R: What does train() do when it calculates ridge regression? [closed]

Question

I am running ridge regression on the Boston dataset. There are many write-ups online for how to do ridge regression.

I will write up the two methods and then pose my question

Initialize with the dataset

library('mlbench')
data(BostonHousing)

First method: According to the Stanford Open course on statistics

library('glmnet')
library('dplyr')

#initialize data matrix for glmnnet
z <- colnames(BostonHousing)
z <- z[-10]
x <- BostonHousing %>% select(one_of(z)) %>% data.matrix()

#create ridge regression with every possible lambda
fit <- glmnet(x,BostonHousing$medv, alpha=0)

#use 10-fold cross validation to choose the lambda with the lowest MSE
cv_fit <- cv.glmnet(x,BostonHousing$medv, alpha = 0)

#create a ridge regression model with the best lambda
fit <- cv_fit$glmnet.fit

#calculate training MSE for best ridge regression model
min(cv_fit$cvm)

Second method: According to this tutorial

library('caret')

#take a random sample of half of the data
split <- createDataPartition(y=BostonHousing$medv, p = 0.5, list = FALSE)

#create training and test sets
train <- BostonHousing[split,]
test <- BostonHousing[-split,]


#calculate ridge regression on every lambda with the training set
ridge <- train(medv ~., data = train, method='ridge',
               lambda = 4,preProcess=c('scale', 'center'))

#use the model to predict values of the test set
ridge.pred <- predict(ridge, test)

#mse for the test error
mean(ridge.pred - test$medv)^2

#select lambda
fitControl <- trainControl(method = "cv", number = 10)
lambdaGrid <- expand.grid(lambda = 10^seq(10, -2, length=100))

#do ridge regression with the best lambda
ridge <- train(medv~., data = train, method='ridge',
               trControl = fitControl,
               #                tuneGrid = lambdaGrid
               preProcess=c('center', 'scale')
)

#predict the test set using the model from the training set
ridge.pred <- predict(ridge, test)

#calculate test mse
sqrt(mean(ridge.pred - test$medv)^2)

I have a few questions, I hope that's alright.

1- Assuming I use the first method, can I estimate the test error of the ridge model with k-fold cross validation?

It only gives me the training error and I'd like to approximate test error.

2- The second approach uses a validation set. Is that desirable in situations with small sample sizes?

The BostonHousing data is 506 rows by 14 variables.

3- Here is the output in the second method

ridge

Ridge Regression 

254 samples
 10 predictor

Pre-processing: centered (10), scaled (10) 
Resampling: Cross-Validated (10 fold) 
Summary of sample sizes: 230, 229, 228, 229, 229, 229, ... 
Resampling results across tuning parameters:

  lambda  RMSE       Rsquared   MAE      
  0e+00   0.5963179  0.6835195  0.4131819
  1e-04   0.5963073  0.6835296  0.4131761
  1e-01   0.5920124  0.6891727  0.4120725

RMSE was used to select the optimal model using
 the smallest value.
The final value used for the model was lambda = 0.1.

Why is ridge regression using resampling? How did they get a lambda of 0.1 when the first method got a lambda of 0.0501?

Answer 1

First question: In case that you tuned you ridge regression with cross validation, you have used all your training data to find the best lambda and the error will therefore be biased downwards if you test it on training data.

Second question: Yes you can still do that. Try the same procedure with different random seeds, that is, perform repeated cross validation. So, split the model in training and test set, tune lambda with CV on training set, determine performance on test set, repeat with different split.

Third question: Did you scale your data in the first method? There might also be different ways to determine lambda because there is no exact method, resulting is different outcomes.