MS 4373 Assignment 4

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

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

K-fold cross validation involves randomly dividing the set of observations into k groups of about equal size (n/k). Some folds are treated as the validation set while the rest are treated as the training set on which the model is performed. The test error is then estimated by averaging the k resulting MSE estimates.

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

i) The validation set approach?

The validation estimate of the test error rate can be highly variable depending on the observations that end up in the validation set and the ones that end up in the training set. The validation approach also only uses a subset of the observations (the training set) to fit the model.

ii) LOOCV?

LOOCV requires fitting the statistical learning method n times, which can be computationally expensive. LOOCV also runs the risk of computational problems, especially when the sample size is extremely large.

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(ISLR)
attach(Default)

set.seed(1) #set the random seed

model <- glm(default ~ income + balance, data = Default, family = "binomial") #income and balance predict default
summary(model)

## 
## 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.

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

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

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

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

prob <- predict(model_train, newdata = Default[-train, ], type = "response")
predict <- rep("No", length(prob))
predict[prob > 0.5] <- "Yes"

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

mean(predict != Default[-train, ]$default)

## [1] 0.0254

The test error rate using the test 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.

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

## [1] 0.0274

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

## [1] 0.0244

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

## [1] 0.0244

The resulting test errors deomstrate how variable the estimates are as different observations are allocated into the training and validation sets.

(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 and dummy variable for student leads to a reduction in the test error rate.

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

## [1] 0.0278

Adding the student variable did not improve the test error rate by much.

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)

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

## 
## 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

The glm() estimate of the standard error for income is 4.348e-01 and for balance, it is 2.274e-04.

(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)
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

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

With the bootstrap method, the estimated standard error for income is 4.866284e-06 and 2.298949e-04 for balance. The standard error estimates are very close between the glm and bootstrap method.

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 mu hat.

library(MASS)
attach(Boston)

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

## [1] 22.53281

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

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

## [1] 0.4088611

The estimate of the standard error of mu hat is 0.4088, which is small in comparison to the mean of medv.

(c) Now estimate the standard error of mu hat 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 two estimates of the satandard error are very close, with .4111 compared to .4088.

(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).

t.test(Boston$medv)

## 
##  One Sample t-test
## 
## data:  Boston$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 confidence interval from the bootstrap method is very close to the interval from the t test.

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

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

## [1] 21.2

(f) We now would like to estimate the standard error of ^??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 of 21.2 matches the estimate obtained in part e, with a standard error of 0.377, which is relatively small compared to median value.

(g) Based on this data set, provide an estimate for the tenth percentile of medv in Boston suburbs. Call this quantity mu hat 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 mu hat 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 bootstrap estimate matches the results found in part g. The standard error obtained using bootstrapping is 0.493, which is small compared to the tenth percentile.

MS 4373 Assignment 4

Katelyn Huynh

March 19, 2021

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

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

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

i) The validation set approach?

The validation estimate of the test error rate can be highly variable depending on the observations that end up in the validation set and the ones that end up in the training set. The validation approach also only uses a subset of the observations (the training set) to fit the model.

ii) LOOCV?

LOOCV requires fitting the statistical learning method n times, which can be computationally expensive. LOOCV also runs the risk of computational problems, especially when the sample size is extremely large.

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

(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.

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

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.

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

The test error rate using the test 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.

The resulting test errors deomstrate how variable the estimates are as different observations are allocated into the training and validation sets.

Adding the student variable did not improve the test error rate by much.

(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 glm() estimate of the standard error for income is 4.348e-01 and for balance, it is 2.274e-04.

(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.

(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.

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

With the bootstrap method, the estimated standard error for income is 4.866284e-06 and 2.298949e-04 for balance. The standard error estimates are very close between the glm and bootstrap method.

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 mu hat.

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

The estimate of the standard error of mu hat is 0.4088, which is small in comparison to the mean of medv.

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

The two estimates of the satandard error are very close, with .4111 compared to .4088.

(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 confidence interval from the bootstrap method is very close to the interval from the t test.

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

(f) We now would like to estimate the standard error of ^??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 estimated median value of 21.2 matches the estimate obtained in part e, with a standard error of 0.377, which is relatively small compared to median value.

(g) Based on this data set, provide an estimate for the tenth percentile of medv in Boston suburbs. Call this quantity mu hat 0.1. (You can use the quantile() function.)

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

The bootstrap estimate matches the results found in part g. The standard error obtained using bootstrapping is 0.493, which is small compared to the tenth percentile.