We now review k-fold cross-validation.
(a) Explain how k-fold cross-validation is implemented.
The dataset is first divided into K different parts (normally 5 or 10). The first part is separated to become the test dataset and the remaining K-1 parts are used for the training dataset. After training, the error or accuracy rate is tested on the 1st K part. This process is repeated with each K part until all of the k parts have been tested. The MSE’s are averaged to get an estimated error rate for new observations.
(b) What are the advantages and disadvantages of k-fold crossvalidation relative to:
i. The validation set approach?
The k-fold cross validation approach is much more stable with less variable mean squared errors than the validation set approach. The k-fold approach also lends itself to use more observations in the training data to get better estimates of the error rate. The main disadvantage of the k-fold cross validation approach is that it is not as easy to implement compared to the validation set approach.
ii. LOOCV?
The main advantages of k-fold cross validation versus LOOCV are that the process is less computationally intensive and there is less variance in the MSE’s. The disadvantage is that k-fold cross validation has higher bias than LOOCV when k < n.
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.
library(ISLR)
summary(Default)## default student balance income
## No :9667 No :7056 Min. : 0.0 Min. : 772
## Yes: 333 Yes:2944 1st Qu.: 481.7 1st Qu.:21340
## Median : 823.6 Median :34553
## Mean : 835.4 Mean :33517
## 3rd Qu.:1166.3 3rd Qu.:43808
## Max. :2654.3 Max. :73554attach(Default)(a) Fit a logistic regression model that uses income and balance to predict default.
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.
set.seed(0)
train = sample(10000, 5000)ii. Fit a multiple logistic regression model using only the training observations.
glm.fit <- glm(default ~ income + balance, data = Default, family = binomial,
subset = train)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.
glm.pred <- rep("No", 5000)
glm.probs <- predict(glm.fit, Default[-train, ], type = "response")
glm.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(glm.pred != default[-train])## [1] 0.0256(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.
3 more trials resulted in an average error rate of 0.0261.
Default_error_rate <- function() {
train = sample(10000, 5000)
glm.fit <- glm(default ~ income + balance, data = Default, family = binomial,
subset = train)
glm.pred <- rep("No", 5000)
glm.probs <- predict(glm.fit, Default[-train, ], type = "response")
glm.pred[glm.probs > 0.5] <- "Yes"
return(mean(glm.pred != Default[-train, ]$default))
}
Default_error_rate()## [1] 0.0264Default_error_rate()## [1] 0.025Default_error_rate()## [1] 0.0268(d) Now consider a logistic regression model that predicts the probability of default using income, balance, and a dummy variable for student. Estimate the test error for this model using the validation set approach. Comment on whether or not including a dummy variable for student leads to a reduction in the test error rate.
The inclusion of the dummy variable did not reduce the test error rate.
Default_error_rate <- function() {
train = sample(10000, 5000)
glm.fit <- glm(default ~ income + balance + student, data = Default,
family = binomial, subset = train)
glm.pred <- rep("No", 5000)
glm.probs <- predict(glm.fit, Default[-train, ], type = "response")
glm.pred[glm.probs > 0.5] <- "Yes"
return(mean(glm.pred != Default[-train, ]$default))
}
Default_error_rate()## [1] 0.0268We 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.
library(ISLR)
attach(Default)## The following objects are masked from Default (pos = 3):
##
## balance, default, income, studentsummary(Default)## default student balance income
## No :9667 No :7056 Min. : 0.0 Min. : 772
## Yes: 333 Yes:2944 1st Qu.: 481.7 1st Qu.:21340
## Median : 823.6 Median :34553
## Mean : 835.4 Mean :33517
## 3rd Qu.:1166.3 3rd Qu.:43808
## Max. :2654.3 Max. :73554(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)
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.
boot.fn = function(data, index)
return(coef(glm(default ~ income + balance, data = data, family = binomial,
subset = index)))(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 standard errors obtained by the logistic regression model and the bootstrap function are very close. In this case, the bootstrap function closely mimics the estimates from the logistic regression model. In other cases, the bootstrap function is used to estimate parameters which do not easily fit a model or formula.
We will now consider the Boston housing data set, from the MASS library.
library(MASS)
set.seed(1)
attach(Boston)
summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08205 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv
## Min. : 1.73 Min. : 5.00
## 1st Qu.: 6.95 1st Qu.:17.02
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00(a) Based on this data set, provide an estimate for the population mean of medv. Call this estimate \(\hat{\mu}\).
mean(medv)## [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.
sd(medv) / sqrt(length(medv))## [1] 0.4088611(c) Now estimate the standard error of \(\hat{\mu}\) using the bootstrap. How does this compare to your answer from (b)?
The mean estimate matched out to 5 decimal places while the standard error matched to 2 decimal places if the formula estimate is rounded.
boot.fn <- function(data, index)
return(mean(data[index]))
library(boot)
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 [\(\hat{\mu}\) − 2SE(\(\hat{\mu}\)), \(\hat{\mu}\) + 2SE(\(\hat{\mu}\))].
c(22.53281 - 2 * 0.4106622, 22.53281 + 2 * 0.4106622)## [1] 21.71149 23.35413t.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
## 22.53281(e) Based on this data set, provide an estimate, \(\hat{\mu}_{med}\), for the median value of medv in the population.
median(medv)## [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.
The bootstrap estimate of the median was exactly the same as the formula (21.2), and the standard error was estimated at 0.377.
boot.fn = function(data, index)
return(median(data[index]))
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 0.3770241(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.)
quantile(medv, c(0.1))## 10%
## 12.75(h) Use the bootstrap to estimate the standard error of \(\hat{\mu}_{0.1}\). Comment on your findings.
The tenth percentile estimate was identical to the calculation from the quantile function (12.75) and the standard error was 0.4926.
boot.fn = function(data, index)
return(quantile(data[index], c(0.1)))
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 0.4925766