We now review k-fold cross-validation. (a) Explain how k-fold cross-validation is implemented. Observations are divided randomly into K groups or folds. The first fold is the validation set and the rest are computed using fit k-1. This is done over and over again using K times and each time a new group of observations is used as the validation set. (b) What are the advantages and disadvantages of k-fold crossvalidation relative to: i. The validation set approach? Advantages: it is simple and easy to implement Disadvantages: the validation MSE can vary alot and also only a set of observations are used to fit the model(only the training set is used not the validation set) ii. LOOCV? Advantages: has less bias, produces a less variable MSE Disadvantage: It is 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.
library(ISLR)
library(ISLR2)##
## Attaching package: 'ISLR2'## The following objects are masked from 'package:ISLR':
##
## Auto, Creditattach(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. :73554dim(Default)## [1] 10000 4logR = glm(default ~ balance + income, data=Default, family='binomial')
summary(logR)##
## 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: i. Split the sample set into a training set and a validation set.
#i
default.num = sample(dim(Default)[1],dim(Default)[1]*.75)
default.train = subset(Default[default.num,])
val.default = subset(Default[-default.num,])ii. Fit a multiple logistic regression model using only the training observations.
logr.default = glm(default ~ balance + income, family = binomial, data = default.train)
summary(logr.default)##
## Call:
## glm(formula = default ~ balance + income, family = binomial,
## data = default.train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.2325 -0.1441 -0.0578 -0.0214 3.7190
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.162e+01 4.991e-01 -23.279 < 2e-16 ***
## balance 5.679e-03 2.611e-04 21.749 < 2e-16 ***
## income 2.319e-05 5.715e-06 4.057 4.96e-05 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 2225.7 on 7499 degrees of freedom
## Residual deviance: 1194.7 on 7497 degrees of freedom
## AIC: 1200.7
##
## 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.
log.prob.default = predict(logr.default, val.default, type = "response")
log.pred.default = rep("No", length(log.prob.default))
log.pred.default[log.prob.default > 0.5] ="Yes"
table(log.pred.default, val.default$default)##
## log.pred.default No Yes
## No 2407 54
## Yes 15 24iv. Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified.
mean(log.pred.default != val.default$default)## [1] 0.0276(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. The second split with the .8 train and .6 validate had the highest error rate.
.65 train and .4 validate Error rate is .026
default.num1 = sample(dim(Default)[1],dim(Default)[1]*.65)
default.train1 = subset(Default[default.num1,])
val.default1 = subset(Default[-default.num1,])
logr.default1 = glm(default ~ balance + income, family = binomial, data = default.train1)
log.prob.default1 = predict(logr.default1, val.default1, type = "response")
log.pred.default1 = rep("No", length(log.prob.default1))
log.pred.default1[log.prob.default1 > 0.4] ="Yes"
table(log.pred.default1, val.default1$default)##
## log.pred.default1 No Yes
## No 3363 67
## Yes 26 44mean(log.pred.default1 != val.default1$default)## [1] 0.02657143.8 train and .6 validate Error rate is .031
default.num2 = sample(dim(Default)[1],dim(Default)[1]*.8)
default.train2 = subset(Default[default.num2,])
val.default2 = subset(Default[-default.num2,])
logr.default2 = glm(default ~ balance + income, family = binomial, data = default.train2)
log.prob.default2 = predict(logr.default2, val.default2, type = "response")
log.pred.default2 = rep("No", length(log.prob.default2))
log.pred.default2[log.prob.default2 > 0.6] ="Yes"
table(log.pred.default2, val.default2$default)##
## log.pred.default2 No Yes
## No 1928 51
## Yes 7 14mean(log.pred.default2 != val.default2$default)## [1] 0.029.4 train and .4 validate Error rate is .028
default.num3 = sample(dim(Default)[1],dim(Default)[1]*.4)
default.train3 = subset(Default[default.num3,])
val.default3 = subset(Default[-default.num3,])
logr.default3 = glm(default ~ balance + income, family = binomial, data = default.train3)
log.prob.default3 = predict(logr.default3, val.default3, type = "response")
log.pred.default3 = rep("No", length(log.prob.default3))
log.pred.default3[log.prob.default3 > 0.4] ="Yes"
table(log.pred.default3, val.default3$default)##
## log.pred.default3 No Yes
## No 5765 113
## Yes 46 76mean(log.pred.default3 != val.default3$default)## [1] 0.0265(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.
