3.) We now review k-fold cross-validation.

  1. Explain how k-fold cross-validation is implemented.

The k-fold cross validation is implemented by taking the number of observations (n) and randomly dividing them into k groups of roughly equal size. These groups act as a validation set, and the remainder acts as a training set. The test error is then estimated by averaging the k resulting MSE estimates.

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

An advantage k-fold cross validation has over the validation set is the validation estimate of the test error rate can be highly variable, depending on precisely which observations are included in the training set and which observations are included in the validation set. And the validation set error rate may tend to overestimate the test error rate for the model fit on the entire data set.

A disadvantage of k-fold cross validation has compared to the validation set is the validation set approach is much simpler and easier to implement.

  1. LOOCV?

One advantage the k-fold cross validation has over LOOCV is LOOCV requires fitting the statistical learning method n times. This has the potential to complicate models when n reaches very high numbers. Also, k-fold cross validation often gives more accurate estimates of the test error rate than LOOCV.

A disadvantage that k-fold cross validation has to LOOCV is LOOCV tends to have less bias than k-fold cross validation.

There is a bias-variance trade-off associated with the choice of k in k-fold cross-validation. Typically using k=5 or k=10 results in test error rate estimates that don’t suffer from excessively high bias or from high variance.

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.

  1. Fit a logistic regression model that uses income and balance to predict default.
library(ISLR)
## Warning: package 'ISLR' was built under R version 4.1.3
Default = Default
set.seed(20)
fit.def = glm(default ~ income + balance, data = Default, family = "binomial")
summary(fit.def)
## 
## 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
  1. Using the validation set approach, estimate the test error of this model. In order to do this, you must perform the following steps:

  1. Split the sample set into a training set and a validation set.

  2. Fit a multiple logistic regression model using only the training observations.

  3. 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.

  4. Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified.

  #i
def.train = sample(dim(Default)[1], dim(Default)[1] / 2)

  #ii
fit.def.train = glm(default ~ income + balance, data = Default, family = "binomial", subset = def.train)
summary(fit.def.train)
## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default, subset = def.train)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.6003  -0.1285  -0.0481  -0.0170   3.7909  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.222e+01  6.764e-01 -18.074  < 2e-16 ***
## income       2.440e-05  7.382e-06   3.305  0.00095 ***
## balance      5.999e-03  3.537e-04  16.958  < 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: 1382.01  on 4999  degrees of freedom
## Residual deviance:  716.55  on 4997  degrees of freedom
## AIC: 722.55
## 
## Number of Fisher Scoring iterations: 8
  #iii
probs = predict(fit.def, newdata = Default[-def.train, ], type = "response")
pred.glm = rep("No", length(probs))
pred.glm[probs > 0.5] = "Yes"

  #iv
mean(pred.glm != Default[-def.train, ]$default)
## [1] 0.0288
#2.86% test error rate with validation set approach.
  1. Repeat the process in (b) three times, using three different splits of the observations into a training set and a validation set. Comment on the results obtained.
train = sample(dim(Default)[1], dim(Default)[1] / 2)
fit.glm = glm(default ~ income + balance, data = Default, family = "binomial", subset = train)
probs = predict(fit.glm, newdata = Default[-train, ], type = "response")
pred.glm = rep("No", length(probs))
pred.glm[probs > 0.5] = "Yes"
mean(pred.glm != Default[-train, ]$default)
## [1] 0.0256
train = sample(dim(Default)[1], dim(Default)[1] / 3)
fit.glm = glm(default ~ income + balance, data = Default, family = "binomial", subset = train)
probs = predict(fit.glm, newdata = Default[-train, ], type = "response")
pred.glm = rep("No", length(probs))
pred.glm[probs > 0.5] = "Yes"
mean(pred.glm != Default[-train, ]$default)
## [1] 0.02684866
train = sample(dim(Default)[1], dim(Default)[1] / 4)
fit.glm = glm(default ~ income + balance, data = Default, family = "binomial", subset = train)
probs = predict(fit.glm, newdata = Default[-train, ], type = "response")
pred.glm = rep("No", length(probs))
pred.glm[probs > 0.5] = "Yes"
mean(pred.glm != Default[-train, ]$default)
## [1] 0.02733333

The validation estimate of the test error rate can be variable, depending on the split determining which observations are included in the training set and which observations are included in the validation set.

