We now review k-fold cross-validation.
(a) Explain how k-fold cross-validation is implemented.
With k-fold Cross Validation, we divide the data set into K different parts (e.g. K = 5, or K = 10, etc). We then remove the first part, fit the model on the remaining K-1 parts, and see how good the predictions are on the left out part (i.e. compute the MSE on the first part). We then repeat this K different times taking out a different part each time By averaging the K different MSE’s we get an estimated validation (test) error rate for new observations.
(b) What are the advantages and disadvantages of k-fold crossvalidation relative to:
i. The validation set approach?
The advantages of the validation set approach is that it is simple and easy to implement.
The disadvantages are that the validation MSE can be highly variable. Also, statistical methods tend to perform worse when trained on fewer observations, and only a subset of observations are used to fit the model to the training data.
ii. LOOCV?
An advantage of LOOCV is that it has less bias than the k-fold crossvalidation method. However, the bias-variance tradeoff, makes LOOCV have the disadvantage of having higher variance.
Another disadvantage is that LOOCV can be computationally intensive, since we fit the model n times.
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)
set.seed(1)
log.reg <- glm(default~income+balance, data=Default, family=binomial)
summary(log.reg)##
## 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. 5.4 Exercises 199
train <- sample(dim(Default)[1], dim(Default)[1] / 2)ii. Fit a multiple logistic regression model using only the training observations.
log.reg2 <- glm(default~income+balance, data=Default, family=binomial, subset = train)
summary(log.reg2)##
## 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.
prob <- predict(log.reg2, newdata = Default[-train, ], type = "response")
prediction <- rep("No", length(prob))
prediction[prob > 0.5] <- "Yes"iv. Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified.
mean(prediction != Default[-train, ]$default)## [1] 0.0254(c) Repeat the process in (b) three times, using three different splits of the observations into a training set and a validation set. Comment on the results obtained.
train <- sample(dim(Default)[1], dim(Default)[1] / 2)
log.reg2 <- glm(default~income+balance, data=Default, family=binomial, subset = train)
prob <- predict(log.reg2, newdata = Default[-train, ], type = "response")
prediction <- rep("No", length(prob))
prediction[prob > 0.5] <- "Yes"
mean(prediction != Default[-train, ]$default)## [1] 0.0274train <- sample(dim(Default)[1], dim(Default)[1] / 2)
log.reg2 <- glm(default~income+balance, data=Default, family=binomial, subset = train)
prob <- predict(log.reg2, newdata = Default[-train, ], type = "response")
prediction <- rep("No", length(prob))
prediction[prob > 0.5] <- "Yes"
mean(prediction != Default[-train, ]$default)## [1] 0.0244train <- sample(dim(Default)[1], dim(Default)[1] / 2)
log.reg2 <- glm(default~income+balance, data=Default, family=binomial, subset = train)
prob <- predict(log.reg2, newdata = Default[-train, ], type = "response")
prediction <- rep("No", length(prob))
prediction[prob > 0.5] <- "Yes"
mean(prediction != Default[-train, ]$default)## [1] 0.0244The results did have variation, depending on which observations were included in the set. However, they each are close to 2.54%, which was our first estimate of the test error rate.
(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)
log.reg2 <- glm(default~income+balance+student, data=Default, family=binomial, subset = train)
prob <- predict(log.reg2, newdata = Default[-train, ], type = "response")
prediction <- rep("No", length(prob))
prediction[prob > 0.5] <- "Yes"
mean(prediction != Default[-train, ]$default)## [1] 0.0278Adding the student variable to the model did not lead to a significant change in the validation set estimate of 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)
log.reg <- glm(default ~ income + balance, data = Default, family = binomial)
summary(log.reg)##
## 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 estimates of the standard errors for the intercept, income, and balance variables are 4.348e-01, 4.985e-06, and 2.274e-04, respectively.
(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 \(\beta_0\), \(\beta_1\), and \(\beta_2\) are 4.344722e-01, 4.866284e-06, and 2.298949e-04, respectively.
(d) Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function.
The estimated standard errors obtained using either method are very similar.
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 \(\hat{\mu}\).
library(MASS)
attach(Boston)
mu_hat <- mean(medv)
mu_hat## [1] 22.53281(b) Provide an estimate of the standard error of \(\hat{\mu}\). 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(nrow(Boston))
se_hat## [1] 0.4088611(c) Now estimate the standard error of \(\hat{\mu}\) 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 estimates of the standard errors are very close to one another.
(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 \([\hat{\mu} - 2SE(\hat{\mu}), \hat{\mu}+2SE(\hat{\mu})]\).
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
## 22.53281conf_int <- c(22.53 - 2 * .4106622, 22.53 + 2 * .4106622)
conf_int## [1] 21.70868 23.35132The confidence intervals are very close to one another using either method.
(e) Based on this data set, provide an estimate, \(\hat{\mu}_{med}\), 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 \(\hat{\mu}_{med}\). 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.
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 median value is 21.2, which is equal to the value obtained using median(medv). The standard error of 0.3770241 is relatively small compared to median value.
(g) Based on this data set, provide an estimate for the tenth percentile of medv in Boston suburbs. Call this quantity \(\hat{\mu}_{0.1}\). (You can use the quantile() function.)
tenthp <- quantile(medv, c(0.1))
tenthp## 10%
## 12.75(h) Use the bootstrap to estimate the standard error of \(\hat{\mu}_{0.1}\). 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 estimated tenth percentile value is 12.75, which is again equal to the value obtained before. Also, the standard error is small compared to the percentile value at only 0.4925766.