We now review k-fold cross-validation.
(a) Explain how k-fold cross-validation is
implemented.
K-fold cross validation is implemented by randomly dividing the set of observations into k groups, or folds, of approximately equal size. The first fold is treated as a validation set, and the method is fit on the remaining k − 1 folds. The mean squared error, MSE1, is then computed on the observations in the held-out fold. This procedure is repeated k times; each time, a different group of observations is treated as a validation set. This process results in k estimates of the test error. The k-fold CV estimate is computed by averaging these values.
(b) What are the advantages and disadvantages of k-fold
crossvalidation relative to:
i. The validation set approach?
Advantages of the validation set approach are that it can be applied to almost any statistical learning method, it is easy to implement, and it gives more accurate estimates of the test error rate than LOOCV. A disadvantage is that the validation set approach can lead to overestimates of the test error rate.
ii. LOOCV?
An advantage of LOOCV is that it gives approximately unbiased estimates of the test error. A disadvantage of LOOCV is that it is computationally expensive and can pose computational problems.
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)
set.seed(1)
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(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.
train <- sample(dim(Default)[1], dim(Default)[1] / 2)ii. Fit a multiple logistic regression model using only the training observations.
fit.glm <- glm(default ~ income + balance, data = Default, family = "binomial", subset = train)
summary(fit.glm)##
## Call:
## glm(formula = default ~ income + balance, family = "binomial",
## data = Default, subset = train)
##
## 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: 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.
probs <- predict(fit.glm, newdata = Default[-train, ], type = "response")
pred.glm <- rep("No", length(probs))
pred.glm[probs > 0.5] <- "Yes"iv. Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified.
mean(pred.glm != Default[-train, ]$default)## [1] 0.0254The validation set error rate is 2.54%.
(c) Repeat the process in (b) three times, using three different splits of the observations into a training set and a validation set. Comment on the results obtained.
train <- sample(dim(Default)[1], dim(Default)[1] / 2)
fit.glm <- glm(default ~ income + balance, data = Default, family = "binomial", subset = train)
probs <- predict(fit.glm, newdata = Default[-train, ], type = "response")
pred.glm <- rep("No", length(probs))
pred.glm[probs > 0.5] <- "Yes"
mean(pred.glm != Default[-train, ]$default)## [1] 0.0274The validation set error rate is 2.82%.
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, ]$default)## [1] 0.0244The validation set error rate is 2.38%.
train <- sample(dim(Default)[1], dim(Default)[1] / 2)
fit.glm <- glm(default ~ income + balance, data = Default, family = "binomial", subset = train)
probs <- predict(fit.glm, newdata = Default[-train, ], type = "response")
pred.glm <- rep("No", length(probs))
pred.glm[probs > 0.5] <- "Yes"
mean(pred.glm != Default[-train, ]$default)## [1] 0.0244The validation set error rate is 2.54%.
The validation set error rate is different for each of the three splits. Dependent on which observations are in the validation set, the error rate varies.
(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.
train <- sample(dim(Default)[1], dim(Default)[1] / 2)
fit.glm <- glm(default ~ income + balance + student, data = Default, family = "binomial", subset = train)
pred.glm <- rep("No", length(probs))
probs <- predict(fit.glm, newdata = Default[-train, ], type = "response")
pred.glm[probs > 0.5] <- "Yes"
mean(pred.glm != Default[-train, ]$default)## [1] 0.0278Including a dummy variable for student did not lead to a reduction in 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)
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: 8The estimated standard errors for income and balance are 4.985e-06 and 2.274e-04.
(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) {
fit <- glm(default ~ income + balance, data = data, family = "binomial", subset = index)
return (coef(fit))
}(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-04The estimated standard errors of the logistic regression coefficients for income and balance are 4.988746e-06 and 2.347490e-04.
(d) Comment on the estimated standard errors obtained using
the glm() function and using your bootstrap function.
Both sets of standard errors seem pretty similar.
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 ˆμ.
library(MASS)
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.
se.hat <- sd(medv) / sqrt(dim(Boston)[1])
se.hat## [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 <- 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 0.007650791 0.4106622The bootstrap standard error of mu.hat of 0.4106622 is close to the standard error of 0.4088611 in (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
[ˆμ − 2SE(ˆμ), ˆμ + 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(22.53281 - 2 * 0.4106622, 22.53281 + 2 * 0.4106622)
CI.mu.hat## [1] 21.71149 23.35413The confidence intervals are almost identical.
(e) 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(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 0.3770241The estimated standard error of the mean using the bootstrap is 21.2, the same as (e). The standard error is 0.3770241.
(g) 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.)
percent.hat <- quantile(medv, c(0.1))
percent.hat## 10%
## 12.75(h) 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 0.0186 0.4925766The percentile value is 12.75, same as (g) and the standard error of ˆμ0.1 is 0.4925766.