STA 6543 Assignment 4

1. (Exercise 5.4.3) We now review k-fold cross-validation.

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

The data is divided into \(k\) subsets (folds). The \(i^{th}\) subset \((i=1,2,...,k)\) is left out and the model is trained on the remaining \(k-1\) subsets and then tested on the \(i^{th}\) subset. This is an iterative process and each fold is left out once to be used to assess the model which was trained on all of the other folds. In the end, all of the observations have been used for both training and testing the model.

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

   1. The validation set approach?

   Advantages: All of the data are used in both training and testing of the model which helps to prevent over/underfitting. Also, \(k\)-fold CV typically has lower variability in test error rates than the validation set approach.

   Disadvantages: Greater computational cost.

   2. LOOCV?

   Advantages: Much lower computational cost for large sample size and lower variance for \(k < n\).

   Disadvantages: Introduces greater bias than LOOCV.

2. (Exercise 5.4.5) 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)
data('Default')

set.seed(1)
model.p2 <- glm(default ~ income + balance, data = Default, family = "binomial")
summary(model.p2)

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.

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

   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.

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

library(caret)

# (i) split data
set.seed(1)
split.p2 <- createDataPartition(Default$default, p = 0.8, list = FALSE, times = 1)
train.p2 <- Default[split.p2, ]
test.p2 <- Default[-split.p2, ]

# (ii) fit model with training data
model.train.p2 <- glm(default ~ income + balance, data = train.p2, family = 'binomial')

# (iii) get predictions for default status
probs.p2 <- predict(model.train.p2, test.p2, type = "response")
preds.p2 <- rep("No", length(probs.p2))
preds.p2[probs.p2 >= 0.5] <- "Yes"

# (iv) compute validation set error
table(test.p2$default, preds.p2)

     preds.p2
        No  Yes
  No  1930    3
  Yes   51   15

error.p2 <- mean(test.p2$default != preds.p2)
cat("The validation set error for the first data split is given by:", error.p2, "\n")

The validation set error for the first data split is given by: 0.02701351 

3. 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.

######################### 2ND DATA SPLIT #########################
# (i) split data 70% train/30% test
set.seed(1)
split.2nd <- createDataPartition(Default$default, p = 0.7, list = FALSE, times = 1)
train.2nd <- Default[split.2nd, ]
test.2nd <- Default[-split.2nd, ]

# (ii) fit model with training data
model.train.2nd <- glm(default ~ income + balance, data = train.2nd, family = 'binomial')

# (iii) get predictions for default status
probs.2nd <- predict(model.train.2nd, test.2nd, type = "response")
preds.2nd <- rep("No", length(probs.2nd))
preds.2nd[probs.2nd >= 0.5] <- "Yes"

# (iv) compute validation set error
table(test.2nd$default, preds.2nd)

     preds.2nd
        No  Yes
  No  2893    7
  Yes   71   28

error.2nd <- mean(test.2nd$default != preds.2nd)
cat("The validation set error for the second data split is given by:", error.2nd, "\n")

The validation set error for the second data split is given by: 0.02600867 

######################### 3RD DATA SPLIT #########################
# (i) split data 60% train/40% test
split.3rd <- createDataPartition(Default$default, p = 0.6, list = FALSE, times = 1)
train.3rd <- Default[split.3rd, ]
test.3rd <- Default[-split.3rd, ]

# (ii) fit model with training data
model.train.3rd <- glm(default ~ income + balance, data = train.3rd, family = 'binomial')

# (iii) get predictions for default status
probs.3rd <- predict(model.train.3rd, test.3rd, type = "response")
preds.3rd <- rep("No", length(probs.3rd))
preds.3rd[probs.3rd >= 0.5] <- "Yes"

# (iv) compute validation set error
table(test.3rd$default, preds.3rd)

     preds.3rd
        No  Yes
  No  3855   11
  Yes   84   49

error.3rd <- mean(test.3rd$default != preds.3rd)
cat("The validation set error for the third data split is given by:", error.3rd, "\n")

