Q10
This question should be answered using the Weekly data set, which is part of the ISLR package. This data is similar in nature to the Smarket data from this chapter’s lab, except that it contains 1,089 weekly returns for 21 years, from the beginning of 1990 to the end of 2010.
10.a
Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
library(ISLR)
summary(Weekly)
cor(Weekly)
attach(Weekly)
plot(Volume)
The correlations between the "lag" variables and today's returns are close to zero.The only substantial correlation is between "Year" and "Volume", when we plot "Volume", we see that it is increasing over time.
10.b
Use the full data set to perform a logistic regression with Direction as the response and the five lag variables plus Volume as predictors. Use the summary function to print the results. Do any of the predictors appear to be statistically significant? If so, which ones?
fit.glm<-glm(Direction~Lag1+Lag2+Lag3+Lag4+Lag5+Volume, data=Weekly, family=binomial)
summary(fit.glm)
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) 0.26686 0.08593 3.106 0.0019 **
Lag1 -0.04127 0.02641 -1.563 0.1181
Lag2 0.05844 0.02686 2.175 0.0296 *
Lag3 -0.01606 0.02666 -0.602 0.5469
Lag4 -0.02779 0.02646 -1.050 0.2937
Lag5 -0.01447 0.02638 -0.549 0.5833
Volume -0.02274 0.03690 -0.616 0.5377
It would seem that "Lag2" is the only predictor statistically significant as its p-value is less than 0.05
10.c
Compute the confusion matrix and overall fraction of correct predictions. Explain what the confusion matrix is telling you about the types of mistakes made by logistic regression.
probs<-predict(fit.glm, type="response")
pred.glm<-rep("Down", length(prbos))
pred.glm[probs > 0.5] <- "Up"
table(pred.glm, Direction)
Direction
pred.glm Down Up
Down 54 48
Up 430 557
We may conclude that the percentage of correct predictions on the training data is which is equal to . In other words, is the training error rate, which is often overly optimistic.
- True positive rate: for weeks when the market goes up, the model is right of the time (557/(48+557))
- True Negative rate: for weeks when the market goes down, the model is right only of the time (54/(54+430))
10.d
Now fit the logistic regression model using a training data period from 1990 to 2008, with Lag2 as the only predictor. Compute the confusion matrix and the overall fraction of correct predictions for the held out data (that is, the data from 2009 and 2010).
train <- (Year < 2009)
Weekly.20092010 <- Weekly[!train,]
Direction.20092010 <- Direction[!train]
fit.glm2 <- glm(Direction~Lag2, data=Weekly, family=binomial, subset=train)
summary(fit.glm2)
probs2 <- predict(fit.glm2, Weekly.20092010, type = "response")
pred.glm2 <- rep("Down", length(probs2))
pred.glm2[probs2 > 0.5] <- "Up"
table(pred.glm2, Direction.20092010)
Direction
pred.glm Down Up
Down 9 5
Up 34 56
In this case, we may conclude that the percentage of correct predictions on the test data is . In other words, is the test error rate.
- For weeks goes up, the model is right
- For weeks goes down, the model is right only
10.e
Repeat (d) using LDA.
library(MASS)
fit.lda <- lda(Direction ~ Lag2, data=Weekly, subset=train)
fit.lda
Prior probabilities of groups:
Down Up
0.4477157 0.5522843
Group means:
Lag2
Down -0.03568254
Up 0.26036581
Coefficients of linear discriminants:
LD1
Lag2 0.4414162
pred.lda <- predict(fit.lda, Weekly.20092010, type = "response")
table(pred.lda$class, Direction.20092010)
Direction.20092010
Down Up
Down 9 5
Up 34 56
The percentage of correct predictions on the test data is .
- For weeks goes up, the model is right
- For weeks goes down, the model is right only
10.f
Repeat (d) using QDA.
fit.qda <- qda(Direction ~ Lag2, data=Weekly, subset=train)
fit.qda
Prior probabilities of groups:
Down Up
0.4477157 0.5522843
Group means:
Lag2
Down -0.03568254
Up 0.26036581
Coefficients of linear discriminants:
LD1
Lag2 0.4414162
pred.qda <- predict(fit.qda, Weekly.20092010)
table(pred.qda$class, Direction.20092010)
Direction.20092010
Down Up
Down 0 0
Up 43 61
The percentage of correct predictions on the test data is .In other words is the test error rate.
