Doing ridge regression in R I have discovered

  • linearRidge in the ridge package - which fits a model, reports coefficients and p values but nothing to measure the overall goodness of fit

  • lm.ridge in the MASS package - which reports coefficients and GCV but no p values for parameters

How can I get all of these things (goodness of fit, coefficients and p values) from the same ridge regression? I'm new to R so not familiar with facilities that may be available e.g. for computing $r^2$ from the data and fitted coefficients.


2 Answers 2


Although I recommended lm.ridge to you in response to an earlier question, you might consider the glmnet package as a better way to get started with ridge regression in R. It has the advantage that you can then follow along with the examples in Chapter 6 of An Introduction to Statistical Learning. The package also has functions for cross-validation of your model. Going through this process will help you understand the issues involved in choosing the penalty value for your ridge regression. It's not as easy as getting an automatic answer, but at the end you will have a better understanding of how the final model was built and be better able to explain (or defend) it to others. And glmnet also allows combinations of ridge and LASSO regression that some favor.

If your interest is in using your model for prediction, then measures of prediction error from cross-validation or bootstrapping are much more important than p-values for individual coefficients. For prediction you want to keep all predictors in the model (if there aren't too many), provided you have minimized overfitting with an approach like ridge regression, rather than omit those with nominally "insignificant" p-values. The p-values reported by linearRidge in the ridge package are based on an algorithm in a paper by the package's author that does not seem to have received much attention, and the package is presently orphaned as the author's email reported to the R repository at CRAN no longer works. So I'd be a bit hesitant to rely on those p-values or that unmaintained package.

If you do need coefficient p-values for some reason, bootstrapping would be a better approach. Once you have an algorithm for choosing the coefficients in a ridge regression, you make multiple bootstrap samples (with replacement) of the same size from your original data, then repeat the entire process to get the regression coefficients for each bootstrap sample with the boot package. You will have to write a function to report the regression coefficients, but that's reasonably straightforward. The distributions of those regression coefficients among the bootstrap analyses provides an estimate of their confidence limits, with the boot.ci function in the boot package. That way you incorporate variability arising from all steps of the modeling process into the coefficient error estimates.

But even those bootstrapped p-values could be misleading, as they would ignore the tradeoffs among coefficients of collinear predictors, which you evidently face based on your earlier question. With even just 2 collinear predictors, their individual regression coefficients are likely to vary widely among bootstrap samples so that their individual p-values may appear insignificant. Nevertheless, their joint contributions to the regression might be much more stable and thus their combination very significant in practical terms. So think really hard about whether you really are interested in p-values for individual coefficients.


For the ridge package you could easily calculate either AIC or BIC or adjusted R2 as measures of goodness of fit, if one uses in these formulae the correct effective degrees of freedom for ridge regression, which work out as the trace of the hat matrix.

Ridge regression models are in fact fit simply as a regular linear regression but with the covariate matrix row augmented with a matrix with sqrt(lambda) [or sqrt(lambdas) in case of adaptive ridge regression] along the diagonal (and p zeros added to the outcome variable y). So given that ridge regression just comes down to doing a linear regression with an augmented covariate matrix you can keep on using many of the features of regular linear model fits. The original paper that the ridge package is based on is worth reading: Significance testing in ridge regression for genetic data

Main question is how to tune your lambda regularization parameter. Below I tuned the optimal lambda for ridge or adaptive ridge regression on the same data used to fit the final ridge or adaptive ridge regression model. In practice, it might be safer to split your data in a training and validation set and use the training set to tune lambda and the validation set to do inference. Or in the case of adaptive ridge regression do 3 splits and use 1 to fit your initial linear model, 1 to tune lambda for the adaptive ridge regression (using the coefficients of the 1st split to define your adaptive weights) and the third to do inference using the optimized lambda & the linear model coefficients derived from the other sections of your data. There are many different strategies to tune lambda for ridge and adaptive ridge regresion though, see this talk. Of course ridge regression will tend to preserve collinear variables and select them together, unlike e.g. LASSO or nonnegative least squares. This is something to keep in mind of course.

The coefficients of regular ridge regression are also heavily biased so this will of course also severely affect the p values. This is less the case with adaptive ridge.

data=data.frame(apply(data,2,function(col) scale(col))) # we standardize all columns
lmfit = lm(y~.,data)
# Coefficients:
#                  Estimate Std. Error t value Pr(>|t|)  
# (Intercept)     9.769e-16  2.767e-02   0.000   1.0000  
# GNP             2.427e+00  9.961e-01   2.437   0.0376 *
# Unemployed      3.159e-01  2.619e-01   1.206   0.2585  
# Armed.Forces    7.197e-02  9.966e-02   0.722   0.4885  
# Population     -1.120e+00  4.343e-01  -2.578   0.0298 *
# Year           -6.259e-01  1.299e+00  -0.482   0.6414  
# Employed        7.527e-02  4.244e-01   0.177   0.8631  
lmcoefs = coef(lmfit)

