Assignment 4

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

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

It is used to evaluate the performance of a machine learning model on unseen data. It shuffles the entire dataset and divides it into k sized folds then runs the model for said k folds.

2. What are the advantages and disadvantages of k-fold cross validation relative to:

1. The validation set approach? The validation set approach has a lower variance since it averages the results across multiple splits. The disadvantage of this approach is that it is computationally expensive since it can require running the model k times.

2. LOOCV? Advantages of this approach is that it has a better balance between the bias-variance tradeoff. Another advantage is that this is computationally less expensive. The disadvantage is that this does have a higher bias and can slightly overestimate the true test error.

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)

## Warning: package 'ISLR2' was built under R version 4.3.3

library(boot)

data(Default)

set.seed(42)

log_model <- glm(default ~ income + balance, data = Default, family = "binomial")
summary(log_model)

## 
## 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

2. 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_indices <- sample(nrow(Default), 0.8 * nrow(Default))
train_set <- Default[train_indices, ]
validation_set <- Default[-train_indices, ]

2. Fit a multiple logistic regression model using only the train ing observations.

model_train <- glm(default ~ income + balance, data = train_set, family = binomial)

3. 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.

pred_probs <- predict(model_train, newdata = validation_set, type = "response")

pred_default <- ifelse(pred_probs > 0.5, "Yes", "No")

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

validation_error <- mean(pred_default != validation_set$default)
print(validation_error)

## [1] 0.0255

3. Repeat the process in (b) three times, using three different splits of the observations into a training set and a validation set. Com ment on the results obtained.

get_validation_error <- function(seed_val) {
  set.seed(seed_val)
  
  model_train <- glm(default ~ income + balance, data = train_set, family = binomial)
  
  pred_probs <- predict(model_train, newdata = validation_set, type = "response")
  pred_default <- ifelse(pred_probs > 0.5, "Yes", "No")
  
  return(mean(pred_default != validation_set$default))
}

seeds <- c(123, 456, 789)

error_rates <- sapply(seeds, get_validation_error)

results <- data.frame(Seed = seeds, ValidationError = error_rates)
print(results)

##   Seed ValidationError
## 1  123          0.0255
## 2  456          0.0255
## 3  789          0.0255

4. Now consider a logistic regression model that predicts the prob ability of default using income, balance, and a dummy variable for student. Estimate the test error for this model using the val idation set approach. Comment on whether or not including a dummyvariable for student leads to a reduction in the test error rate.

set.seed(42)

log_model_student <- glm(default ~ income + balance + student, data = Default, family = "binomial")
summary(log_model_student)

## 
## Call:
## glm(formula = default ~ income + balance + student, family = "binomial", 
##     data = Default)
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.087e+01  4.923e-01 -22.080  < 2e-16 ***
## income       3.033e-06  8.203e-06   0.370  0.71152    
## balance      5.737e-03  2.319e-04  24.738  < 2e-16 ***
## studentYes  -6.468e-01  2.363e-01  -2.738  0.00619 ** 
## ---
## 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: 1571.5  on 9996  degrees of freedom
## AIC: 1579.5
## 
## Number of Fisher Scoring iterations: 8

pred_probs1 <- predict(log_model_student, newdata = validation_set, type = "response")
pred_default1 <- ifelse(pred_probs > 0.5, "Yes", "No")

validation_error1 <- mean(pred_default1 != validation_set$default)
print(validation_error1)

## [1] 0.0255

The test error rate is the same as all other test error rates even with the inclusion of the student dummyvariable.

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.

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(42)

log_model <- glm(default ~ income + balance, data = Default, family = "binomial")
summary(log_model)

## 
## 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

2. 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.

set.seed(42)
boot.fn <- function(data, index) {
  coef(glm(default ~ income + balance, data = data, subset = index, family = "binomial"))
}
boot.fn(Default, 1:8000)

##   (Intercept)        income       balance 
## -1.179560e+01  2.226489e-05  5.786903e-03

3. 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(42)
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 -2.292405e-02 4.435269e-01
## t2*  2.080898e-05  2.737444e-08 5.073444e-06
## t3*  5.647103e-03  1.176249e-05 2.299133e-04

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

The standard errors using the boot function are slightly higher than the ones obtained using the glm function.

9. 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 ˆµ.

data(Boston)
mu_hat <- mean(Boston$medv)
mu_hat

## [1] 22.53281

2. 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.

sd <- sd(Boston$medv)
n <- length(Boston$medv)

se_mu_hat <- sd / sqrt(n)
se_mu_hat

## [1] 0.4088611

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

mean.fn <- function(data, index)
  mean(data[index])

boot(Boston$medv, mean.fn, R = 1000)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = mean.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original       bias    std. error
## t1* 22.53281 -0.002375494   0.4185115

The standard error using the boot function is higher than the standard error that was manually calculated.

4. Based on your bootstrap estimate from (c), provide a 95% con fidence 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), ˆµ + 2SE(ˆµ)].

mu_hat - 2*.4185115

## [1] 21.69578

mu_hat + 2*.4185115

## [1] 23.36983

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

The confidence interval calculated manually is slightly lower than that of the one provided using the t.test function.

5. Based on this data set, provide an estimate, ˆµmed, for the median value of medv in the population.

mu_hat_med <- median(Boston$medv)
mu_hat_med

## [1] 21.2

6. 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.

med.fn <- function(data, index)
  median(data[index])

boot(Boston$medv, med.fn, R = 1000)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = med.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*     21.2 -0.0137   0.3906596

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

mu_hat_01 <- quantile(Boston$medv, 0.10)
mu_hat_01

##   10% 
## 12.75

8. Use the bootstrap to estimate the standard error of ˆµ0.1. Com ment on your findings.

percentile <- function(data, index) 
  quantile(data[index], 0.10)

boot(data = Boston$medv, percentile, R = 1000)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = percentile, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*    12.75  0.0201   0.4950819