1. We now review k-fold cross-validation.
  1. Explain how k-fold cross-validation is implemented.

First, the set of observations is randomly divided into k groups, or folds, of approximately equal size. From these, The first fold is treated as a validation set, and the method is fit on the remaining k − 1 folds. The mean squared error is then computed on the observations in the held-out fold. This procedure is repeated k times, with a different group of observations is treated as a validation set for each iteration. As a result, we are left with k estimates of the test error; these values are then averaged to arrive at the k-fold CV estimate.

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

Though he validation set approach is conceptually simple and easy to implement, it has a disadvantage of leading to overestimates of the test error rate, since in this approach the training set used to fit the statistical learning method contains only half the observations of the entire data set. Additionally, its estimate of the test error rate can be highly variable.

(b.ii) What are the advantages and disadvantages of k-fold cross-validation relative to LOOCV?

A disadvantage of LOOCV is that it requires fitting the statistical learning method n times which has the potential to be computationally expensive. In addition, since the mean of many highly correlated quantities has higher variance than does the mean of many quantities that are not as highly correlated, the test error estimate resulting from LOOCV tends to have higher variance than does the test error estimate resulting from k-fold CV. This is because When performing LOOCV, we are essentially averaging the outputs of n fitted models that are trained on an almost identical set of observations. As a result, these outputs are highly correlated with each other.

Alternatively to the validation set approach, LOOCV will give approximately unbiased estimates of the test error, since each training set contains n − 1 observations, which is almost as many as the number of observations in the full data set. Therefore, from the perspective of bias reduction, LOOCV is to be preferred to k-fold CV.

Ultimately, there is a bias-variance trade-off associated with the choice of k in k-fold cross-validation. In turn, k-fold cross-validation using k = 5 or k = 10 is typically performed, as these values have been shown empirically to yield test error rate estimates that suffer neither from excessively high bias nor from very high variance.

  1. 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)
set.seed(4)
fit_g1 <- glm(default ~ income + balance, data = Default, family = "binomial")
summary(fit_g1)
## 
## 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:

(b.i) Split the sample set into a training set and a validation set.

set.seed(1)
training_data <- sample(nrow(Default), nrow(Default) / 2)

(b.ii) Fit a multiple logistic regression model using only the training observations.

fit_g2 <- glm(default ~ income + balance, data = Default, family = "binomial", subset = training_data)
summary(fit_g2)
## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default, subset = training_data)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.5830  -0.1428  -0.0573  -0.0213   3.3395  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.194e+01  6.178e-01 -19.333  < 2e-16 ***
## income       3.262e-05  7.024e-06   4.644 3.41e-06 ***
## balance      5.689e-03  3.158e-04  18.014  < 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: 1523.8  on 4999  degrees of freedom
## Residual deviance:  803.3  on 4997  degrees of freedom
## AIC: 809.3
## 
## Number of Fisher Scoring iterations: 8

(b.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.

post_prob <- predict(fit_g2, newdata = Default[-training_data,], type = "response")
g2_prediction <- rep("No", length(post_prob))
g2_prediction[post_prob > 0.5] <- "Yes"

(b.iv) Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified.

As shown below, the resultant validation set error is 2.54%.

mean(g2_prediction != Default[-training_data,]$default)
## [1] 0.0254
  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.

Comments: as mentioned within the textbook, the variability of the validation set error associated with this approach can be observed. This variability results from the changes of which observations are included in the training and validation set during each iteration of the below for loop.

for(i in 1:3){
  set.seed(i)
  tmp_training_data <- sample(nrow(Default), nrow(Default) / 2)
  tmp_fit <- glm(default ~ income + balance, data = Default, family = "binomial", subset = tmp_training_data)
  tmp_post_prob <- predict(tmp_fit, newdata = Default[-tmp_training_data,], type = "response")
  tmp_prediction <- rep("No", length(tmp_post_prob))
  tmp_prediction[tmp_post_prob > 0.5] <- "Yes"
  print(mean(tmp_prediction != Default[-tmp_training_data,]$default))
}
## [1] 0.0254
## [1] 0.0238
## [1] 0.0264
  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.