There was not a significant increase in the error rate. I used a .65 train and .6 validate set. The test error was .029 which is close to the error rates from previous tests.
student1 = ifelse(student == "Yes", 1, 0)
Default1 = data.frame(Default, student1)def.logr.num = sample(dim(Default1)[1],dim(Default1)[1]*.65)
def.logr.train = subset(Default1[def.logr.num,])
def.logr.val = subset(Default1[-def.logr.num,])
logr.d = glm(default ~ income + balance + student1, data = Default1, family = binomial)
logr.prob.def = predict(logr.d, def.logr.val, type = "response")
logr.pred.def = rep("No", length(logr.prob.def))
logr.pred.def[logr.prob.def > 0.6] ="Yes"
table(logr.pred.def, def.logr.val$default)##
## logr.pred.def No Yes
## No 3367 101
## Yes 3 29mean(logr.pred.def != def.logr.val$default)## [1] 0.02971429We 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. Income = 4.98 Balance = 2.27
set.seed(3)
mlrdefault = glm(default ~ income + balance, data = Default, family = binomial)
summary(mlrdefault)##
## 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) + coef(glm(default ~ balance + income, data = data, subset = index, family = binomial))(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, 250)##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = Default, statistic = boot.fn, R = 250)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* -1.154047e+01 -2.418584e-02 4.421693e-01
## t2* 5.647103e-03 1.876376e-05 2.416242e-04
## t3* 2.080898e-05 -2.107241e-07 4.868477e-06(d)Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function. The estimated standard errors obtained using glm() were smaller than those obtained by using bootstrap but not by much. There was not a significant difference between the two methoods.
We will now consider the Boston housing data set, from the ISLR2 library. (a) Based on this data set, provide an estimate for the population mean of medv. Call this estimate ˆμ.
summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08204 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 lstat
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 1.73
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.: 6.95
## Median : 5.000 Median :330.0 Median :19.05 Median :11.36
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :12.65
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:16.95
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :37.97
## medv
## Min. : 5.00
## 1st Qu.:17.02
## Median :21.20
## Mean :22.53
## 3rd Qu.:25.00
## Max. :50.00attach(Boston)
mu = mean(medv)
mu## [1] 22.53281(b) Provide an estimate of the standard error of ˆμ. 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.
stde.mu = sd(medv)/sqrt(length(medv))
stde.mu## [1] 0.4088611(c) Now estimate the standard error of ˆμ using the bootstrap. How does this compare to your answer from (b)? My anwser from b was .408 rounded to .41. With bootstrap I got .416 rounded to .42. There was only a slight difference in the SE
boot.b = function(data, index)return(mean(data[index]))
boot.b2 = boot(medv, boot.b, 250)
boot.b2##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = medv, statistic = boot.b, R = 250)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 22.53281 -0.003516996 0.4011974(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$medv). Hint: You can approximate a 95% confidence interval using the formula [ˆμ − 2SE(ˆμ), ˆμ + 2SE(ˆμ)].
t.test(Boston$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
## 22.53281set.seed(3)
ci.boot = c(22.53281 - 2 * .4161404, 22.53281 + 2 * .4161404)
ci.boot## [1] 21.70053 23.36509(e) Based on this data set, provide an estimate, ˆμmed, for the median value of medv in the population.
median.mu = median(medv)
median.mu## [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. The standard error for ˆμmed is .38.
boot.bm = function(data, index)return(median(data[index]))
boot.bm3 = boot(medv, boot.bm, 250)
boot.bm3##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = medv, statistic = boot.bm, R = 250)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 21.2 0.001 0.3758912(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.)
mu.1 = quantile(medv, c(0.1))
mu.1## 10%
## 12.75(h) Use the bootstrap to estimate the standard error of ˆμ0.1. Comment on your findings. The standard error is .48
boot.bme = function(data, index)return(quantile(data[index], c(0.1)))
boot.bme4 = boot(medv, boot.bme, 250)
boot.bme4##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = medv, statistic = boot.bme, R = 250)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 12.75 0.0532 0.4792373