MS 4373 Assignment 8

Q5. We have seen that we can fit an SVM with a non-linear kernel in order to perform classification using a non-linear decision boundary. We will now see that we can also obtain a non-linear decision boundary by performing logistic regression using non-linear transformations of the features.

(a) Generate a data set with n=500 and p=2, such that the observations belong to two classes with a quadratic decision boundary between them.

set.seed(1)
x1 <- runif(500) - 0.5
x2 <- runif(500) - 0.5
y <- 1 * (x1^2 - x2^2 > 0)

(b) Plot the observations, colored according to their class labels. Your plot should display X1 on the x-axis and X2 on the y-axis.

plot(x1[y == 0], x2[y == 0], col = "maroon", xlab = "X1", ylab = "X2")
points(x1[y == 1], x2[y == 1], col = "cornflowerblue")

(c) Fit a logistic regression model to the data, using X1 and X2 as predictors.

df <- data.frame(x1 = x1, x2 = x2, y = as.factor(y))
logit <- glm(y ~ x1 + x2, data = df, family = "binomial")
summary(logit)

## 
## Call:
## glm(formula = y ~ x1 + x2, family = "binomial", data = df)
## 
## Deviance Residuals: 
##    Min      1Q  Median      3Q     Max  
## -1.179  -1.139  -1.112   1.206   1.257  
## 
## Coefficients:
##              Estimate Std. Error z value Pr(>|z|)
## (Intercept) -0.087260   0.089579  -0.974    0.330
## x1           0.196199   0.316864   0.619    0.536
## x2          -0.002854   0.305712  -0.009    0.993
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 692.18  on 499  degrees of freedom
## Residual deviance: 691.79  on 497  degrees of freedom
## AIC: 697.79
## 
## Number of Fisher Scoring iterations: 3

(d) Apply this model to training data in order to obtain a predicted class label for each training observation. Plot the observations, colored according to the predicted class labels. The decision boundary should be linear.

logit.prob <- predict(logit, newdata = df, type = 'response')
logit.pred <- ifelse(logit.prob > 0.5, 1, 0)
plot(df$x1, df$x2, , col = logit.pred + 2)

(e) Now ﬁt a logistic regression model to the data using non-linear functions of X1 and X2 as predictors (e.g. X21 , X1×X2, log(X2), and so forth).

nonlin <- glm(y ~ poly(x1, 2) + poly(x2, 2), data = df, family = 'binomial')

## Warning: glm.fit: algorithm did not converge

## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurred

summary(nonlin)

## 
## Call:
## glm(formula = y ~ poly(x1, 2) + poly(x2, 2), family = "binomial", 
##     data = df)
## 
## Deviance Residuals: 
##        Min          1Q      Median          3Q         Max  
## -1.079e-03  -2.000e-08  -2.000e-08   2.000e-08   1.297e-03  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)
## (Intercept)     -94.48    2963.78  -0.032    0.975
## poly(x1, 2)1   3442.52  104411.28   0.033    0.974
## poly(x1, 2)2  30110.74  858421.66   0.035    0.972
## poly(x2, 2)1    162.82   26961.99   0.006    0.995
## poly(x2, 2)2 -31383.76  895267.48  -0.035    0.972
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 6.9218e+02  on 499  degrees of freedom
## Residual deviance: 4.2881e-06  on 495  degrees of freedom
## AIC: 10
## 
## Number of Fisher Scoring iterations: 25

(f) Apply this model to the training data in order to obtain a predicted class label for each training observation. Plot the observations, colored according to the predicted class labels. The decision boundary should be obviously non-linear. If it is not, then repeat (a)-(e) until you come up with an example in which the predicted class labels are obviously non-linear.

nonlin.prob <- predict(nonlin, newdata = df, type = 'response')
nonlin.pred <- ifelse(nonlin.prob > 0.5, 1, 0)
plot(df$x1, df$x2, col = nonlin.pred + 2)

(g) Fit a support vector classiﬁer to the data with X1 and X2 as predictors. Obtain a class prediction for each training observation. Plot the observations, colored according to the predicted class labels.

library(e1071)
svm1 <- svm(y ~ ., data = df, kernel = 'linear', cost = 0.01)
plot(svm1, df)

(h) Fit a SVM using a non-linear kernel to the data. Obtain a class prediction for each training observation. Plot the observations, colored according to the predicted class labels.

svm.nl <- svm(y ~ ., data = df, kernel = 'radial', gamma = 1)
plot(svm.nl, data = df)

(i) Comment on your results.

