search for: hkb

...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 estimate of the ridge constant, together with GCV values. I found on the web other studies where people simply choose among one of these quantities. It will be perfect to me to do the same, but how? Which are the decisional criteria, if there are criteria? HK...

Given the following data, I am plotting log.det ~ norm.beta, where the points depend on a parameter, lambda (but there is no functional form). I want to find the (x,y) positions along this curve corresponding to two special values of lambda lambda.HKB <- 0.004275357 lambda.LW <- 0.03229531 and draw reference lines at ~ -45 degrees (or normal to the curve) thru these points. How can I do this? A complete example is below > pd lambda log.det norm.beta 0.000 0.000 -12.92710 3.806801 0.005 0.005 -14.41144 2.819460 0.010 0...

...tes10.txt", header=TRUE) > diabetes10 > library(MASS) > select(lm.ridge(y=diabetes10 ~ age+sex+bmi+map+tc , diabetes10, > lambda = seq(0,0.1,0.0001))) > > First of all, i am confused about the lamda values, > Second of all, my output is: > > modified HKB estimator is -1.334073e-29 > modified L-W estimator is -5.610557e-28 > smallest value of GCV at 1e-04 > > > I have no idea what that is telling me and where I am supposed to work out > which variables have been selected. > > Someone help me out please! ----- Fran...

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 advisable to use the results of regression corresponding to lambda= 10; so the next thing I do is: reg=lm....

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 use the results of regression corresponding to lambda= 10; so the next thing I do is: reg=lm....

Hello R-list, I am trying to calculate a ridge regression using first the *lm.ridge()* function from the MASS package and then applying the obtained Hoerl Kennard Baldwin (HKB) estimator as a penalty scalar to the *ols()* function provided by Frank Harrell in his Design package. It looks like this: > rrk1<-lm.ridge(lnbcpc ~ lntex + lnbeerp + lnwinep + lntemp + pop, subset(aa, Jahr>=1957 & Jahr<=1966)) > f <- ols(lnbcpc ~ lntex + lnbeerp + lnwine...

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 means that it is advisable to use the results of regression corresponding to lambda= 10; so the next thing I do is: best &l...

...s 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) modified HKB estimator is 0.004275 modified L-W estimator is 0.0323 smallest value of GCV at 0.003 > ##let's choose 0.03 for lambda > coef(gr)[gr$lam == 0.03,] GNP.deflator GNP Unemployed Armed.Forces Population Year -1620.429355 0.021060 0.007994 -0.013146...