The validation set error for the third data split is given by: 0.02375594 

######################### 4TH DATA SPLIT #########################
# (i) split data 50% train/50% test
split.4th <- createDataPartition(Default$default, p = 0.5, list = FALSE, times = 1)
train.4th <- Default[split.4th, ]
test.4th <- Default[-split.4th, ]

# (ii) fit model with training data
model.train.4th <- glm(default ~ income + balance, data = train.4th, family = 'binomial')

# (iii) get predictions for default status
probs.4th <- predict(model.train.4th, test.4th, type = "response")
preds.4th <- rep("No", length(probs.4th))
preds.4th[probs.4th >= 0.5] <- "Yes"

# (iv) compute validation set error
table(test.4th$default, preds.4th)

     preds.4th
        No  Yes
  No  4813   20
  Yes  111   55

error.4th <- mean(test.4th$default != preds.4th)
cat("The validation set error for the fourth data split is given by:", error.4th, "\n")

The validation set error for the fourth data split is given by: 0.02620524 

The test error seems to decrease and then increase again as the amount of data used to train the model decreases from 80% to 50%. This is not unexpected, since we know that, in general, as model flexibility increases, training error decreases and testing error decreases, reaches some minimum point, then increases again.

4. 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.

# code dummy variable for student
Default$studentDummy <- ifelse(Default$student == "Yes", 1, 0)

# fit logreg model with dummy
model.new.p2 <- glm(default ~ income + balance + studentDummy, data = Default,
                    family = 'binomial')
summary(model.new.p2)

Call:
glm(formula = default ~ income + balance + studentDummy, 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 ***
studentDummy -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

set.seed(1)
split.new <- createDataPartition(Default$default, p = 0.8, list = FALSE, times = 1)
train.new <- Default[split.new, ]
test.new <- Default[-split.new, ]

model.new.train <- glm(default ~ income + balance + studentDummy, 
                       data = train.new, family = 'binomial')

probs.new <- predict(model.new.train, test.new, type = "response")
preds.new <- rep("No", length(probs.new))
preds.new[probs.new >= 0.5] <- "Yes"

table(test.new$default, preds.new)

     preds.new
        No  Yes
  No  1928    5
  Yes   50   16

error.new <- mean(test.new$default != preds.new)
cat("The test error for the model with the dummy student variable is:", error.new, "\n")

The test error for the model with the dummy student variable is: 0.02751376 

Including a dummy variable for student led to a model with a higher test error rate than the model which does not include the dummy variable using the exact same data split.

3. (Exercise 5.4.6) 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(20)
p3.fit <- glm(default ~ income + balance, data = Default, family = 'binomial')
summary(p3.fit)

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 estimated standard errors for the coefficients associated with income and balance are 4.985e-06 and 2.274 e-04, respectively.

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.

boot.fn <- function(df, x) {
  boot.fit <- glm(df[,1] ~ df[,4] + df[,3], data = df, subset = x,
                  family = 'binomial')
  names(boot.fit$coefficients) <- c('Intercept', 'Income', 'Balance')
  get.coefs <- coef(boot.fit)
  return(get.coefs[2:3])
}

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.

library(boot)
boot(data = Default, boot.fn, R = 1000)

ORDINARY NONPARAMETRIC BOOTSTRAP


Call:
boot(data = Default, statistic = boot.fn, R = 1000)


Bootstrap Statistics :
        original       bias     std. error
t1* 2.080898e-05 9.312739e-09 5.114083e-06
t2* 5.647103e-03 2.283201e-05 2.231546e-04

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

The estimated standard errors are as follows:

Methods	Income	Balance
glm()	4.985e-06	2.274e-04
bootstrap	5.114083e-06	2.231546e-04

Clearly the estimates are very close in value.

4. (Exercise 5.4.9) 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 \(\hat{\mu}\).

library(ISLR2)
data("Boston")
mu.hat <- mean(Boston$medv)
mu.hat

[1] 22.53281

Based on this data set, the average median value of owner-occupied homes among suburbs of Boston is approximately \(\hat{\mu} = \$22,532.81\).