Non-linear kernel SVMs perform well at identifying non-linear boundaries. The logistic regression models also perform well. However, SVM with linear kernel and logistic regression without an interaction term do not perform as well when finding non-linear decision boundaries.

Q7. In this problem, you will use support vector approaches in order to predict whether a given car gets high or low gas mileage based on the Auto data set.

(a) Create a binary variable that takes on a 1 for cars with gas mileage above the median, and a 0 for cars with gas mileage below the median.

library(ISLR)
var <- ifelse(Auto$mpg > median(Auto$mpg), 1, 0)
Auto$mpglevel <- as.factor(var)

(b) Fit a support vector classiﬁer to the data with various values of cost, in order to predict whether a car gets high or low gas mileage. Report the cross-validation errors associated with different values of this parameter. Comment on your results.

library(e1071)
set.seed(1)
tune.out = tune(svm, mpglevel ~ ., data = Auto, kernel = "linear", ranges = list(cost = c(0.01, 0.1, 1, 5, 10, 100)))
summary(tune.out)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     1
## 
## - best performance: 0.01025641 
## 
## - Detailed performance results:
##    cost      error dispersion
## 1 1e-02 0.07653846 0.03617137
## 2 1e-01 0.04596154 0.03378238
## 3 1e+00 0.01025641 0.01792836
## 4 5e+00 0.02051282 0.02648194
## 5 1e+01 0.02051282 0.02648194
## 6 1e+02 0.03076923 0.03151981

A cost of 1 results in the lowest error rate of 0.01.

(c) Now repeat (b), this time using SVMs with radial and polynomial basis kernels, with different values of gamma and degree and cost. Comment on your results.

set.seed(1)
tune.out = tune(svm, mpglevel ~ ., data = Auto, kernel = "polynomial", ranges = list(cost = c(0.1, 1, 5, 10), degree = c(2, 3, 4)))
summary(tune.out)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost degree
##    10      2
## 
## - best performance: 0.5130128 
## 
## - Detailed performance results:
##    cost degree     error dispersion
## 1   0.1      2 0.5511538 0.04366593
## 2   1.0      2 0.5511538 0.04366593
## 3   5.0      2 0.5511538 0.04366593
## 4  10.0      2 0.5130128 0.08963366
## 5   0.1      3 0.5511538 0.04366593
## 6   1.0      3 0.5511538 0.04366593
## 7   5.0      3 0.5511538 0.04366593
## 8  10.0      3 0.5511538 0.04366593
## 9   0.1      4 0.5511538 0.04366593
## 10  1.0      4 0.5511538 0.04366593
## 11  5.0      4 0.5511538 0.04366593
## 12 10.0      4 0.5511538 0.04366593

set.seed(1)
tune.out = tune(svm, mpglevel ~ ., data = Auto, kernel = "radial", ranges = list(cost = c(0.1, 1, 5, 10), gamma = c(0.01, 0.1, 1, 5, 10, 100)))
summary(tune.out)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost gamma
##    10  0.01
## 
## - best performance: 0.02557692 
## 
## - Detailed performance results:
##    cost gamma      error dispersion
## 1   0.1 1e-02 0.08929487 0.04382379
## 2   1.0 1e-02 0.07403846 0.03522110
## 3   5.0 1e-02 0.04852564 0.03303346
## 4  10.0 1e-02 0.02557692 0.02093679
## 5   0.1 1e-01 0.07903846 0.03874545
## 6   1.0 1e-01 0.05371795 0.03525162
## 7   5.0 1e-01 0.02820513 0.03299190
## 8  10.0 1e-01 0.03076923 0.03375798
## 9   0.1 1e+00 0.55115385 0.04366593
## 10  1.0 1e+00 0.06384615 0.04375618
## 11  5.0 1e+00 0.05884615 0.04020934
## 12 10.0 1e+00 0.05884615 0.04020934
## 13  0.1 5e+00 0.55115385 0.04366593
## 14  1.0 5e+00 0.49493590 0.04724924
## 15  5.0 5e+00 0.48217949 0.05470903
## 16 10.0 5e+00 0.48217949 0.05470903
## 17  0.1 1e+01 0.55115385 0.04366593
## 18  1.0 1e+01 0.51794872 0.05063697
## 19  5.0 1e+01 0.51794872 0.04917316
## 20 10.0 1e+01 0.51794872 0.04917316
## 21  0.1 1e+02 0.55115385 0.04366593
## 22  1.0 1e+02 0.55115385 0.04366593
## 23  5.0 1e+02 0.55115385 0.04366593
## 24 10.0 1e+02 0.55115385 0.04366593

Adjusting the values of degree increases the error, especially in comparison to the model in part b. The smaller the gamma value, the smaller the error rate will be.