- For weeks goes up, the model is right
- For weeks goes down, the model is right only
We may note, that QDA achieves a correctness of even though the model chooses “Up” the whole time.
10.g
Repeat (d) using KNN with K = 1.
library(class)
train.X <- as.matrix(Lag2[train])
test.X <- as.matrix(Lag2[!train])
train.Direction <- Direction[train]
set.seed(1)
pred.knn <- knn(train.X, test.X, train.Direction, k = 1)
table(pred.knn, Direction.20092010)
Direction.20092010
pred.knn Down Up
Down 21 30
Up 22 31
The percentage of correct predictions on the test data is . In other words is the test error rate.
- For weeks goes up, the model is right
- For weeks goes down, the model is right only
10.h
Which of these methods appears to provide the best results on this data?
If we compare the test error rate, we see that logistic regression and LDA have the minimum error rates, followed by QDA and KNN.
10.i
Experiment with different combinations of predictors, including possible transformations and interactions, for each of the methods. Report the variables, method, and associated confusion matrix that appears to provide the best results on the held out data. Note that you should also experiment with values for K in the KNN classifier.
# Logistic regression with Lag2:Lag1
fit.glm3 <- glm(Direction ~ Lag2:Lag1, data = Weekly, family = binomial, subset = train)
probs3 <- predict(fit.glm3, Weekly.20092010, type = "response")
pred.glm3 <- rep("Down", length(probs3))
pred.glm3[probs3 > 0.5] = "Up"
table(pred.glm3, Direction.20092010)
mean(pred.glm3 == Direction.20092010)
0.5865385
# LDA with Lag2 interaction with Lag1
fit.lda2 <- lda(Direction ~ Lag2:Lag1, data = Weekly, subset = train)
pred.lda2 <- predict(fit.lda2, Weekly.20092010)
mean(pred.lda2$class == Direction.20092010)
0.5769231
# QDA with sqrt(abs(Lag2))
fit.qda2 <- qda(Direction ~ Lag2 + sqrt(abs(Lag2)), data = Weekly, subset = train)
pred.qda2 <- predict(fit.qda2, Weekly.20092010)
table(pred.qda2$class, Direction.20092010)
mean(pred.qda2$class == Direction.20092010)
0.5769231
# KNN k =10
pred.knn2 <- knn(train.X, test.X, train.Direction, k = 10)
table(pred.knn2, Direction.20092010)
mean(pred.knn2 == Direction.20092010)
# KNN k = 100
pred.knn3 <- knn(train.X, test.X, train.Direction, k = 100)
table(pred.knn3, Direction.20092010)
mean(pred.knn3 == Direction.20092010)
0.5576923
Out of these combinations, the original logistic regression and LDA have the best performance in terms of test error rates.
11
In this problem, you will develop a model to predict whether a given car gets high or low gas mileage based on the Auto data set.
11.a
Create a binary variable, mpg01, that contains a 1 if mpg contains a value above its median, and a 0 if mpg contains a value below its median. You can compute the median using the median() function. Note you may find it helpful to use the data.frame() function to create a single data set containing both mpg01 and the other Auto variables.
attach(Auto)
mpg01 <- rep(0, length(mpg))
mpg01[mpg > median(mpg)] <- 1
Auto <- data.frame(Auto, mpg01)
11.b
Explore the data graphically in order to investigate the association between mpg01 and the other features. Which of the other features seem most likely to be useful in predicting mpg01? Scatterplots and boxplots may be useful tools to answer this question. Describe your findings.
