Assignment 4

---

Libraries

library(ISLR)
library(boot)
library(MASS)

Exercises

Exercise 3

We now review k-fold cross-validation.

• (a) Explain how k-fold cross-validation is implemented.
  • This approach involves randomly dividing the set of observations into k groups, or folds, of approximately equal size. The first fold is treated as a validation set, and the method is fit on the remaining k − 1 folds. The mean squared error, MSE1, is then computed on the observations in the held-out fold. This procedure is repeated k times; each time, a different group of observations is treated as a validation set. This process results in k estimates of the test error, MSE1, MSE2,…, MSEk. The k-fold CV estimate is computed by averaging these values. In practice, one typically performs k-fold CV using k = 5 or k = 10.
• (b) What are the advantages and disadvantages of k-fold cross validation relative to: i. The validation set approach? ii. LOOCV?
  • Validation Set: The validation estimate of the test error rate can be highly variable, depending on precisely which observations are included in the training set and which observations are included in the validation set. In the validation approach, only a subset of the observations—those that are included in the training set rather than in the validation set—are used to fit the model. Since statistical methods tend to perform worse when trained on fewer observations, this suggests that the validation set error rate may tend to overestimate the test error rate for the model fit on the entire data set.
  • LOOCV: LOOCV requires fitting the statistical learning method n times. This has the potential to be computationally expensive (except for linear models fit by least squares. But cross-validation is a very general approach that can be applied to almost any statistical learning method. Some statistical learning methods have computationally intensive fitting procedures, and so performing LOOCV may pose computational problems, especially if n is extremely large.
  • k-fold cross validation: It often gives more accurate estimates of the test error rate than does LOOCV. However, performing k-fold CV for, say, k = 5 or k = 10 leads to an intermediate level of bias, since each training set contains (k − 1)n/k observations—fewer than in the LOOCV approach, but substantially more than in the validation set approach. Therefore, from the perspective of bias reduction, it is clear that LOOCV is to be preferred to k-fold CV.

Exercise 5

(a) 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.

#Exercise 5-a
attach(Default) #Attach data to use variable names

set.seed(1)
log_mod1 = glm(default ~ income + balance, data = Default, family = "binomial") #Fit logistic regression model
summary(log_mod1)

## 
## 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:

• Split the sample set into a training set and a validation set.

# Exercise 5-b Split Sample (50-50)
train_1 = sample(dim(Default)[1], 0.50*dim(Default)[1])
test_1 = Default[-train_1, ]

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

# Exercise 5-b Fit Multiple Regression
log_mod2 = glm(default ~ income + balance, data = Default, family = "binomial", subset = train_1)
summary(log_mod2)

## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default, subset = train_1)
## 
## 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

• 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.

# Exercise 5-b Posterior Probability
probs_1 = predict(log_mod2, newdata = test_1, type = "response")
pred.log_mod2 = rep("No", length(probs_1))
pred.log_mod2[probs_1 > 0.5] = "Yes"

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

result_1 = mean(pred.log_mod2 != Default[-train_1, ]$default)

percent = "%"
text1 = "Split @ 50%:"
paste(text1, round(result_1*100, digits = 2) , percent)

## [1] "Split @ 50%: 2.54 %"

• Interpretation: The above value represents the test error rate with the validation set approach.

(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.

# Exercise 5-c Three Times
train_2 = sample(dim(Default)[1], 0.75*dim(Default)[1])
test_2 = Default[-train_2, ]
log_mod3 = glm(default ~ income + balance, data = Default, family = "binomial", subset = train_2)
probs_2 = predict(log_mod3, newdata = test_2, type = "response")
pred.log_mod3 = rep("No", length(probs_2))
pred.log_mod3[probs_2 > 0.5] = "Yes"
result_2 = mean(pred.log_mod3 != Default[-train_2, ]$default)

train_3 = sample(dim(Default)[1], 0.25*dim(Default)[1])
test_3 = Default[-train_3, ]
log_mod4 = glm(default ~ income + balance, data = Default, family = "binomial", subset = train_3)
probs_3 = predict(log_mod4, newdata = test_3, type = "response")
pred.log_mod4 = rep("No", length(probs_3))
pred.log_mod4[probs_3 > 0.5] = "Yes"
result_3 = mean(pred.log_mod4 != Default[-train_3, ]$default)