(d) Make some plots to justify (C) and (D)

svm.linear = svm(mpglevel ~ ., data = Auto, kernel = "linear", cost = 1)
svm.poly = svm(mpglevel ~ ., data = Auto, kernel = "polynomial", cost = 10, 
    degree = 2)
svm.radial = svm(mpglevel ~ ., data = Auto, kernel = "radial", cost = 10, gamma = 0.01)
plotpairs = function(fit) {
    for (name in names(Auto)[!(names(Auto) %in% c("mpg", "mpglevel", "name"))]) {
        plot(fit, Auto, as.formula(paste("mpg~", name, sep = "")))
    }
}
plotpairs(svm.linear)

plotpairs(svm.poly)

plotpairs(svm.radial)

Q8. This problem involves the OJ data set which is part of the ISLR package.

(a) Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.

library(ISLR)
set.seed(1)
train = sample(dim(OJ)[1], 800)
OJ.train = OJ[train, ]
OJ.test = OJ[-train, ]

(b) Fit a support vector classifier to the training data using cost=0.01, with Purchase as the response and the other variables as predictors. Use the summary() function to produce summary statistics, and describe the results obtained.

library(e1071)
svm.linear = svm(Purchase ~ ., kernel = "linear", data = OJ.train, cost = 0.01)
summary(svm.linear)

## 
## Call:
## svm(formula = Purchase ~ ., data = OJ.train, kernel = "linear", cost = 0.01)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  0.01 
## 
## Number of Support Vectors:  435
## 
##  ( 219 216 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

(c) What are the training and test error rates?

train.pred = predict(svm.linear, OJ.train)
table(OJ.train$Purchase, train.pred)

##     train.pred
##       CH  MM
##   CH 420  65
##   MM  75 240

(75 + 65)/(420 + 65 + 75 + 240)

## [1] 0.175

test.pred = predict(svm.linear, OJ.test)
table(OJ.test$Purchase, test.pred)

##     test.pred
##       CH  MM
##   CH 153  15
##   MM  33  69

(33 + 15)/(153 + 15 + 33 + 69)

## [1] 0.1777778

The training error rate is 0.175 while the test error rate is 0.178.

(d) Use the tune() function to select an optimal cost. Consider values in the range 0.01 to 10.

set.seed(1554)
tune.out = tune(svm, Purchase ~ ., data = OJ.train, kernel = "linear", ranges = list(cost = 10^seq(-2, 1, by = 0.25)))
summary(tune.out)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##       cost
##  0.3162278
## 
## - best performance: 0.17125 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.17750 0.06635343
## 2   0.01778279 0.17750 0.05916080
## 3   0.03162278 0.17500 0.06095308
## 4   0.05623413 0.17375 0.06755913
## 5   0.10000000 0.17625 0.06755913
## 6   0.17782794 0.17625 0.06573569
## 7   0.31622777 0.17125 0.06483151
## 8   0.56234133 0.17375 0.06573569
## 9   1.00000000 0.17250 0.06258328
## 10  1.77827941 0.17500 0.06997023
## 11  3.16227766 0.17250 0.06661456
## 12  5.62341325 0.17625 0.07155272
## 13 10.00000000 0.17875 0.07072295

(e) Compute the training and test error rates using this new value for cost.

svm.linear = svm(Purchase ~ ., kernel = "linear", data = OJ.train, cost = tune.out$best.parameters$cost)
train.pred = predict(svm.linear, OJ.train)
table(OJ.train$Purchase, train.pred)

##     train.pred
##       CH  MM
##   CH 423  62
##   MM  71 244

(71 + 62)/(423 + 62 + 71 + 244)

## [1] 0.16625

test.pred = predict(svm.linear, OJ.test)
table(OJ.test$Purchase, test.pred)

##     test.pred
##       CH  MM
##   CH 155  13
##   MM  29  73

(29 + 13)/(155 + 13 + 29 + 73)

## [1] 0.1555556

The training error rate is 0.166 and the test error is 0.1555

Repeat parts (b) through (e) using a support vector machine with a radial kernel. Use the default value for gamma.

