Assignment #4

3. We now review k-fold cross-validation.

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

This method involves randomly dividing the observations into k groups, or folds, of approximately equal sizes. Initially, one fold serves as the validation set while the method is trained on the remaining k − 1 folds. The mean squared error (MSE1) is calculated based on the data in the held-out fold. This process repeats k times, with each iteration using a different set of observations as the validation set. As a result, k estimates of the test error are obtained. The k-fold cross-validation estimate is derived by averaging these values.

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

1. The validation set approach?

Advantages: The validation set approach is straightforward in concept and simple to implement. Disadvantages: The validation MSE may exhibit significant variability, and the model is trained only on a subset of observations (training data).

2. LOOCV?

Advantages: LOOCV exhibits reduced bias. Unlike the validation approach, which generates varying MSE values due to randomness in the splitting process, LOOCV consistently produces the same results when applied repeatedly, as each split is based on one observation at a time. Disadvantage: LOOCV demands significant computational resources.

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

## 
## 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.

## i
train <- sample(dim(Default)[1], dim(Default)[1] / 2)

## ii
fit.glm <- glm(default ~ income + balance, data = Default, family = "binomial", subset = train)
summary(fit.glm)

## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default, subset = train)
## 
## 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
probs <- predict(fit.glm, newdata = Default[-train, ], type = "response")
pred.glm <- rep("No", length(probs))
pred.glm[probs > 0.5] <- "Yes"

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

## [1] 0.0254

The validation set approach gave us a 2.54% test error rate.

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.

train <- sample(dim(Default)[1], dim(Default)[1] / 2)
fit.glm <- glm(default ~ income + balance, data = Default, family = "binomial", subset = train)
probs <- predict(fit.glm, newdata = Default[-train, ], type = "response")
pred.glm <- rep("No", length(probs))
pred.glm[probs > 0.5] <- "Yes"
mean(pred.glm != Default[-train, ]$default)

## [1] 0.0274

train <- sample(dim(Default)[1], dim(Default)[1] / 2)
fit.glm <- glm(default ~ income + balance, data = Default, family = "binomial", subset = train)
probs <- predict(fit.glm, newdata = Default[-train, ], type = "response")
pred.glm <- rep("No", length(probs))
pred.glm[probs > 0.5] <- "Yes"
mean(pred.glm != Default[-train, ]$default)

## [1] 0.0244

train <- sample(dim(Default)[1], dim(Default)[1] / 2)
fit.glm <- glm(default ~ income + balance, data = Default, family = "binomial", subset = train)
probs <- predict(fit.glm, newdata = Default[-train, ], type = "response")
pred.glm <- rep("No", length(probs))
pred.glm[probs > 0.5] <- "Yes"
mean(pred.glm != Default[-train, ]$default)

## [1] 0.0244

Each of the three different splits produced a different resulting error rate and this shows the rate varies by which observations are in the training/validation sets.

4. Now consider a logistic regression model that predicts the prob- ability 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(dim(Default)[1], dim(Default)[1] / 2)
fit.glm <- glm(default ~ income + balance + student, data = Default, family = "binomial", subset = train)
pred.glm <- rep("No", length(probs))
probs <- predict(fit.glm, newdata = Default[-train, ], type = "response")
pred.glm[probs > 0.5] <- "Yes"
mean(pred.glm != Default[-train, ]$default)

## [1] 0.0278

Including a dummy variable for student did not lead to a reduction in the test error rate.

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(1)
attach(Default)

## The following objects are masked from Default (pos = 3):
## 
##     balance, default, income, student

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

## 
## 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 glm() estimates of the standard errors for the coefficients intercept, income and balance are 4.348e-01, 4.985e-06 and 2.274e-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(data, index) {
    fit <- glm(default ~ income + balance, data = data, family = "binomial", subset = index)
    return (coef(fit))
}

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(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 -3.945460e-02 4.344722e-01
## t2*  2.080898e-05  1.680317e-07 4.866284e-06
## t3*  5.647103e-03  1.855765e-05 2.298949e-04

The bootstrap estimates of the standard errors for the coefficients B0, B1 and B2 are respectively 0.4245, 4.866 x 10^(-6) and 2.298 x 10^(-4).

4. Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function. The estimated standard errors obtained by each of the methods are very similar to one another.

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

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

library(MASS)
attach(Boston)
myboston <- Boston
mu_hat <- mean(myboston$medv)
mu_hat

## [1] 22.53281

μˆ = 22.53

2. 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.

se_mu = sd(myboston$medv)/sqrt(length(myboston$medv))
se_mu

## [1] 0.4088611

Standard error of μˆ = 0.40886

3. 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)
              return(mean(data[index]))

boot_res<- boot(myboston$medv,boot_fn, R = 1000)

boot_res

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = myboston$medv, statistic = boot_fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original      bias    std. error
## t1* 22.53281 0.007650791   0.4106622

Standard error from Bootstrap for μˆ is 0.4045557, this is very similar but slightly lower than that from (b),which is 0.40886.

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 [μˆ − 2SE(μˆ), μˆ + 2SE(μˆ)].

lower_bd <- mu_hat - (2*0.4045557)
upper_db <- mu_hat + (2*0.4045557)
lower_bd

## [1] 21.72369

upper_db

## [1] 23.34192

t.test(myboston$medv)

## 
##  One Sample t-test
## 
## data:  myboston$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

Confidence intervals calculated using the bootstrap estimate of SE and the one sample t-test are about the same to one another.

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

median(myboston$medv)

## [1] 21.2

6. 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.

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

We get an estimated median value of 21.2 which is the same as the value obtained in (e), with a standard error of 0.3770 which is relatively small, indicating the med estimate for populating is fairly accurate.

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

quantile(myboston$medv,0.1)

##   10% 
## 12.75

μˆ0.1=12.75

8. Use the bootstrap to estimate the standard error of μˆ0.1. Comment on your findings.

set.seed(1)
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  0.0339   0.4767526

Tenth percentile estimated value is 12.75, which is again the same as the value obtained in (g), with a standard error of 0.4768 which is relatively small compared to the percentile value.