Assignment 5
Nicholas Mata jnm705
2025-04-08
2
For parts (a) through (c), indicate which of i. through iv. is correct. Justify your answer.
(a) The lasso, relative to least squares, is:
i. More flexible and hence will give improved prediction accuracy when its increase in bias is less than its decrease in variance.
ii. More flexible and hence will give improved prediction accuracy when its increase in variance is less than its decrease in bias.
iii. Less flexible and hence will give improved prediction accuracy when its increase in bias is less than its decrease in variance.
iv. Less flexible and hence will give improved prediction accuracy when its increase in variance is less than its decrease in bias.
- is correct. The less flexible LASSO restricts the model, decreasing its variance and increasing its bias. If the bias increase is less than the reduction in variance (if the reduction in varianve outweighs the increase in bias), then the prediction accuracy will increase.
(b) Repeat (a) for ridge regression relative to least squares.
also iii. Ridge regression is also a less flexible model so the same reasoning as above applies.
(c) Repeat (a) for non-linear methods relative to least squares.
Non-linear methods are more flexible and will introduce more variance while reducing bias. The answer would be ii. If the bias reduction outweighs the increased variance, the model will have a higher prediction accuracy.
9
In this exercise, we will predict the number of applications received using the other variables in the College data set.
library(ISLR2)
str(College)## 'data.frame': 777 obs. of 18 variables:
## $ Private : Factor w/ 2 levels "No","Yes": 2 2 2 2 2 2 2 2 2 2 ...
## $ Apps : num 1660 2186 1428 417 193 ...
## $ Accept : num 1232 1924 1097 349 146 ...
## $ Enroll : num 721 512 336 137 55 158 103 489 227 172 ...
## $ Top10perc : num 23 16 22 60 16 38 17 37 30 21 ...
## $ Top25perc : num 52 29 50 89 44 62 45 68 63 44 ...
## $ F.Undergrad: num 2885 2683 1036 510 249 ...
## $ P.Undergrad: num 537 1227 99 63 869 ...
## $ Outstate : num 7440 12280 11250 12960 7560 ...
## $ Room.Board : num 3300 6450 3750 5450 4120 ...
## $ Books : num 450 750 400 450 800 500 500 450 300 660 ...
## $ Personal : num 2200 1500 1165 875 1500 ...
## $ PhD : num 70 29 53 92 76 67 90 89 79 40 ...
## $ Terminal : num 78 30 66 97 72 73 93 100 84 41 ...
## $ S.F.Ratio : num 18.1 12.2 12.9 7.7 11.9 9.4 11.5 13.7 11.3 11.5 ...
## $ perc.alumni: num 12 16 30 37 2 11 26 37 23 15 ...
## $ Expend : num 7041 10527 8735 19016 10922 ...
## $ Grad.Rate : num 60 56 54 59 15 55 63 73 80 52 ...sum(is.na(College$Apps))## [1] 0(a) Split the data set into a training set and a test set.
set.seed(13)
index = sample(nrow(College), 0.5 * nrow(College))
train_college = College[index,]
test_college = College[-index,](b) Fit a linear model using least squares on the training set, and report the test error obtained.
linear_model = lm(Apps ~ ., data = train_college)
probabilities = predict(linear_model, newdata = test_college)
test_mse = mean((test_college$Apps - probabilities)^2)
test_mse## [1] 1062514sqrt(test_mse)## [1] 1030.783(c) Fit a ridge regression model on the training set, with λ chosen by cross-validation. Report the test error obtained.
library(glmnet)## Loading required package: Matrix## Loaded glmnet 4.1-8set.seed(13)
X_train = model.matrix(Apps ~ ., data = train_college)[,-1]
y_train = train_college$Apps
X_test = model.matrix(Apps ~ ., data = test_college)[,-1]
y_test = test_college$Apps
cv_ridge = cv.glmnet(X_train, y_train, alpha = 0)
best_lambda = cv_ridge$lambda.min
ridge_model = glmnet(X_train, y_train, alpha = 0, lambda = best_lambda)
ridge_probs = predict(ridge_model, s = best_lambda, newx = X_test)
ridge_mse = mean((ridge_probs - y_test)^2)
ridge_mse## [1] 1174478sqrt(ridge_mse)## [1] 1083.733(d) Fit a lasso model on the training set, with λ chosen by cross- validation. Report the test error obtained, along with the number of non-zero coefficient estimates.