cor(Auto[, -9])
mpg cylinders displacement horsepower weight acceleration year origin mpg01
mpg 1.0000000 -0.7776175 -0.8051269 -0.7784268 -0.8322442 0.4233285 0.5805410 0.5652088 0.8369392
cylinders -0.7776175 1.0000000 0.9508233 0.8429834 0.8975273 -0.5046834 -0.3456474 -0.5689316 -0.7591939
displacement -0.8051269 0.9508233 1.0000000 0.8972570 0.9329944 -0.5438005 -0.3698552 -0.6145351 -0.7534766
horsepower -0.7784268 0.8429834 0.8972570 1.0000000 0.8645377 -0.6891955 -0.4163615 -0.4551715 -0.6670526
weight -0.8322442 0.8975273 0.9329944 0.8645377 1.0000000 -0.4168392 -0.3091199 -0.5850054 -0.7577566
acceleration 0.4233285 -0.5046834 -0.5438005 -0.6891955 -0.4168392 1.0000000 0.2903161 0.2127458 0.3468215
year 0.5805410 -0.3456474 -0.3698552 -0.4163615 -0.3091199 0.2903161 1.0000000 0.1815277 0.4299042
origin 0.5652088 -0.5689316 -0.6145351 -0.4551715 -0.5850054 0.2127458 0.1815277 1.0000000 0.5136984
mpg01 0.8369392 -0.7591939 -0.7534766 -0.6670526 -0.7577566 0.3468215 0.4299042 0.5136984 1.0000000
pairs(Auto)
boxplot(cylinders ~ mpg01, data=Auto, main="cylinders vs mpg01")
boxplot(displacement ~ mpg01, data=Auto, main="displacement vs mpg01")
boxplot(horsepower ~ mpg01, data=Auto, main="horsepower vs mpg01")
boxplot(weight ~ mpg01, data=Auto, main="weight vs mpg01")
boxplot(acceleration ~ mpg01, data=Auto, main="acceleration vs mpg01")
boxplot(year ~ mpg01, data=Auto, main="year vs mpg01")
boxplot(origin ~ mpg01, data=Auto, main="origin vs mpg01")
We may conclude that there exists some association between "mpg01" and "cyclinders","weight","displacement" and "horsepower".
11.c
Split the data into a training set and a test set.
train <- (year %%2 == 0)
Auto.train <- Auto[train,]
Auto.test <- Auto[!train,]
mpg01.test <- mpg01[!train]
11.d
Perform LDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
library(MASS)
fit.lda <- lda(mpg01 ~ cylinders+weight+displacement+horsepower, data=Auto, subset=train)
fit.lda
pred.lda <- predict(fit.lda, Auto.test)
table(pred.lda$class, mpg01.test)
mean(pred.lda$class != mpg01.test)
[1] 0.1263736
11.e
Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
fit.qda <- qda(mpg01 ~ cylinders+weight+displacement+horsepower, data=Auto, subset=train)
fit.qda
pred.qda <- predict(fit.qda, Auto.test)
table(pred.qda$class, mpg01.test)
mean(pred.lda$class != mpg01.test)
[1] 0.1263736
11.f
Perform logistic regression on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
fit.glm <- glm(mpg01 ~ cylinders+weight+displacement+horsepower, data=Auto, family=binomial, subset=train)
summary(fit.glm)
probs <- predict(fit.glm, Auto.test, type = "response")
pred.glm <- rep(0, length(probs))
pred.glm[probs > 0.5] = 1
table(pred.glm, mpg01.test)
mean(pred.glm != mpg01.test)
[1] 0.1208791
11.g
Perform KNN on the training data, with several values of K, in order to predict mpg01. Use only the variables that seemed most associated with mpg01 in (b). What test errors do you obtain? Which value of K seems to perform the best on this data set?
train.X <- cbind(cylinders, weight, displacement, horsepower)[train, ]
test.X <- cbind(cylinders, weight, displacement, horsepower)[!train, ]
train.mpg01 <- mpg01[train]
set.seed(1)
pred.knn <- knn(train.X, test.X, train.mpg01, k = 1)
table(pred.knn, mpg01.test)
mean(pred.knn!=mpg01.test)
0.1538462
pred.knn <- knn(train.X, test.X, train.mpg01, k = 10)
table(pred.knn, mpg01.test)
mean(pred.knn!=mpg01.test)
[1] 0.1648352
pred.knn <- knn(train.X, test.X, train.mpg01, k = 100)
table(pred.knn, mpg01.test)
mean(pred.knn!=mpg01.test)
0.1428571
We have concluded that we have a test error of for K=100. So, a K value of 100 seems to perform the best
12
This problem involves writing functions.
12.a
Write a function, Power(), that prints out the result of raising 2 to the 3rd power. In other words, your function should compute 23 and print out the results.
Power2 <- function() {
2^3
}
Power()
12.b
Create a new function, Power2(), that allows you to pass any two numbers, x and a, and prints out the value of x^a. You can do this by beginning your function with the line.
Power2=function(x,a){
x^a
}
Power2(3,8)
12.c
Using the Power2() function that you just wrote, compute , , and .
