Conceptual Questions
3.) We now review k-fold cross-validation.
(a) Explain how k-fold cross-validation is implemented.
The k-fold cross-validation is implemented by obtaining a certain number of chosen observations and these observations are then split to number of k that is chosen.The mean squared are calculated in each of the k fold and the average of all the mean squared are calculated in the end to identify the resulting mean squared error of all the k folds.
(b) What are the advantages and disadvantages of k-fold cross-validation relative to:
i. The validation set approach?
An advantage of k-fold cross-validation to validation set approach is that the variability may be lower for k-fold in comparison to validation set. The reason is that the observations that is used for train and test set may matter in calculating the test error rate, and since validation set approach will set aside only one subset of the data, the variability is higher when introduced to new data. The k-fold cross-validation uses different training and test set that essentially attempts to avoid this issue.
A disadvantage of k-fold cross-validation to validation set approach is that the model have to run in the number of k-fold that is chosen which means that it may take more computational power to do so.
ii. LOOCV?
An advantage of k-fold cross-validation to LOOCV approach is that k-fold cross validation will take less computational power because LOOCV will fit the model in the form of n - 1 observations rather than splitting the train set into the number of k-fold.
A disadvantage of k-fold cross-validation to LOOCV is that k-fold cross validation may perform less in regards of providing an unbiased result. The reason to this is that since LOOCV is fitted on n - 1 observations then it may produced an unbiased estimate in comparison to k-fold cross-validation.
Applied Questions
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)
attach(Default)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)
##
## 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.
ii. 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.
The test error rate of the 50/50 split is 2.54%
set.seed(1)
#splitting to 50/50 train and validation set
train.part<-sample(nrow(Default),nrow(Default)*0.5)
default.train<-Default[train.part,]
default.test<-Default[-train.part,]
glm.fit <- glm(default ~ income + balance, data = default.train, family =binomial)
glm.pred <- predict(glm.fit, default.test, type="response")
glm.probs = rep("No", dim(Default)[1])
glm.probs[glm.pred > 0.5] = "Yes"
mean(glm.probs != default.test$default)## [1] 0.0254(c) Repeat the process in (b) three times, using three different splits of the observations into a training set and a validation set. Comment on the results obtained.
The test error rate of the 90/10 split is 2.9%
set.seed(1)
#splitting to 90/10 train and validation set
train.part<-sample(nrow(Default),nrow(Default)*0.9)
default.train<-Default[train.part,]
default.test<-Default[-train.part,]
glm.fit <- glm(default ~ income + balance, data = default.train, family =binomial)
glm.pred <- predict(glm.fit, default.test, type="response")
glm.probs = rep("No", dim(Default)[1])
glm.probs[glm.pred > 0.5] = "Yes"
mean(glm.probs != default.test$default)## [1] 0.029The test error rate of the 80/20 split is 2.6%
set.seed(1)
#splitting to 80/20 train and validation set
train.part<-sample(nrow(Default),nrow(Default)*0.8)
default.train<-Default[train.part,]
default.test<-Default[-train.part,]
glm.fit <- glm(default ~ income + balance, data = default.train, family =binomial)
glm.pred <- predict(glm.fit, default.test, type="response")
glm.probs = rep("No", dim(Default)[1])
glm.probs[glm.pred > 0.5] = "Yes"
mean(glm.probs != default.test$default)## [1] 0.026The test error rate of the 70/30 split is 2.65%
set.seed(1)
#splitting to 70/30 train and validation set
train.part<-sample(nrow(Default),nrow(Default)*0.7)
default.train<-Default[train.part,]
default.test<-Default[-train.part,]
glm.fit <- glm(default ~ income + balance, data = default.train, family =binomial)
glm.pred <- predict(glm.fit, default.test, type="response")
glm.probs = rep("No", dim(Default)[1])
glm.probs[glm.pred > 0.5] = "Yes"
mean(glm.probs != default.test$default)## [1] 0.0265The different sets of samples on train and validation set produced a different test error rate in each, which are 2.54% on a 50/50 split, 2.9% on a 90/10 split, 2.6% on a 80/20 split, and 2.65% on a 70/30 split.
(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.
