# Practical 3 Solutions

Make sure to at least finish Practical 1 before continuing with this practical! It might be tempting to skip some stuff if you’re behind, but you may end up more lost$$...$$we will happily answer questions from any previous practical each session.

We continue analysing the fat data in package faraway in this practical. Run the following commands to load the package and obtain our reduced dataset fat1 of interest. We also load the glmnet package for performing ridge regression.

library(faraway)
fat1 <- fat[,-c(2, 3, 8)]
library(glmnet)

### Q1 Ridge Regression

1. Define a vector y containing all the values of the response variable brozek.
y = fat1$brozek 1. Stack the predictors column-wise into a matrix, and define this matrix as x. x = model.matrix(brozek~., fat1)[,-1] 1. Fit a model of the predictors for the response using ridge regression over a grid of 200 $$\lambda$$ values. Hint: type ?glmnet and look at argument nlambda. ridge = glmnet(x, y, alpha = 0, nlambda = 200) 1. Plot the regularisation paths of the ridge regresssion coefficients as a function of log $$\lambda$$. plot(ridge, xvar = 'lambda') ### Q2 Cross-Validation 1. Using cv.glmnet with 5-fold CV and random seed equal to 2, find the values of min-CV $$\lambda$$ and 1-se $$\lambda$$. set.seed(2) ridge.cv = cv.glmnet(x, y, alpha=0, nfolds = 5) ridge.cv$lambda.min
##  0.6294393
ridge.cv$lambda.1se ##  1.454086 1. Plot the CV $$\lambda$$ graph next to the regularisation-path plot indicating with vertical lines the two $$\lambda$$ values. par(mfrow=c(1,2)) plot(ridge.cv) plot(ridge, xvar = 'lambda') abline(v = log(ridge.cv$lambda.min), lty = 3)
abline(v = log(ridge.cv$lambda.1se), lty = 3) 1. Which predictor has the highest absolute coefficient? abs(coef(ridge.cv)) ## 15 x 1 sparse Matrix of class "dgCMatrix" ## s1 ## (Intercept) 11.084963486 ## age 0.101081788 ## weight 0.002854877 ## height 0.141616551 ## adipos 0.325547676 ## neck 0.239066118 ## chest 0.118435588 ## abdom 0.324829169 ## hip 0.033194250 ## thigh 0.127019937 ## knee 0.028284305 ## ankle 0.095625648 ## biceps 0.042553472 ## forearm 0.197620295 ## wrist 1.265884058 # it is wrist for 1-se lambda abs(coef(ridge.cv, s ='lambda.min')) ## 15 x 1 sparse Matrix of class "dgCMatrix" ## s1 ## (Intercept) 9.174745514 ## age 0.099911854 ## weight 0.012172773 ## height 0.108068586 ## adipos 0.301584522 ## neck 0.354687819 ## chest 0.089843263 ## abdom 0.462355382 ## hip 0.028571052 ## thigh 0.156802210 ## knee 0.007910553 ## ankle 0.035339905 ## biceps 0.048198960 ## forearm 0.288207387 ## wrist 1.525865471 # it is also wrist for min-CV lambda 1. Type plot(ridge.cv) into the R console. What is plotted? plot(ridge.cv) # We see the estimated CV MSE for each value of log lambda, along with # the 1 standard error interval bars of those estimates. # The vertical lines show the min-CV lambda and 1-se lambda values. ### Q3 Comparison of Predictive Performance Using 50 repetitions with random seed set to 2, compare whether the model with min-CV $$\lambda$$ or 1-se $$\lambda$$ may be preferable in terms of predictive performance. In this case use 5-fold CV within cv.glmnet() and also a 50% splitting for training and testing. Produce boxplots of the correlation values obtained from each cycle of your for loop. Finally, calculate the mean correlations from the two models. Under which approach does ridge seem to perform better? repetitions = 50 cor.1 = c() cor.2 = c() set.seed(2) for(i in 1:repetitions){ # Step (i) data splitting training.obs = sample(1:252, 126) y.train = fat1$brozek[training.obs]
x.train = model.matrix(brozek~., fat1[training.obs, ])[,-1]
y.test = fat1\$brozek[-training.obs]
x.test = model.matrix(brozek~., fat1[-training.obs, ])[,-1]

# Step (ii) training phase
ridge.train = cv.glmnet(x.train, y.train, alpha = 0, nfolds = 5)

# Step (iii) generating predictions
predict.1 = predict(ridge.train, x.test, s = 'lambda.min')
predict.2 = predict(ridge.train, x.test, s = 'lambda.1se')

# Step (iv) evaluating predictive performance
cor.1[i] = cor(y.test, predict.1)
cor.2[i] = cor(y.test, predict.2)
}

boxplot(cor.1, cor.2, names = c('Ridge min-CV lambda','Ridge 1-se lambda'),
ylab = 'Test correlation', col = 3) mean(cor.1)
##  0.8204005
mean(cor.2)
##  0.796162
# Ridge regression based on min-CV lambda seems to be better.