HW4: STA 6543-Campbell

Ch5 (pg. 197): 3, 5, 6, 9

Problem 3

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

The data is randomly divided into k groups. Ideally these groups are all the same size, but approximately the same will suffice. A model is then fitted k times using one of the k-folds (subsets) as the validation set and remaining subsets comprise the training set. The mean error rate is then taken, which is used as the estimate of test error for the model.

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

1. The validation set approach?

K-fold cross-validation uses more of the data set than the validation set approach, so it provides a more accurate estimate of the test error rate. One drawback of k-fold CV compared to the validation set approach is that it is more computationally time consuming when the dataset is large.

2. LOOCV? K-fold CV is more computationally efficient because the model only needs to run as many times as there are folds, where as LOOCV has to run n times (n = number of observations). Additionally, when using k=5 or k=10 folds, k-fold CV has shown to obtain test error rate estimates that balance the bias-variance trade-off. This means it will provide more accurate estimates compared to LOOCV.

Problem 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.

set.seed(17)
default_data = Default
logit_default = glm(default ~ income + balance, data = default_data, family = "binomial")
summary(logit_default)

Call:
glm(formula = default ~ income + balance, family = "binomial", 
    data = default_data)

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

b) 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_index = sample(1:nrow(default_data), 0.7*nrow(default_data))
train_default = default_data[train_index, ]
test_default = default_data[-train_index, ]

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

logit_default2 = glm(default ~ income + balance, data = train_default, family = "binomial")
summary(logit_default2)

Call:
glm(formula = default ~ income + balance, family = "binomial", 
    data = train_default)

Deviance Residuals: 
    Min       1Q   Median       3Q      Max  
-2.4704  -0.1478  -0.0590  -0.0219   3.6997  

Coefficients:
              Estimate Std. Error z value Pr(>|z|)    
(Intercept) -1.145e+01  5.156e-01 -22.218  < 2e-16 ***
income       2.273e-05  5.955e-06   3.817 0.000135 ***
balance      5.568e-03  2.673e-04  20.835  < 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: 2064.0  on 6999  degrees of freedom
Residual deviance: 1128.5  on 6997  degrees of freedom
AIC: 1134.5

Number of Fisher Scoring iterations: 8

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.

test_probs = predict(logit_default2, test_default, type = "response")
test_preds = if_else(test_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.

table(test_preds, test_default$default)

          
test_preds   No  Yes
       No  2893   63
       Yes   10   34

(error = mean(test_preds != test_default$default))

[1] 0.02433333

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 created a for loop to repeat the validation approach of estimating the error rate of the logistic model with three different splits. The error rates found with these three different splits were 0.0273, 0.0217, and 0.031. Compared to the error rate found in the original split (0.0243), they are all similar.

errors = c()
seeds = c(23, 55, 91)
for(i in 1:3){
  set.seed(seeds[i])
  temp_train_index = sample(1:nrow(default_data), 0.7*nrow(default_data))
  temp_train = default_data[temp_train_index, ]
  temp_test = default_data[-temp_train_index, ]
  temp_logit = glm(default ~ income + balance, data = temp_train, family = "binomial")
  temp_test_probs = predict(temp_logit, temp_test, type = "response")
  temp_test_preds = if_else(temp_test_probs > 0.5, "Yes", "No")
  errors[i] = mean(temp_test_preds != temp_test$default)
}
errors[i+1] = error

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.

The dummy variable studentYes is found to be significant with a p-value of less than 0.05. However, including a dummy variable does not lead to a reduction in error rate. In fact, the error rates for the validation set for each train/test split used earlier when including student are all is greater than the error rate when student is not included.

new_errors = c()
seeds = c(23, 55, 91, 17)
for(i in 1:4){
  set.seed(seeds[i])
  temp_train_index = sample(1:nrow(default_data), 0.7*nrow(default_data))
  temp_train = default_data[temp_train_index, ]
  temp_test = default_data[-temp_train_index, ]
  temp_logit = glm(default ~ ., data = temp_train, family = "binomial")
  summary(temp_logit)$coefficients
  temp_test_probs = predict(temp_logit, temp_test, type = "response")
  temp_test_preds = if_else(temp_test_probs > 0.5, "Yes", "No")
  new_errors[i] = mean(temp_test_preds != temp_test$default)
}

print("Errors not including student"); errors

[1] "Errors not including student"
[1] 0.02733333 0.02166667 0.03100000 0.02433333

print("Errors including student"); new_errors

[1] "Errors including student"
[1] 0.02766667 0.02233333 0.03200000 0.02500000

Problem 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.

