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 the observations and separating them into “folds” at random. A model is trained on k−1 of the folds and tested on the remaining fold. This process is repeated k times, meaning each k fold is used as the testing data. The test error is then cross-validated 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 advantages of the k-fold cross-validation in comparison to the validation set approach are consistency and accuracy. Depending on the observations used in the training set compared to the ones in the validation set, the validation estimate of the test error rate can vary drastically. Second, only a subset of the observations are used to fit the model. The validation set approach can over-estimate the test error when compared to a model that is fit on the entire data set, since most models will improve with a larger data set.
The disadvantages of the k-fold cross-validation in comparison to the validation set approach are complexity and timeliness. The validation set approach is easier to understand, making it easier to perform and explain to others. Additionally, the validation set approach only requires a model to be trained and tested once. Large data sets or large k values would require far more effort and time put in than a simple validation set.
ii. LOOCV?
The advantages of the k-fold cross-validation in comparison to the LOOCV approach are being less computationally demanding and potentially less variance. K-fold cross-validation requires the model to be fitted only k times, while LOOCV would fit the model n times. Also, the bias-variance trade off gives the k-fold method less variance but a higher chance of bias than the LOOCV method.
The disadvantages of the k-fold cross-validation in comparison to the LOOCV approach are potentially being more computationally demanding and potentially more bias. In some cases, when n is small, LOOCV can actually require less computational power than k-fold cross-validation. Secondly, LOOCV has lower bias but higher variance than the k-fold cross-validation method.
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)
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: 8b. 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] 0.0274train <- 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, ]$default)## [1] 0.0244train <- sample(dim(Default)[1], dim(Default)[1] / 2)
fit.glm <- glm(default ~ income + balance, data = Default, family = "binomial", subset = train)
probs <- predict(fit.glm, newdata = Default[-train, ], type = "response")
pred.glm <- rep("No", length(probs))
pred.glm[probs > 0.5] <- "Yes"
mean(pred.glm != Default[-train, ]$default)## [1] 0.0244The first validation set error rate is the highest of the four, including the first test ran in part b, at 2.74%. The last two rates match at a 2.44% error rate. These were the lowest of the four rates, giving a variance of 0.30%. While this variance is low, it still shows the values used in the test and training sets have some influence on the 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)
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] 0.0278The test error rate for this model is 2.78%. This is actually higher than the previous range of error rates provided in part c, so the addition of a dummy variable for student does not appear to reduce the 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)## 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 estimated standard errors are \(β0=4.348*10^{-1}\) for the intercept, \(β1= 4.985*10^{-6}\) for income, and \(β2=2.274*10^{-4}\) for balance.
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 standard errors of the coefficients are \(β0=4.345*10^{-1}\) for the intercept, \(β1=4.866*10^{-6}\) for income, and \(β2=2.299x10^{-4}\) for balance.
d. Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function.
The estimated standard errors obtained by the glm function and the bootstrap function are very similar. The intercepts are basically identical, while the glm predicted a slightly higher income and a slightly lower balance coefficient than the bootstrap function.
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 \(\hat{μ}\).
library(MASS)
attach(Boston)
mu.hat <- mean(medv)
mu.hat## [1] 22.53281The estimated population mean is \(\hat{μ}= 22.533\)
b. Provide an estimate of the standard error of \(\hat{μ}\). 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] 0.4088611The estimate of the standard error of \(\hat{μ}\) is \(4.089*10^{-1}\)
c. Now estimate the standard error of \(\hat{μ}\) 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 estimate of the standard error of \(\hat{μ}\) is \(4.107*10^{-1}\) . This is slightly higher than the calculated standard error of \(4.089*10^{-1}\) from part 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).
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 * 0.4119, 22.53 + 2 * 0.4119)
CI.mu.hat## [1] 21.7062 23.3538The bootstrap 95% confidence interval of 21.7062- 23.3538 is nearly identical to the 95% confidence interval provided by the t test of 21.72953 - 23.33608.
e. Based on this data set, provide an estimate, \(\hat{μ}\) med, for the median value of medv in the population.
med.hat <- median(medv)
med.hat## [1] 21.2f. We now would like to estimate the standard error of \(\hat{μ}\) 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.
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 medians are identical at 21.2, and the standard error is very small at 0.3770.
g. Based on this data set, provide an estimate for the tenth percentile of medv in Boston suburbs. Call this quantity \(\hat{μ}\) 0.1. (You can use the quantile() function.)
percent.hat <- quantile(medv, c(0.1))
percent.hat## 10%
## 12.75h. Use the bootstrap to estimate the standard error of \(\hat{μ}\) 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 of 12.75 is identical to the tenth percentile calculated in g. The standard error of the tenth percentile is 0.4926, which is pretty low compared to percentile value itself.