(a) Explain how k-fold cross-validation is implemented K-fold cross-validation is a widely used technique in machine learning and statistical modeling that helps to assess the performance and generalization capability of a model. It involves dividing a dataset into k equally sized folds, where k is a positive integer greater than 1.Split the data into k equally sized subsets or folds. Each fold contains an equal number of samples from the dataset.For each fold, use the other k-1 folds as the training data to fit the model. This means that k different models are trained on k-1 folds of data.Evaluate the performance of the trained model on the remaining fold that was not used during training. This gives an estimate of how well the model generalizes to new data.Repeat the training and evaluation process for each of the k folds. This means that each fold is used once for evaluation and k-1 times for training.Compute the average performance over the k iterations. This gives an estimate of the overall performance of the model.
(b) What are the advantages and disadvantages of k-fold crossvalidation relative to:i. The validation set approach? ii. LOOCV? i.relative to the validation set approach: Advantages:K-fold cross-validation allows for a more efficient use of data, as all available data is used for training and testing the model. It provides a more reliable estimate of the model’s performance than the validation set approach.It can be used for both small and large datasets. Disadvantages: K-fold cross-validation can be computationally expensive, especially when the dataset is large or the model is complex.The performance estimate obtained by k-fold cross-validation can be sensitive to the value of k chosen.K-fold cross-validation may not work well for imbalanced datasets where the classes are not equally represented in each fold. ii relative to LOOCV: Advantages: K-fold cross-validation is less computationally expensive than LOOCV, especially when the dataset is large.K-fold cross-validation can provide a more reliable estimate of the model’s performance than LOOCV, as it reduces the variance in the estimate of the model’s performance. It can be used for both small and large datasets. Disadvantages: LOOCV uses all the available data for training except for one sample, which may lead to a more accurate estimate of the model’s performance.LOOCV may work better for small datasets, as k-fold cross-validation may result in too few samples for training the model.
#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(4)
glm.Default <- glm(default~income+balance, data=Default,family=binomial )
summary(glm.Default)##
## 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(4)
train <- sample(nrow(Default), size = nrow(Default) * 0.5)
train.default=Default[train,]
test.default=Default[-train,]
dim(train.default)## [1] 5000 4dim(test.default)## [1] 5000 4ii. Fit a multiple logistic regression model using only the training observations.
set.seed(4)
glm.Default <- glm(default~income+balance, data=Default,family=binomial,subset = train)
summary(glm.Default)##
## Call:
## glm(formula = default ~ income + balance, family = binomial,
## data = Default, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.9240 -0.1468 -0.0558 -0.0191 3.5650
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.135e+01 6.096e-01 -18.613 <2e-16 ***
## income 9.504e-06 7.000e-06 1.358 0.175
## balance 5.772e-03 3.281e-04 17.591 <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: 1477.13 on 4999 degrees of freedom
## Residual deviance: 792.65 on 4997 degrees of freedom
## AIC: 798.65
##
## 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.
library(ISLR)
attach(Default)
glm.Default <- glm(default~income+ balance, data=Default,family=binomial,subset = train)
glm.probs=predict(glm.Default, test.default, type="response")
glm.pred<- rep("No", 5000)
glm.pred[glm.probs>0.5]="Yes"
default.test=default[-train]
table(glm.pred, default.test)## default.test
## glm.pred No Yes
## No 4815 107
## Yes 21 57iv. Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified.
mean(glm.pred==default.test)## [1] 0.9744mean(glm.pred!=default.test)## [1] 0.0256The validation set approach gave us a 2.56% test error rate.
(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.
# Try 1 , (0.6/0.4)#
set.seed(4)
train <- sample(nrow(Default), size = nrow(Default) * 0.6)
train.default=Default[train,]
test.default=Default[-train,]
default.test=default[-train]
glm.Default <- glm(default~income+ balance, data=Default,family=binomial,subset = train)
glm.probs=predict(glm.Default, test.default, type="response")
glm.pred<- rep("No", length(default.test))
glm.pred[glm.probs>0.5]="Yes"
table(glm.pred, default.test)## default.test
## glm.pred No Yes
## No 3857 82
## Yes 18 43mean(glm.pred==default.test)## [1] 0.975mean(glm.pred!=default.test)## [1] 0.025The validation set approach gave us a 2.25% test error rate.
# Try 2 (0.25/0.75)#
set.seed(4)
train <- sample(nrow(Default), size = nrow(Default) * 0.75)
train.default=Default[train,]
test.default=Default[-train,]
default.test=default[-train]
glm.Default <- glm(default~income+ balance, data=Default,family=binomial,subset = train)
glm.probs=predict(glm.Default, test.default, type="response")
glm.pred<- rep("No", length(default.test))
glm.pred[glm.probs>0.5]="Yes"
table(glm.pred, default.test)## default.test
## glm.pred No Yes
## No 2413 42
## Yes 17 28mean(glm.pred==default.test)## [1] 0.9764mean(glm.pred!=default.test)## [1] 0.0236The validation set approach gave us a 2.36% test error rate.