The test error rate of the 50/50 split with the student variable is 2.6%
set.seed(1)
#splitting to 50/50 train and validation set with the student variable
train.part<-sample(nrow(Default),nrow(Default)*0.5)
default.train<-Default[train.part,]
default.test<-Default[-train.part,]
glm.fit <- glm(default ~ income + balance + student, data = default.train, family =binomial)
glm.pred <- predict(glm.fit, default.test, type="response")
glm.probs = rep("No", dim(Default)[1])
glm.probs[glm.pred > 0.5] = "Yes"
mean(glm.probs != default.test$default)## [1] 0.026The test error rate of the 90/10 split with the student variable is 3.1%
set.seed(1)
#splitting to 90/10 train and validation set with the student variable
train.part<-sample(nrow(Default),nrow(Default)*0.9)
default.train<-Default[train.part,]
default.test<-Default[-train.part,]
glm.fit <- glm(default ~ income + balance + student, data = default.train, family =binomial)
glm.pred <- predict(glm.fit, default.test, type="response")
glm.probs = rep("No", dim(Default)[1])
glm.probs[glm.pred > 0.5] = "Yes"
mean(glm.probs != default.test$default)## [1] 0.031The test error rate of the 80/20 split with the student variable is 2.75%
set.seed(1)
#splitting to 80/20 train and validation set with the student variable
train.part<-sample(nrow(Default),nrow(Default)*0.8)
default.train<-Default[train.part,]
default.test<-Default[-train.part,]
glm.fit <- glm(default ~ income + balance + student, data = default.train, family =binomial)
glm.pred <- predict(glm.fit, default.test, type="response")
glm.probs = rep("No", dim(Default)[1])
glm.probs[glm.pred > 0.5] = "Yes"
mean(glm.probs != default.test$default)## [1] 0.0275The test error rate of the 70/30 split with the student variable is 2.77%
set.seed(1)
#splitting to 70/30 train and validation set with the student variable
train.part<-sample(nrow(Default),nrow(Default)*0.7)
default.train<-Default[train.part,]
default.test<-Default[-train.part,]
glm.fit <- glm(default ~ income + balance + student, data = default.train, family =binomial)
glm.pred <- predict(glm.fit, default.test, type="response")
glm.probs = rep("No", dim(Default)[1])
glm.probs[glm.pred > 0.5] = "Yes"
mean(glm.probs != default.test$default)## [1] 0.0277The model that includes that student variable did not reduce the test error rate in comparison to the test error rate of the 50/50 split, the model without the student variable test error rate = 2.54% vs the model with the student variable test error rate = 2.6%.
The model that includes that student variable did not reduced the test error rate in comparison to the test error rate of the 90/10 split the model without the student variable test error rate = 2.9% vs the model with the student variable test error rate = 3.1%.
The model that includes that student variable did not reduce the test error rate in comparison to the test error rate of the 80/20 split, the model without the student variable test error rate = 2.6% vs the model with the student variable test error rate = 2.75%.
The model that includes that student variable did not reduced the test error rate in comparison to the test error rate of the 70/30 split, the model without the student variable test error rate = 2.65% vs the model with the student variable test error rate = 2.77%.
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.
(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.
The estimated standard errors for the coefficients associated with income is 4.985e-06 and 2.274e-04 for balance.
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)
##
## 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.
set.seed(1)
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-04(d) Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function.
The estimated standard errors obtained by the bootstrap function are quite similar to the glm function. The glm function obtained coefficients associated with income = 4.985e-06 and balance = 2.274e-04. The bootstrap function obtained coefficients associated with income = 4.866284e-06 and balance = 2.298949e-04.**
9.) We will now consider the Boston housing data set, from the MASS 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)?
The estimate of the standard error of ˆμ obtained by bootstrap is quite similar to standard error of ˆμ from (b), bootstrap = 0.4106622 vs 0.4088611
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 0.4106622(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), ˆμ + 2SE(ˆμ)].
The mean of medv obtained by the t.test is similar to the result obtained by the bootstrap.
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
## 22.53281d.mu.hat <- c(22.53 - 2 * 0.4107, 22.53 + 2 * 0.4107)
d.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.
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.
The median obtained by the bootstrap is the same as the median obtained in (e).
set.seed(1)
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 0.02295 0.3778075(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.)
perc.hat <- quantile(medv, c(0.1))
perc.hat## 10%
## 12.75(h) Use the bootstrap to estimate the standard error of ˆμ0.1. Comment on your findings. The median obtained by the bootstrap is the same as the median obtained in (g).
set.seed(1)
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 0.0339 0.4767526