Comment: as shown below, including a dummy variable for student does not appear to lead to a reduction in the test error rate. In fact, when the following code was ran for seeds 1-3, the test error rate in each case increased.

set.seed(3)
tmp_training_data <- sample(nrow(Default), nrow(Default) / 2)
tmp_fit <- glm(default ~ income + balance + student, data = Default, family = "binomial", subset = tmp_training_data)
tmp_post_prob <- predict(tmp_fit, newdata = Default[-tmp_training_data,], type = "response")
tmp_prediction <- rep("No", length(tmp_post_prob))
tmp_prediction[tmp_post_prob > 0.5] <- "Yes"
mean(tmp_prediction != Default[-tmp_training_data,]$default)
## [1] 0.0272
  1. 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.

The estimated standard errors for the coefficients are shown below:

fit_q6a <- glm(default ~ income + balance, data = Default, family = "binomial")
summary(fit_q6a)
## 
## 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. 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) {
    tmp_fit <- glm(default ~ income + balance, data = data, family = "binomial", subset = index)
    return (coef(tmp_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.

The outputted estimate of the standard errors of the logistic regression coefficients are shown below:

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 -4.138024e-02 4.381705e-01
## t2*  2.080898e-05  2.071209e-07 4.860557e-06
## t3*  5.647103e-03  1.975295e-05 2.310331e-04
  1. Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function.

As can be seen from the above two outputs, there was consistency between the estimated standard errors obtained from glm() and boot.fn(), and ultimately were similar.

  1. We will now consider the Boston housing data set, from the MASS library.
  1. Based on this data set, provide an estimate for the population mean of medv. Call this estimate \(\hat\mu\).
library(MASS)
mu_hat = mean(Boston$medv)
mu_hat
## [1] 22.53281
  1. Provide an estimate of the standard error of \(\hat\mu\). Interpret this result.

Interpretation: the standard error of \(\hat\mu\) shown below is arrived at by dividing the sample standard deviation by the square root of the amount of observations.

mu_hat_err = sd(Boston$medv) / sqrt(nrow(Boston))
mu_hat_err
## [1] 0.4088611
  1. Now estimate the standard error of \(\hat\mu\) using the bootstrap. How does this compare to your answer from (b)?

Comment: as outputted below, the bootstrap estimated standard error of 0.4186383 follows suit with 0.4088611 outputted in part (b).

set.seed(4)
boot.fn <- function(data, index){
  mean(data[index])
}
boot(Boston$medv, boot.fn, 1000)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original      bias    std. error
## t1* 22.53281 -0.01136423   0.4186383
  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).

As subsequently portrayed below, the confidence interval given by t.test() is similar to that of the bootstrap estimate.

t.test(Boston$medv)
## 
##  One Sample t-test
## 
## data:  Boston$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
c_interval = c(22.53281 - (2 * 0.4186383), 22.53281 + (2 * 0.4186383))
c_interval
## [1] 21.69553 23.37009
  1. Based on this data set, provide an estimate, \(\hat\mu_{med}\), for the median value of medv in the population.
mu_hat_med <- median(Boston$medv)
mu_hat_med
## [1] 21.2
  1. We now would like to estimate the standard error of \(\hat\mu_{med}\). 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.

Comments: the estimated median below matches that outputted in part (e) above. Further, we have a relatively small standard error with respect to the median value.

set.seed(4)
boot.fn <- function(data, index){
    median(data[index])
}
boot(Boston$medv, boot.fn, 1000)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original   bias    std. error
## t1*     21.2 -0.00455   0.3830777
  1. Based on this data set, provide an estimate for the tenth percentile of medv in Boston suburbs. Call this quantity \(\hat\mu_{0.1}\). (You can use the quantile() function.)
mu_hat_tenth_perc <- quantile(Boston$medv, c(0.1))
mu_hat_tenth_perc
##   10% 
## 12.75
  1. Use the bootstrap to estimate the standard error of \(\hat\mu_{0.1}\). Comment on your findings.

Comment: as shown below, an estimated tenth percentile value of 12.75 was obtained, which matches that outputted in part (g). A corresponding standard error of 0.5048959 was given which was slightly higher than that associated with the mean and median values above, though still relatively reasonable.

set.seed(4)
boot.fn<-function(data, index){
  quantile(data[index], c(0.1))
}
boot(Boston$medv ,boot.fn, 1000)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*    12.75  -0.005   0.5048959