# Try 3 (0.1/0.9)#
set.seed(4)
train <- sample(nrow(Default), size = nrow(Default) * 0.9)
train.default=Default[train,]
test.default=Default[-train,]
default.test=default[-train]
glm.Default <- glm(default~income+ balance, data=Default,family=binomial,subset = train)
glm.probs=predict(glm.Default, test.default, type="response")
glm.pred<- rep("No", length(default.test))
glm.pred[glm.probs>0.5]="Yes"
table(glm.pred, default.test)## default.test
## glm.pred No Yes
## No 966 17
## Yes 7 10mean(glm.pred==default.test)## [1] 0.976mean(glm.pred!=default.test)## [1] 0.024The validation set approach gave us a 2.4% 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
set.seed(4)
train <- sample(nrow(Default), size = nrow(Default) * 0.5)
train.default=Default[train,]
test.default=Default[-train,]
default.test=default[-train]
glm.Default <- glm(default~income+ balance+ student, data=Default,
family=binomial,subset = train)
glm.probs=predict(glm.Default, test.default, type="response")
glm.pred<- rep("No", length(default.test))
glm.pred[glm.probs>0.5]="Yes"
table(glm.pred, default.test)## default.test
## glm.pred No Yes
## No 4816 111
## Yes 20 53mean(glm.pred==default.test)## [1] 0.9738mean(glm.pred!=default.test)## [1] 0.0262The validation set approach gave us a 2.56% test error rate without “student” as a dummy variable, and gave us a 2.62% test error rate with “student” as a dummy variable.Dummy variable for “student” DID NOT lead to a reduction in the test error rate.
##Problem 6## #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. DDo 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(4)
glm.Default <- glm(default~income+balance, data=Default,family=binomial )
summary(glm.Default)##
## 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 glm() estimates of the standard errors for the coefficients intercept, income and balance are respectively 4.348e-01, 4.985e-06 and 2.274e-04.
(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))))}
dim(Default)## [1] 10000 4boot.fn(Default,1:10000)## (Intercept) income balance
## -1.154047e+01 2.080898e-05 5.647103e-03(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,500)##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = Default, statistic = boot.fn, R = 500)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* -1.154047e+01 1.835069e-02 4.355763e-01
## t2* 2.080898e-05 -3.497509e-07 4.726585e-06
## t3* 5.647103e-03 -4.977180e-06 2.360959e-04The bootstrap estimates of the standard errors for the coefficients B0, B1 and B2 are respectively :4.355763e-01, 4.726585e-06, 2.360959e-04
(d) Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function. The glm() estimates of the standard errors for the coefficients intercept, income and balance are respectively 4.348e-01, 4.985e-06 and 2.274e-04.
The bootstrap estimates of the standard errors for the coefficients B0, B1 and B2 are respectively :4.355763e-01, 4.726585e-06, 2.360959e-04
The estimated standard errors obtained by the two methods are pretty close.
##9. 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.
library(ISLR2)
attach(Boston)
mean.M =mean(medv)
mean.M## [1] 22.53281estimate for the population mean of medv ˆµ= 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.
stderr.mean = sd(medv)/sqrt(length(medv))
stderr.mean## [1] 0.4088611the standard error of the sample mean is 0.4088611
(c) Now estimate the standard error of ˆµ using the bootstrap. How does this compare to your answer from (b)?
boot.fn=function(data, index){return(mean(data[index]))}
library(boot)
boot(medv,boot.fn,500)##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = medv, statistic = boot.fn, R = 500)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 22.53281 -0.02674308 0.3910375standard error of ˆµ using the bootstrap is 0.4099959, is similar to the estimate found in (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).
Hint: You can approximate a 95 % confidence interval using the
formula [ˆµ − 2SE(ˆµ), µˆ + 2SE(ˆµ)].
CI.Boston <- c(22.53-2*0.41, 22.53+2*0.41)
CI.Boston## [1] 21.71 23.35t.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
## 22.53281A 95 % confidence interval using bootstrap is between 21.71 and 23.35, using t test we have a CI of 22.53. The two result are approximately the same.
(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.fn2=function(data, index){return(median(data[index]))}
library(boot)
boot(medv,boot.fn2,500)##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = medv, statistic = boot.fn2, R = 500)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 21.2 0.0096 0.3914899We get an estimated median value of 21.2 which is equal to the value obtained in (e), with a standard error of 0.3874 which is relatively small compared to median value.
(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.fn3=function(data, index){return(quantile(data[index], c(0.1)))}
library(boot)
boot(medv,boot.fn3,500)##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = medv, statistic = boot.fn3, R = 500)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 12.75 0.0196 0.4838948Estimated quantile value of 12.75 from bootstap which is same as the value in (g), and the standard error of 0.5135083.