Problem 3

We now review k-fold cross-validation.
(a) Explain how k-fold cross-validation is implemented.

k-fold cross-validation is implemented by dividing observations into k groups, using the first fold as the validation set, and fitting the method on the remaining folds.

(b) What are the advantages and disadvantages of k-fold cross-validation relative to:
i. The validation set approach?

The validation set is very simple involving just a training and validation set, meaning error rates can be overestimated.

ii. LOOCV?

LOOCV can be very computationally expensive with a large n value, as it is essentially using n as k. k-folds with a much smaller k value (around 5-10) will still be quite effective with much less computation.

Problem 5

In Chapter 4, we used logistic regression to predict the probability of default using income and balance on the Default data set. We will now estimate the test error of this logistic regression model using the validation set approach. Do not forget to set a random seed before beginning your analysis.

library(ISLR2)
attach(Default)

(a) Fit a logistic regression model that uses income and balance to predict default.

set.seed(1)
glm.fit=glm(default~income+balance, data=Default, family=binomial)

(b) Using the validation set approach, estimate the test error of this model. In order to do this, you must perform the following steps:
i. Split the sample set into a training set and a validation set.
ii. Fit a multiple logistic regression model using only the training observations.
iii. Obtain a prediction of default status for each individual in the validation set by computing the posterior probability of default for that individual, and classifying the individual to the default category if the posterior probability is greater than 0.5.
iv. Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified.

train=sample(dim(Default)[1], dim(Default)[1]/2)
glm.fit=glm(default~income+balance, data=Default, family=binomial, subset=train)
glm.pred=rep("No", dim(Default)[1]/2)
glm.probs=predict(glm.fit, Default[-train,], type="response")
glm.pred[glm.probs>0.5]="Yes"
mean(glm.pred!=Default[-train,]$default)
## [1] 0.0254

(c) Repeat the process in (b) three times, using three different splits of the observations into a training set and a validation set. Comment on the results obtained.

train=sample(dim(Default)[1], dim(Default)[1]/2)
glm.fit=glm(default~income+balance, data=Default, family=binomial, subset=train)
glm.pred=rep("No", dim(Default)[1]/2)
glm.probs=predict(glm.fit, Default[-train,], type="response")
glm.pred[glm.probs>0.5]="Yes"
mean(glm.pred!=Default[-train,]$default)
## [1] 0.0274
train=sample(dim(Default)[1], dim(Default)[1]/2)
glm.fit=glm(default~income+balance, data=Default, family=binomial, subset=train)
glm.pred=rep("No", dim(Default)[1]/2)
glm.probs=predict(glm.fit, Default[-train,], type="response")
glm.pred[glm.probs>0.5]="Yes"
mean(glm.pred!=Default[-train,]$default)
## [1] 0.0244
train=sample(dim(Default)[1], dim(Default)[1]/2)
glm.fit=glm(default~income+balance, data=Default, family=binomial, subset=train)
glm.pred=rep("No", dim(Default)[1]/2)
glm.probs=predict(glm.fit, Default[-train,], type="response")
glm.pred[glm.probs>0.5]="Yes"
mean(glm.pred!=Default[-train,]$default)
## [1] 0.0244

The error rate appears to vary a little bit between 0.024 and 0.028.

(d) Now consider a logistic regression model that predicts the probability of default using income, balance, and a dummy variable for student. Estimate the test error for this model using the validation set approach. Comment on whether or not including a dummy variable for student leads to a reduction in the test error rate.

train=sample(dim(Default)[1], dim(Default)[1]/2)
glm.fit=glm(default~income+balance+student, data=Default, family=binomial, subset=train)
glm.pred=rep("No", dim(Default)[1]/2)
glm.probs=predict(glm.fit, Default[-train,], type="response")
glm.pred[glm.probs>0.5]="Yes"
mean(glm.pred!=Default[-train,]$default)
## [1] 0.0278

Adding in a student dummy variable does not lead to a reduction in test error rate, in fact it appears to be slightly higher.

