Assignment #5
MK
2024-03-22
- For parts (a) through (c), indicate which of i. through iv. is correct. Justify your answer.
- More flexible and hence will give improved prediction ac- curacy when its increase in bias is less than its decrease in variance.
- More flexible and hence will give improved prediction accu- racy when its increase in variance is less than its decrease in bias.
- Less flexible and hence will give improved prediction accu- racy when its increase in bias is less than its decrease in variance.
- Less flexible and hence will give improved prediction accu- racy when its increase in variance is less than its decrease in bias.
The lasso, relative to least squares, is: iii is the correct response. Lasso is better than least squares because it can balance bias and variance well. When least squares gives too much variation, Lasso can reduce it by accepting a bit more bias, resulting in more accurate predictions. Also, Lasso makes it easier to pick the important variables, which makes it easier to understand compared to methods like ridge regression.
Repeat (a) for ridge regression relative to least squares: iii is the correct choice. Ridge regression and lasso outperform least squares by balancing bias and variance. As λ increases, ridge regression’s flexibility decreases, reducing variance but increasing bias. Understanding this λ-variance-bias relationship is key. Least squares coefficients can lead to significant variance with minor changes in training data, while ridge regression can trade off a small bias increase for substantial variance reduction, performing well under such conditions. Ridge regression is particularly effective when least squares estimates have high variance. One notable distinction is that lasso conducts variable selection, making interpretation easier.
Repeat (a) for non-linear methods relative to least squares: ii is the correct option. Non-linear models offer greater flexibility and have lower bias compared to least squares models.
- In this exercise, we will predict the number of applications received using the other variables in the College data set.
library(ISLR)- Split the data set into a training set and a test set.
attach(College)
x=model.matrix(Apps~.,College)[,-1]
y=College$Apps
set.seed(10)
train=sample(1:nrow(x), nrow(x)/2)
test=(-train)
College.train = College[train, ]
College.test = College[test, ]
y.test=y[test]- Fit a linear model using least squares on the training set, and report the test error obtained.
pls.fit<-lm(Apps~., data=College, subset=train)
summary(pls.fit)##
## Call:
## lm(formula = Apps ~ ., data = College, subset = train)
##
## Residuals:
## Min 1Q Median 3Q Max
## -5139.5 -473.3 -21.1 353.2 7402.7
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -629.36179 639.35741 -0.984 0.325579
## PrivateYes -647.56836 192.17056 -3.370 0.000832 ***
## Accept 1.68912 0.05038 33.530 < 2e-16 ***
## Enroll -1.02383 0.27721 -3.693 0.000255 ***
## Top10perc 48.19124 8.10714 5.944 6.42e-09 ***
## Top25perc -10.51538 6.44952 -1.630 0.103865
## F.Undergrad 0.01992 0.05364 0.371 0.710574
## P.Undergrad 0.04213 0.05348 0.788 0.431373
## Outstate -0.09489 0.02674 -3.549 0.000436 ***
## Room.Board 0.14549 0.07243 2.009 0.045277 *
## Books 0.06660 0.31115 0.214 0.830623
## Personal 0.05663 0.09453 0.599 0.549475
## PhD -10.11489 7.11588 -1.421 0.156027
## Terminal -2.29300 8.03546 -0.285 0.775528
## S.F.Ratio 22.07117 18.70991 1.180 0.238897
## perc.alumni 2.08121 6.00673 0.346 0.729179
## Expend 0.07654 0.01672 4.577 6.45e-06 ***
## Grad.Rate 9.99706 4.49821 2.222 0.026857 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1092 on 370 degrees of freedom
## Multiple R-squared: 0.9395, Adjusted R-squared: 0.9367
## F-statistic: 338 on 17 and 370 DF, p-value: < 2.2e-16pred.app<-predict(pls.fit, College.test)
test.error<-mean((College.test$Apps-pred.app)^2)
test.error## [1] 1020100MSE for linear model using the least squares is 1,020,100
- Fit a ridge regression model on the training set, with λ chosen by cross-validation. Report the test error obtained.