# function to augment covariate matrix with matrix with sqrt(lambda) along diagonal to fit ridge penalized regression
dataaug=function (lambda, data) { p=ncol(data)-1 # nr of covariates; data contains y in first column
                                  data.frame(rbind(as.matrix(data[,-1]),diag(sqrt(lambda),p)),yaugm=c(data$y,rep(0,p))) } 
# function to calculate optimal penalization factor lambda of ridge regression on basis of BIC value of regression model
BICval_ridge = function (lambda, data) BIC(lm(yaugm~.,data=dataaug(lambda, data)))
lambda_ridge = optimize(BICval_ridge, interval=c(0,10), data=data)$minimum
lambda_ridge # ridge lambda optimized based on BIC value = 5.575865e-05
ridgefit = lm(yaugm~.,data=dataaug(lambda_ridge, data))
# Coefficients:
#               Estimate Std. Error t value Pr(>|t|)   
# (Intercept)  -0.000388   0.018316  -0.021  0.98338   
# GNP           2.411782   0.769161   3.136  0.00680 **
# Unemployed    0.312543   0.202167   1.546  0.14295   
# Armed.Forces  0.071125   0.077050   0.923  0.37057   
# Population   -1.115754   0.336651  -3.314  0.00472 **
# Year         -0.608894   1.002436  -0.607  0.55266   
# Employed      0.072204   0.328229   0.220  0.82885   

# function to calculate optimal penalization factor lambda of adaptive ridge regression 
# (with gamma=2) on basis of BIC value of regression model
BICval_adridge = function (lambda, data, init.coefs) { dat = dataaug(lambda*(1/(abs(init.coefs)+1E-5)^2), data)
                                                       BIC(lm(yaugm~.,data=dat)) }
lamvals = 10^seq(-12,-1,length.out=100)
BICvals = sapply(lamvals,function (lam) BICval_adridge (lam, data, lmcoefs))
firstderivBICvals = function (lambda) splinefun(x=lamvals, y=BICvals)(lambda, deriv=1)
plot(lamvals, BICvals, type="l", ylab="BIC", xlab="Adaptive ridge lambda", log="x")
lambda_adridge = lamvals[which.min(lamvals*firstderivBICvals(lamvals))] # we place optimal lambda at middle flat part of BIC as coefficients are most stable there
lambda_adridge # 4.641589e-08
abline(v=lambda_adridge, col="red")

enter image description here

adridgefit = lm(yaugm~.,data=dataaug(lambda_adridge, data))
# Coefficients:
#                 Estimate Std. Error t value Pr(>|t|)   
# (Intercept)   -1.121e-05  1.828e-02  -0.001  0.99952   
# GNP            2.427e+00  7.716e-01   3.146  0.00666 **
# Unemployed     3.159e-01  2.029e-01   1.557  0.14026   
# Armed.Forces   7.197e-02  7.719e-02   0.932  0.36590   
# Population    -1.120e+00  3.364e-01  -3.328  0.00459 **
# Year          -6.259e-01  1.006e+00  -0.622  0.54326   
# Employed       7.527e-02  3.287e-01   0.229  0.82197  

Normally in ridge regression the effective degrees of freedom are defined as the trace of the hat matrix - you'd have to check if the BICs calculated as above use these correct degrees of freedom though, otherwise you'd have to calculate them yourself.

  • 1
    $\begingroup$ Why should we believe the p-values produced by lm, given that $\lambda$ has been tuned and the data have been augmented? This seems to lie at the heart of the original question. $\endgroup$
    – whuber
    Commented May 17, 2019 at 16:14
  • $\begingroup$ Well as I mentioned one could tune the lambda on an independent subset of your data & do inference on the remainder. But this paper, bmcbioinformatics.biomedcentral.com/articles/10.1186/…, which the ridge package is based on shows that for normal ridge regression the p values are fairly reliable in comparison to permutation based p values. Effective degrees of freedom need to be calculated based on the trace of the hat matrix though, but otherwise everything is just based on a regression with an augmented covariate matrix as that's what your ridge model is. $\endgroup$ Commented May 18, 2019 at 20:56
  • $\begingroup$ The uncertainty in the exact lambda value that you pick is usually probably also not very large (e.g. coefficients are nearly identical across 3 orders of magnitude in adaptive ridge fit above) so I'm not sure taking that uncertainty into account would change p values much. $\endgroup$ Commented May 18, 2019 at 20:58
  • 1
    $\begingroup$ Trickier is the the uncertainty in the linear model coefficients that enter the augmented matrix in the adaptive penalty weights in adaptive ridge. But it would be straightforward I think to take that into account if you wanted to by resampling coefficients from the original linear model fit multivariate normal distribution (ie plugging the variance covariance matrix from the origin lm fit into MASS's mvrnorm) and augmenting the covariate matrix many times with these resampled coefficients and fitting your adaptive ridge fit many times. Ie using parametric bootstrapping. I'll update my answer $\endgroup$ Commented May 18, 2019 at 21:02
  • $\begingroup$ Thank you for that thoughtful analysis! $\endgroup$
    – whuber
    Commented May 18, 2019 at 21:29

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