Assignment 4

library(tidyverse)
library(openintro)
library(ISLR)
library(ISLR2)

Exercise 3

We now review \(k\)-fold cross-validation.

\((a)\) Explain how \(k\)-fold cross-validation is implemented.

• The \(k\)-fold cross validation is implemented by taking the \(n\) observations and randomly splitting it into \(k\) non-overlapping groups of length of (approximately) \(n/k\). These groups acts as a validation set, and the remainder (of length \(n−n/k\)) acts as a training set. The test error is then estimated by averaging the \(k\) resulting MSE estimates.

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

\((i.)\) The validation set approach?

• The validation set approach has two main drawbacks compared to \(k\)-fold cross-validation. First, the validation estimate of the test error rate can be highly variable (depending on precisely which observations are included in the training set and which observations are included in the validation set). Second, only a subset of the observations are used to fit the model. Since statistical methods tend to perform worse when trained on fewer observations, this suggests that the validation set error rate may tend to overestimate the test error rate for the model fit on the entire data set.

\((ii.)\) LOOCV?

• The LOOCV cross-validation approach is a special case of \(k\)-fold cross-validation in which \(k=n\). This approach has two drawbacks compared to \(k\)-fold cross-validation. First, it requires fitting the potentially computationally expensive model \(n\) times compared to \(k\)-fold cross-validation which requires the model to be fitted only \(k\) times. Second, the LOOCV cross-validation approach may give approximately unbiased estimates of the test error, since each training set contains \(n−1\) observations; however, this approach has higher variance than \(k\)-fold cross-validation (since the outputs of \(n\) fitted models trained on an almost identical set of observations are averaged, these outputs are highly correlated, and the mean of highly correlated quantities has higher variance than less correlated ones). So, there is a bias-variance trade-off associated with the choice of \(k\) in \(k\)-fold cross-validation; typically using \(k=5\) or \(k=10\) yield test error rate estimates that suffer neither from excessively high bias nor from very high variance.

Exercise 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.

attach(Default)
set.seed(1)
model1 <- glm(default ~ income + balance, data = Default, family = "binomial")
summary(model1)

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

\((i.)\) Split the sample set into a training set and a validation set.

set.seed(1)
sample <- sample(nrow(Default), nrow(Default)/2)
train <- Default[sample,]
val <- Default[-sample,]

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

model2 <- glm(default ~ income + balance, data = train, family = "binomial")
summary(model2)

## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = train)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.5830  -0.1428  -0.0573  -0.0213   3.3395  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.194e+01  6.178e-01 -19.333  < 2e-16 ***
## income       3.262e-05  7.024e-06   4.644 3.41e-06 ***
## balance      5.689e-03  3.158e-04  18.014  < 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: 1523.8  on 4999  degrees of freedom
## Residual deviance:  803.3  on 4997  degrees of freedom
## AIC: 809.3
## 
## Number of Fisher Scoring iterations: 8

\((iii.)\) 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.

probs <- predict(model2, newdata = val, type = "response")
pred.glm <- rep("No", length(probs))
pred.glm[probs > 0.5] <- "Yes"

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

mean(pred.glm != val$default)

## [1] 0.0254

The test error rate with the validation approach is 2.54%.

\((c)\) 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.

set.seed(1)
sample <- sample(nrow(Default), nrow(Default)*0.7) #70/30 split
train <- Default[sample,]
val <- Default[-sample,]
model3 <- glm(default ~ income + balance, data = train, family = "binomial")
probs <- predict(model3, newdata = val, type = "response")
pred.glm <- rep("No", length(probs))
pred.glm[probs > 0.5] <- "Yes"
mean(pred.glm != val$default)

## [1] 0.02666667

sample <- sample(nrow(Default), nrow(Default)*0.8) #80/20 split
train <- Default[sample,]
val <- Default[-sample,]
model4 <- glm(default ~ income + balance, data = train, family = "binomial")
probs <- predict(model4, newdata = val, type = "response")
pred.glm <- rep("No", length(probs))
pred.glm[probs > 0.5] <- "Yes"
mean(pred.glm != val$default)