chooseCRANmirror(ind=1)
install.packages('glmnet')##
## The downloaded binary packages are in
## /var/folders/qq/bfgnq9g96sz_3lrzwxn_zp3c0000gn/T//RtmpphC54P/downloaded_packageslibrary(glmnet)## Loading required package: Matrix## Loaded glmnet 4.1-8grid=10^seq(10,-2,length=100)
ridge.mod=glmnet(x[train,],y[train],alpha=0,lambda=grid)
summary(ridge.mod)## Length Class Mode
## a0 100 -none- numeric
## beta 1700 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 5 -none- call
## nobs 1 -none- numericcv.college.out=cv.glmnet(x[train,],y[train] ,alpha=0)
bestlam=cv.college.out$lambda.min
bestlam## [1] 411.3927ridge.pred=predict(ridge.mod,s=bestlam,newx=x[test,])
mean((ridge.pred-y.test)^2)## [1] 985020.1The ridge regression model performs better on this dataset than ordinary least squares (OLS). Ridge regression is especially useful when OLS struggles with high variability in the data. However, interpreting the relationship between variables becomes tricky with ridge regression because it includes all variables but adjusts their importance, making it harder to understand their impact on the outcome.
- Fit a lasso model on the training set, with λ chosen by cross- validation. Report the test error obtained, along with the num- ber of non-zero coefficient estimates.
lasso.mod=glmnet(x[train,],y[train],alpha=1,lambda=grid)
summary(lasso.mod)## Length Class Mode
## a0 100 -none- numeric
## beta 1700 dgCMatrix S4
## df 100 -none- numeric
## dim 2 -none- numeric
## lambda 100 -none- numeric
## dev.ratio 100 -none- numeric
## nulldev 1 -none- numeric
## npasses 1 -none- numeric
## jerr 1 -none- numeric
## offset 1 -none- logical
## call 5 -none- call
## nobs 1 -none- numericcv.out=cv.glmnet(x[train,],y[train],alpha=1)
bestlam=cv.out$lambda.min
bestlam## [1] 24.66235lasso.pred=predict(lasso.mod,s=bestlam,newx=x[test,])
mean((lasso.pred-y.test)^2)## [1] 1008145out=glmnet(x,y,alpha=1,lambda = grid)
lasso.coef=predict(out,type="coefficients",s=bestlam)[1:18,]
lasso.coef[lasso.coef!=0]## (Intercept) PrivateYes Accept Enroll Top10perc
## -6.324960e+02 -4.087012e+02 1.436837e+00 -1.410240e-01 3.143012e+01
## Top25perc P.Undergrad Outstate Room.Board Personal
## -8.606536e-01 1.480293e-02 -5.342495e-02 1.205819e-01 4.379135e-05
## PhD Terminal S.F.Ratio perc.alumni Expend
## -5.121245e+00 -3.371192e+00 2.717231e+00 -1.039648e+00 6.838161e-02
## Grad.Rate
## 4.700317e+00MSE for the lasso model is 1,008,145
- 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.
