Question 3

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

k-fold cross-validation is implemented by splitting a dataset into x folds to split and test on different x portions of the data at a time. Data will essentially take turns being a training data set vs a testing data set.

  1. What are the advantages and disadvantages of k-fold crossvalidation relative to:

  1. The validation set approach? Adv. to validation set is less overfitting on the training dataset. Disadv. is that it is more computationally expensive since you have to fit on x folds instead of once.

  2. LOOCV? Adv of k-fold here is that it’s less computational, since LOOCV is strictly testing on a single data point at a time. Disadv of k-fold is that results can vary depending on the random fold at the time of training.

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

library(ISLR)
## Warning: package 'ISLR' was built under R version 4.6.1
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)
## 
## 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.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.
set.seed(1)
train <- sample(nrow(Default), nrow(Default) / 2)
Default.train <- Default[train,]
Default.val <- Default[-train,]
glm.fit2 <- glm(default ~ income + balance, data = Default.train, family = binomial)
glm.probs <- predict(glm.fit2, Default.val, type = "response")
glm.pred <- rep("No", nrow(Default.val))
glm.pred[glm.probs > 0.5] <- "Yes"
mean(glm.pred != Default.val$default)
## [1] 0.0254
  1. Repeat the process in (b) three times, using three different splits of the observations into a training set and a validation set. Comment on the results obtained. - The different splits are all different from each other with the different seeds. Seed 2 is at .0238, seed 3 at .0264 and seed 4 at .0256
set.seed(2)
train <- sample(nrow(Default), nrow(Default) / 2)
glm.fit3 <- glm(default ~ income + balance, data = Default[train, ], family = binomial)
glm.probs <- predict(glm.fit3, Default[-train,], type = "response")
glm.pred <- rep("No", nrow(Default[-train,]))
glm.pred[glm.probs > 0.5] <- "Yes"
mean(glm.pred != Default[-train, ]$default)
## [1] 0.0238
set.seed(3)
train <- sample(nrow(Default), nrow(Default) / 2)
glm.fit4 <- glm(default ~ income + balance, data = Default[train, ], family = binomial)
glm.probs <- predict(glm.fit4, Default[-train,], type = "response")
glm.pred <- rep("No", nrow(Default[-train,]))
glm.pred[glm.probs > 0.5] <- "Yes"
mean(glm.pred != Default[-train, ]$default)
## [1] 0.0264
set.seed(4)
train <- sample(nrow(Default), nrow(Default) / 2)
glm.fit5 <- glm(default ~ income + balance, data = Default[train, ], family = binomial)
glm.probs <- predict(glm.fit5, Default[-train,], type = "response")
glm.pred <- rep("No", nrow(Default[-train,]))
glm.pred[glm.probs > 0.5] <- "Yes"
mean(glm.pred != Default[-train, ]$default)
## [1] 0.0256
  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. - Incorporating student as a dummy variable does not lead to a reduction in test error at: .026.
set.seed(1)
train <- sample(nrow(Default), nrow(Default) / 2)
glm.fit6 <- glm(default ~ income + balance + student, data = Default[train,], family = binomial)
glm.probs <- predict(glm.fit6, Default[-train, ], type = "response")
glm.pred <- rep("No", nrow(Default[-train, ]))
glm.pred[glm.probs > 0.5] <- "Yes"
mean(glm.pred != Default[-train, ]$default)
## [1] 0.026

Question 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.
library(ISLR2)
## Warning: package 'ISLR2' was built under R version 4.6.1
## 
## Attaching package: 'ISLR2'
## The following objects are masked from 'package:ISLR':
## 
##     Auto, Credit
library(boot)
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)
## 
## 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) {
  glm.fit <- glm(default ~ income + balance, data = data, subset = index, family = binomial)
  return(coef(glm.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.
set.seed(1)
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 -3.945460e-02 4.344722e-01
## t2*  2.080898e-05  1.680317e-07 4.866284e-06
## t3*  5.647103e-03  1.855765e-05 2.298949e-04
  1. Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function.

Question 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)
library(boot)
μ <- mean(Boston$medv)
summary(μ)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   22.53   22.53   22.53   22.53   22.53   22.53
  1. Provide an estimate of the standard error of ˆμ. Interpret this result. Hint: We can compute the standard error of the sample mean by dividing the sample standard deviation by the square root of the number of observations. - The standard of error is .4089, which means our sample mean will typically vary .41 units from the true pop. mean
se.mu <- sd(Boston$medv) / sqrt(nrow(Boston))
se.mu
## [1] 0.4088611
  1. Now estimate the standard error of ˆμ using the bootstrap. How does this compare to your answer from (b)? - The standard error using bootstrap is .414 which is slightly higher and slightly worse than our standard error of .4089 when dividing the st.dev by sqrt of n.
boot.fn.mean <- function(data, index) {
  return(mean(data$medv[index]))
}

boot(Boston, boot.fn.mean, R = 1000)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston, statistic = boot.fn.mean, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original    bias    std. error
## t1* 22.53281 0.0109415    0.414028
  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$medv). Hint: You can approximate a 95 % confidence interval using the formula [ˆμ − 2SE(ˆμ), ˆμ + 2SE(ˆμ)].
μ - 2 * 0.414028
## [1] 21.70475
μ + 2 * 0.414028
## [1] 23.36086
t.test(Boston$medv)
## 
##  One Sample t-test
## 
## data:  Boston$medv
## t = 55.111, df = 505, p-value < 2.2e-16
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
##  21.72953 23.33608
## sample estimates:
## mean of x 
##  22.53281
  1. Based on this data set, provide an estimate, ˆμmed, for the median value of medv in the population.
u.med <- median(Boston$medv)
summary(u.med)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##    21.2    21.2    21.2    21.2    21.2    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. - The st error is .3778 which is lower than the st. error of our mean of .414.
set.seed(1)
boot.fn.median <- function(data, index) {
  return(median(data$medv[index]))
}
boot(Boston, boot.fn.median, R = 1000)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston, statistic = boot.fn.median, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*     21.2 0.02295   0.3778075
  1. Based on this data set, provide an estimate for the tenth percentile of medv in Boston census tracts. Call this quantity ˆμ0.1. (You can use the quantile() function.)
u.0.1 <- quantile(Boston$medv, 0.1)
u.0.1
##   10% 
## 12.75
  1. Use the bootstrap to estimate the standard error of ˆμ0.1. Comment on your findings. - st.error of the tenth percentile is .4768 which is higher than both the st. error of the median and the mean, meaning it’s worse.
set.seed(1)
boot.fn.tenth <- function(data, index) {
  return(quantile(data$medv[index], 0.1))
}
boot(Boston, boot.fn.tenth, R = 1000)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston, statistic = boot.fn.tenth, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*    12.75  0.0339   0.4767526