Assignment 4

Chapter 05 (page 197): 3, 5, 6, 9

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

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

• k-fold cross validation is determined by the randomly dividing the amount of observations into k groups in approx equal sizes. The first group or fold is treated as the validation set and the method is fit on the remaining k - 1 folds. You repeat this k number of times on a different group of observations. After completion this results in k estimates of the test error. the cross validation piece is calculated by the average of all the k folds.

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

i. The validation set approach?

• Advantage: Easy to implement
• Disadvantage: The test error rate can be highly variable, depending on which observations are included in the validation set. This means that only a subset of the observations are used to fit the model - so over estimation of the test error rate can occur.

ii. LOOCV?

• Advantage: Less bias. You will use almost all of the data set. Using the validation approach it produces different MSE when applied repeatedly due to randomness in the splitting process, while performing LOOCV multiple times will always yield the same results, because we split based on 1 obs. each time.
• Disadvantage: Not as easy to implement

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.

library(ISLR)

## Warning: package 'ISLR' was built under R version 4.0.5

data("Default")
summary(Default)

##  default    student       balance           income     
##  No :9667   No :7056   Min.   :   0.0   Min.   :  772  
##  Yes: 333   Yes:2944   1st Qu.: 481.7   1st Qu.:21340  
##                        Median : 823.6   Median :34553  
##                        Mean   : 835.4   Mean   :33517  
##                        3rd Qu.:1166.3   3rd Qu.:43808  
##                        Max.   :2654.3   Max.   :73554

lr1 = glm(default ~ balance + income, data = Default, family = binomial)

summary(lr1)

## 
## Call:
## glm(formula = default ~ balance + income, 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 ***
## balance      5.647e-03  2.274e-04  24.836  < 2e-16 ***
## income       2.081e-05  4.985e-06   4.174 2.99e-05 ***
## ---
## 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.

trainDefault = sample(dim(Default)[1], dim(Default)[1]*0.50)
testDefault = Default[-trainDefault, ]

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

lr1 = glm(default ~ balance + income, data = Default, family = binomial, subset = trainDefault)

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.

log.prob_def = predict(lr1, testDefault, type = "response")
log.pred_def = rep("No", dim(Default)[1]*0.50)
log.pred_def[log.prob_def > 0.5] = "Yes"
table(log.pred_def, testDefault$default)

##             
## log.pred_def   No  Yes
##          No  4815  106
##          Yes   21   58

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

mean(log.pred_def !=testDefault$default)

## [1] 0.0254

• Test rate error: 2.86%

(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.

#i
trainDefault = sample(dim(Default)[1], dim(Default)[1]*0.50)
testDefault = Default[-trainDefault, ]

#ii
lr2 = glm(default ~ balance + income, data = Default, family = binomial, subset = trainDefault)

#iii
log.prob_def = predict(lr2, testDefault, type = "response")
log.pred_def = rep("No", dim(Default)[1]*0.50)
log.pred_def[log.prob_def > 0.5] = "Yes"
table(log.pred_def, testDefault$default)

##             
## log.pred_def   No  Yes
##          No  4811  120
##          Yes   20   49

mean(log.pred_def !=testDefault$default)

## [1] 0.028

#i
trainDefault = sample(dim(Default)[1], dim(Default)[1]*0.50)
testDefault = Default[-trainDefault, ]

#ii
lr3 = glm(default ~ balance + income, data = Default, family = binomial, subset = trainDefault)

#iii
log.prob_def = predict(lr3, testDefault, type = "response")
log.pred_def = rep("No", dim(Default)[1]*0.50)
log.pred_def[log.prob_def > 0.5] = "Yes"
table(log.pred_def, testDefault$default)

##             
## log.pred_def   No  Yes
##          No  4810  109
##          Yes   25   56

mean(log.pred_def !=testDefault$default)

## [1] 0.0268

#i
trainDefault = sample(dim(Default)[1], dim(Default)[1]*0.50)
testDefault = Default[-trainDefault, ]

