library(ISLR)
library(boot)
library(MASS)We now review k-fold cross-validation.
(a) Explain how k-fold cross-validation is implemented.
The k-fold cross validation is implemented by taking a set of nth observations and randomly splitting them into k non-overlapping groups. 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.
(b) What are the advantages and disadvantages of k-fold cross-validation relative to:
i. The validation set approach? The validation set approach is by seperating the training data into two sets. However, there are two disadvantages: The estimate of the test error rate can be highly variable depending on which observations are included in the training and validation sets; the validation set error rate may tend to overestimate the test error rate for the model fit on the entire data set.
ii. LOOCV?
LOOCV is a special case of k-fold cross-validation with k = n. Thus, LOOCV is the most computationally intense (disadvantage) method since the model must be repeatedly fit n times. Also, LOOCV has higher variance, but lower bias, than k-fold CV.
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.
data(Default)
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 ...summary(Default)## default student balance income
## No :9667 No :7056 Min. : 0.0 Min. : 772
## Yes: 333 Yes:2944 1st Qu.: 481.7 1st Qu.:21340
## Median : 823.6 Median :34553
## Mean : 835.4 Mean :33517
## 3rd Qu.:1166.3 3rd Qu.:43808
## Max. :2654.3 Max. :73554(a) Fit a logistic regression model that uses income and balance to predict default.
logmodel = glm(default ~ balance + income, data = Default, family = binomial)
summary(logmodel)##
## Call:
## glm(formula = default ~ balance + income, 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 ***
## balance 5.647e-03 2.274e-04 24.836 < 2e-16 ***
## income 2.081e-05 4.985e-06 4.174 2.99e-05 ***
## ---
## 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)
row.number = sample(1:nrow(Default), 0.8*nrow(Default))
defaultrain = Default[row.number,]
defaultest = Default[-row.number,]ii. Fit a multiple logistic regression model using only the training observations.
glmfit <- glm(default ~ income + balance, data = defaultrain, family = "binomial")
summary(glmfit)##
## Call:
## glm(formula = default ~ income + balance, family = "binomial",
## data = defaultrain)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.4758 -0.1413 -0.0563 -0.0210 3.4620
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.168e+01 4.893e-01 -23.879 < 2e-16 ***
## income 2.547e-05 5.631e-06 4.523 6.1e-06 ***
## balance 5.613e-03 2.531e-04 22.176 < 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: 2313.6 on 7999 degrees of freedom
## Residual deviance: 1239.2 on 7997 degrees of freedom
## AIC: 1245.2
##
## Number of Fisher Scoring iterations: 8iii. 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.
glmfit.pred = predict(glmfit, defaultest, type = "response")
glmfit.pred = ifelse(glmfit.pred >0.5, "Yes", "No")iv. Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified.
mean(glmfit.pred != defaultest$default)## [1] 0.026(c) Repeat the process in (b) three times, using three different splits of the observations into a training set and a validation set. Comment on the results obtained.
glmvalid = function() {
row.number = sample(1:nrow(Default), 0.8*nrow(Default))
defaultrain = Default[row.number,]
defaultest = Default[-row.number,]
glmfit <- glm(default ~ income + balance, data = defaultrain, family = "binomial")
glmfit.pred = predict(glmfit, defaultest, type = "response")
glmfit.pred = ifelse(glmfit.pred >0.5, "Yes", "No")
return(mean(glmfit.pred != defaultest$default))
}glmvalid()## [1] 0.024glmvalid()## [1] 0.0265glmvalid()## [1] 0.0315(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.
glmstudentfit = update(glmfit,.~. +student)
summary(glmstudentfit)##
## Call:
## glm(formula = default ~ income + balance + student, family = "binomial",
## data = defaultrain)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.4736 -0.1389 -0.0543 -0.0202 3.5171
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.094e+01 5.517e-01 -19.834 < 2e-16 ***
## income 5.877e-06 9.172e-06 0.641 0.52170
## balance 5.713e-03 2.585e-04 22.100 < 2e-16 ***
## studentYes -7.282e-01 2.683e-01 -2.714 0.00664 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 2313.6 on 7999 degrees of freedom
## Residual deviance: 1231.9 on 7996 degrees of freedom
## AIC: 1239.9
##
## Number of Fisher Scoring iterations: 8glmstudentfit.pred = predict(glmstudentfit, defaultest, type = "response")
glmstudentfit.pred = ifelse(glmstudentfit.pred > 0.5, "Yes", "No")
mean(glmstudentfit.pred != defaultest$default)## [1] 0.0275We 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.
glmfit <- glm(default ~ income + balance, data = Default, family = "binomial")
summary(glmfit)##
## 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) {
glmfit <- glm(default ~ income + balance, data = data, family = "binomial", subset = index)
return (coef(glmfit))
}(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.
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 -3.914317e-02 4.355389e-01
## t2* 2.080898e-05 1.631784e-07 4.852930e-06
## t3* 5.647103e-03 1.843950e-05 2.304681e-04d) Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function.
The estimated standard errors that were obtained were very similar appearing all the way down to the thrid digit: 4.348e-01 vs 0.4364, 4.985e-06 vs 4.861752e-06, 2.274e-04 vs 2.308681e-04
detach(Default)We will now consider the Boston housing data set, from the MASS library.
Based on this data set, provide an estimate for the population mean of medv. Call this estimate ˆµ.
data(Boston)
attach(Boston)
mu.hat <- mean(medv)
mu.hat## [1] 22.53281(b) 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.
medv.sd=sd(medv)/sqrt(dim(Boston)[1])
medv.sd## [1] 0.4088611(c) Now estimate the standard error of ˆµ using the bootstrap. How does this compare to your answer from (b)?
set.seed(1)
boot.fn <- function(data, index) {
mu.hat <- mean(data[index])
return (mu.hat)
}
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(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 [ˆµ − 2SE(ˆµ), µˆ + 2SE(ˆµ)].
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
## 22.53281CI.mu.hat <- c(22.53 - 2 * 0.4107, 22.53 + 2 * 0.4107)
CI.mu.hat## [1] 21.7086 23.3514(e) Based on this data set, provide an estimate, ˆµmed, for the median value of medv in the population.
medv.med = median(medv)
medv.med## [1] 21.2(f) 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 -0.0386 0.3770241(g) Based on this data set, provide an estimate for the tenth percentile of medv in Boston suburbs. Call this quantity ˆµ0.1. (You can use the quantile() function.)
medv.tenth = quantile(medv, c(0.1))
medv.tenth## 10%
## 12.75(h) Use the bootstrap to estimate the standard error of ˆµ0.1. Comment on your findings.
boot.fn= function(data, index) return(quantile(data[index], c(0.1)))
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