set.seed(410)
svm.radial = svm(Purchase ~ ., data = OJ.train, kernel = "radial")
summary(svm.radial)

## 
## Call:
## svm(formula = Purchase ~ ., data = OJ.train, kernel = "radial")
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  1 
## 
## Number of Support Vectors:  373
## 
##  ( 188 185 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

train.pred = predict(svm.radial, OJ.train)
table(OJ.train$Purchase, train.pred)

##     train.pred
##       CH  MM
##   CH 441  44
##   MM  77 238

(77 + 44)/(441 + 44 + 77 + 238)

## [1] 0.15125

test.pred = predict(svm.radial, OJ.test)
table(OJ.test$Purchase, test.pred)

##     test.pred
##       CH  MM
##   CH 151  17
##   MM  33  69

(33 + 17)/(151 + 17 + 33 + 69)

## [1] 0.1851852

The training error for radical is 0.15 and the test error rate is 0.185.

set.seed(1)
tune.out = tune(svm, Purchase ~ ., data = OJ.train, kernel = "radial", ranges = list(cost = 10^seq(-2, 1, by = 0.25)))
summary(tune.out)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##       cost
##  0.5623413
## 
## - best performance: 0.16875 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.39375 0.04007372
## 2   0.01778279 0.39375 0.04007372
## 3   0.03162278 0.35750 0.05927806
## 4   0.05623413 0.19500 0.02443813
## 5   0.10000000 0.18625 0.02853482
## 6   0.17782794 0.18250 0.03291403
## 7   0.31622777 0.17875 0.03230175
## 8   0.56234133 0.16875 0.02651650
## 9   1.00000000 0.17125 0.02128673
## 10  1.77827941 0.17625 0.02079162
## 11  3.16227766 0.17750 0.02266912
## 12  5.62341325 0.18000 0.02220485
## 13 10.00000000 0.18625 0.02853482

svm.radial = svm(Purchase ~ ., data = OJ.train, kernel = "radial", cost = tune.out$best.parameters$cost)
train.pred = predict(svm.radial, OJ.train)
table(OJ.train$Purchase, train.pred)

##     train.pred
##       CH  MM
##   CH 437  48
##   MM  71 244

(71 + 48)/(437 + 48 + 71 + 244)

## [1] 0.14875

test.pred = predict(svm.radial, OJ.test)
table(OJ.test$Purchase, test.pred)

##     test.pred
##       CH  MM
##   CH 150  18
##   MM  30  72

(30 + 18)/(150 + 18 + 30 + 72)

## [1] 0.1777778

Now, after tuning, the training error is 0.149 and the test error is 0.178.

(g) Repeat parts (b) through (e) using a support vector machine with a polynomial kernel. Set degree=2.

set.seed(1)
svm.poly = svm(Purchase ~ ., data = OJ.train, kernel = "poly", degree = 2)
summary(svm.poly)

## 
## Call:
## svm(formula = Purchase ~ ., data = OJ.train, kernel = "poly", degree = 2)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  polynomial 
##        cost:  1 
##      degree:  2 
##      coef.0:  0 
## 
## Number of Support Vectors:  447
## 
##  ( 225 222 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

train.pred = predict(svm.poly, OJ.train)
table(OJ.train$Purchase, train.pred)

##     train.pred
##       CH  MM
##   CH 449  36
##   MM 110 205

(110 + 36)/(449 + 36 + 110 + 205)

## [1] 0.1825

test.pred = predict(svm.poly, OJ.test)
table(OJ.test$Purchase, test.pred)

##     test.pred
##       CH  MM
##   CH 153  15
##   MM  45  57

(45 + 15)/(153 + 15 + 45 + 57)

## [1] 0.2222222

The training and testing error rates using a svm with degree 2 is 0.18 and 0.22 respectively.

set.seed(1)
tune.out = tune(svm, Purchase ~ ., data = OJ.train, kernel = "poly", degree = 2, 
    ranges = list(cost = 10^seq(-2, 1, by = 0.25)))
summary(tune.out)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##      cost
##  3.162278
## 
## - best performance: 0.1775 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.39125 0.04210189
## 2   0.01778279 0.37125 0.03537988
## 3   0.03162278 0.36500 0.03476109
## 4   0.05623413 0.33750 0.04714045
## 5   0.10000000 0.32125 0.05001736
## 6   0.17782794 0.24500 0.04758034
## 7   0.31622777 0.19875 0.03972562
## 8   0.56234133 0.20500 0.03961621
## 9   1.00000000 0.20250 0.04116363
## 10  1.77827941 0.18500 0.04199868
## 11  3.16227766 0.17750 0.03670453
## 12  5.62341325 0.18375 0.03064696
## 13 10.00000000 0.18125 0.02779513

svm.poly = svm(Purchase ~ ., data = OJ.train, kernel = "poly", degree = 2, cost = tune.out$best.parameters$cost)
train.pred = predict(svm.poly, OJ.train)
table(OJ.train$Purchase, train.pred)