Problem 6

We continue to consider the use of a logistic regression model to predict the probability of default using income and balance on the Default data set. In particular, we will now compute estimates for the standard errors of the income and balance logistic regression coefficients in two different ways: (1) using the bootstrap, and (2) using the standard formula for computing the standard errors in the glm() function. Do not forget to set a random seed before beginning your analysis.
(a) Using the summary() and glm() functions, determine the estimated standard errors for the coefficients associated with income and balance in a multiple logistic regression model that uses both predictors.

set.seed(1)
glm.fit=glm(default~income+balance, data=Default, family=binomial)
summary(glm.fit)
## 
## Call:
## glm(formula = default ~ income + balance, family = binomial, 
##     data = Default)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.4725  -0.1444  -0.0574  -0.0211   3.7245  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.154e+01  4.348e-01 -26.545  < 2e-16 ***
## income       2.081e-05  4.985e-06   4.174 2.99e-05 ***
## balance      5.647e-03  2.274e-04  24.836  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 2920.6  on 9999  degrees of freedom
## Residual deviance: 1579.0  on 9997  degrees of freedom
## AIC: 1585
## 
## Number of Fisher Scoring iterations: 8

(b) Write a function, boot.fn(), that takes as input the Default data set as well as an index of the observations, and that outputs the coefficient estimates for income and balance in the multiple logistic regression model.

boot.fn=function(data, index){
  fit=glm(default~income+balance, data=data, family=binomial, subset=index)
  return(coef(fit))
}

(c) Use the boot() function together with your boot.fn() function to estimate the standard errors of the logistic regression coefficients for income and balance.

library(boot)
boot(Default, boot.fn, 100)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Default, statistic = boot.fn, R = 100)
## 
## 
## Bootstrap Statistics :
##          original        bias     std. error
## t1* -1.154047e+01  8.556378e-03 4.122015e-01
## t2*  2.080898e-05 -3.993598e-07 4.186088e-06
## t3*  5.647103e-03 -4.116657e-06 2.226242e-04
detach(Default)

(d) Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function.

The estimated standard errors obtained by the two methods are fairly similar.

Problem 9

We will now consider the Boston housing data set, from the ISLR2 library.
(a) Based on this data set, provide an estimate for the population mean of medv . Call this estimate \(\hat{μ}\).

library(MASS)
## 
## Attaching package: 'MASS'
## The following object is masked from 'package:ISLR2':
## 
##     Boston
attach(Boston)
mu.hat=mean(medv)
mu.hat
## [1] 22.53281

(b) Provide an estimate of the standard error of \(\hat{μ}\). Interpret this result.
Hint: We can compute the standard error of the sample mean by dividing the sample standard deviation by the square root of the number of observations.

se.hat=sd(medv)/sqrt(dim(Boston)[1])
se.hat
## [1] 0.4088611

(c) Now estimate the standard error of \(\hat{μ}\) using the bootstrap. How does this compare to your answer from (b)?

set.seed(1)
boot.fn=function(data, index){
    mu=mean(data[index])
    return (mu)
}
boot(medv, boot.fn, 1000)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original      bias    std. error
## t1* 22.53281 0.007650791   0.4106622

The results are slightly different, though very similar.

(d) Based on your bootstrap estimate from (c), provide a 95 % confidence interval for the mean of medv . Compare it to the results obtained using t.test(Boston$medv).
Hint: You can approximate a 95 % confidence interval using the formula \([\hat{μ} − 2SE(\hat{μ}), \hat{μ} + 2SE(\hat{μ})]\).

t.test(medv)
## 
##  One Sample t-test
## 
## data:  medv
## t = 55.111, df = 505, p-value < 2.2e-16
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
##  21.72953 23.33608
## sample estimates:
## mean of x 
##  22.53281
CI.mu.hat<-c(mu.hat-2*0.4107, mu.hat+2*0.4107)
CI.mu.hat
## [1] 21.71141 23.35421

The confidence interval from the bootstrap is very similar to the one found using t.test.

(e) Based on this data set, provide an estimate, \(\hat{μ}_{med}\), for the median value of medv in the population.

median(medv)
## [1] 21.2

(f) We now would like to estimate the standard error of \(\hat{μ}_{med}\). Unfortunately, there is no simple formula for computing the standard error of the median. Instead, estimate the standard error of the median using the bootstrap. Comment on your findings.

boot.fn=function(data, index){
    mu=median(data[index])
    return (mu)
}
boot(medv, boot.fn, 1000)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*     21.2 -0.0386   0.3770241

We get a median of 21.2 which matches the value found in (e) with a standard error of 0.377

(g) Based on this data set, provide an estimate for the tenth percentile of medv in Boston census tracts. Call this quantity \(\hat{μ}_{0.1}\). (You can use the quantile() function.)

quantile(medv,.1)
##   10% 
## 12.75

(h) Use the bootstrap to estimate the standard error of \(\hat{μ}_{0.1}\). Comment on your findings.

boot.fn=function(data, index){
    mu=quantile(data[index],0.1)
    return (mu)
}
boot(medv, boot.fn, 1000)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*    12.75  0.0186   0.4925766

The standard error for the quantile is 0.4925