Power2(10, 3)
Power2(8, 17)
Power2(131, 3)
12.d
Now create a new function, Power3(), that actually returns the result x^a as an R object, rather than simply printing it to the screen. That is, if you store the value x^a in an object called result within your function, then you can simply return() this result, using the following line:
Power3 <- function(x , a) {
result <- x^a
return(result)
}
12.e
Now using the Power3() function, create a plot of . The x-axis should display a range of integers from 1 to 10, and the y-axis should display . Label the axes appropriately, and use an appropriate title for the figure. Consider displaying either the x-axis, the y-axis, or both on the log-scale. You can do this by using log=‘‘x’’, log=‘‘y’, or log=‘‘xy’’ as arguments to the plot() function.
x<-1:10
plot(x， Power(x, 2), xlab="Log of x", ylab="Log of x^2", main="Log of x^2 vs Log of x")
12.f
Create a function, PlotPower(), that allows you to create a plot of x against x^a for a fixed a and for a range of values of x. For instance, if you call
PlotPower <- function(x, a) {
plot(x, Power3(x, a))
}
PlotPower(1:10, 3)
13
Using the Boston data set, fit classification models in order to predict whether a given suburb has a crime rate above or below the median. Explore logistic regression, LDA, and KNN models using various sub- sets of the predictors. Describe your findings.
attach(Boston)
crim01 <- rep(0, length(crim))
crim01[crim > median(crim)] <- 1
Auto <- data.frame(Boston, crim01)
train <- 1:(length(crim)/2)
test <- (length(crim)/2) + 1 : length(crim)
Boston.train <- Boston[train,]
Boston.test <- Boston[test,]
crim01.test <- crim01[test]
## logistic regression
fit.glm <- glm(crim01 ~ . - crim01 - crim, data=Boston, family=binomial, subset=train)
#summary(fit.glm)
probs <- predict(fit.glm, crim01.test, type = "response")
pred.glm <- rep(0, length(probs))
pred.glm[probs > 0.5] = 1
table(pred.glm, crim01.test)
mean(pred.glm != crim01.test)
[1] 0.1818182
We may conclude that, for this logistic regression, we have a test error rate of .
## LDA
library(MASS)
fit.lda <- lda(crim01 ~ . - crim01 - crim, data=Boston, subset=train)
#fit.lda
pred.lda <- predict(fit.lda, Boston.test)
table(pred.lda$class, crim01.test)
mean(pred.lda$class != crim01.test)
[1] 0.1343874
We may conclude that, for this LDA, we have a test error rate of .
## QDA
fit.qda <- qda(crim01 ~ . - crim01 - crim, data=Boston, subset=train)
#fit.lda <- lda(crim01 ~ . - crim01 - crim - chas - nox, data = Boston, subset = train)
pred.qda <- predict(fit.qda, Boston.test)
table(pred.qda$class, crim01.test)
mean(pred.lda$class != crim01.test)
0.1343874
#0.1501976
## KNN
### k=1
train.X <- cbind(zn, indus, chas, nox, rm, age, dis, rad, tax, ptratio, black, lstat, medv)[train, ]
test.X <- cbind(zn, indus, chas, nox, rm, age, dis, rad, tax, ptratio, black, lstat, medv)[!train, ]
train.crim01 <- crim01[train]
set.seed(1)
pred.knn <- knn(train.X, test.X, train.crim01, k = 1)
table(pred.knn, crim01.test)
mean(pred.knn!= crim01.test)
0.458498
### k=10
train.X <- cbind(zn, indus, chas, nox, rm, age, dis, rad, tax, ptratio, black, lstat, medv)[train, ]
test.X <- cbind(zn, indus, chas, nox, rm, age, dis, rad, tax, ptratio, black, lstat, medv)[!train, ]
train.crim01 <- crim01[train]
set.seed(1)
pred.knn <- knn(train.X, test.X, train.crim01, k = 10)
table(pred.knn, crim01.test)
mean(pred.knn!= crim01.test)
0.1185771
### k=10
train.X <- cbind(zn, indus, chas, nox, rm, age, dis, rad, tax, ptratio, black, lstat, medv)[train, ]
test.X <- cbind(zn, indus, chas, nox, rm, age, dis, rad, tax, ptratio, black, lstat, medv)[!train, ]
train.crim01 <- crim01[train]
set.seed(1)
pred.knn <- knn(train.X, test.X, train.crim01, k = 100)
table(pred.knn, crim01.test)
mean(pred.knn!= crim01.test)
0.4901186