We now review k-fold cross-validation. (a) Explain how k-fold cross-validation is implemented.
A dataset is randomly separated into k groups of equal size. One of the subsets, or folds, is used as the initial validation set while the rest are used for training. Once the MSE is found, the process is done again using a different fold as the validation set until each fold has served as the validation set. The k fold cross validation estimate is the average of the MSEs found.
(b) What are the advantages and disadvantages of k-fold crossvalidation relative to: i. The validation set approach? K-fold CV, in comparison to the validation set approach, although computationally more demanding, the result being an average of results reduces variance and has the model be evaluated on multiple subsets of the data.
ii. LOOCV? An advantage of k-fold CV over LOOCV is k-fold uses less computational power. K-fold also gives more accurate estimates of the test error rate.
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(ISLR2)attach(Default)str(Default)## 'data.frame': 10000 obs. of 4 variables:
## $ default: Factor w/ 2 levels "No","Yes": 1 1 1 1 1 1 1 1 1 1 ...
## $ student: Factor w/ 2 levels "No","Yes": 1 2 1 1 1 2 1 2 1 1 ...
## $ balance: num 730 817 1074 529 786 ...
## $ income : num 44362 12106 31767 35704 38463 ...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(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.
set.seed(1)
train = sample(nrow(Default), 0.5*nrow(Default))
default_train = Default[train, ]
default_test = Default[-train, ]ii. Fit a multiple logistic regression model using only the training observations.
glm_fit = glm(default ~ income + balance, data = default_train, family = binomial)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.
default_probabilities = predict(glm_fit, default_test)
default_predictions = ifelse(default_probabilities > 0.5, "Yes", "No")iv. Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified.
table(default_predictions, default_test$default)##
## default_predictions No Yes
## No 4833 122
## Yes 10 35mean(default_predictions != default_test$default) * 100## [1] 2.642.64% misclassification percentage
(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.
set.seed(1)
train = sample(nrow(Default), 0.7*nrow(Default))
default_train = Default[train, ]
default_test = Default[-train, ]
glm_fit = glm(default ~ income + balance, data = default_train, family = binomial)
default_probabilities = predict(glm_fit, default_test)
default_predictions = ifelse(default_probabilities > 0.5, "Yes", "No")
table(default_predictions, default_test$default)##
## default_predictions No Yes
## No 2897 89
## Yes 1 13mean(default_predictions != default_test$default) * 100## [1] 370/30 split has a 3% misclassified percentage.
set.seed(1)
train = sample(nrow(Default), 0.9*nrow(Default))
default_train = Default[train, ]
default_test = Default[-train, ]
glm_fit = glm(default ~ income + balance, data = default_train, family = binomial)
default_probabilities = predict(glm_fit, default_test)
default_predictions = ifelse(default_probabilities > 0.5, "Yes", "No")
table(default_predictions, default_test$default)##
## default_predictions No Yes
## No 962 33
## Yes 1 4mean(default_predictions != default_test$default) * 100## [1] 3.490/10 split has a 3.4% misclassified percentage.
set.seed(1)
train = sample(nrow(Default), 0.95*nrow(Default))
default_train = Default[train, ]
default_test = Default[-train, ]
glm_fit = glm(default ~ income + balance, data = default_train, family = binomial)
default_probabilities = predict(glm_fit, default_test)
default_predictions = ifelse(default_probabilities > 0.5, "Yes", "No")
table(default_predictions, default_test$default)##
## default_predictions No Yes
## No 477 19
## Yes 1 3mean(default_predictions != default_test$default) * 100## [1] 495/5 split has a 4% misclassified percentage.
There was more misclassification as the amount of data used in training went up.