install.packages('pls')##
## The downloaded binary packages are in
## /var/folders/qq/bfgnq9g96sz_3lrzwxn_zp3c0000gn/T//RtmpphC54P/downloaded_packageslibrary(pls)##
## Attaching package: 'pls'## The following object is masked from 'package:stats':
##
## loadingspcr.college=pcr(Apps~., data=College.train,scale=TRUE,validation="CV")
summary(pcr.college)## Data: X dimension: 388 17
## Y dimension: 388 1
## Fit method: svdpc
## Number of components considered: 17
##
## VALIDATION: RMSEP
## Cross-validated using 10 random segments.
## (Intercept) 1 comps 2 comps 3 comps 4 comps 5 comps 6 comps
## CV 4347 4345 2371 2391 2104 1949 1898
## adjCV 4347 4345 2368 2396 2085 1939 1891
## 7 comps 8 comps 9 comps 10 comps 11 comps 12 comps 13 comps
## CV 1899 1880 1864 1861 1870 1873 1891
## adjCV 1893 1862 1857 1853 1862 1865 1885
## 14 comps 15 comps 16 comps 17 comps
## CV 1903 1727 1295 1260
## adjCV 1975 1669 1283 1249
##
## TRAINING: % variance explained
## 1 comps 2 comps 3 comps 4 comps 5 comps 6 comps 7 comps 8 comps
## X 32.6794 56.94 64.38 70.61 76.27 80.97 84.48 87.54
## Apps 0.9148 71.17 71.36 79.85 81.49 82.73 82.79 83.70
## 9 comps 10 comps 11 comps 12 comps 13 comps 14 comps 15 comps
## X 90.50 92.89 94.96 96.81 97.97 98.73 99.39
## Apps 83.86 84.08 84.11 84.11 84.16 84.28 93.08
## 16 comps 17 comps
## X 99.86 100.00
## Apps 93.71 93.95validationplot(pcr.college, val.type="MSEP")
pcr.pred=predict(pcr.college,x[test,],ncomp=10)
mean((pcr.pred-y.test)^2)## [1] 1422699Just by looking at the graph of the model, it seems the MSE is lowest with around 10 components. The summary confirms that using 10 components gives the lowest error and explains most of the variance in both predictors (92.89%) and the response variable (84.08%).
- 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.
pls.college=plsr(Apps~., data=College.train,scale=TRUE, validation="CV")
validationplot(pls.college, val.type="MSEP")
summary(pls.college)## Data: X dimension: 388 17
## Y dimension: 388 1
## Fit method: kernelpls
## Number of components considered: 17
##
## VALIDATION: RMSEP
## Cross-validated using 10 random segments.
## (Intercept) 1 comps 2 comps 3 comps 4 comps 5 comps 6 comps
## CV 4347 2178 1872 1734 1615 1453 1359
## adjCV 4347 2171 1867 1726 1586 1427 1341
## 7 comps 8 comps 9 comps 10 comps 11 comps 12 comps 13 comps
## CV 1347 1340 1329 1317 1310 1305 1305
## adjCV 1330 1324 1314 1302 1296 1291 1291
## 14 comps 15 comps 16 comps 17 comps
## CV 1305 1307 1307 1307
## adjCV 1291 1292 1293 1293
##
## TRAINING: % variance explained
## 1 comps 2 comps 3 comps 4 comps 5 comps 6 comps 7 comps 8 comps
## X 24.27 38.72 62.64 65.26 69.01 73.96 78.86 82.18
## Apps 76.96 84.31 86.80 91.48 93.37 93.75 93.81 93.84
## 9 comps 10 comps 11 comps 12 comps 13 comps 14 comps 15 comps
## X 85.35 87.42 89.18 91.41 92.70 94.58 97.16
## Apps 93.88 93.91 93.93 93.94 93.95 93.95 93.95
## 16 comps 17 comps
## X 98.15 100.00
## Apps 93.95 93.95pls.pred=predict(pls.college,x[test,],ncomp=9)
mean((pls.pred-y.test)^2)## [1] 1049868The PLS model resulted with 11 components because this had the lowest CV at 1310 with high variance explained at 89.18.
- 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 ap- proaches?
test.avg = mean(College.test[, "Apps"])
lm.test.r2 = 1 - mean((College.test[, "Apps"] - pred.app)^2) /mean((College.test[, "Apps"] - test.avg)^2)
ridge.test.r2 = 1 - mean((College.test[, "Apps"] - ridge.pred)^2) /mean((College.test[, "Apps"] - test.avg)^2)
lasso.test.r2 = 1 - mean((College.test[, "Apps"] - lasso.pred)^2) /mean((College.test[, "Apps"] - test.avg)^2)
pcr.test.r2 = 1 - mean((pcr.pred-y.test)^2) /mean((College.test[, "Apps"] - test.avg)^2)
pls.test.r2 = 1 - mean((pls.pred-y.test)^2) /mean((College.test[, "Apps"] - test.avg)^2)
barplot(c(lm.test.r2, ridge.test.r2, lasso.test.r2, pcr.test.r2, pls.test.r2), names.arg=c("OLS", "Ridge", "Lasso", "PCR", "PLS"), main="Test R-squared")
Judging by the test r2 values, these models aren’t too bad in terms of
accuracy. Except for the PCR model, all others show high accuracy.
- We will now try to predict per capita crime rate in the Boston data set.