## [1] 0.0265

sample <- sample(nrow(Default), nrow(Default)*0.6) #60/40 split
train <- Default[sample,]
val <- Default[-sample,]
model5 <- glm(default ~ income + balance, data = train, family = "binomial")
probs <- predict(model5, newdata = val, type = "response")
pred.glm <- rep("No", length(probs))
pred.glm[probs > 0.5] <- "Yes"
mean(pred.glm != val$default)

## [1] 0.027

The validation error rates marginally change across the different splits tested. Based on the above simulations, a 80/20 split seems to perform the best.

\((d)\) 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(1)
sample <- sample(nrow(Default), nrow(Default)*0.8) #80/20 split
train <- Default[sample,]
val <- Default[-sample,]
model6 <- glm(default ~ income + balance + student, data = train, 
               family = "binomial")
pred.glm <- rep("No", length(probs))
probs <- predict(model6, newdata = val, type = "response")
pred.glm[probs > 0.5] <- "Yes"
mean(pred.glm != val$default)

## [1] 0.0275

It doesn’t seem that adding the “student” dummy variable leads to a reduction in the validation set estimate of the test error rate.

Exercise 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(1)
model7 <- glm(default ~ income + balance, data = Default, family = "binomial")
summary(model7)

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

## Standard Error for income: 4.985167e-06

## Standard Error for balance: 0.0002273731

\((b)\) 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("income","balance")])
}

\((c)\) 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)

## 
## Attaching package: 'boot'

## The following object is masked from 'package:openintro':
## 
##     salinity

bootstrap <- boot(Default, boot.fn, 1000)

##       income      balance 
## 2.080898e-05 5.647103e-03

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

The estimated standard errors obtained through the bootstrap function are much larger than the errors obtained through glm(). This could sugest that the glm() estimates might be underestimating the real variablility of the coefficients, potentially due to violation of model assumptions.

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

library(MASS)

## 
## Attaching package: 'MASS'

## The following object is masked from 'package:ISLR2':
## 
##     Boston

## The following objects are masked from 'package:openintro':
## 
##     housing, mammals

## The following object is masked from 'package:dplyr':
## 
##     select

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

## [1] 22.53281

\((b)\) 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 <- sd(medv) / sqrt(dim(Boston)[1])
se.hat

## [1] 0.4088611

\((c)\) Now estimate the standard error of \(\hat{\mu}\) 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 bootstrap estimated standard error of \(\hat{\mu}\) of 0.4106 is very close to the estimate found in \((b)\) of 0.4088. As the two estimates are close, this suggests that the parametric assumptions of the original standard error from \((b)\) are likely to be valid.

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

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.mu.hat <- c(22.53 - 2 * 0.4119, 22.53 + 2 * 0.4119)
CI.mu.hat

## [1] 21.7062 23.3538

The bootstrap confidence interval is very close to the one provided by the t.test() function.

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

med.hat <- median(medv)
med.hat

## [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.

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 -0.0386   0.3770241

The estimated median value is 21.2 which is equal to the value obtained in \((e)\), with a standard error of 0.377 which suggests that if repeated sampling was conducted from the population and the median was calculated, the value would vary by 0.377 on average. This SE is the smallest so far, possibly due to the median being less sensitive to outliers than the mean.

\((g)\) 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.)

percent.hat <- quantile(medv, c(0.1))
percent.hat

##   10% 
## 12.75

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

boot.fn <- function(data, index) {
    mu <- quantile(data[index], c(0.1))
    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*    12.75  0.0186   0.4925766

The estimated tenth percentile value is 12.75 which is again equal to the value obtained in \((g)\), with a standard error of 0.4925. This suggests that if repeated sampling was conducted from the population and the 10th percentile was calculated, the value would vary by 0.4925 on average. This is slightly higher than the SE calculated with the median, since lower percentiles can be more affected by skewness and outliers compared to the median. Out of all the metrics compared, the median provided the optimal SE.