HW4
Vu Hong Ha
2023-05-24
#Question 13 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.
library(pacman)
p_load(class, tidyverse,MASS, corrplot, kableExtra, ISLR2)
data("Weekly")
head(Weekly)## Year Lag1 Lag2 Lag3 Lag4 Lag5 Volume Today Direction
## 1 1990 0.816 1.572 -3.936 -0.229 -3.484 0.1549760 -0.270 Down
## 2 1990 -0.270 0.816 1.572 -3.936 -0.229 0.1485740 -2.576 Down
## 3 1990 -2.576 -0.270 0.816 1.572 -3.936 0.1598375 3.514 Up
## 4 1990 3.514 -2.576 -0.270 0.816 1.572 0.1616300 0.712 Up
## 5 1990 0.712 3.514 -2.576 -0.270 0.816 0.1537280 1.178 Up
## 6 1990 1.178 0.712 3.514 -2.576 -0.270 0.1544440 -1.372 Down(a) Produce some numerical and graphical summaries of the Weekly data
summary(Weekly)## Year Lag1 Lag2 Lag3
## Min. :1990 Min. :-18.1950 Min. :-18.1950 Min. :-18.1950
## 1st Qu.:1995 1st Qu.: -1.1540 1st Qu.: -1.1540 1st Qu.: -1.1580
## Median :2000 Median : 0.2410 Median : 0.2410 Median : 0.2410
## Mean :2000 Mean : 0.1506 Mean : 0.1511 Mean : 0.1472
## 3rd Qu.:2005 3rd Qu.: 1.4050 3rd Qu.: 1.4090 3rd Qu.: 1.4090
## Max. :2010 Max. : 12.0260 Max. : 12.0260 Max. : 12.0260
## Lag4 Lag5 Volume Today
## Min. :-18.1950 Min. :-18.1950 Min. :0.08747 Min. :-18.1950
## 1st Qu.: -1.1580 1st Qu.: -1.1660 1st Qu.:0.33202 1st Qu.: -1.1540
## Median : 0.2380 Median : 0.2340 Median :1.00268 Median : 0.2410
## Mean : 0.1458 Mean : 0.1399 Mean :1.57462 Mean : 0.1499
## 3rd Qu.: 1.4090 3rd Qu.: 1.4050 3rd Qu.:2.05373 3rd Qu.: 1.4050
## Max. : 12.0260 Max. : 12.0260 Max. :9.32821 Max. : 12.0260
## Direction
## Down:484
## Up :605
##
##
##
## corrplot(cor(Weekly[,-9]), method="square")
(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.
attach(Weekly)
Weekly.fit<-glm(Direction~Lag1+Lag2+Lag3+Lag4+Lag5+Volume, data=Weekly, family=binomial)
summary(Weekly.fit)##
## Call:
## glm(formula = Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 +
## Volume, family = binomial, data = Weekly)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.6949 -1.2565 0.9913 1.0849 1.4579
##
## 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
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1496.2 on 1088 degrees of freedom
## Residual deviance: 1486.4 on 1082 degrees of freedom
## AIC: 1500.4
##
## Number of Fisher Scoring iterations: 4(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.
Wk.probs <- predict(Weekly.fit, type = "response")
Wk.probs[1:10]## 1 2 3 4 5 6 7 8
## 0.6086249 0.6010314 0.5875699 0.4816416 0.6169013 0.5684190 0.5786097 0.5151972
## 9 10
## 0.5715200 0.5554287contrasts(Direction)## Up
## Down 0
## Up 1Wk.pred <- rep("Down", length(Wk.probs))
Wk.pred[Wk.probs > .5] = "Up"
table(Wk.pred, Direction)## Direction
## Wk.pred Down Up
## Down 54 48
## Up 430 557(54 + 557) / length(Wk.probs)## [1] 0.5610652mean(Wk.pred == Direction)## [1] 0.5610652(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.test <- Weekly[!train, ]
dim(Weekly.test)## [1] 104 9Direction.test <- Direction[!train]
Weekly.fit<-glm(Direction~Lag2, data=Weekly,family=binomial, subset=train)
logWeekly.prob= predict(Weekly.fit, Weekly.test, type = "response")
logWeekly.pred = rep("Down", length(logWeekly.prob))
logWeekly.pred[logWeekly.prob > 0.5] = "Up"
table(logWeekly.pred, Direction.test)## Direction.test
## logWeekly.pred Down Up
## Down 9 5
## Up 34 56mean(logWeekly.pred == Direction.test)## [1] 0.625(e) Repeat (d) using LDA
library(MASS)
lda.fit <- lda(Direction ~ Lag2, data = Weekly,
subset = train)
lda.fit## Call:
## lda(Direction ~ Lag2, data = Weekly, subset = train)
##
## 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 0.4414162plot(lda.fit)
lda.pred <- predict(lda.fit, Weekly.test)
