3. We now review k-fold cross-validation.

(a) Explain how k-fold cross-validation is implemented.

The k-fold cross validation involves dividing the amount of observations into k groups (or folds). The first fold is considered a validation set and the remainders are used at training sets to fit the model. This process are repeated k times, with k=number of times we want. Each time, a different group will be treated as a validation set. Test error is then computed by averaging the all the MSE estimates from all the folds.

(b) What are the advantages and disadvantages of k-fold crossvalidation relative to:

i. The validation set approach?

Advantages:The validation set approach is conceptually simple and is easy to implement

Disadvantages: The validation MSE can be highly variable (depends on which observations are included in the training/validation set). Also in validation set approach, only one subset of observations are used to fit the model, as a training set, and may overestimate the test error rate.

ii. LOOCV?

Advantages: LOOCV have less bias. We repeatedly fit the statistical learning method using training data that contains n-1 obs so almost all the data set is used. LOOCV produces a less variable MSE.

Disadvantage: LOOCV is computationally intensive

5. 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)
str(Default)
## 'data.frame':    10000 obs. of  4 variables:
##  $ default: Factor w/ 2 levels "No","Yes": 1 1 1 1 1 1 1 1 1 1 ...
##  $ student: Factor w/ 2 levels "No","Yes": 1 2 1 1 1 2 1 2 1 1 ...
##  $ balance: num  730 817 1074 529 786 ...
##  $ income : num  44362 12106 31767 35704 38463 ...

(a) Fit a logistic regression model that uses income and balance to predict 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: 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(1)
tr_ind <- sample(nrow(Default), 0.8*nrow(Default), replace = F)
train <- Default[tr_ind,]
test <- Default[-tr_ind,]

ii. Fit a multiple logistic regression model using only the training observations.

fit.glm <- glm(default ~ income + balance, data=train, family="binomial")
summary(fit.glm)
## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = train)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.4758  -0.1413  -0.0563  -0.0210   3.4620  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.168e+01  4.893e-01 -23.879  < 2e-16 ***
## income       2.547e-05  5.631e-06   4.523  6.1e-06 ***
## balance      5.613e-03  2.531e-04  22.176  < 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: 2313.6  on 7999  degrees of freedom
## Residual deviance: 1239.2  on 7997  degrees of freedom
## AIC: 1245.2
## 
## Number of Fisher Scoring iterations: 8

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.prob = predict.glm(fit.glm, newdata = test, type = "response") #create a prediction
glm.pred = rep("No",length(glm.prob))                              #use these for confusion Matrix or table
glm.pred[glm.prob > 0.5] = "Yes"
caret::confusionMatrix(as.factor(glm.pred), as.factor(test$default)) 
## Confusion Matrix and Statistics
## 
##           Reference
## Prediction   No  Yes
##        No  1928   50
##        Yes    2   20
##                                          
##                Accuracy : 0.974          
##                  95% CI : (0.966, 0.9805)
##     No Information Rate : 0.965          
##     P-Value [Acc > NIR] : 0.01371        
##                                          
##                   Kappa : 0.4252         
##                                          
##  Mcnemar's Test P-Value : 7.138e-11      
##                                          
##             Sensitivity : 0.9990         
##             Specificity : 0.2857         
##          Pos Pred Value : 0.9747         
##          Neg Pred Value : 0.9091         
##              Prevalence : 0.9650         
##          Detection Rate : 0.9640         
##    Detection Prevalence : 0.9890         
##       Balanced Accuracy : 0.6423         
##                                          
##        'Positive' Class : No             
##

iv. Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified

mean(glm.pred != test$default) 
## [1] 0.026

(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 error rate is different when we change the split of data (70/30, 60/40, and 50/50). At first they decrease but after the split of 60/40, the error rate starts to increase.

set.seed(1)
tr_ind <- sample(nrow(Default), 0.7*nrow(Default), replace = F)
train <- Default[tr_ind,]
test <- Default[-tr_ind,]

fit.glm <- glm(default ~ income + balance, data=train, family="binomial")

glm.prob = predict.glm(fit.glm, newdata = test, type = "response") 
glm.pred = rep("No",length(glm.prob))                             
glm.pred[glm.prob > 0.5] = "Yes"

mean(glm.pred != test$default) 
## [1] 0.02666667
set.seed(1)
tr_ind <- sample(nrow(Default), 0.6*nrow(Default), replace = F)
train <- Default[tr_ind,]
test <- Default[-tr_ind,]

