3.We now review k-fold cross-validation.
(a) Explain how k-fold cross-validation is implemented.
In the k-fold cross validation method the observations are randomly divided into k groups/folds of about the same size. The first fold is the validation set, while the other k folds are used to fit the model.Then, the MSE is computed using the observations in the held-out fold. This process is then repeated k times using a different fold of observations as the validation set each time. This leaves us with k estimates of the test error that are then used to compute the k-fold CV estimate by taking the average of these values.
(b) What are the advantages and disadvantages of k-fold cross validation relative to:
i. The validation set approach? ii. LOOCV?
In comparison to the k-fold CV approach the validation set approach can have a lot more variability in the test error rate. In the validation approach, only one subset of observations are used to fit the model. This leads to an overestimate of the test error rate for the model fit on the whole data set. However, the advantages that the validation set approach has over the k-fold CV is that it is easy to use.
The LOOCV can be computationally intensive in comparison to the k-fold validation approach since the LOOCV approach requires that the method be fit n times. An advantage of the LOOCV approach is it gives unbiased estimates of the test error since each training set contains n-1. =The test error is then based off of the left over observations. However, since the the mean of the outputs of the LOOCV method are highly correlated, they also have a higher variance than the k-fold CV approach. With the k-fold approach you can avoid this issue by using k=5 or k=10 which yield test error rates estimates that have neither high bias nor high variance. 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 4.0.5attach(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(10000,5000)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",5000)
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, ]$default)## [1] 0.0254Our validation set error rate is 2.54%
(c) Repeat the process in (b) three times, using three different splits of the observations into a training set and a validation set. Comment on the results obtained.
set.seed(1)
train <- sample(nrow(Default),nrow(Default)*0.6)
fit.glm3 <- glm(default ~ income + balance, data = Default, family = "binomial", subset = train)
probs <- predict(fit.glm3, newdata = Default[-train, ], type = "response")
pred.glm <- rep("No", length(probs))
pred.glm[probs > 0.5] <- "Yes"
mean(pred.glm != Default[-train, ]$default)## [1] 0.025set.seed(1)
train <- sample(nrow(Default),nrow(Default)*0.7)
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, ]$default)## [1] 0.02666667set.seed(1)
train <- sample(nrow(Default),nrow(Default)*0.8)
fit.glm <- glm(default ~ income + balance, data = Default, family = "binomial", subset = train)
probs <- predict(fit.glm, newdata = Default[-train, ], type = "response")
pred.glm <- rep("No", length(probs))
pred.glm[probs > 0.5] <- "Yes"
mean(pred.glm != Default[-train, ]$default)## [1] 0.026From the results of the test error for the validation estimates we’ll notice that the rates vary for the different samples.
(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)
train <- sample(nrow(Default),nrow(Default)*0.5)
fit.glm <- glm(default ~ income + balance + student, 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] 0.026Adding a “student” dummy variable did not lead to a reduction in the test error rate of the validation set.
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 computes 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(2)
attach(Default)## The following objects are masked from Default (pos = 3):
##
## balance, default, income, studentfit.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 glm() estimates of the standard errors for the coefficients β0, β1 and β2 are respectively 4.348e-01, 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 -3.956457e-02 4.228903e-01
## t2* 2.080898e-05 -4.605498e-07 4.924565e-06
## t3* 5.647103e-03 3.048004e-05 2.228767e-04The bootstrap estimates of the standard errors for the coefficients β0, β1 and β2 are respectively 0.4229, 4.925 x 10^(-6) and 2.229 x 10^(-4).
(d) Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function.
The estimated standard errors results from the two different methods are similar.
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 μ^.
library(MASS)
attach(Boston)
mu.hat <- mean(Boston$medv)
mu.hat## [1] 22.53281(b) Provide an estimate of the standard error of μ^. Interpret this result.
se.hat <- sd(Boston$medv) / sqrt(dim(Boston)[1])
se.hat## [1] 0.4088611(c) Now estimate the standard error of μ^ using the bootstrap. How does this compare to your answer from (b) ?
set.seed(2)
boot.fn <- function(data, index) {
mu <- mean(data[index])
return (mu)
}
boot(Boston$medv, boot.fn, 1000)##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = Boston$medv, statistic = boot.fn, R = 1000)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 22.53281 -0.008232806 0.407299The bootstrap estimated standard error of μ^ of 0.4073 is very close to the estimate found in (b) of 0.4089.
(d) Based on your bootstrap estimate from (c), provide a 95% confidence interval for the mean of “medv”. Compare it to the results obtained using t.test(Boston$medv).
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
## 22.53281CI.mu.hat <- c(22.53281 - 2 * 0.407299, 22.53281 + 2 * 0.407299)
CI.mu.hat## [1] 21.71821 23.34741The bootstrap confidence interval is similar to the one given in the t.test(Boston$medv) function.
(e) Based on this data set, provide an estimate, μ^med, 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 μ^med. 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.
set.seed(2)
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 -0.02465 0.3879954We get an estimated median value of 21.2 which is equal to the value obtained in (e), with a standard error of ~0.388.
(g) Based on this data set, provide an estimate for the tenth percentile of “medv” in Boston suburbs. Call this quantity μ^0.1
percent.hat <- quantile(medv, c(0.1))
percent.hat## 10%
## 12.75(h) Use the bootstrap to estimate the standard error of μ^0.1. Comment on your findings
set.seed(2)
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 0.00275 0.5159129We get an estimated tenth percentile value of 12.75 which is again equal to the value obtained in (g), with a standard error of 0.5159.