Problem 3

We now review k-fold cross-validation.

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

K-fold cross validation is implemented by breaking the dataset into k parts and then running cross validation on those parts rather than each individual entry in the dataset. We average k MSEs (as opposed to n-1) to get an estimated test error rate.

(b) What are the advantages and disadvantages of k-fold crossvalidation relative to:
i. The validation set approach?

Advantages
K-fold CV is more stable than The Validation Approach.
K-fold CV generally has a much lower variability in the MSEs than The Validation Approach.

Disadvantages
K-fold CV is computationally more intense than The validation approach.

ii. LOOCV?

Advantages
K-fold CV is less computationally intensive than LOOCV, while still having similar stability.
K-fold CV has less variance than LOOCV.

Disadvantages
When k<n K-fold CV has more bias than LOOCV.

Problem 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.

(a) Fit a logistic regression model that uses income and balance to predict default.
logist.Defualt <- glm(default ~ income + balance, data = Default, family = binomial)
summary(logist.Defualt)
## 
## 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)
set.training <- sample(nrow(Default), nrow(Default)/2)
training.default <- Default[set.training, ]
#head(training.default)
test.default <- Default[-set.training, ]
ii. Fit a multiple logistic regression model using only the training observations.
va.Defualt <- glm(default ~ income + balance, data = training.default, family = binomial)
summary(va.Defualt)
## 
## Call:
## glm(formula = default ~ income + balance, family = binomial, 
##     data = training.default)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.5830  -0.1428  -0.0573  -0.0213   3.3395  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.194e+01  6.178e-01 -19.333  < 2e-16 ***
## income       3.262e-05  7.024e-06   4.644 3.41e-06 ***
## balance      5.689e-03  3.158e-04  18.014  < 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: 1523.8  on 4999  degrees of freedom
## Residual deviance:  803.3  on 4997  degrees of freedom
## AIC: 809.3
## 
## 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.
prob.Default.glm <- predict(va.Defualt, newdata = test.default, type = 'response')
pred.Default.glm <- rep('No', length(prob.Default.glm))
pred.Default.glm[prob.Default.glm > 0.5] <- 'Yes'

table(pred.Default.glm, test.default$default, dnn = c('Predicted Default', "Actual Default"))
##                  Actual Default
## Predicted Default   No  Yes
##               No  4824  108
##               Yes   19   49

The majority of predictions that default status is “No” and they are relatively accurate.

iv. Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified.
mean(pred.Default.glm != Default[-set.training,]$default)
## [1] 0.0254

The validation error is 0.00254, which is very small. This means that this model is fairly accurate at predicting the default status.

(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.
logi.error <- rep(0, 3)

for (i in 1:3) {
    set.seed(i+1)
    set.training2 <- sample(nrow(Default), nrow(Default)/2)
    training.default2 <- Default[set.training2, ]
    test.default2 <- Default[-set.training2, ]
    
    va.Defualt2 <- glm(default ~ income + balance, data = training.default2, family = binomial)
    
    prob.Default.glm2 <- predict(va.Defualt2, newdata = test.default2, type = 'response')
    pred.Default.glm2 <- rep('No', length(prob.Default.glm2))
    pred.Default.glm2[prob.Default.glm2 > 0.5] <- 'Yes'
   
    logi.error[i] <-  mean(pred.Default.glm2 != Default[-set.training2, ]$default)
  }

print(logi.error)
## [1] 0.0238 0.0264 0.0256

All 3 MSEs produced by the different splits are different, but very similar to the original MSE we found of 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.
Default$sdummy <- ifelse(Default$student == 'No', 0, 1)

dummy.error <- rep(0, 4)

for (i in 1:4) {
  
set.seed(i)
set.training3 <- sample(nrow(Default), nrow(Default)/2)
training.default3 <- Default[set.training3, ]
test.default3 <- Default[-set.training3, ]

va.Defualt3 <- glm(default ~ income + balance + sdummy, data = training.default3, family = binomial)

prob.Default.glm3 <- predict(va.Defualt3, newdata = test.default3, type = 'response')
pred.Default.glm3 <- rep('No', length(prob.Default.glm3))
pred.Default.glm3[prob.Default.glm3 > 0.5] <- 'Yes'
   
dummy.error[i] <- mean(pred.Default.glm3 != Default[-set.training3, ]$default)
}

print(dummy.error)
## [1] 0.0260 0.0246 0.0272 0.0262

The MSEs do not change noticably with the inclusion of the dummy variable for student. If anything they are negligibly higher than the logistic model without the dummy variable for student.

