hw5
Subash Shrestha
4/2/2021
Questions 2,9,11
Question 2:
For parts(a) through (c), indicate which of i.through iv. is correct. Justify your answer.
a) The lasso, relative to least square 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.
Correct Answer for (a) is: iii
Explanation: Since Lasso method is less flexible as compared to the Least square method, it will give improved prediction accuracy when its increase in bias is less than its decrease in variance. When lease square method estimates have high variance, the lasso will yield a reduction in variance for the expense of increase in bias. It will help accurate prediction.
b) The ridge regression, relative to lease squares is:
Correct Answer for (b) is: iii
Explanation:As lambda increases, the flexibility of ridge regression fit decreases leading to decreased variance but increased bias. The slight change in training data leads to produce large change in variance. The lasso performs variance selection and makes it easier to interpret.
c) The non-linear methods, relative to least squares is:
correct Answer for (c) is: ii Non-linear method is more flexible than the least square method so it will yield improved prediction accuracy when its increase in variance is less than its decrease in bias.
Question 9
In this exercise, we will predict the number of applications received using the other variables in the college data set.
- split the data set into a training set and a test set.
library(ISLR)## Warning: package 'ISLR' was built under R version 4.0.2attach(College)
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 ...set.seed(1)
trainingindex <- sample(nrow(College), 0.75 * nrow(College))
head(trainingindex)## [1] 679 129 509 471 299 270train <- College[trainingindex, ]
test <- College[-trainingindex, ]
dim(College)## [1] 777 18dim(train)## [1] 582 18dim(test)## [1] 195 18- Fit a linear model using lease square on the training set, and report the test error obtained
lm.fit <- lm(Apps~., data = train )
summary(lm.fit)##
## Call:
## lm(formula = Apps ~ ., data = train)
##
## Residuals:
## Min 1Q Median 3Q Max
## -5773.1 -425.2 4.5 327.9 7496.3
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -5.784e+02 4.707e+02 -1.229 0.21962
## PrivateYes -4.673e+02 1.571e+02 -2.975 0.00305 **
## Accept 1.712e+00 4.567e-02 37.497 < 2e-16 ***
## Enroll -1.197e+00 2.151e-01 -5.564 4.08e-08 ***
## Top10perc 5.298e+01 6.158e+00 8.603 < 2e-16 ***
## Top25perc -1.528e+01 4.866e+00 -3.141 0.00177 **
## F.Undergrad 7.085e-02 3.760e-02 1.884 0.06002 .
## P.Undergrad 5.771e-02 3.530e-02 1.635 0.10266
## Outstate -8.143e-02 2.077e-02 -3.920 9.95e-05 ***
## Room.Board 1.609e-01 5.361e-02 3.002 0.00280 **
## Books 2.338e-01 2.634e-01 0.887 0.37521
## Personal 6.611e-03 6.850e-02 0.097 0.92315
## PhD -1.114e+01 5.149e+00 -2.163 0.03093 *
## Terminal 9.186e-01 5.709e+00 0.161 0.87223
## S.F.Ratio 1.689e+01 1.542e+01 1.096 0.27368
## perc.alumni 2.256e+00 4.635e+00 0.487 0.62667
## Expend 5.567e-02 1.300e-02 4.284 2.16e-05 ***
## Grad.Rate 6.427e+00 3.307e+00 1.944 0.05243 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1009 on 564 degrees of freedom
## Multiple R-squared: 0.9336, Adjusted R-squared: 0.9316
## F-statistic: 466.7 on 17 and 564 DF, p-value: < 2.2e-16lm.pred = predict( lm.fit, newdata = test )
lm.error = mean((test$Apps - lm.pred)^2 )
lm.error## [1] 1384604The test error obtained from linear model using least square method is 1020100
c) Fit a ridge regression model on the training set, with lambda chosen by cross-validation. Report the test error obtained
library(glmnet)## Warning: package 'glmnet' was built under R version 4.0.2## Loading required package: Matrix## Loaded glmnet 4.1-1xtrain = model.matrix(Apps~., data = train[,-1])
ytrain = train$Apps
xtest = model.matrix(Apps~., data = test [,-1])
ytest = test$Apps
set.seed(1)ridge.fit = cv.glmnet(xtrain, ytrain, alpha = 0 )
plot(ridge.fit)
ridge.lambda = ridge.fit$lambda.min
ridge.lambda## [1] 364.6228ridge.pred = predict(ridge.fit, s = ridge.lambda, newx = xtest)
ridge.error = mean((ridge.pred - ytest)^2)
ridge.error## [1] 1260111- Fit a lasso model on the training set, with lambda chosen by cross- validation. Report the test error obtained, along with the number of non-zero coefficient estimates.