set.seed(13)
X_train = model.matrix(Apps ~ ., data = train_college)[,-1]
y_train = train_college$Apps
X_test = model.matrix(Apps ~ ., data = test_college)[,-1]
y_test = test_college$Apps
cv_lasso = cv.glmnet(X_train, y_train, alpha = 1)
best_lambda = cv_lasso$lambda.min
lasso_model = glmnet(X_train, y_train, alpha = 1, lambda = best_lambda)
lasso_probs = predict(lasso_model, s = best_lambda, newx = X_test)
lasso_mse = mean((lasso_probs - y_test)^2)
lasso_mse## [1] 1055797sqrt(lasso_mse)## [1] 1027.52(e) Fit a PCR model on the training set, with M chosen by cross-validation. Report the test error obtained, along with the value of M selected by cross-validation.
library(pls)##
## Attaching package: 'pls'## The following object is masked from 'package:stats':
##
## loadingsset.seed(13)
pcr_model = pcr(Apps ~ ., data = train_college, scale = TRUE, validation = "CV")
#validationplot(pcr_model, val.type = "MSEP")
cv_results = RMSEP(pcr_model)
best_m = which.min(cv_results$val[1, , -1])
probabilities = predict(pcr_model, newdata = test_college, ncomp = best_m)
pcr_mse = mean((probabilities - test_college$Apps)^2)
pcr_mse## [1] 1062514sqrt(pcr_mse)## [1] 1030.783(f) Fit a PLS model on the training set, with M chosen by cross-validation. Report the test error obtained, along with the value of M selected by cross-validation.
set.seed(13)
pls_model = plsr(Apps ~ ., data = train_college, scale = TRUE, validation = "CV")
cv_results = RMSEP(pls_model)
best_m = which.min(cv_results$val[1, , -1])
probabilities = predict(pls_model, newdata = test_college, ncomp = best_m)
pls_mse = mean((probabilities - test_college$Apps)^2)
pls_mse## [1] 1062514sqrt(pls_mse)## [1] 1030.783(g) Comment on the results obtained. How accurately can we predict the number of college applications received? Is there much difference among the test errors resulting from these five approaches?
Lasso 1027.52
OLS 1030.78
PCR 1030.78
PLS 1030.78
Ridge 1083.73
We can predict with moderate accuracy, around 34% error. The test errors across the board are similar between the 5 methods, suggesting the dataset is fairly robust the modeling method.
11
We will now try to predict per capita crime rate in the Boston data set.
str(Boston)## 'data.frame': 506 obs. of 13 variables:
## $ crim : num 0.00632 0.02731 0.02729 0.03237 0.06905 ...
## $ zn : num 18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
## $ indus : num 2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
## $ chas : int 0 0 0 0 0 0 0 0 0 0 ...
## $ nox : num 0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
## $ rm : num 6.58 6.42 7.18 7 7.15 ...
## $ age : num 65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
## $ dis : num 4.09 4.97 4.97 6.06 6.06 ...
## $ rad : int 1 2 2 3 3 3 5 5 5 5 ...
## $ tax : num 296 242 242 222 222 222 311 311 311 311 ...
## $ ptratio: num 15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
## $ lstat : num 4.98 9.14 4.03 2.94 5.33 ...
## $ medv : num 24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...sum(is.na(Boston$crim))## [1] 0(a) Try out some of the regression methods explored in this chapter, such as best subset selection, the lasso, ridge regression, and PCR. Present and discuss results for the approaches that you consider.
library(leaps)
regfit_full = regsubsets(crim ~ ., data = Boston, nvmax = 12)
summary(regfit_full)## Subset selection object
## Call: regsubsets.formula(crim ~ ., data = Boston, nvmax = 12)
## 12 Variables (and intercept)
## Forced in Forced out
## zn FALSE FALSE
## indus FALSE FALSE
## chas FALSE FALSE
## nox FALSE FALSE
## rm FALSE FALSE
## age FALSE FALSE
## dis FALSE FALSE
## rad FALSE FALSE
## tax FALSE FALSE
## ptratio FALSE FALSE
## lstat FALSE FALSE
## medv FALSE FALSE
## 1 subsets of each size up to 12
## Selection Algorithm: exhaustive
## zn indus chas nox rm age dis rad tax ptratio lstat medv
## 1 ( 1 ) " " " " " " " " " " " " " " "*" " " " " " " " "
## 2 ( 1 ) " " " " " " " " " " " " " " "*" " " " " "*" " "
## 3 ( 1 ) " " " " " " " " " " " " " " "*" " " " " "*" "*"
## 4 ( 1 ) "*" " " " " " " " " " " "*" "*" " " " " " " "*"
## 5 ( 1 ) "*" "*" " " " " " " " " "*" "*" " " " " " " "*"
## 6 ( 1 ) "*" " " " " "*" " " " " "*" "*" " " "*" " " "*"
## 7 ( 1 ) "*" " " " " "*" " " " " "*" "*" " " "*" "*" "*"
## 8 ( 1 ) "*" "*" " " "*" " " " " "*" "*" " " "*" "*" "*"
## 9 ( 1 ) "*" "*" " " "*" "*" " " "*" "*" " " "*" "*" "*"
## 10 ( 1 ) "*" " " "*" "*" "*" " " "*" "*" "*" "*" "*" "*"
## 11 ( 1 ) "*" "*" "*" "*" "*" " " "*" "*" "*" "*" "*" "*"
## 12 ( 1 ) "*" "*" "*" "*" "*" "*" "*" "*" "*" "*" "*" "*"reg_summary = summary(regfit_full)
reg_summary$rsq## [1] 0.3912567 0.4207965 0.4244920 0.4334892 0.4378328 0.4421077 0.4451514
## [8] 0.4470236 0.4482891 0.4487917 0.4493353 0.4493378plot(regfit_full, scale = "r2")
plot(regfit_full, scale = "adjr2")
plot(regfit_full, scale = "Cp")
plot(regfit_full, scale = "bic")
2 variables will be used to predict crim based on best
subset method. rad and lstat. The 2 variable
model has comparable rsquare and BIC values to the rest of the models,
while also being simpler.