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. 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.
logist.Defualt.6 <- glm(default ~ income + balance, data = Default, family = binomial)
summary(logist.Defualt.6)
## 
## 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){
 return(coef(glm(default ~ income + balance, 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.
set.seed(1)
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 -3.945460e-02 4.344722e-01
## t2*  2.080898e-05  1.680317e-07 4.866284e-06
## t3*  5.647103e-03  1.855765e-05 2.298949e-04
(d) Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function.

The standard errors are almost indistinguishable between the functions.

detach(Default)

Problem 9

We will now consider the Boston housing data set, from the MASS library.

(a) Based on this data set, provide an estimate for the population mean of medv. Call this estimate \(\hat{\mu}\).
mu.hat <- mean(medv)
mu.hat 
## [1] 22.53281
(b) Provide an estimate of the standard error of \(\hat{\mu}\). 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.
mu.hat.se <- sd(medv)/sqrt(nrow(Boston))
mu.hat.se
## [1] 0.4088611

The estimated standard error is 0.4088611

(c) Now estimate the standard error of \(\hat{\mu}\) using the bootstrap. How does this compare to your answer from (b)?
boot.fn2 <- function(data,index){
 return(mean(data[index]))
 }

set.seed(1)
btstrp <- boot(medv,boot.fn2, 1000)

Estimated Standard Error: 0.4088611
Bootstrap Estimated SE: 0.4106622
The standard errors are similar. The bootstrap error is slightly higher than the estimated error.

**(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\(medv)*.** ##### *Hint: You can approximate a 95% confidence interval using theformula [\)$ − 2SE(\(\hat{\mu}\)), \(\hat{\mu}\) + 2SE(\(\hat{\mu}\))].
t.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 
##  22.53281

95% Confidence Interval: [21.72953, 23.33608]
Based on the t-test we could assume 95% confidence that the actual mean for medv is between 21.72953 and 23.33608.

ci.mu <- c(btstrp$t0 - 2 * 0.4106622, btstrp$t0 + 2 * 0.4106622)
ci.mu
## [1] 21.71148 23.35413

95% Confidence Interval: [21.71148, 23.35413]
Based on the bootstrap confidence interval we could assume 95% confidence that the actual mean for medv is between 21.71148 and 23.35413.

The two confidence intervals are very similar.

(e) Based on this data set, provide an estimate, \(\hat{\mu}_{med}\), for the median value of medv in the population.
median(medv)
## [1] 21.2

The median estimate for medv is 21.2.

(f) We now would like to estimate the standard error of \(\hat{\mu}_{med}\). 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]))
 }

set.seed(1)
btstrp2 <- boot(medv,boot.fn3, 1000)

btstrp2
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn3, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*     21.2 0.02295   0.3778075

Median Estimate: 21.2
Bootstrap Median Estimate: 21.2

The two estimates are exactly the same.

(g) Based on this data set, provide an estimate for the tenth percentile of medv in Boston suburbs. Call this quantity \(\hat{\mu}_{0.1}\). (You can use the quantile() function.)
perc.10 <- quantile(medv, .10)
perc.10
##   10% 
## 12.75

The estimate for the 10th percentile is 12.75.

(h) Use the bootstrap to estimate the standard error of \(\hat{\mu}_{0.1}\). Comment on your findings.
boot.fn4 <- function(data,index){
 return(quantile(data[index], 0.1))
}

boot(medv, boot.fn4, 1000)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn4, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*    12.75 0.01455   0.4823468

The 10th percentile using the bootstrap estimate is 12.75. This is the exact same value we got using the original estimate.

detach(Boston)