set.seed(1)
lasso.fit = cv.glmnet(xtrain, ytrain, aplha =1)
plot(lasso.fit)
Now we can see the relationship between log lambda and MSE. we are going to find the best lambda that reduces the error.
lasso.lambda = lasso.fit$lambda.min
lasso.lambda## [1] 1.945882We will use this lambda to predict traiing model on test data
lasso.pred = predict (lasso.fit, s = lasso.lambda, newx = xtest )
lasso.error = mean((lasso.pred - ytest )^2)
lasso.error## [1] 1394834The lasso model is not as strong at predicting as the ridge regression on this data
lasso.coe = predict (lasso.fit, type = "coefficients", s = lasso.lambda)[1:18, ]
lasso.coe## (Intercept) (Intercept) Accept Enroll Top10perc
## -1.030320e+03 0.000000e+00 1.693638e+00 -1.100616e+00 5.104012e+01
## Top25perc F.Undergrad P.Undergrad Outstate Room.Board
## -1.369969e+01 7.851307e-02 6.036558e-02 -9.861002e-02 1.475536e-01
## Books Personal PhD Terminal S.F.Ratio
## 2.185146e-01 0.000000e+00 -8.390364e+00 1.349648e+00 2.457249e+01
## perc.alumni Expend Grad.Rate
## 7.684156e-01 5.858007e-02 5.324356e+00We can see that lasso chose variables that are helpful in predicting test dataset. It is more simplier that ridge regression but produces less error.
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.
#install.packages("pls")
library(pls)## Warning: package 'pls' was built under R version 4.0.2##
## Attaching package: 'pls'## The following object is masked from 'package:stats':
##
## loadingspcr.fit = pcr(Apps~., data = train, scale = TRUE, validation = "CV")
validationplot(pcr.fit, val.type = "MSEP")
summary(pcr.fit)## Data: X dimension: 582 17
## Y dimension: 582 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 3862 3773 2100 2106 1812 1678 1674
## adjCV 3862 3772 2097 2104 1800 1666 1668
## 7 comps 8 comps 9 comps 10 comps 11 comps 12 comps 13 comps
## CV 1674 1625 1580 1579 1582 1586 1590
## adjCV 1669 1613 1574 1573 1576 1580 1584
## 14 comps 15 comps 16 comps 17 comps
## CV 1585 1480 1189 1123
## adjCV 1580 1463 1179 1115
##
## TRAINING: % variance explained
## 1 comps 2 comps 3 comps 4 comps 5 comps 6 comps 7 comps 8 comps
## X 32.159 57.17 64.41 70.20 75.53 80.48 84.24 87.56
## Apps 5.226 71.83 71.84 80.02 83.01 83.07 83.21 84.46
## 9 comps 10 comps 11 comps 12 comps 13 comps 14 comps 15 comps
## X 90.54 92.81 94.92 96.73 97.81 98.69 99.35
## Apps 85.00 85.22 85.22 85.23 85.36 85.45 89.93
## 16 comps 17 comps
## X 99.82 100.00
## Apps 92.84 93.36pcr.pred = predict(pcr.fit, test, ncomp = 10 )
pcr.error = mean((pcr.pred - test$Apps)^2)
pcr.error## [1] 1952693- 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- valiation.
pls.fit = plsr(Apps~., data = train, scale = TRUE, validation = "CV")
validationplot(pls.fit, val.type = "MSEP")
summary(pls.fit)## Data: X dimension: 582 17
## Y dimension: 582 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 3862 1916 1708 1496 1415 1261 1181
## adjCV 3862 1912 1706 1489 1396 1237 1170
## 7 comps 8 comps 9 comps 10 comps 11 comps 12 comps 13 comps
## CV 1166 1152 1146 1144 1144 1140 1138
## adjCV 1157 1144 1137 1135 1135 1131 1129
## 14 comps 15 comps 16 comps 17 comps
## CV 1137 1137 1137 1137
## adjCV 1129 1129 1129 1129
##
## TRAINING: % variance explained
## 1 comps 2 comps 3 comps 4 comps 5 comps 6 comps 7 comps 8 comps
## X 25.67 47.09 62.54 65.0 67.54 72.28 76.80 80.63
## Apps 76.80 82.71 87.20 90.8 92.79 93.05 93.14 93.22
## 9 comps 10 comps 11 comps 12 comps 13 comps 14 comps 15 comps
## X 82.71 85.53 88.01 90.95 93.07 95.18 96.86
## Apps 93.30 93.32 93.34 93.35 93.36 93.36 93.36
## 16 comps 17 comps
## X 98.00 100.00
## Apps 93.36 93.36pls.pred = predict(pls.fit, test, ncomp = 7)
pls.error = mean((pls.pred - test$Apps )^2)
pls.error## [1] 1333314- Comment on the result obtained. How accurately can we predict the numbers of collge application received? Is there much difference among the test errors resulting from these five approaches?
