Problem 5.3

We now review k-fold cross-validation.

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

K fold CV is essentially taking a training dataset and dividing it into k equally sized folds. For each of these folds, a model is trained using all other folds and tested on that specific fold. This results in k different estimates of test error, and thus the total error is the average of all fold test errors.

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

K fold CV has a significantly lower variability than validation set, as having the dataset split once can lead to drastically different results (for example, observations of a specific class may not occur in the test split).

However, the validation set approach is less computationally intensive, particularly in very large datasets where training a single model may take up a large amount of resources.

  1. LOOCV?

LOOCV will have a higher variance as the testing dataset will be restricted to one value, and thus K fold CV can often have a more accurate test error due to the bias variance tradeoff; LOOCV will have a low bias due to having more data, but the presence of a single response variable will lead to a much higher variance. K fold CV is also much less computationally demanding, as there are significantly fewer models being trained.

Problem 5.5

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.

  1. Fit a logistic regression model that uses income and balance to predict default.
library(ISLR2)
set.seed(123)
lr <- glm(default ~ income + balance, family = 'binomial', data=Default)
  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.
tr_indices <- createDataPartition(Default$default, p=0.8, list=F)

train_def <- Default[tr_indices,]
test_def <- Default[-tr_indices,]
  1. Fit a multiple logistic regression model using only the training observations.
split_lr <- glm(default ~ income + balance, family = 'binomial', data=Default)
  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.
split_preds <- predict(split_lr, newdata=test_def)
split_preds <- factor(ifelse(split_preds >= 0.5, 'Yes', 'No'))
  1. Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified.
misclassification <- mean(test_def$default != split_preds)
misclassification
## [1] 0.03051526

We see the misclassification rate to be approximately 2.7%.

  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.
for (i in 1:3){
  # will repeat above code for 3 different seeds
  set.seed(i)
  
  tr_indices <- createDataPartition(Default$default, p=0.8, list=F)
  train_def <- Default[tr_indices,]
  test_def <- Default[-tr_indices,]
  
  split_lr <- glm(default ~ income + balance, family = 'binomial', data=Default)
  
  split_preds <- predict(split_lr, newdata=test_def)
  split_preds <- factor(ifelse(split_preds >= 0.5, 'Yes', 'No'))
  
  misclassification <- mean(test_def$default != split_preds)

  cat("Misclassification for attempt ", i, " is: ", misclassification * 100, '%\n')
}
## Misclassification for attempt  1  is:  2.901451 %
## Misclassification for attempt  2  is:  2.501251 %
## Misclassification for attempt  3  is:  2.801401 %

After repeating it three times we see misclassification rates in a similar range, between 2.5% and 2.9%.

  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.

We will estimate the test error using validation set approach, and will repeat it 25 times.

misclassification_rates <- sapply(1:25, function(i){
  set.seed(i)
  
  tr_indices <- createDataPartition(Default$default, p=0.8, list=F)
  train_def <- Default[tr_indices,]
  test_def <- Default[-tr_indices,]
  
  split_lr <- glm(default ~ income + balance + student, family = 'binomial', data=Default)
  
  split_preds <- predict(split_lr, newdata=test_def)
  split_preds <- factor(ifelse(split_preds >= 0.5, 'Yes', 'No'))
  
  misclassification <- mean(test_def$default != split_preds)
  
  return(misclassification)
})

cat("The average misclassification rate is", mean(misclassification_rates) * 100, '%')
## The average misclassification rate is 2.747374 %

We see that the performance is relatively similar, with a misclassification rate of 2.74%. There does not appear to be a large improvement in performance using student.

Problem 5.6

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.

set.seed(341)
  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.
mlr <- glm(default ~ income + balance, data=Default, family = 'binomial')
summary(mlr)
## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default)
## 
## 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

The income coefficient is 2.081e-05 with a standard error of 4.985e-06, while the balance coefficient is 5.647e-03 with a standard error of 2.274e-04.

  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, indices){
  sample_data <- data[indices,]  # sample with replacement

  sample_data$default <- as.factor(sample_data$default)
  
  mlr <- glm(default ~ income + balance, data=sample_data, family = 'binomial')
  return(mlr$coefficients)
}
  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.
boot_data <- boot(Default, boot.fn, 1000)
boot_data
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Default, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##          original        bias     std. error
## t1* -1.154047e+01 -3.348636e-02 4.274945e-01
## t2*  2.080898e-05  1.364367e-07 4.883897e-06
## t3*  5.647103e-03  1.860915e-05 2.314111e-04
summary(mlr) # for comparison
## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default)
## 
## 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. Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function.

