Questions #3, 5, 6, 9

0.0.1 k-fold Cross Validation

Q3. We now review k-fold cross-validation.

  1. Explain how k-fold cross-validation is implemented.

K-fold cross-validation involves randomly dividing the set of observations into k groups (folds), of approximately equal size. The first fold is treated as a validation set, and the remainder is treated as a training set. The k-fold CV estimate is estimated by averaging the k resulting MSE estimates.

  1. What are the advantages and disadvantages of k-fold cross-validation relative to:
    1. The validation set approach?

The validation set approach is simple as you randomly divide the available set of observations into two parts: training set and validation set (or hold-out set). This approach can be highly variable depending on which observations are included in the training set and which are included in the validation set; may tend to overestimate the test error rate for the model fit on the entire data set.

    1. LOOCV?

LOOCV is similar to validation set approach as it attempts to address the method’s drawbacks/disadvantages by using a single observation for the validation set and the remaining observations for the training set. It has far less bias and does not overestimate the test error rate as much as the validation set does, and yields the same result when performing LOOCV multiple times. However, it is a poor estimate because it is highly variable since it is based upon a single observation and can potentially be expensive to implement because the model has to be fit n times.

0.0.2 Logistic Regression to Predict Probability of Default

Q5. 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.

  1. Fit a logistic regression model that uses income and balance to predict default.
library(ISLR)
attach(Default)
set.seed(1)
glm.fit = glm(default ~ income + balance,
              data = Default,
              family = "binomial")
summary(glm.fit)
## 
## 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
  1. Using the validation set approach, estimate the test error of this model. In order to do this, you must perform the following steps:
    1. Split the sample set into a training set and a validation set.
train = sample(dim(Default)[1], dim(Default)[1] / 2)
    1. Fit a multiple logistic regression model using only the training observations.
glm.fit = glm(default ~ income + balance,
              data = Default,
              family = "binomial",
              subset = train)
summary(glm.fit)
## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default, subset = train)
## 
## 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
    1. 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.probs = predict(glm.fit,
                    newdata = Default[-train, ],
                    type = "response")
glm.pred = rep("No", length(glm.probs))
glm.pred[glm.probs > 0.5] = "Yes"
    1. Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified.
mean(glm.pred != Default[-train, ]$default)
## [1] 0.0254
  1. 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 three different splits each produce a different resulting error rate which shows that the rate varies by which observations are in the training/validation sets.

#1
train = sample(dim(Default)[1], dim(Default)[1] / 2)
glm.fit = glm(default ~ income + balance,
              data = Default,
              family = "binomial",
              subset = train)
glm.probs = predict(glm.fit,
                    newdata = Default[-train, ],
                    type = "response")
glm.pred = rep("No", length(glm.probs))
glm.pred[glm.probs > 0.5] = "Yes"
mean(glm.pred != Default[-train, ]$default)
## [1] 0.0274
#2
train = sample(dim(Default)[1], dim(Default)[1] / 2)
fit.glm = glm(default ~ income + balance,
              data = Default,
              family = "binomial",
              subset = train)
glm.probs = predict(fit.glm,
                    newdata = Default[-train, ],
                    type = "response")
glm.pred = rep("No", length(glm.probs))
glm.pred[glm.probs > 0.5] = "Yes"
mean(glm.pred != Default[-train, ]$default)
## [1] 0.0244
#3
train = sample(dim(Default)[1], dim(Default)[1] / 2)
fit.glm = glm(default ~ income + balance,
              data = Default,
              family = "binomial",
              subset = train)
glm.probs = predict(fit.glm,
                    newdata = Default[-train, ],
                    type = "response")
glm.pred = rep("No", length(glm.probs))
glm.pred[glm.probs > 0.5] = "Yes"
mean(glm.pred != Default[-train, ]$default)
## [1] 0.0244
  1. 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.

Including the dummy variable did not result in a reduction in the test rate.

train = sample(dim(Default)[1], dim(Default)[1] / 2)
glm.fit = glm(default ~ income + balance + student,
              data = Default,
              family = "binomial",
              subset = train)
glm.probs = predict(fit.glm,
                    newdata = Default[-train, ],
                    type = "response")
glm.pred = rep("No", length(glm.probs))
glm.pred[glm.probs > 0.5] = "Yes"
mean(glm.pred != Default[-train, ]$default)
## [1] 0.0266

0.0.3 Logistic Regression with Bootstrap and Standard Errors

Q6. 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.

  1. 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.
set.seed(1)
train = sample(dim(Default)[1], dim(Default)[1] / 2)
glm.fit <- glm(default ~ income + balance,
               data = Default,
               family = "binomial")
summary(glm.fit)
## 
## 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
  1. 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) {
    fit = glm(default ~ income + balance,
              data = data,
              family = "binomial",
              subset = index)
    return (coef(fit))
}
  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.

The bootstrap estimates of the standard errors of the coefficients are β1 and β2 are respectively 4.858 x 10^(-6) and 2.300 x 10^(-4).

library(boot)
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 -3.912114e-02 4.347403e-01
## t2*  2.080898e-05  1.585717e-07 4.858722e-06
## t3*  5.647103e-03  1.856917e-05 2.300758e-04
  1. Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function.

The estimated standard errors obtained are very similar/close.

detach(Default)

0.0.4 Estimates with μˆ.

Q9. We will now consider the Boston housing data set, from the ISLR2 library.

  1. Based on this data set, provide an estimate for the population mean of medv. Call this estimate (mu-hat).
library(ISLR2)
## Warning: package 'ISLR2' was built under R version 4.1.3
## 
## Attaching package: 'ISLR2'
## The following objects are masked from 'package:ISLR':
## 
##     Auto, Credit
attach(Boston)
mu.hat = mean(medv)
mu.hat
## [1] 22.53281
  1. Provide an estimate of the standard error of (mu-hat). 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.
se.hat = sd(medv) / sqrt(dim(Boston)[1])
se.hat
## [1] 0.4088611
  1. Now estimate the standard error of (mu-hat) using the bootstrap. How does this compare to your answer from (b)?

The bootstrap estimated standard error from mu.hat is 0.4106622, which is similar to the estimate found from se.hat of 0.4088611.

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
  1. 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 [μˆ − 2SE(μˆ), μˆ + 2SE(μˆ)].

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

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 * 0.4106622, 22.53 + 2 * 0.4106622)
CI.mu.hat
## [1] 21.70868 23.35132
  1. Based on this data set, provide an estimate, (mu-hat)med, for the median value of medv in the population.
med.hat = median(medv)
med.hat
## [1] 21.2
  1. We now would like to estimate the standard error of (mu-hat)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.

Estimated median value of 21.2 is equal to the value obtained in (e); with a standard error of 0.3770241 which is relatively small.

boot.fn = function(data, index) {
    mu = median(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*     21.2 -0.0386   0.3770241
  1. Based on this data set, provide an estimate for the tenth percentile of medv in Boston suburbs. Call this quantity (mu-hat)0.1. (You can use the quantile() function.)
percent.hat = quantile(medv, c(0.1))
percent.hat
##   10% 
## 12.75
  1. Use the bootstrap to estimate the standard error of (mu-hat)0.1. Comment on your findings.

Estimated 12.75 which is equal to the value obtained in (g); with a standard error of 0.4925766 which is relatively small.

boot.fn = function(data, index) {
    mu = quantile(data[index], c(0.1))
    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*    12.75  0.0186   0.4925766
detach(Boston)