barplot(c(lm.error, ridge.error, lasso.error, pcr.error, pls.error),col = "blue", xlab = "Regression Methods", ylab = "Test Error", main = "Test error for all models", names.arg = c("OLS", "Ridge", "Lasso", "PCR", "PLS"))
In order to mention how accurately we can predict, we have to compare r squared values for each methods
avg.apps = mean(test$Apps)
avg.apps## [1] 3268.051lm.r2 = 1 - mean((lm.pred - test$ Apps)^2) / mean((avg.apps-test$Apps)^2)
lm.r2## [1] 0.9086432ridge.r2 = 1 - mean((ridge.pred - test$Apps)^2) / mean ((avg.apps - test$Apps)^2)
ridge.r2## [1] 0.9168573lasso.r2 = 1- mean((lasso.pred - test$Apps)^2) / mean((avg.apps - test$Apps)^2)
lasso.r2## [1] 0.9079682pcr.r2 = 1 - mean((pcr.pred - test$Apps)^2) / mean((avg.apps - test$Apps)^2)
pcr.r2## [1] 0.8711605pls.r2 = 1 - mean((pls.pred - test$Apps)^2) / mean((avg.apps - test$Apps)^2)
pls.r2## [1] 0.9120273barplot(c(lm.r2,ridge.r2, lasso.r2, pcr.r2, pls.r2), col = "grey", xlab = "Regression Methods", ylab = "Test R-squared", main = "Test R squared for all methods", names.arg = c("OLS","Ridge", "Lasso", "PCR", "PLS"))
we can see that with exception of PCR, all other models gave higher accuracy.
Question 11:
We will now try to predict per capita crime rate in the Boston data set.
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.
set.seed(1)
library(MASS)## Warning: package 'MASS' was built under R version 4.0.2data(Boston)
attach(Boston)Best subset selection
library(leaps)## Warning: package 'leaps' was built under R version 4.0.2predict.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)
}
}
rmse.cv = sqrt(apply(cv.errors, 2, mean))
plot(rmse.cv, pch = 19, type ="b")
summary(best.fit)## Subset selection object
## Call: regsubsets.formula(crim ~ ., data = Boston[folds != i, ], nvmax = p)
## 13 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
## black FALSE FALSE
## lstat FALSE FALSE
## medv FALSE FALSE
## 1 subsets of each size up to 13
## Selection Algorithm: exhaustive
## zn indus chas nox rm age dis rad tax ptratio black 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 ) "*" "*" "*" "*" "*" " " "*" "*" "*" "*" "*" "*" "*"
## 13 ( 1 ) "*" "*" "*" "*" "*" "*" "*" "*" "*" "*" "*" "*" "*"which.min(rmse.cv)## [1] 9boston.bsm.error = (rmse.cv[which.min(rmse.cv)])^2
boston.bsm.error## [1] 42.81453The CV estimate for the test MSE is 42.81453, the lowest MSE.
Lasso model
boston.x = model.matrix(crim~., data = Boston)[,-1]
boston.y = Boston$crim
boston.lasso = cv.glmnet(boston.x, boston.y, aplha =1, type.measure = "mse")
plot(boston.lasso)
coef(boston.lasso)## 14 x 1 sparse Matrix of class "dgCMatrix"
## 1
## (Intercept) 2.176491
## zn .
## indus .
## chas .
## nox .
## rm .
## age .
## dis .
## rad 0.150484
## tax .
## ptratio .
## black .
## lstat .
## medv .boston.lasso.error = (boston.lasso$cvm[boston.lasso$lambda == boston.lasso$lambda.1se])
boston.lasso.error## [1] 62.74783Including only one variable rad, MSE is 55.0202
Ridge Regression
boston.ridge = cv.glmnet(boston.x, boston.y, type.measure = "mse", alpha = 0 )
plot(boston.ridge)
coef(boston.ridge)## 14 x 1 sparse Matrix of class "dgCMatrix"
## 1
## (Intercept) 1.523899548
## zn -0.002949852
## indus 0.029276741
## chas -0.166526006
## nox 1.874769661
## 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 -0.023480540boston.ridge.error = boston.ridge$cvm[boston.ridge$lambda == boston.ridge$lambda.1se]
boston.ridge.error## [1] 58.8156MSE = 58.66963
PCR
library(pls)
boston.pcr = pcr(crim~., data = Boston, scale = TRUE, validation = "CV")
summary(boston.pcr)## 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 45.4b)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.
Best subset selection had MSE of 42.81453 which was lowest and best.
c)Does your chosen model involve all of the features in the data set? Why or why not?
The Best subset selection was choosen even though it did not include all the variables. variables included were zn, indus, nox, dis, rad, ptratio, black, lstat, and medv. if all variable were included, more variation will be observed but we are trying to more prediction accuracy wil low variance and low MSE.