Assignment #4

Question 3

We now review k-fold cross-validation.

(a)

Explain how k-fold cross-validation is implemented.

k-fold cross-validation is used to estimate how well a model will perform on new data. The full data set is randomly divided into equal sized folds, where one fold is held to be used as validation. The model is then fitted onto the remaining k-folds to predict the fold that was held and the error rate is recorded.

##(b) > What are the advantages and disadvantages of k-fold cross validation relative to:

i.

The validation set approach?

The validation set approach can be highly variable because the data is split into one training half and one testing half. The estimate of error can change from one split to the other. This approach is also less biased as the model is only trained on half the data set.

However, this approach fits the model k times rather than once which can take longer and to do, and is more complex since it is divided into folds.

##ii.

LOOCV?

LOOCV fits model exactly n

Question 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.

library(ISLR2)
set.seed(240)

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

summary(glm.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

(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.

set.seed(240)

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

ii.

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

glm.fit <- 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.

glm.probs <- predict(glm.fit, newdata = Default[-train, ], type = "response")
glm.pred <- ifelse(glm.probs > 0.5, "Yes", "No")

iv.

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

mean(glm.pred != Default[-train, ]$default)

(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.

# split 1
set.seed(100)
train <- sample(nrow(Default), nrow(Default) / 2)
fit <- glm(default ~ income + balance, data = Default, family = binomial, subset = train)
pred <- ifelse(predict(fit, Default[-train, ], type = "response") > 0.5, "Yes", "No")
mean(pred != Default[-train, ]$default)

# split 2
set.seed(200)
train <- sample(nrow(Default), nrow(Default) / 2)
fit <- glm(default ~ income + balance, data = Default, family = binomial, subset = train)
pred <- ifelse(predict(fit, Default[-train, ], type = "response") > 0.5, "Yes", "No")
mean(pred != Default[-train, ]$default)

# split 3
set.seed(300)
train <- sample(nrow(Default), nrow(Default) / 2)
fit <- glm(default ~ income + balance, data = Default, family = binomial, subset = train)
pred <- ifelse(predict(fit, Default[-train, ], type = "response") > 0.5, "Yes", "No")
mean(pred != Default[-train, ]$default)

The three estimates are close to each other but not exact since the model is trained on a difference random half of observations and uses the other half to validate. The test error depends on the particular split but each set is in a consistent range of 2.6%-2.9%.

(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.

set.seed(240)
train <- sample(nrow(Default), nrow(Default) / 2)

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

pred <- ifelse(predict(fit, Default[-train, ], type = "response") > 0.5, "Yes", "No")
mean(pred != Default[-train, ]$default)

No, including a dummy variable for student does not lead to a reduction in the test error since the error rate is 2.78%, which lands in the previous range of 2.6%-2.9%.

Question 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(240)

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

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

(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)

set.seed(240)
boot(Default, boot.fn, R = 1000)

(d)

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

Comparing the glm() function and the bootstrap function, the standard errors are very close to one another, meaning that the model provided by the glm() function is true since te boostrap function does not depend on a formula, it is considered to be more reliable.

Question 9

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

(a)

Based on this data set, provide an estimate for the population mean of medv. Call this estimate ˆμ.

mu.hat <- mean(Boston$medv)
mu.hat

(b)

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.

n  <- length(Boston$medv)
se <- sd(Boston$medv)/sqrt(n)
se

A standard error of 0.409 shows that the sample mean of mu hat is scattered around the true population mean by $409 since medv is in the $1000’s. This shows how precise the estimate of 22.53 is since 0.409 is very small.

(c)

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

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

set.seed(240)
boot(Boston$medv, boot.fn, R = 1000)

Compared to the previous standard of error of mu hat, there is a small difference between 0.4088611 and 0.4100267 which means that the assumptions are reasonable.

(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 [ˆμ − 2SE(ˆμ), ˆμ + 2SE(ˆμ)].

mu.hat  <- mean(Boston$medv)
boot.se <- 0.4156
c(mu.hat - 2 * boot.se, mu.hat + 2 * boot.se)
t.test(Boston$medv)

The two intervals also have little difference as the bootstrap function produces almost the same interval from the results of a t-test. This shows that the standard errors are reliable.

(e)

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

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

(f)

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.

boot.fn.med <- function(data, index) 
{
return(median(data[index]))
}

set.seed(240)
boot(Boston$medv, boot.fn.med, R = 1000)

The spread of the standard of error is 0.386 which is small compared to the previous median of 21.2, meaning the median is estimated very well.

(g)

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.0.1.hat <- quantile(Boston$medv, 0.10)
mu.0.1.hat

(h)

Use the bootstrap to estimate the standard error of ˆμ0.1. Comment on your findings.

boot.fn.10 <- function(data, index) 
{
return(quantile(data[index], 0.10))
}

set.seed(240)
boot(Boston$medv, boot.fn.10, R = 1000)

The tenth percentile was estimated at 12.75 with a standard of error of 0.5102 in comparison to the previous error of 0.3806, which is noticeable the largest difference across all errors so far.