set.seed(13)
index = sample(nrow(Boston), 0.5 * nrow(Boston))
train_boston = Boston[index,]
test_boston = Boston[-index,]
# linear model with chosen best subset
linear_model = lm(crim ~ rad + lstat, data = train_boston)
probabilities = predict(linear_model, newdata = test_boston)
test_mse = mean((test_boston$crim - probabilities)^2)
test_mse## [1] 33.22543sqrt(test_mse)## [1] 5.76415# Ridge
library(glmnet)
set.seed(13)
X_train = model.matrix(crim ~ ., data = train_boston)[,-1]
y_train = train_boston$crim
X_test = model.matrix(crim ~ ., data = test_boston)[,-1]
y_test = test_boston$crim
cv_ridge = cv.glmnet(X_train, y_train, alpha = 0)
best_lambda = cv_ridge$lambda.min
ridge_model = glmnet(X_train, y_train, alpha = 0, lambda = best_lambda)
ridge_probs = predict(ridge_model, s = best_lambda, newx = X_test)
ridge_mse = mean((ridge_probs - y_test)^2)
ridge_mse## [1] 33.08957sqrt(ridge_mse)## [1] 5.752353# LASSO
set.seed(13)
X_train = model.matrix(crim ~ ., data = train_boston)[,-1]
y_train = train_boston$crim
X_test = model.matrix(crim ~ ., data = test_boston)[,-1]
y_test = test_boston$crim
cv_lasso = cv.glmnet(X_train, y_train, alpha = 1)
best_lambda = cv_lasso$lambda.min
lasso_model = glmnet(X_train, y_train, alpha = 1, lambda = best_lambda)
lasso_probs = predict(lasso_model, s = best_lambda, newx = X_test)
lasso_mse = mean((lasso_probs - y_test)^2)
lasso_mse## [1] 32.88823sqrt(lasso_mse)## [1] 5.734826# PCR
library(pls)
set.seed(13)
pcr_model = pcr(crim ~ ., data = train_boston, scale = TRUE, validation = "CV")
#validationplot(pcr_model, val.type = "MSEP")
cv_results = RMSEP(pcr_model)
best_m = which.min(cv_results$val[1, , -1])
probabilities = predict(pcr_model, newdata = test_boston, ncomp = best_m)
pcr_mse = mean((probabilities - test_boston$crim)^2)
pcr_mse## [1] 34.82423sqrt(pcr_mse)## [1] 5.901206RMSE
OLS (two variable model): 5.76415
Ridge: 5.752353
LASSO: 5.734826
PCR: 5.901206
LASSO had the lowest RMSE so it would be chosen as the best method.
(b) Propose a model (or set of models) that seem to perform well on this data set, and justify your answer. Make sure that you are evaluating model performance using validation set error, cross- validation, or some other reasonable alternative, as opposed to using training error.
OLS with appropriate variables, Ridge Regression, and LASSO would be viable for this data set based on their RMSE in comparison with each other and PCR.
(c) Does your chosen model involve all of the features in the data set? Why or why not?
lasso_coef = predict(lasso_model, type = "coefficients", s = best_lambda, newx = X_test)[1:12,]
lasso_coef## (Intercept) zn indus chas nox rm
## 5.24016233 0.04936863 -0.11608797 -0.17313686 0.00000000 0.00000000
## age dis rad tax ptratio lstat
## 0.00000000 -0.80540024 0.52582341 0.00000000 0.00000000 0.13698970The chosen model did not use all of the available features because LASSO filtered out irrelevant variables for the dataset.