train_4 = sample(dim(Default)[1], 0.65*dim(Default)[1])
test_4 = Default[-train_4, ]
log_mod5 = glm(default ~ income + balance, data = Default, family = "binomial", subset = train_4)
probs_4 = predict(log_mod5, newdata = test_4, type = "response")
pred.log_mod5 = rep("No", length(probs_4))
pred.log_mod5[probs_4 > 0.5] = "Yes"
result_4 = mean(pred.log_mod5 != Default[-train_4, ]$default)

## [1] "Split 1 @ 75%: 2.56 %"

## [1] "Split 2 @ 25%: 2.53 %"

## [1] "Split 3 @ 65%: 2.66 %"

(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.

# Exercise 5-d
train_5 = sample(dim(Default)[1], 0.50*dim(Default)[1])
test_5 = Default[-train_5, ]
log_mod6 = glm(default ~ income + balance + student, data = Default, family = "binomial", subset = train_5)
probs_5 = predict(log_mod6, newdata = test_5, type = "response")
pred.log_mod6 = rep("No", length(probs_5))
pred.log_mod6[probs_5 > 0.5] = "Yes"
result_5 = mean(pred.log_mod6 != Default[-train_5, ]$default)

text_compare = "Original Model @ 50% Split with Income + Balance: "
text5 = "Split @ 50% with Income + Balance + Student:"

## [1] "Original Model @ 50% Split with Income + Balance:  2.54 %"

## [1] "Split @ 50% with Income + Balance + Student: 2.42 %"

## [1] "Including the dummy variable resulted in a decreased test error rate."

Exercise 6

(a) 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.

#Exercise 6-a
lr6 = glm(default ~ income + balance, data = Default, family = "binomial")
set.seed(1)
summary(lr6)

## 
## 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

• Estimated standard errors:
  • balance 2.274e-04
  • income: 4.985e-066

(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 multiplelogistic regression model.

# Exercise 6-b
set.seed(1)
boot.fn = function(data, index) return(coef(glm(default ~ income + balance, data = data, family = "binomial", subset = index)))

(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.

• Bootstrap Estimated Errors:
  • t1: 4.344722e-01
  • t2: 4.866284e-06
  • t3: 2.298949e-04

Exercise 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 ˆµ.

# Exercise 9-a
attach(Boston)

mu.hat = mean(medv)
print(mu.hat)

## [1] 22.53281

(b) Provide an estimate of the standard error of ˆµ. Interpret this result.

# Exercise 9-b
standard_error.hat = sd(medv) / sqrt(dim(Boston)[1])
standard_error.hat

## [1] 0.4088611

• This value is relatively small compared to the actual value of the mean(medv).

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

# Exercise 9-c
set.seed(1)
boot.mu.fn = function(data, index) {
    mu = mean(data[index])
    return (mu)
}

bstrap = boot(medv, boot.mu.fn, 1000)
bstrap

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

• The bootstrap estimated standard error of μ̂ of 0.4106622 is very close to the estimate found in (b) of 0.4089.

(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).

# Exercise 9-d
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

c(bstrap$t0 - 2 * 0.4106622, bstrap$t0 + 2 * 0.4106622)

## [1] 21.71148 23.35413

• The bootstrap confidence interval is very close to the one provided by the t.test() function.

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

# Exercise 9-e
medv_median = median(medv)
medv_median

## [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.

# Exercise 9-f
set.seed(1)
boot.med.fn = function(data, index) return(median(data[index]))
bstrap.med = boot(medv, boot.med.fn, 1000)
bstrap.med

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

• The standard error according to the bootstrap is 0.3778075.

(g) Based on this data set, provide an estimate for the tenth percentile of medv in Boston suburbs. Call this quantity ˆµ0.1. (You can use the quantile() function.)

# Exercise 9-g
percent.hat = quantile(medv, c(0.1))
percent.hat

##   10% 
## 12.75

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

# Exercise 9-h
boot.h.fn = function(data, index) return(quantile(data[index], c(0.1)))
boot(medv, boot.h.fn, 1000)

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