names(lda.pred)## [1] "class" "posterior" "x"lda.class <- lda.pred$class
table(lda.class, Direction.test)## Direction.test
## lda.class Down Up
## Down 9 5
## Up 34 56mean(lda.class == Direction.test)## [1] 0.625(f) Repeat (d) using QDA.
qda.fit <- qda(Direction ~ Lag2, data = Weekly,
subset = train)
qda.fit## Call:
## qda(Direction ~ Lag2, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581qda.class <- predict(qda.fit, Weekly.test)$class
table(qda.class, Direction.test)## Direction.test
## qda.class Down Up
## Down 0 0
## Up 43 61mean(qda.class == Direction.test)## [1] 0.5865385(g) Repeat (d) using KNN with K = 1
Wk.train=as.matrix(Lag2[train])
Wk.test=as.matrix(Lag2[!train])
train.Direction =Direction[train]
set.seed(1)
knn.pred=knn(Wk.train,Wk.test,train.Direction,k=1)
table(knn.pred,Direction.test)## Direction.test
## knn.pred Down Up
## Down 21 30
## Up 22 31mean(knn.pred == Direction.test)## [1] 0.5(h) Repeat (d) using naive Bayes
library(e1071)
nb.fit <- naiveBayes(Direction ~ Lag2, data = Weekly,
subset = train)
nb.fit##
## Naive Bayes Classifier for Discrete Predictors
##
## Call:
## naiveBayes.default(x = X, y = Y, laplace = laplace)
##
## A-priori probabilities:
## Y
## Down Up
## 0.4477157 0.5522843
##
## Conditional probabilities:
## Lag2
## Y [,1] [,2]
## Down -0.03568254 2.199504
## Up 0.26036581 2.317485nb.class <- predict(nb.fit, Weekly.test)
table(nb.class, Direction.test)## Direction.test
## nb.class Down Up
## Down 0 0
## Up 43 61mean(nb.class == Direction.test)## [1] 0.5865385(i) Which of these methods appears to provide the best results on this data?
print("The methods that have the highest accuracy rates are the Logistic Regression and Linear Discriminant Analysis; both having rates of 62.5%.")## [1] "The methods that have the highest accuracy rates are the Logistic Regression and Linear Discriminant Analysis; both having rates of 62.5%."(j) 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.
We combine 2 variables: log 1 and log 2 to test
GLM method
Weekly.fit<-glm(Direction~ Lag1 + Lag2, data=Weekly,family=binomial, subset=train)
logWeekly.prob= predict(Weekly.fit, Weekly.test, type = "response")
logWeekly.pred = rep("Down", length(logWeekly.prob))
logWeekly.pred[logWeekly.prob > 0.5] = "Up"
table(logWeekly.pred, Direction.test)## Direction.test
## logWeekly.pred Down Up
## Down 7 8
## Up 36 53mean(logWeekly.pred == Direction.test)## [1] 0.5769231lda
lda.fit <- lda(Direction ~ Lag1 + Lag2, data = Weekly,
subset = train)
lda.fit## Call:
## lda(Direction ~ Lag1 + Lag2, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag1 Lag2
## Down 0.289444444 -0.03568254
## Up -0.009213235 0.26036581
##
## Coefficients of linear discriminants:
## LD1
## Lag1 -0.3013148
## Lag2 0.2982579plot(lda.fit)
lda.pred <- predict(lda.fit, Weekly.test)
names(lda.pred)## [1] "class" "posterior" "x"lda.class <- lda.pred$class
table(lda.class, Direction.test)## Direction.test
## lda.class Down Up
## Down 7 8
## Up 36 53mean(lda.class == Direction.test)## [1] 0.5769231qda
qda.fit <- qda(Direction ~ Lag1 + Lag2, data = Weekly,
subset = train)
qda.fit## Call:
## qda(Direction ~ Lag1 + Lag2, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag1 Lag2
## Down 0.289444444 -0.03568254
## Up -0.009213235 0.26036581qda.class <- predict(qda.fit, Weekly.test)$class
table(qda.class, Direction.test)## Direction.test
## qda.class Down Up
## Down 7 10
## Up 36 51mean(qda.class == Direction.test)## [1] 0.5576923knn (k=10)
Wk.train=as.matrix(Lag2[train])
Wk.test=as.matrix(Lag2[!train])
train.Direction =Direction[train]
set.seed(1)
knn.pred=knn(Wk.train,Wk.test,train.Direction,k=10)
table(knn.pred,Direction.test)## Direction.test
## knn.pred Down Up
## Down 17 21
## Up 26 40mean(knn.pred == Direction.test)## [1] 0.5480769naive Bayes
nb.fit <- naiveBayes(Direction ~ Lag1 + Lag2, data = Weekly,
subset = train)
nb.fit##
## Naive Bayes Classifier for Discrete Predictors
##
## Call:
## naiveBayes.default(x = X, y = Y, laplace = laplace)
##
## A-priori probabilities:
## Y
## Down Up
## 0.4477157 0.5522843
##
## Conditional probabilities:
## Lag1
## Y [,1] [,2]
## Down 0.289444444 2.211721
## Up -0.009213235 2.308387
##
## Lag2
## Y [,1] [,2]
## Down -0.03568254 2.199504
## Up 0.26036581 2.317485nb.class <- predict(nb.fit, Weekly.test)
table(nb.class, Direction.test)## Direction.test
## nb.class Down Up
## Down 3 8
## Up 40 53mean(nb.class == Direction.test)## [1] 0.5384615