#ii
lr4 = glm(default ~ balance + income, data = Default, family = binomial, subset = trainDefault)

#iii
log.prob_def = predict(lr4, testDefault, type = "response")
log.pred_def = rep("No", dim(Default)[1]*0.50)
log.pred_def[log.prob_def > 0.5] = "Yes"
table(log.pred_def, testDefault$default)

##             
## log.pred_def   No  Yes
##          No  4822  106
##          Yes   24   48

mean(log.pred_def !=testDefault$default)

## [1] 0.026

• Test error rate lr2: 2.5%
• Test error rate lr3: 2.6%
• Test error rate lr4: 2.7%
• Each test rate error is different. This shows you thta each time the model is run different obeservations are used.

(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.

#i
trainDefault = sample(dim(Default)[1], dim(Default)[1]*0.50)
testDefault = Default[-trainDefault, ]

#ii
lr5 = glm(default ~ balance + income + student, data = Default, family = binomial, subset = trainDefault)

#iii
log.prob_def = predict(lr5, testDefault, type = "response")
log.pred_def = rep("No", dim(Default)[1]*0.50)
log.pred_def[log.prob_def > 0.5] = "Yes"
table(log.pred_def, testDefault$default)

##             
## log.pred_def   No  Yes
##          No  4822  100
##          Yes   29   49

mean(log.pred_def !=testDefault$default)

## [1] 0.0258

• Test Error rate: 2.6%
• Adding in the Student variable didn’t add any value/impact to the test error rate.

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.

lr6 = glm(default ~ balance + income, data = Default, family = binomial)

summary(lr6)

## 
## Call:
## glm(formula = default ~ balance + income, 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 ***
## balance      5.647e-03  2.274e-04  24.836  < 2e-16 ***
## income       2.081e-05  4.985e-06   4.174 2.99e-05 ***
## ---
## 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-06

(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 ~ balance + income, 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.

library(boot)

boot(Default, boot.fn, 100)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Default, statistic = boot.fn, R = 100)
## 
## 
## Bootstrap Statistics :
##          original        bias     std. error
## t1* -1.154047e+01 -3.334939e-02 4.537593e-01
## t2*  5.647103e-03  1.442870e-05 2.248760e-04
## t3*  2.080898e-05  2.447808e-07 5.056124e-06

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

• The bootstrap estimates of the standard errors for the coefficients:
• B0: 4.334877e-01
• B1: 2.298971e-04
• B2: 4.093430e-06

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.

library(MASS)
attach(Boston)
mu.hat <- mean(medv)
mu.hat

## [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.

stand_error_hat <- sd(medv) / sqrt(dim(Boston)[1])
stand_error_hat

## [1] 0.4088611

(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) {
    mu <- mean(data[index])
    return (mu)
}
boot(medv, boot.fn, 1000)

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

• Bootstrap estimated standard error: 0.4106622
• Standard error(b): 0.4088611
• Both estimates are pretty similar

(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)].

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

CI.mu.hat <- c(22.53 - 2 * 0.4119, 22.53 + 2 * 0.4119)
CI.mu.hat

## [1] 21.7062 23.3538

• the t-test Confidence interval is very similar to the bootstrap confidence interval

(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.3 = function(data, index) return(median(data[index]))
boot3 = boot(medv, boot.3, 1000)
boot3

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.3, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*     21.2 -0.0386   0.3770241

• Standard error: 0.3770241
• Estimated mean value: 21.2
• The bootstrap estimated mean was also 21.2

(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.)

ten_perc= quantile(medv, c(0.1))
ten_perc

##   10% 
## 12.75

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

boot.4 = function(data, index) return(quantile(data[index], c(0.1)))
boot4 = boot(medv, boot.4, 1000)
boot4

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.4, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*    12.75  0.0186   0.4925766

• Standard error: 0.4925766
• Estimated quantile value: 12.75
• the same value for the estimated quantile is also found in (g)