Home work 5
Karim I
2022-10-31
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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. (b) Repeat (a) for ridge regression relative to least squares. (c) Repeat (a) for non-linear methods relative to least squares.
less flexible, and hence more accurate, when the increase in bias is less than the decrease in variance. The Lasso solution might result in a decrease in variance with a limited increase in bias. This has the potential to improve forecast accuracy. Lasso’s advantage demonstrates that least squares is rooted in bias-variance and also conducts variable selection, which can aid in the interpretation of other approaches.
less flexible, and hence more accurate, when the increase in bias is less than the decrease in variance. The bias variance trade-off underpins both Lasso’s advantage and Ridge regression. The variable diminishes as lambda grows, whereas the ridge regression fit lowers but increases its bias. When there is a little change in the training data, the least squares coefficient yields a significant change and a greater number for variance. Ridge regression can still perform by trading off a modest increase in bias for a significant decrease in variance.
More flexible, and hence more accurate, when the rise in variance is less than the decrease in bias.
- In this exercise, we will predict the number of applications received using the other variables in the College data set.
library(ISLR2)
data(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 ...pairs(College)
- Split the data set into a training set and a test set.
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 least squares on the training set, and report the test error obtained.
1384604 is the linear model fit on all variables. While the rs is at 93.63%. We can try to reduce the mistake and raise the r2.
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.err=mean((test$Apps-lm.pred)^2)
lm.err## [1] 1384604- Fit a ridge regression model on the training set, with λ chosen by cross-validation. Report the test error obtained.
xtrain=model.matrix(Apps~., data=train[,-1])
ytrain=train$Apps
xtest=model.matrix(Apps~., data=test[,-1])
ytest=test$Apps
set.seed(1)
library(glmnet)## Loading required package: Matrix## Loaded glmnet 4.1-4ridge.fit=cv.glmnet(xtrain, ytrain, alpha=0)
plot(ridge.fit)
ridge.lambda=ridge.fit$lambda.min
ridge.lambda## [1] 364.6228ridge.pred=predict(ridge.fit, s=ridge.lambda, newx = xtest)
ridge.err=mean((ridge.pred-ytest)^2)
ridge.err## [1] 1260111Based on our findings, we may conclude that rdige regression outperforms OLS. We can see from our data that the ridge regression has a flaw in that it reduces the coefficients to zero for some variables, resulting in a prediction accuracy that is difficult to understand.
- Fit a lasso model on the training set, with λ chosen by crossvalidation. Report the test error obtained, along with the number of non-zero coefficient estimates.
Lasso identified a variable that might be useful in predicting the test dataset. By forecasting accuracy, we can observe that a simpler model did not perform as well as the ridge regression. However, because the variance of the ridge regression was smaller than that of the Lasso, we can see that it created less error. The function of the predictors may have created the ridge-outperformance response.
set.seed(1)
lasso.fit=cv.glmnet(xtrain, ytrain, alpha=1)
plot(lasso.fit)
lasso.lambda=lasso.fit$lambda.min
lasso.lambda## [1] 1.945882lasso.pred=predict(lasso.fit, s=lasso.lambda, newx = xtest)
lasso.err=mean((lasso.pred-ytest)^2)
lasso.err## [1] 1394834lasso.coeff=predict(lasso.fit, type="coefficients", s=lasso.lambda)[1:18,]
lasso.coeff## (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+00- Fit a PCR model on the training set, with M chosen by crossvalidation. Report the test error obtained, along with the value of M selected by cross-validation.
We can see that the PCR model did not beat any of the preceding models since we picked a large number of primary components. With the PCR model, the bias decreased while the variance rose. The U-shape for the mistakes is caused by CV faults. Reducing the number of principal components would not have resulted in a better conclusion since the CV error would have been larger.
library(pls)##
## 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.err=mean((pcr.pred-test$Apps)^2)
pcr.err## [1] 1952693- Fit a PLS model on the training set, with M chosen by crossvalidation. Report the test error obtained, along with the value of M selected by cross-validation.
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.err=mean((pls.pred-test$Apps)^2)
pls.err## [1] 1333314Comment on the results obtained. How accurately can we pre- dict the number of college applications received? Is there much difference among the test errors resulting from these five ap- proaches?
Except for PCR, we may conclude that all have high accuracy rates based on the r2.
avg.apps=mean(test$Apps)
lm.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] 0.8711605We 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. The model’s predictors explain 99.52% of the variance in the applicable PCR model, which has 12 components. The crime rate in 12 components is 44.13%. This is an excellent prediction model in addition to the BSM.
set.seed(1)
library(MASS)##
## Attaching package: 'MASS'## The following object is masked from 'package:ISLR2':
##
## Bostondata(Boston)
attach(Boston)
library(leaps)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)
}
}
rmse.cv = sqrt(apply(cv.errors, 2, mean))
plot(rmse.cv, pch = 19, type = "b")
boston.bsm.err=(rmse.cv[which.min(rmse.cv)])^2
boston.bsm.err## [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 45.4- 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, crossvalidation, or some other reasonable alternative, as opposed to using training error.
The computations above assist us in answering this question. The best subset selection approach identified the model with the lowest cross-validation error, and this model had an MSE of 44.32201.
- Does your chosen model involve all of the features in the data set? Why or why not? The optimal subset selection contains just indus, zn, nox, rad, dis, ptratio,lstat, and medv as variables, for a total of nine variables. If the model did not filter out the other variables, we may have greater variation. As a result, the more variables we have, the better our prediction accuracy will be, as we are seeking for higher prediction accuracy and lower MSE.