3. We now review k-fold cross-validation. (a) Explain how k-fold cross-validation is implemented. First, the data set in k parts called folds. We remove the first, fit the model with the remaining folds and see how well it can predict the held out fold. This step is then repeated for each fold. The overall MSE is obtained by averaging every MSE calculated in the previous step.

(b) What are the advantages and disadvantages of k-fold cross-validation relative to: i. The validation set approach? The validation set approach is more variable since it only use a random subset as training data. On the other hand is relatively simple and easy to apply.

ii. LOOCV? K-fold cross validation is less computational expensive, especially for large sample. On the other end, LOOCV has less bias since it always yield to the same results.

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.

library(ISLR)
attach(Default)
set.seed(1)

(a) Fit a logistic regression model that uses income and balance to predict default.

m1 <- glm(default ~ income+balance,data = Default, family = "binomial")

(b) Using the validation set approach, estimate the test error of this model. In order to do this, you must perform the following steps: i. Split the sample set into a training set and a validation set.

train <- sample(nrow(Default),nrow(Default)/2)

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

m2 <- glm(default ~ income + balance, data = Default, family = "binomial", subset = train)

iii. 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.

probabilities = predict(m2, newdata= Default[-train], type = "response")
predictions = ifelse(probabilities > 0.5,"No", "Yes")

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

mean(predictions == Default[-train]$default)
## [1] 0.0266

(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.

train <- sample(nrow(Default),nrow(Default)/3)
m2 <- glm(default ~ income + balance, data = Default, family = "binomial", subset = train)
probabilities = predict(m2, newdata= Default[-train], type = "response")
predictions = ifelse(probabilities > 0.5,"No", "Yes")
mean(predictions == Default[-train]$default)
## [1] 0.0266
train <- sample(nrow(Default),nrow(Default)/7)
m2 <- glm(default ~ income + balance, data = default, family = "binomial", subset = train)
probabilities = predict(m2, newdata= Default[-train], type = "response")
predictions = ifelse(probabilities > 0.5,"No", "Yes")
mean(predictions == Default[-train]$default)
## [1] 0.027
train <- sample(nrow(Default),nrow(Default)/10)
m2 <- glm(default ~ income + balance, data = Default, family = "binomial", subset = train)
probabilities = predict(m2, newdata= Default[-train], type = "response")
predictions = ifelse(probabilities > 0.5,"No", "Yes")
mean(predictions == Default[-train]$default)
## [1] 0.0268

Despite the size difference, the three splits produces similar results. The validation set approach seems to be fairly consistent.

(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.

train <- sample(nrow(Default),nrow(Default)/8)
m3 <- glm(default ~ income + balance + student, data = Default, family = "binomial", subset = train)
probabilities = predict(m3, newdata= Default[-train], type = "response")
predictions = ifelse(probabilities > 0.5,"No", "Yes")
mean(predictions == Default[-train]$default)
## [1] 0.0268

The student variable appears to not impact test accuracy.

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.

set.seed(2)
summary(m2)
## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default, subset = train)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -1.1883  -0.1213  -0.0431  -0.0161   3.4364  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.269e+01  1.600e+00  -7.934 2.12e-15 ***
## income       2.536e-05  1.653e-05   1.535    0.125    
## balance      6.335e-03  8.364e-04   7.575 3.60e-14 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 296.77  on 999  degrees of freedom
## Residual deviance: 150.20  on 997  degrees of freedom
## AIC: 156.2
## 
## Number of Fisher Scoring iterations: 8

(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) {m4 <- glm(default ~ income + balance, data = data, family = "binomial", subset = index)
    return (coef(m4))}

The estimated standard errors related to the logistic model are: - intercept: 5.358e-01 - income: 6.207e-06 - balance: 2.783e-04 (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.

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 -3.956457e-02 4.228903e-01
## t2*  2.080898e-05 -4.605498e-07 4.924565e-06
## t3*  5.647103e-03  3.048004e-05 2.228767e-04

The bootstrap estimated the following standard errors: - intercept: 4.449820e-01 - income: 4.835222e-06 - balance: 2.326538e-04 (d) Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function. The logistic regression model has lower estimated standard errors

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 μˆ.

library(MASS)
attach(Boston)
mu.hat = mean(medv)
mu.hat
## [1] 22.53281

(b) Provide an estimate of the standard error of μˆ. 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.

standard.error = sd(medv)/sqrt(nrow(Boston))
standard.error
## [1] 0.4088611

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

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

The estimated standard errors obtained are very similar (0.4088611 vs 0.4106622)

(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). Hint: You can approximate a 95 % confidence interval using the formula [μˆ − 2SE(μˆ), μˆ + 2SE(μˆ)].

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 = c(mu.hat - 2*standard.error,mu.hat + 2*standard.error)
CI
## [1] 21.71508 23.35053

The confidence interval obtained with the one sample t-test appears to be less restrictive.

(e) Based on this data set, provide an estimate, μˆmed, for the median value of medv in the population.

mu.med = median(medv)
mu.med
## [1] 21.2

(f) We now would like to estimate the standard error of μˆmed. 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(1)
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 0.02295   0.3778075

(g) Based on this data set, provide an estimate for the tenth percentile of medv in Boston suburbs. Call this quantity μˆ0.1. (You can use the quantile() function.)

quantile(medv,c(0.1))
##   10% 
## 12.75

(h) Use the bootstrap to estimate the standard error of μˆ0.1. Comment on your findings.

boot.fn= function(data, index) return(quantile(data[index], c(0.1)))
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 0.01455   0.4823468

The estimated standard error obtained of median with the bootstrap is larger than the one related to the mean. This might suggests that the mean is better measure of central tendency.