QUESTIONS

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

Explain how k-fold cross-validation is implemented.

#Answer The implementation of the k-fold cross validation is as follows: 1. Divide the dataset randomly into k folds of equal size (k=5 or 10 preferred). 2. The first fold will be the validation dataset, and the statistical learning method will be trained using the remaining folds. 3. The MSE is then calculated using the validation dataset.The process is repeated K times each with a different validation dataset and a corresponding MSE. 4.An estimate of the test error which is the average of the MSEs is then calculated.

What are the advantages and disadvantages of k-fold cross validation relative to:

The validation set approach? #ANSWER 1.An advantage is that it doesn’t overestimate the test error rate.

A disadvantage is that has a slower training time as multiple folds are trained.

LOOCV? #ANSWER Advantage: It is less time intensive than LOOCV. Disadvantage: If has less bias than LOOCV hence it is a poorer choice for bias reduction.

Question 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.

Fit a logistic regression model that uses income and balance to predict default.

set.seed(1)
default.glm <- glm(default~income+balance, data=Default, family = "binomial")
summary(default.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

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.

set.seed(2)
split <- sample.split(Default, SplitRatio = 0.8)
Defaulttrain <- subset(Default, split==TRUE)
Defaulttest<- subset(Default, split==FALSE)

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

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

summary(train.glm)

## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = Defaulttrain)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.2167  -0.1416  -0.0550  -0.0198   3.7482  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.171e+01  5.201e-01 -22.505  < 2e-16 ***
## income       2.172e-05  5.949e-06   3.651 0.000261 ***
## balance      5.714e-03  2.733e-04  20.907  < 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: 2062.7  on 7499  degrees of freedom
## Residual deviance: 1121.0  on 7497  degrees of freedom
## AIC: 1127
## 
## Number of Fisher Scoring iterations: 8

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.

Default.probs <- predict(train.glm, Defaulttest, type="response")
Default.pred <- rep("No", 2500)
Default.pred[Default.probs>0.5] <- "Yes"

table(Default.pred, Defaulttest$default)

##             
## Default.pred   No  Yes
##          No  2384   70
##          Yes   14   32

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

mean(Default.pred!=Defaulttest$default)

## [1] 0.0336

The validation test error rate is 0.0336.

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.

SOLUTION

#1. 50-50 split

set.seed(3)
split1 <- sample.split(Default, SplitRatio = 0.5)
Defaulttrain1 <- subset(Default, split1==TRUE)
Defaulttest1<- subset(Default, split1==FALSE)

## Logistic model

train1.glm <- glm(default~income+balance, data=Defaulttrain1, family="binomial")

summary(train1.glm)

## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = Defaulttrain1)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.1769  -0.1387  -0.0524  -0.0184   3.7739  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.190e+01  6.448e-01 -18.456  < 2e-16 ***
## income       2.153e-05  7.357e-06   2.926  0.00343 ** 
## balance      5.863e-03  3.405e-04  17.220  < 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: 1402.61  on 4999  degrees of freedom
## Residual deviance:  741.39  on 4997  degrees of freedom
## AIC: 747.39
## 
## Number of Fisher Scoring iterations: 8

Default.probs1 <- predict(train1.glm, Defaulttest1, type="response")
Default.pred1 <- rep("No", 5000)
Default.pred1[Default.probs1>0.5] <- "Yes"

table(Default.pred1, Defaulttest1$default)

##              
## Default.pred1   No  Yes
##           No  4798  122
##           Yes   27   53

#
mean(Default.pred1!=Defaulttest1$default)

## [1] 0.0298

The validation test error is 0.0298.

#2. 70-30 split

set.seed(4)
split2 <- sample.split(Default, SplitRatio = 0.7)
Defaulttrain2 <- subset(Default, split2==TRUE)
Defaulttest2<- subset(Default, split2==FALSE)

## Logistic model

train2.glm <- glm(default~income+balance, data=Defaulttrain2, family="binomial")

summary(train2.glm)

## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = Defaulttrain2)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.4353  -0.1470  -0.0604  -0.0229   3.6905  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.131e+01  5.898e-01 -19.185  < 2e-16 ***
## income       2.045e-05  6.806e-06   3.005  0.00265 ** 
## balance      5.520e-03  3.065e-04  18.011  < 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: 1543.58  on 4999  degrees of freedom
## Residual deviance:  834.82  on 4997  degrees of freedom
## AIC: 840.82
## 
## Number of Fisher Scoring iterations: 8

Default.probs2 <- predict(train2.glm, Defaulttest2, type="response")
Default.pred2 <- rep("No", 5000)
Default.pred2[Default.probs2>0.5] <- "Yes"

table(Default.pred2, Defaulttest2$default)

##              
## Default.pred2   No  Yes
##           No  4832  106
##           Yes   14   48

#
mean(Default.pred2!=Defaulttest2$default)

## [1] 0.024

#3. 60-40 split

set.seed(9)
split3 <- sample.split(Default, SplitRatio = 0.6)
Defaulttrain3 <- subset(Default, split3==TRUE)
Defaulttest3<- subset(Default, split3==FALSE)
## Logistic model

train3.glm <- glm(default~income+balance, data=Defaulttrain3, family="binomial")

summary(train3.glm)

## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = Defaulttrain3)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.4085  -0.1487  -0.0623  -0.0236   3.6705  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.123e+01  5.891e-01 -19.055  < 2e-16 ***
## income       2.009e-05  6.778e-06   2.964  0.00304 ** 
## balance      5.462e-03  3.057e-04  17.863  < 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: 1517.14  on 4999  degrees of freedom
## Residual deviance:  836.79  on 4997  degrees of freedom
## AIC: 842.79
## 
## Number of Fisher Scoring iterations: 8

Default.probs3 <- predict(train3.glm, Defaulttest3, type="response")
Default.pred3 <- rep("No", 5000)
Default.pred3[Default.probs3>0.5] <- "Yes"

table(Default.pred3, Defaulttest3$default)

##              
## Default.pred3   No  Yes
##           No  4829  104
##           Yes   13   54

#
mean(Default.pred3!=Defaulttest3$default)

## [1] 0.0234

There isn’t a big difference in the test error rate among the different splits.

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.

Answer

set.seed(3)
splits <- sample.split(Default, SplitRatio = 0.8)
train <- subset(Default, splits==TRUE)
test<- subset(Default, splits==FALSE)

## Logistic model

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

summary(glm.fit)

## 
## Call:
## glm(formula = default ~ income + balance + student, family = "binomial", 
##     data = train)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.1417  -0.1394  -0.0533  -0.0192   3.7559  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.110e+01  5.865e-01 -18.932   <2e-16 ***
## income       5.910e-06  9.598e-06   0.616   0.5381    
## balance      5.791e-03  2.779e-04  20.840   <2e-16 ***
## studentYes  -5.852e-01  2.777e-01  -2.107   0.0351 *  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 2062.7  on 7499  degrees of freedom
## Residual deviance: 1116.6  on 7496  degrees of freedom
## AIC: 1124.6
## 
## Number of Fisher Scoring iterations: 8

probs <- predict(glm.fit,test, type="response")
pred <- rep("No", 2500)
pred[probs>0.5] <- "Yes"

table(pred, test$default)

##      
## pred    No  Yes
##   No  2382   70
##   Yes   16   32

#
mean(pred!=test$default)

## [1] 0.0344

There isn’t a difference in the test error rate when the dummy variable was included.

Exercise 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) usingthe standard formula for computing the standard errors in the glm() function. Do not forget to set a random seed before beginning your analysis.

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.

summary(glm(default~income+balance, data=Default, family="binomial"))$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

The standard error for the coefficient associated with income is 4.985e^-6 The standard error for the coefficient associated with balance is 2.274e^-4

Write a function, boot.fn(), that takes as input the Default dataset 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=1:nrow(data)){
  coef(glm(default~income+balance, data=data, subset=index,family="binomial"))[-1]
}

Use the boot() function together with your boot.fn() function to estimate the standard errors of the logistic regression coefficients for income and balance.

set.seed(1)
# The boot function
boot(Default, boot.fn, R=1000)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Default, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##         original       bias     std. error
## t1* 2.080898e-05 1.680317e-07 4.866284e-06
## t2* 5.647103e-03 1.855765e-05 2.298949e-04

Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function #Answer i.From the output of the glm function: The standard error for the coefficient associated with income is 4.985e^-6 The standard error for the coefficient associated with balance is 2.274e^-4

From the output of the boostrap function: The standard error for the coefficient associated with income is 4.886284e^-6 The standard error for the coefficient associated with balance is 2.2989e^-4

These estimates are pretty similar but we see the estimates for the bootstrap standard error is a little bit bigger than the one for the glm function() indicating the glm() function had the best fit.

#Exercise 9 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.

u <-mean(medv)
u

## [1] 22.53281

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.

s.e <- sd(medv)/sqrt(nrow(Boston))
s.e

## [1] 0.4088611

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

se.fn <- function(data, index){
mean(data[index])
  
}
set.seed(1)
boot(Boston$medv, se.fn, R=1000)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = se.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original      bias    std. error
## t1* 22.53281 0.007650791   0.4106622

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)].

cfI <- c(u-2*0.4106622, u+2*0.4106622)
cfI

## [1] 21.71148 23.35413

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

The confidence interval using the t-test function is wider indicating instability.

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

U.med <- median(medv)
U.med

## [1] 21.2

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.

st.med <- function(data, index){
   median(data[index])
}
set.seed(1)
boot(Boston$medv, st.med, R=1000)

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

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(medv, 0.1)
u0.1

##   10% 
## 12.75

Use the bootstrap to estimate the standard error of ˆµ0.1. Comment on your finding

quantile.fn <- function(data, index){
  quantile(data[index], 0.1)
}

set.seed(1)
boot(Boston$medv, quantile.fn, R=1000)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = quantile.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*    12.75  0.0339   0.4767526

From the output, we see that there is little bias in the estimate for the 10th percentile.

Homework 4

Obehi Ikpea

2023-03-08

QUESTIONS

Question 5

SOLUTION

Answer

Exercise 6