have run a ridge regression on a data set 'final' 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

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. > >

I encounter such problem with lmer under FreeBSD, but not under Windows. Anyone knows why? Thanks. > example(lmer) lmer> (fm1 <- lmer(Reaction ~ Days + (Days|Subject), sleepstudy)) Error in UseMethod("as.logical") : no applicable method for "as.logical" > traceback() 9: as.logical(EMverbose) 8: as.logical(EMverbose) 7: lmerControl() 6:

I am using gam in mgcv (1.3-22) and trying to use gcv to help with model selection. However, I'm a little confused by the process of assessing GCV scores based on their magnitude (or on relative changes in magnitude). Differences in GCV scores often seem "obvious" with my poisson gams but with negative binomial, the decision seems less clear. My data represent a similar pattern as

I am trying to understand the code for lm.ridge from the MASS package. Here is the part I am having trouble understanding: if(Inter <- attr(Terms, "intercept")) { Xm <- colMeans(X[, -Inter]) Ym <- mean(Y) p <- p - 1 X <- X[, -Inter] - rep(Xm, rep(n, p)) Y <- Y - Ym } else Ym <- Xm <- NA Xscale <- drop(rep(1/n, n) %*% X^2)^0.5 X <- X/rep(Xscale, rep.int(n,

Hi, I've been using gamm4 to build GAMMs for exploring environmental influences on genetic ancestry. Things have gone well and I have 2 very straightforward questions: 1. I've used method=REML. Am I correct that this is an alternative method for estimating the smooth functions in GAMMs rather than GCV that is often used for GAMs? I've read up on REML and it makes sense, but I'm

Hello, I have a question regarding model selection and dropping of terms for GAMs fitted with package mgcv. I am following the approach suggested in Wood (2001), Wood and Augustin (2002). I fitted a saturated model, and I find from the plots that for two of the covariates, 1. The confidence interval includes 0 almost everywhere 2. The degrees of freedom are NOT close to 1 3. The partial

Hi, I had a questions about doing ridge regression in R. Why is it that when I try this on datasets with more predictors than samples (p>n) using lambda=0, it still finds coefficients for all predictors? I thought when lambda=0, it should be like ordinary regression and therefore not find coefficients for all due to singularity? I would greatly appreciate your help. Thank you, --James K.

Hi all, I have run a ridge regression as follows: reg=lm.ridge(final$l~final$lag1+final$lag2+final$g+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

hello to everybody: I would to know with ranges of GCV or UBRE values can be considered as adequate to consider a GAM as correct Thanks in advance -- David Nogu?s Bravo Functional Ecology and Biodiversity Department Pyrenean Institute of Ecology Spanish Research Council Av. Monta?ana 1005 Zaragoza - CP 50059 976716030 - 976716019 (fax)

I received an assignment that I have to do in R, but I'm absolutely not very good at it. The task is the following: http://www.nabble.com/file/p22804957/question8.jpg To do this, we also get the following pieces of code (not in correct order): http://www.nabble.com/file/p22804957/hints.jpg I'm terrible at this and I'm completely stuck. The model I chose can be found in here:

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

For an application of ridge regression, I need to get the covariance matrices of the estimated regression coefficients in addition to the coefficients for all values of the ridge contstant, lambda. I've studied the code in MASS:::lm.ridge, but don't see how to do this because the code is vectorized using one svd calculation. The relevant lines from lm.ridge, using X, Y are:

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

It seems to me that summary for ridge coxph() prints summary but returns NULL. It is not a big issue because one can calculate statistics directly from a coxph.object. However, for some reason the score test is not calculated for ridge coxph(), i.e score nor rscore components are not included in the coxph object when ridge is specified. Please find the code below. I use 2.9.2 R with 2.35-4 version

Hi, I have optimized the shrinkage parameter (GCV)for ridge and got my r2 value is 70% . to check the sensitivity of the result, I did permutation test. I permuted the response vector and run for 1000 times and draw a distribution. But now, I get r2 values highest 98% and some of them more than 70 %. Is it expected from such type of test? *I was under impression that, r2 with real data set

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