The coefficient for income (2.081e-05) has a standard error of 4.985e-06, and the coefficient for balance (5.647e-03) has a standard error of 2.274e-04.

logit_default = glm(default ~ income + balance, data = default_data, family = "binomial")
summary(logit_default)$coefficients

                 Estimate   Std. Error    z value      Pr(>|z|)
(Intercept) -1.154047e+01 4.347564e-01 -26.544680 2.958355e-155
income       2.080898e-05 4.985167e-06   4.174178  2.990638e-05
balance      5.647103e-03 2.273731e-04  24.836280 3.638120e-136

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 = 1:nrow(data)) {
  temp_logit = glm(default ~ income + balance, data = data, subset = index, family = "binomial")
  return(temp_logit$coefficients)
}

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.

boot(default_data, boot.fn, 1000)

ORDINARY NONPARAMETRIC BOOTSTRAP


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


Bootstrap Statistics :
         original        bias     std. error
t1* -1.154047e+01 -3.771940e-02 4.190856e-01
t2*  2.080898e-05  1.161211e-07 4.759188e-06
t3*  5.647103e-03  1.488493e-05 2.223174e-04

View(default_data)

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

The estimated standard errors obtained using the glm function are 4.985167e-06 for income and 2.273731e-04 for balance. The standard error obtained using the bootstrap function are closely equivalent -> 4.759188e-06 for income and 2.223174e-04 for balance.

Problem 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 \(\hat{\mu}\).

set.seed(17)
boston_data = Boston
(mu_hat = mean(boston_data$medv))

[1] 22.53281

b) Provide an estimate of the standard error of \(\hat{\mu}\). Interpret this result.

The standard error estimated is 0.4088611. Since medv represents median value of homes (in $1000s), this means the sample mean of this data set (22.53281 -> 22,533) could be off by about $410 in either direction.

sd(boston_data$medv) / sqrt(nrow(boston_data))

[1] 0.4088611

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

The standard error estimated using bootstrap is 0.4168157, which is slightly larger than the standard error estimated in part b (0.4088611).

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

ORDINARY NONPARAMETRIC BOOTSTRAP


Call:
boot(data = boston_data$medv, statistic = boot.fn, R = 1000)


Bootstrap Statistics :
    original      bias    std. error
t1* 22.53281 -0.01082431   0.4168157

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).

The 95% confidence interval based on the bootstrap estimate from (c) is (21.71585, 23.34977). This confidence interval is slightly wider than the confidence interval obtained using t.test(Boston$medv) which is (21.72953, 23.33608). Note: though not identical, these confidence intervals are very close to each other

t.test(boston_data$medv)

    One Sample t-test

data:  boston_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 

# I am choosing to use 1.96 instead of 2 while calculating my 95% confidence interval to be more precise
(conf_interval = c(mu_hat - 1.96*0.4168157, mu_hat + 1.96*0.4168157))

[1] 21.71585 23.34977

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

(med_hat = median(boston_data$medv))

[1] 21.2

f) 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.

The standard error of \(\hat{\mu}_{med}\) is 0.3935471. This is slightly less than the standard error for \(\hat{\mu}\) which in theory means the sample median is more accurate to the population median.

boot.fn <- function(data, index) {
  return(median(data[index]))
}
boot(boston_data$medv, boot.fn, 1000)

g) Based on this data set, provide an estimate for the tenth percentile of medv in Boston suburbs. Call this quantity \(\hat{\mu}_{0.1}\).

(mu_tenth_percentile = quantile(boston_data$medv, 0.1))

  10% 
12.75 

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

The standard error of \(\hat{\mu}_{0.1}\) is 0.4896878.

boot.fn <- function(data, index) {
  return(quantile(data[index], 0.1))
}
boot(boston_data$medv, boot.fn, 1000)