Assignment 4

Problem 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 amount of observations, n, and randomly splitting them into k, non-overlapping groups of length of approximately n/k. These groups are the validation set, and the remainder is the training set. The test error is then estimated by averaging the k resulting mean squared error estimates. (b) What are the advantages and disadvantages of k-fold crossvalidation relative to: i. The validation set approach? Advantages: Validation set approach is conceptually simple & easily implemented. Disadvantages: Validation MSE can be highly variable & only a subset of observations are used to fit the model. ii. LOOCV? Advantages: Less bias & validation approach produces different MSE when repeatedly applied due to randomness during process of splitting, while performing LOOCV multiple times will always produce same results, because split based on one observation every time. Disadvantage is that it is computationally intensive.

Problem 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)

data("Default")

summary(Default)

##  default    student       balance           income     
##  No :9667   No :7056   Min.   :   0.0   Min.   :  772  
##  Yes: 333   Yes:2944   1st Qu.: 481.7   1st Qu.:21340  
##                        Median : 823.6   Median :34553  
##                        Mean   : 835.4   Mean   :33517  
##                        3rd Qu.:1166.3   3rd Qu.:43808  
##                        Max.   :2654.3   Max.   :73554

logmodel1 = glm(default ~ balance + income, data = Default, family = binomial)

summary(logmodel1)

## 
## Call:
## glm(formula = default ~ balance + income, 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 ***
## balance      5.647e-03  2.274e-04  24.836  < 2e-16 ***
## income       2.081e-05  4.985e-06   4.174 2.99e-05 ***
## ---
## 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.

trainDefault = sample(dim(Default)[1], dim(Default)[1]*0.50)
testDefault = Default[-trainDefault, ]

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

LOGmodel2 = glm(default ~ balance + income, data = Default, family = binomial, subset = trainDefault)

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.

log.prob_def = predict(LOGmodel2, testDefault, type = "response")
log.pred_def = rep("No", dim(Default)[1]*0.50)
log.pred_def[log.prob_def > 0.5] = "Yes"
table(log.pred_def, testDefault$default)

##             
## log.pred_def   No  Yes
##          No  4821  116
##          Yes   12   51

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

mean(log.pred_def !=testDefault$default)

## [1] 0.0256

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.The three different splits each produce a different resulting error rate which shows that the rate varies by which observations are in the training/validation sets.

#i
trainDefault = sample(dim(Default)[1], dim(Default)[1]*0.50)
testDefault = Default[-trainDefault, ]

#ii
LOGmodel2 = glm(default ~ balance + income, data = Default, family = binomial, subset = trainDefault)

#iii
log.prob_def = predict(LOGmodel2, testDefault, type = "response")
log.pred_def = rep("No", dim(Default)[1]*0.50)
log.pred_def[log.prob_def > 0.5] = "Yes"
table(log.pred_def, testDefault$default)

##             
## log.pred_def   No  Yes
##          No  4822  106
##          Yes   19   53

#iv
mean(log.pred_def !=testDefault$default)

## [1] 0.025

#i
trainDefault = sample(dim(Default)[1], dim(Default)[1]*0.50)
testDefault = Default[-trainDefault, ]

#ii
LOGmodel2 = glm(default ~ balance + income, data = Default, family = binomial, subset = trainDefault)

#iii
log.prob_def = predict(LOGmodel2, testDefault, type = "response")
log.pred_def = rep("No", dim(Default)[1]*0.50)
log.pred_def[log.prob_def > 0.5] = "Yes"
table(log.pred_def, testDefault$default)

##             
## log.pred_def   No  Yes
##          No  4821  105
##          Yes   16   58

#iv
mean(log.pred_def !=testDefault$default)

## [1] 0.0242

#i
trainDefault = sample(dim(Default)[1], dim(Default)[1]*0.50)
testDefault = Default[-trainDefault, ]

#ii
LOGmodel2 = glm(default ~ balance + income, data = Default, family = binomial, subset = trainDefault)

#iii
log.prob_def = predict(LOGmodel2, testDefault, type = "response")
log.pred_def = rep("No", dim(Default)[1]*0.50)
log.pred_def[log.prob_def > 0.5] = "Yes"
table(log.pred_def, testDefault$default)

##             
## log.pred_def   No  Yes
##          No  4817  108
##          Yes   15   60

#iv
mean(log.pred_def !=testDefault$default)

## [1] 0.0246

4. 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. Including the dummy variable for student leads DID NOT seem to result in a reduction in the test error rate.

#i
trainDefault = sample(dim(Default)[1], dim(Default)[1]*0.50)
testDefault = Default[-trainDefault, ]

#ii
LOGmodel2 = glm(default ~ balance + income + student, data = Default, family = binomial, subset = trainDefault)

#iii
log.prob_def = predict(LOGmodel2, testDefault, type = "response")
log.pred_def = rep("No", dim(Default)[1]*0.50)
log.pred_def[log.prob_def > 0.5] = "Yes"
table(log.pred_def, testDefault$default)

##             
## log.pred_def   No  Yes
##          No  4822  109
##          Yes   16   53

#iv
mean(log.pred_def !=testDefault$default)

## [1] 0.025

Problem 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. The glm() estimates of the standard errors for the coefficients intercept, income and balance are respectively 4.348e-01, 4.985e-06 and 2.274e-04.

set.seed(1)
attach(Default)

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

## 
## 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

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. The bootstrap estimates of the standard errors for the coefficients B0, B1 and B2 are respectively 4.492830e-01, 4.988746e-06, and 2.347490e-04.

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

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 two methods are very similar.

Problem 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 ˆμ.

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

## [1] 22.53281

2. 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.

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

## [1] 0.4088611

3. Now estimate the standard error of ˆμ using the bootstrap. How does this compare to your answer from (b)? The bootstrap estimated standard error of (mu-hat) of 0.4106622 is similar to the estimate found in (b) of 0.4088611.

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

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 [ˆμ − 2SE(ˆμ), μˆ + 2SE(ˆμ)]. The bootstrap confidence interval is very close to the one provided by the t.test() function.

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

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

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

## [1] 21.2

6. 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. We get an estimated median value of 21.2 which is equal to the value obtained in (e), with a standard error of 0.3770241 which is relatively small compared to median value.

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

7. Based on this data set, provide an estimate for the tenth percentile of medv in Boston suburbs. Call this quantity ˆμ0.1. (You can use the quantile() function.)

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

##   10% 
## 12.75

8. Use the bootstrap to estimate the standard error of ˆμ0.1. Comment on your findings. We get an estimated tenth percentile value of 12.75 which is again equal to the value obtained in (g), with a standard error of 0.4925766 which is relatively small compared to percentile value.

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