Libraries
library(ISLR2)
library(boot)
library(MASS)##
## Attaching package: 'MASS'## The following object is masked from 'package:ISLR2':
##
## BostonWe now review k-fold cross-validation.
(a) Explain how k-fold cross-validation is implemented.
"This approach involves randomly dividing the set of observations into k groups, or folds, of approximately equal size. The first fold is treated as a validation set, and the method is fit on the remaining k − 1 folds. The mean squared error, MSE1, is then computed on the observations in the held-out fold. This procedure is repeated k times; each time, a different group of observations is treated as a validation set. This process results in k estimates of the test error. The k-fold CV estimate is computed by averaging these values."(b) What are the advantages and disadvantages of k-fold cross-validation relative to:
"The validation set approach is conceptually simple and is easy to implement to its advantage, whereas the disadvantage is that the MSE can be highly variable and Only a subset of observations are used to fit the model (training data).""LOOCV tends to have less bias. Where The validation approach produces different MSE when applied repeatedly due to randomness in the splitting process, performing the LOOCV multiple times will always yield the same results, because we split based on 1 obs. each time. To its disadvantage, however, is that it can be computationally intensive."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.
data("Default")
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. :73554logmodel1 = glm(default ~ balance + income, data = Default, family = binomial)
summary(logmodel1)##
## Call:
## glm(formula = default ~ balance + income, 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 ***
## balance 5.647e-03 2.274e-04 24.836 < 2e-16 ***
## income 2.081e-05 4.985e-06 4.174 2.99e-05 ***
## ---
## 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:
Split the sample set into a training set and a validation set.
Fit a multiple logistic regression model using only the training observations.
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.
Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified.
#i
trainDefault = sample(dim(Default)[1], dim(Default)[1]*0.50)
testDefault = Default[-trainDefault, ]
#ii
LOGmodel2 = glm(default ~ balance + income, data = Default, family = binomial, subset = trainDefault)
#iii
log.prob_def = predict(LOGmodel2, testDefault, type = "response")
log.pred_def = rep("No", dim(Default)[1]*0.50)
log.pred_def[log.prob_def > 0.5] = "Yes"
table(log.pred_def, testDefault$default)##
## log.pred_def No Yes
## No 4800 108
## Yes 24 68#iv
mean(log.pred_def !=testDefault$default)## [1] 0.0264(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.
1
#i
trainDefault = sample(dim(Default)[1], dim(Default)[1]*0.50)
testDefault = Default[-trainDefault, ]
#ii
LOGmodel2 = glm(default ~ balance + income, data = Default, family = binomial, subset = trainDefault)
#iii
log.prob_def = predict(LOGmodel2, testDefault, type = "response")
log.pred_def = rep("No", dim(Default)[1]*0.50)
log.pred_def[log.prob_def > 0.5] = "Yes"
table(log.pred_def, testDefault$default)##
## log.pred_def No Yes
## No 4814 119
## Yes 15 52#iv
mean(log.pred_def !=testDefault$default)## [1] 0.02682
#i
trainDefault = sample(dim(Default)[1], dim(Default)[1]*0.50)
testDefault = Default[-trainDefault, ]
#ii
LOGmodel2 = glm(default ~ balance + income, data = Default, family = binomial, subset = trainDefault)
#iii
log.prob_def = predict(LOGmodel2, testDefault, type = "response")
log.pred_def = rep("No", dim(Default)[1]*0.50)
log.pred_def[log.prob_def > 0.5] = "Yes"
table(log.pred_def, testDefault$default)##
## log.pred_def No Yes
## No 4805 128
## Yes 12 55#iv
mean(log.pred_def !=testDefault$default)## [1] 0.0283
#i
trainDefault = sample(dim(Default)[1], dim(Default)[1]*0.50)
testDefault = Default[-trainDefault, ]
#ii
LOGmodel2 = glm(default ~ balance + income, data = Default, family = binomial, subset = trainDefault)
#iii
log.prob_def = predict(LOGmodel2, testDefault, type = "response")
log.pred_def = rep("No", dim(Default)[1]*0.50)
log.pred_def[log.prob_def > 0.5] = "Yes"
table(log.pred_def, testDefault$default)##
## log.pred_def No Yes
## No 4826 108
## Yes 17 49#iv
mean(log.pred_def !=testDefault$default)## [1] 0.025(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.