fit.glm <- glm(default ~ income + balance, data=train, family="binomial")

glm.prob = predict.glm(fit.glm, newdata = test, type = "response") 
glm.pred = rep("No",length(glm.prob))                             
glm.pred[glm.prob > 0.5] = "Yes"

mean(glm.pred != test$default)
## [1] 0.025
set.seed(1)
tr_ind <- sample(nrow(Default), 0.5*nrow(Default), replace = F)
train <- Default[tr_ind,]
test <- Default[-tr_ind,]

fit.glm <- glm(default ~ income + balance, data=train, family="binomial")

glm.prob = predict.glm(fit.glm, newdata = test, type = "response") 
glm.pred = rep("No",length(glm.prob))                             
glm.pred[glm.prob > 0.5] = "Yes"

mean(glm.pred != test$default)
## [1] 0.0254

(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.

Adding the dummy variable for student increase the test error rate, but just slightly. New test error rate is 2.75% while the old one is 0.026

set.seed(1)
tr_ind <- sample(nrow(Default), 0.8*nrow(Default), replace = F)
train <- Default[tr_ind,]
test <- Default[-tr_ind,]

fit.glm <- glm(default ~ income + balance + student, data=train, family="binomial")

glm.prob = predict.glm(fit.glm, newdata = test, type = "response") 
glm.pred = rep("No",length(glm.prob))                             
glm.pred[glm.prob > 0.5] = "Yes"

mean(glm.pred != test$default)
## [1] 0.0275

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. 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.

The estimate Std. Error for the coefficients are: 0.04348, 4.985 x 10-6, and 2.274 x 10-4

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: 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.

library(boot)

boot.fn=function(data =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.

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 -3.890536e-02 4.347841e-01
## t2*  2.080898e-05  1.566550e-07 4.864757e-06
## t3*  5.647103e-03  1.843880e-05 2.301132e-04

Compare to the coefficients of the ‘glm()’ function, both the boot function and glm function give the same coefficients

summary(fit.glm)$coef
##                  Estimate   Std. Error    z value      Pr(>|z|)
## (Intercept) -1.154047e+01 4.347564e-01 -26.544680 2.958355e-155
## income       2.080898e-05 4.985167e-06   4.174178  2.990638e-05
## balance      5.647103e-03 2.273731e-04  24.836280 3.638120e-136

(d) Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function.

They have almost similar estimated standard errors. For bootstrap, Std error for ‘intercept’ and ‘balance’ are slightly higher while intercept for ‘income’ slightly lower

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(ISLR)
library(MASS)
u = mean(Boston$medv)
u
## [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.

dim(Boston)
## [1] 506  14
se.u <- sd(Boston$medv)/sqrt(dim(Boston)[1])
se.u
## [1] 0.4088611

The standard error is 0.4088

(c) Now estimate the standard error of ˆµ using the bootstrap. How does this compare to your answer from (b)?

set.seed(1)
boot.fn <- function(data, index) {
    u <- mean(data[index])
    return (u)
}
boot(Boston$medv, boot.fn, 1000)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original      bias    std. error
## t1* 22.53281 0.007650791   0.4106622

The standard error using bootstrap is similar to using the formula 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)].

#Confidence interval from t-test
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
#Confidence interval from bootstrap
CI.boot.mean <- c(u-2*se.u, u+2*se.u)
CI.boot.mean
## [1] 21.71508 23.35053

The confidence interval, using t-test and bootstrap, are pretty close to each other.

(e) Based on this data set, provide an estimate, ˆµmed, for the median value of medv in the population.

med <- median(Boston$medv)
med
## [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.

set.seed(1)

boot.fn = function(data, index) {
  med <- median(data[index])
  return (med)
}

SE.boot.med = boot(Boston$medv, boot.fn, R=1000)
SE.boot.med
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*     21.2 0.02295   0.3778075

(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.)

u0.1 <- quantile(Boston$medv,.1)
u0.1
##   10% 
## 12.75

(h) Use the bootstrap to estimate the standard error of ˆµ0.1. Comment on your findings.

set.seed(1)

boot.fn = function(data, index) {
  u0.1 <- quantile(data[index],c(0.1))
  return (u0.1)
}
SE.boot.u0.1 = boot(Boston$medv, boot.fn, R=1000)
SE.boot.u0.1
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*    12.75  0.0339   0.4767526

The tenth percentile from bootstrap is also 12.75, similar to g)