- 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.
install.packages('leaps')##
## The downloaded binary packages are in
## /var/folders/qq/bfgnq9g96sz_3lrzwxn_zp3c0000gn/T//RtmpphC54P/downloaded_packageslibrary(leaps)library(MASS)
set.seed(1)
attach(Boston)
predict.regsubsets = function(object, newdata, id, ...) {
form = as.formula(object$call[[2]])
mat = model.matrix(form, newdata)
coefi = coef(object, id = id)
mat[, names(coefi)] %*% coefi
}
k = 10
p = ncol(Boston) - 1
folds = sample(rep(1:k, length = nrow(Boston)))
cv.errors = matrix(NA, k, p)
for (i in 1:k) {
best.fit = regsubsets(crim ~ ., data = Boston[folds != i, ], nvmax = p)
for (j in 1:p) {
pred = predict(best.fit, Boston[folds == i, ], id = j)
cv.errors[i, j] = mean((Boston$crim[folds == i] - pred)^2)
}
}
mean.cv.errors <- apply(cv.errors, 2, mean)
plot(mean.cv.errors, type = "b", xlab = "# of variables", ylab = "CV error")
which.min(mean.cv.errors)## [1] 9mean.cv.errors[which.min(mean.cv.errors)]## [1] 42.81453x = model.matrix(crim ~ . - 1, data = Boston)
y = Boston$crim
cv.lasso = cv.glmnet(x, y, type.measure = "mse")
plot(cv.lasso)
coef(cv.lasso)## 14 x 1 sparse Matrix of class "dgCMatrix"
## s1
## (Intercept) 2.176491
## zn .
## indus .
## chas .
## nox .
## rm .
## age .
## dis .
## rad 0.150484
## tax .
## ptratio .
## black .
## lstat .
## medv .sqrt(cv.lasso$cvm[cv.lasso$lambda == cv.lasso$lambda.1se])## [1] 7.921353x = model.matrix(crim ~ . - 1, data = Boston)
y = Boston$crim
cv.ridge = cv.glmnet(x, y, type.measure = "mse", alpha = 0)
plot(cv.ridge)
coef(cv.ridge)## 14 x 1 sparse Matrix of class "dgCMatrix"
## s1
## (Intercept) 1.523899542
## zn -0.002949852
## indus 0.029276741
## chas -0.166526007
## nox 1.874769665
## rm -0.142852604
## age 0.006207995
## dis -0.094547258
## rad 0.045932737
## tax 0.002086668
## ptratio 0.071258052
## black -0.002605281
## lstat 0.035745604
## medv -0.023480540sqrt(cv.ridge$cvm[cv.ridge$lambda == cv.ridge$lambda.1se])## [1] 7.669133pcr.crime = pcr(crim ~ ., data = Boston, scale = TRUE, validation = "CV")
summary(pcr.crime)## Data: X dimension: 506 13
## Y dimension: 506 1
## Fit method: svdpc
## Number of components considered: 13
##
## VALIDATION: RMSEP
## Cross-validated using 10 random segments.
## (Intercept) 1 comps 2 comps 3 comps 4 comps 5 comps 6 comps
## CV 8.61 7.175 7.180 6.724 6.731 6.727 6.727
## adjCV 8.61 7.174 7.179 6.721 6.725 6.724 6.724
## 7 comps 8 comps 9 comps 10 comps 11 comps 12 comps 13 comps
## CV 6.722 6.614 6.618 6.607 6.598 6.553 6.488
## adjCV 6.718 6.609 6.613 6.602 6.592 6.546 6.481
##
## TRAINING: % variance explained
## 1 comps 2 comps 3 comps 4 comps 5 comps 6 comps 7 comps 8 comps
## X 47.70 60.36 69.67 76.45 82.99 88.00 91.14 93.45
## crim 30.69 30.87 39.27 39.61 39.61 39.86 40.14 42.47
## 9 comps 10 comps 11 comps 12 comps 13 comps
## X 95.40 97.04 98.46 99.52 100.0
## crim 42.55 42.78 43.04 44.13 45.4Propose 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. Based on the MSE, the best subset selection model had the lowest cross-validation error with the MSE of 42.8.
Does your chosen model involve all of the features in the data set? Why or why not? The selected model is the best subset selection model, featuring 9 predictors and the lowest Mean Squared Error (MSE). By excluding four out of the 13 predictors, we aim to minimize variance. This model is designed to achieve both low variance and MSE, while maintaining good accuracy.