Ridge regression is a method we can use to fit a regression model when multicollinearity is present in the data.

In a nutshell, least squares regression tries to find coefficient estimates that minimize the sum of squared residuals (RSS):

**RSS = Σ(y _{i} – ŷ_{i})2**

where:

**Σ**: A greek symbol that means*sum***y**: The actual response value for the i_{i}^{th}observation**ŷ**: The predicted response value based on the multiple linear regression model_{i}

Conversely, ridge regression seeks to minimize the following:

**RSS + λΣβ _{j}^{2}**

where *j* ranges from 1 to *p* predictor variables and λ ≥ 0.

This second term in the equation is known as a *shrinkage penalty*. In ridge regression, we select a value for λ that produces the lowest possible test MSE (mean squared error).

This tutorial provides a step-by-step example of how to perform ridge regression in R.

**Step 1: Load the Data**

For this example, we’ll use the R built-in dataset called **mtcars**. We’ll use **hp** as the response variable and the following variables as the predictors:

- mpg
- wt
- drat
- qsec

To perform ridge regression, we’ll use functions from the **glmnet** package. This package requires the response variable to be a vector and the set of predictor variables to be of the class **data.matrix**.

The following code shows how to define our data:

**#define response variable
y <- mtcars$hp
#define matrix of predictor variables
x <- data.matrix(mtcars[, c('mpg', 'wt', 'drat', 'qsec')])
**

**Step 2: Fit the Ridge Regression Model**

Next, we’ll use the **glmnet()** function to fit the ridge regression model and specify **alpha=0**.

Note that setting alpha equal to 1 is equivalent to using Lasso Regression and setting alpha to some value between 0 and 1 is equivalent to using an elastic net.

Also note that ridge regression requires the data to be standardized such that each predictor variable has a mean of 0 and a standard deviation of 1.

Fortunately **glmnet()** automatically performs this standardization for you. If you happened to already standardize the variables, you can specify **standardize=False**.

**library(glmnet)
#fit ridge regression model
model <- glmnet(x, y, alpha = 0)
#view summary of model
summary(model)
Length Class Mode
a0 100 -none- numeric
beta 400 dgCMatrix S4
df 100 -none- numeric
dim 2 -none- numeric
lambda 100 -none- numeric
dev.ratio 100 -none- numeric
nulldev 1 -none- numeric
npasses 1 -none- numeric
jerr 1 -none- numeric
offset 1 -none- logical
call 4 -none- call
nobs 1 -none- numeric
**

**Step 3: Choose an Optimal Value for Lambda**

Next, we’ll identify the lambda value that produces the lowest test mean squared error (MSE) by using k-fold cross-validation.

Fortunately, **glmnet** has the function **cv.glmnet()** that automatically performs k-fold cross validation using k = 10 folds.

**#perform k-fold cross-validation to find optimal lambda value
cv_model <- cv.glmnet(x, y, alpha = 0)
#find optimal lambda value that minimizes test MSE
best_lambda <- cv_model$lambda.min
best_lambda
[1] 10.04567
#produce plot of test MSE by lambda value
plot(cv_model)
**

The lambda value that minimizes the test MSE turns out to be **10.04567**.

**Step 4: Analyze Final Model**

Lastly, we can analyze the final model produced by the optimal lambda value.

We can use the following code to obtain the coefficient estimates for this model:

**#find coefficients of best model
best_model <- glmnet(x, y, alpha = 0, lambda = best_lambda)
coef(best_model)
5 x 1 sparse Matrix of class "dgCMatrix"
s0
(Intercept) 475.242646
mpg -3.299732
wt 19.431238
drat -1.222429
qsec -17.949721**

We can also produce a Trace plot to visualize how the coefficient estimates changed as a result of increasing lambda:

**#produce Ridge trace plot
plot(model, xvar = "lambda")**

Lastly, we can calculate the R-squared of the model on the training data:

**#use fitted best model to make predictions
y_predicted <- predict(model, s = best_lambda, newx = x)
#find SST and SSE
sst <- sum((y - mean(y))^2)
sse <- sum((y_predicted - y)^2)
#find R-Squared
rsq <- 1 - sse/sst
rsq
[1] 0.7999513
**

The R-squared turns out to be **0.7999513**. That is, the best model was able to explain **79.99%** of the variation in the response values of the training data.

You can find the complete R code used in this example here.