2. Provide an estimate of the standard error of \(\hat{\mu}\). 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

[1] 0.4088611

The standard error estimate is given by \(\widehat{SE} = 0.4089\). For a sample of owner-occupied homes in 506 Boston suburbs, the \(\hat{\mu}\) estimate will vary approximately by \(\pm \,\,\$408.90\), on average. This is a relatively low standard error, which tells us that \(\hat{\mu} = \$22,532.81\) is a good estimate of the population mean of medv; our sample medv sufficiently represents the true population.

3. Now estimate the standard error of \(\hat{\mu}\) using the bootstrap. How does this compare to your answer from (b)?

set.seed(506)
boot.new <- function(df, x) {
  mu.boot <- mean(df[x])
  return(mu.boot)
}

boot(Boston$medv, boot.new, R = 1000)

ORDINARY NONPARAMETRIC BOOTSTRAP


Call:
boot(data = Boston$medv, statistic = boot.new, R = 1000)


Bootstrap Statistics :
    original      bias    std. error
t1* 22.53281 -0.01338261   0.4163944

The standard error estimate for \(\hat{\mu}\) from bootstrapping is \(\widehat{SE}_{boot} = 0.4164\). This estimate is slightly higher than that obtained in part (b), but still relatively small. We can still conclude that our sample medv sufficiently represents the true population.

4. 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 \([\hat{\mu} - 2SE(\hat{\mu}), \hat{\mu} + 2SE(\hat{\mu})]\).

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 

mu.hat.CI <- function(mu, se) {
  CI <- c(mu - 2*se, mu + 2*se)
  names(CI) <- c("Lower", "Upper")
  return(CI)
}

mu.hat.CI(mu.hat, 0.4163944)

   Lower    Upper 
21.70002 23.36560 

The 95% confidence interval based on the bootstrap estimate of standard error is approximately ($21,700.02, $23,365.60). This interval is only slightly wider (by about $60.00) than the one obtained using t.test(Boston$medv) which is approximately ($21,729.53, $23,336.08).

5. Based on this data set, provide an estimate, \(\hat{\mu}_{med}\), for the median value of medv in the population.

mu.hat.med <- median(Boston$medv)
mu.hat.med

[1] 21.2

The population estimate for the median value of medvis \(\hat{\mu}_{med} = \$21,200.00\).

6. We now would like to estimate the standard error of \(\hat{\mu}_{med}\). 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.

set.seed(506)
boot.4f <- function(df, x) {
  med.boot <- median(df[x])
  return(med.boot)
}

boot(Boston$medv, boot.4f, R = 1000)

ORDINARY NONPARAMETRIC BOOTSTRAP


Call:
boot(data = Boston$medv, statistic = boot.4f, R = 1000)


Bootstrap Statistics :
    original  bias    std. error
t1*     21.2 -0.0372   0.3863877

The standard error estimate for \(\hat{\mu}_{med}\) from bootstrapping is \(\widehat{SE}_{median} = 0.3864\). Again, this is quite small. We can conclude that \(\hat{\mu}_{med} = \$21,200.00\) is a good estimate of the true population median of medv.

7. Based on this data set, provide an estimate for the tenth percentile of medv in Boston census tracts. Call this quantity \(\hat{\mu}_{0.1}\). (You can use the quantile() function.)

hat.mu10 <- quantile(Boston$medv, probs = 0.1)
hat.mu10

  10% 
12.75 

The 10th percentile estimate for the population of medvis \(\hat{\mu}_{0.1} = \$12,750.00\).

8. Use the bootstrap to estimate the standard error of \(\hat{\mu}_{0.1}\). Comment on your findings.

boot.4h <- function(df, x) {
  boot10 <- quantile(df[x], probs = 0.1)
  return(boot10)
}

boot(Boston$medv, boot.4h, R = 1000)

ORDINARY NONPARAMETRIC BOOTSTRAP


Call:
boot(data = Boston$medv, statistic = boot.4h, R = 1000)


Bootstrap Statistics :
    original  bias    std. error
t1*    12.75  0.0134   0.5027208

The standard error estimate for \(\hat{\mu}_{0.1}\) from bootstrapping is \(\widehat{SE}_{0.1} = 0.5027\). A bit higher than previous SE estimates, but still relatively small. We can conclude that \(\hat{\mu}_{0.1} = \$12,750.00\) is a good estimate of the true population 10th percentile of medv.