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

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?