  1. 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.
set.seed(6)
subset = sample(nrow(Default), nrow(Default)*0.7)
default.train = Default[subset,]
default.test = Default[-subset,]

dum = glm(default ~ income + balance + student, family = binomial, data=default.train)
summary(dum)
## 
## Call:
## glm(formula = default ~ income + balance + student, family = binomial, 
##     data = default.train)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.2048  -0.1318  -0.0503  -0.0179   3.7413  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.125e+01  6.069e-01 -18.533   <2e-16 ***
## income       2.394e-06  9.888e-06   0.242   0.8087    
## balance      5.976e-03  2.886e-04  20.709   <2e-16 ***
## studentYes  -7.037e-01  2.841e-01  -2.477   0.0133 *  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 2064.0  on 6999  degrees of freedom
## Residual deviance: 1067.9  on 6996  degrees of freedom
## AIC: 1075.9
## 
## Number of Fisher Scoring iterations: 8
predict.dum = predict(dum, default.test, type="response")
class.dum = ifelse(predict.dum>0.7,"Yes","No")

table(default.test$default, class.dum, dnn=c("Actual","Predicted"))
##       Predicted
## Actual   No  Yes
##    No  2900    3
##    Yes   83   14
round(mean(class.dum!=default.test$default),4)
## [1] 0.0287

Adding the “student” dummy variable doesn’t lead to a reduction in the validation set estimate of the test error rate.

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.

  1. 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(32)
attach(Default)

fit.glm = glm(default ~ income + balance, data = Default, family = "binomial")
summary(fit.glm)
## 
## 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

SE estimates for coefficients are respectively 0.4348, 4.985e-6, and 2.274e-04.

  1. 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))}
  1. 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, 1000)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Default, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##          original        bias     std. error
## t1* -1.154047e+01 -2.093229e-02 4.157888e-01
## t2*  2.080898e-05  7.080066e-08 4.760232e-06
## t3*  5.647103e-03  9.980680e-06 2.161297e-04

SE estimates for coefficients are respectively 0.4158, 4.760e-06, and 2.161e-04.

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

Estimated standard errors obtained by the two methods are very similar.

9.) We will now consider the Boston housing data set, from the ISLR2 library.

  1. Based on this data set, provide an estimate for the population mean of medv. Call this estimate ˆμ.
library(ISLR2)
## Warning: package 'ISLR2' was built under R version 4.1.3
## 
## Attaching package: 'ISLR2'
## The following objects are masked from 'package:ISLR':
## 
##     Auto, Credit
attach(Boston)
mu.hat = mean(medv)
mu.hat
## [1] 22.53281
  1. Provide an estimate of the standard error of ˆμ. Interpret this result.
se.hat = sd(medv) / sqrt(dim(Boston)[1])
se.hat
## [1] 0.4088611
  1. Now estimate the standard error of ˆμ using the bootstrap. How does this compare to your answer from (b)?
set.seed(999)
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 -0.008379249   0.4101731

Bootstrap estimated standard error of μˆ of 0.4101 is very close to the estimate found in (b) of 0.4089.

  1. 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).
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(22.53 - 2 * 0.4101, 22.53 + 2 * 0.4101)
CI.mu.hat
## [1] 21.7098 23.3502

Bootstrap confidence interval is very close to the one provided by the t.test() function.

  1. Based on this data set, provide an estimate, ˆμmed, for the median value of medv in the population.
med.hat = median(medv)
med.hat
## [1] 21.2
  1. We now would like to estimate the standard error of ˆμmed. 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  0.0015   0.3677608

Estimated median value of 21.2 which is equal to the value obtained in (e), with a standard error of 0.3678 which is relatively small compared to median value.

  1. Based on this data set, provide an estimate for the tenth percentile of medv in Boston census tracts. Call this quantity ˆμ0.1.
perc.hat = quantile(medv, c(0.1))
perc.hat
##   10% 
## 12.75
  1. Use the bootstrap to estimate the standard error of ˆμ0.1. Comment on your findings.
boot.fn = function(data, index)
{mu = quantile(data[index], c(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 -0.00335    0.511486

Estimated tenth percentile value of 12.75 which is again equal to the value obtained in (g), with a standard error of 0.5115 which is relatively small compared to percentile value.