#i
trainDefault = sample(dim(Default)[1], dim(Default)[1]*0.50)
testDefault = Default[-trainDefault, ]
#ii
LOGmodel2 = glm(default ~ balance + income + student, data = Default, family = binomial, subset = trainDefault)
#iii
log.prob_def = predict(LOGmodel2, testDefault, type = "response")
log.pred_def = rep("No", dim(Default)[1]*0.50)
log.pred_def[log.prob_def > 0.5] = "Yes"
table(log.pred_def, testDefault$default)##
## log.pred_def No Yes
## No 4808 124
## Yes 20 48#iv
mean(log.pred_def !=testDefault$default)## [1] 0.0288"In our case, we get a 2.7% error rate, which isn't that much differenct than the other splits. There isn't any impact on the error rate even if we add the student variable."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.
LOGmodel3 = glm(default ~ balance + income, data = Default, family = binomial)
summary(LOGmodel3)##
## Call:
## glm(formula = default ~ balance + income, 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 ***
## balance 5.647e-03 2.274e-04 24.836 < 2e-16 ***
## income 2.081e-05 4.985e-06 4.174 2.99e-05 ***
## ---
## 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 ~ balance + income, 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.
boot(Default, boot.fn, 100)##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = Default, statistic = boot.fn, R = 100)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* -1.154047e+01 -1.056755e-01 4.365345e-01
## t2* 5.647103e-03 4.506066e-05 2.439301e-04
## t3* 2.080898e-05 1.049700e-06 4.841117e-06(d) Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function.
"So we get the estimated standard errors for the coefficients, so there for income and balance are 2.294044e-04 and 3.875525e-06."We will now consider the Boston housing data set, from the ISLR2 library.
data("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 ˆµ.
attach(Boston)
mean.medv = mean(medv)
mean.medv## [1] 22.53281(b) Provide an estimate of the standard error of ˆµ. Interpret this result.
stderr.mean = sd(medv)/sqrt(length(medv))
stderr.mean## [1] 0.4088611(c) Now estimate the standard error of ˆµ using the bootstrap. How does this compare to your answer from (b)?
boot.fn2 = function(data, index) return(mean(data[index]))
boot2 = boot(medv, boot.fn2, 100)
boot2##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = medv, statistic = boot.fn2, R = 100)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 22.53281 0.07524901 0.3999444(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
## 22.53281(e) Based on this data set, provide an estimate, ˆµmed, for the median value of medv in the population.
median.medv = median(medv)
median.medv## [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.
boot.fn3 = function(data, index) return(median(data[index]))
boot3 = boot(medv, boot.fn3, 1000)
boot3##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = medv, statistic = boot.fn3, R = 1000)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 21.2 -0.0416 0.3768507(g) Based on this data set, provide an estimate for the tenth percentile of medv in Boston census tracts. Call this quantity ˆµ0.1. (You can use the quantile() function.)
tenth.medv = quantile(medv, c(0.1))
tenth.medv## 10%
## 12.75(h) Use the bootstrap to estimate the standard error of ˆµ0.1. Comment on your findings.
boot.fn4 = function(data, index) return(quantile(data[index], c(0.1)))
boot4 = boot(medv, boot.fn4, 1000)
boot4##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = medv, statistic = boot.fn4, R = 1000)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 12.75 -0.0096 0.508632"We found the estimated quantile value of 12.75 from bootstap, which is same as the value in (g) and the standard error of 0.487."…