##     train.pred
##       CH  MM
##   CH 451  34
##   MM  90 225

(90 + 34)/(451 + 34 + 90 + 225)

## [1] 0.155

test.pred = predict(svm.poly, OJ.test)
table(OJ.test$Purchase, test.pred)

##     test.pred
##       CH  MM
##   CH 154  14
##   MM  41  61

(41 + 14)/(154 + 14 + 41 + 61)

## [1] 0.2037037

MS 4373 Assignment 8

Katelyn Huynh

May 13, 2020

Q5. We have seen that we can fit an SVM with a non-linear kernel in order to perform classification using a non-linear decision boundary. We will now see that we can also obtain a non-linear decision boundary by performing logistic regression using non-linear transformations of the features.

(a) Generate a data set with n=500 and p=2, such that the observations belong to two classes with a quadratic decision boundary between them.

(b) Plot the observations, colored according to their class labels. Your plot should display X1 on the x-axis and X2 on the y-axis.

(c) Fit a logistic regression model to the data, using X1 and X2 as predictors.

(d) Apply this model to training data in order to obtain a predicted class label for each training observation. Plot the observations, colored according to the predicted class labels. The decision boundary should be linear.

(e) Now ﬁt a logistic regression model to the data using non-linear functions of X1 and X2 as predictors (e.g. X21 , X1×X2, log(X2), and so forth).

(g) Fit a support vector classiﬁer to the data with X1 and X2 as predictors. Obtain a class prediction for each training observation. Plot the observations, colored according to the predicted class labels.

(h) Fit a SVM using a non-linear kernel to the data. Obtain a class prediction for each training observation. Plot the observations, colored according to the predicted class labels.

(i) Comment on your results.

Non-linear kernel SVMs perform well at identifying non-linear boundaries. The logistic regression models also perform well. However, SVM with linear kernel and logistic regression without an interaction term do not perform as well when finding non-linear decision boundaries.

Q7. In this problem, you will use support vector approaches in order to predict whether a given car gets high or low gas mileage based on the Auto data set.

(a) Create a binary variable that takes on a 1 for cars with gas mileage above the median, and a 0 for cars with gas mileage below the median.

(b) Fit a support vector classiﬁer to the data with various values of cost, in order to predict whether a car gets high or low gas mileage. Report the cross-validation errors associated with different values of this parameter. Comment on your results.

A cost of 1 results in the lowest error rate of 0.01.

(c) Now repeat (b), this time using SVMs with radial and polynomial basis kernels, with different values of gamma and degree and cost. Comment on your results.

Adjusting the values of degree increases the error, especially in comparison to the model in part b. The smaller the gamma value, the smaller the error rate will be.

(d) Make some plots to justify (C) and (D)

Q8. This problem involves the OJ data set which is part of the ISLR package.

(a) Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.

(b) Fit a support vector classifier to the training data using cost=0.01, with Purchase as the response and the other variables as predictors. Use the summary() function to produce summary statistics, and describe the results obtained.

(c) What are the training and test error rates?

The training error rate is 0.175 while the test error rate is 0.178.

(d) Use the tune() function to select an optimal cost. Consider values in the range 0.01 to 10.

(e) Compute the training and test error rates using this new value for cost.

The training error rate is 0.166 and the test error is 0.1555

Repeat parts (b) through (e) using a support vector machine with a radial kernel. Use the default value for gamma.

The training error for radical is 0.15 and the test error rate is 0.185.

Now, after tuning, the training error is 0.149 and the test error is 0.178.

(g) Repeat parts (b) through (e) using a support vector machine with a polynomial kernel. Set degree=2.

The training and testing error rates using a svm with degree 2 is 0.18 and 0.22 respectively.

After tuning, the training and testing error rates are 0.156 and 0.20 respectively.

(h) Overall, which approach seems to give the best results on this data?

The radial approach appears to be the best option given its error rates.