Using 1000 bootstrapped samples we can compare the result to what we found in part a). We see the standard errors for both coefficients are close to each other. Income has approximately the same coefficient at 2.081e-0.5 for both, with a slightly lower error for boostrap (4.88e-06 compared to 4.96e-06); balance is similar with an approximate coefficient of 5.647e-03 but a slightly higher standard error for bootstrap (2.31e-04 compared to 2.27e-04).

Problem 5.9

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

head(Boston)
##      crim zn indus chas   nox    rm  age    dis rad tax ptratio lstat medv
## 1 0.00632 18  2.31    0 0.538 6.575 65.2 4.0900   1 296    15.3  4.98 24.0
## 2 0.02731  0  7.07    0 0.469 6.421 78.9 4.9671   2 242    17.8  9.14 21.6
## 3 0.02729  0  7.07    0 0.469 7.185 61.1 4.9671   2 242    17.8  4.03 34.7
## 4 0.03237  0  2.18    0 0.458 6.998 45.8 6.0622   3 222    18.7  2.94 33.4
## 5 0.06905  0  2.18    0 0.458 7.147 54.2 6.0622   3 222    18.7  5.33 36.2
## 6 0.02985  0  2.18    0 0.458 6.430 58.7 6.0622   3 222    18.7  5.21 28.7
  1. Based on this data set, provide an estimate for the population mean of medv. Call this estimate mu_hat.
mu_hat <- mean(Boston$medv)
mu_hat
## [1] 22.53281

The estimated mean value of owner occupied homes is around $22,500.

  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 thenumber of observations.

std <- sd(Boston$medv)/sqrt(nrow(Boston))
std
## [1] 0.4088611

The standard error is 0.408, or about $408.

  1. Now estimate the standard error of mu_hat using the bootstrap. How does this compare to your answer from (b)?
boot.fn <- function(data, indices){
  sample_data <- data[indices,]  # sample with replacement
  
  mean <- mean(sample_data$medv)

  return(mean)
}

boot_data <- boot(Boston, boot.fn, 1000)
boot_data
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original     bias    std. error
## t1* 22.53281 0.02163241   0.4211374

Using bootstrap, we find a mean of 22.532 and a standard error of 0.421, which is similar to what we saw in the previous section.

  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 [mu_hat − 2SE(mu_hat), mu_hat + 2SE(mu_hat)].

Using the mean medv and standard error for it, we find the 95% confidence interval to be 21.690 to 23.374. Using the t.test we see:

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

This is very close to our prediction using boostrap, as expected.

  1. Based on this data set, provide an estimate, mu_hat_med, for the median value of medv in the population.
med <- median(Boston$medv)
med
## [1] 21.2

The median value of medv is 21.2, or $21,200.

  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.
boot.fn <- function(data, indices){
  sample_data <- data[indices,]  # sample with replacement
  
  med <- median(sample_data$medv)

  return(med)
}

boot_data <- boot(Boston, boot.fn, 1000)
boot_data
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*     21.2 -0.0332    0.380378

Using bootrap, we find the standard error of the median to be 0.380, or $380.

  1. Based on this data set, provide an estimate for the tenth percentile of medv in Boston census tracts. Call this quantity mu_hat0.1. (You can use the quantile() function.)
mu_hat01 <- quantile(Boston$medv, 0.1)
mu_hat01
##   10% 
## 12.75

The 10th percentile value of medv is 12.75, or $12,750.

  1. Use the bootstrap to estimate the standard error of mu_hat0.1. Comment on your findings.
boot.fn <- function(data, indices){
  sample_data <- data[indices,]  # sample with replacement
  
  mu_hat01 <- quantile(sample_data$medv, 0.1)

  return(mu_hat01)
}

boot_data <- boot(Boston, boot.fn, 1000)
boot_data
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*    12.75  0.0146   0.5014013

Using bootstrap we can estimate the standard error to be 0.501, or $501. This is slightly larger than the standard error of the overall median found in part f).