(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.
set.seed(1)
Default$student_dummy = ifelse(Default$student == "Yes", 1, 0)
train = sample(nrow(Default), 0.5*nrow(Default))
default_train = Default[train, ]
default_test = Default[-train, ]
glm_fit = glm(default ~ income + balance + student_dummy, data = default_train, family = binomial)
default_probabilities = predict(glm_fit, default_test)
default_predictions = ifelse(default_probabilities > 0.5, "Yes", "No")
table(default_predictions, default_test$default)##
## default_predictions No Yes
## No 4832 119
## Yes 11 38mean(default_predictions != default_test$default) * 100## [1] 2.6The misclassification percentage of 2.6% with the addition of a student dummy variable is lower than the 2.64% misclassification without the student dummy variable. The inclusion of a student dummy variable did reduce the test error rate.
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)
##
## 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: 8income standard error: 4.985 x 10^-6
balance standard error: 2.274 x 10^-4
(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) {
coef(glm(default ~ income + balance, data = data, family = binomial, subset = index))
}boot.fn(Default, 1:10000)## (Intercept) income balance
## -1.154047e+01 2.080898e-05 5.647103e-03(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, 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(d) Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function.
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: 8The estimated standard errors from the glm differ enough from those from the bootstrap method that we can say the standard errors from the bootstrap method are more accurate.
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 μˆ.
mu_hat = mean(Boston$medv)
mu_hat## [1] 22.53281(b) Provide an estimate of the standard error of μˆ. Interpret this result. Hint: We can compute the standard error of the sample mean by dividing the sample standard deviation by the square root of the number of observations.
sd(Boston$medv)/sqrt(length(Boston$medv))## [1] 0.4088611This low standard error shows that there is not much variation in
medv.
(c) Now estimate the standard error of μˆ using the bootstrap. How does this compare to your answer from (b)?
boot.fn = function(data, index) {
mean(data$medv[index])
}boot_results = boot(Boston, boot.fn, 1000)
boot_results##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = Boston, statistic = boot.fn, R = 1000)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 22.53281 0.0109415 0.414028boot_se = sd(boot_results$t)The bootstrapping standard error is close to the estimated standard error from (b).
(d) Based on your bootstrap estimate from (c), provide a 95 % confidence interval for the mean of medv. Compare it to the results obtained using t.test(Boston$medv). Hint: You can approximate a 95 % confidence interval using the formula [μˆ − 2SE(μˆ), μˆ + 2SE(μˆ)].
# 95% confidence interval from bootstrap estimate
mu_hat - 2 * boot_se## [1] 21.70475mu_hat + 2 * boot_se## [1] 23.36086# confidence interval from t.test()
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(e) Based on this data set, provide an estimate, μˆmed, for the median value of medv in the population.
mu_hat_med = median(Boston$medv)
mu_hat_med## [1] 21.2(f) We now would like to estimate the standard error of μˆmed. Unfortunately, there is no simple formula for computing the standard error of the median. Instead, estimate the standard error of the median using the bootstrap. Comment on your findings.
boot.fn = function(data, index) {
median(data$medv[index])
}boot_med_results = boot(Boston, boot.fn, 1000)
boot_med_results##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = Boston, statistic = boot.fn, R = 1000)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 21.2 -0.02095 0.3707169sd(boot_med_results$t)## [1] 0.3707169The small standard error of the median implies that our estimate of the median is a good estimate because different samples would result in similar medians.
(g) Based on this data set, provide an estimate for the tenth percentile of medv in Boston census tracts. Call this quantity μˆ0.1. (You can use the quantile() function.)
mu_hat_0.1 = quantile(Boston$medv, 0.1)
mu_hat_0.1## 10%
## 12.75(h) Use the bootstrap to estimate the standard error of μˆ0.1. Comment on your findings.
boot.fn = function(data, index) {
quantile(data$medv[index], 0.1)
}boot(Boston, boot.fn, 1000)##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = Boston, statistic = boot.fn, R = 1000)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 12.75 0.02795 0.5077737The standard error repeatedly comes back around 0.5, indicating that our estimate is reliable.