Chapter 05 (page 197): 3, 5, 6, 9

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

  1. Explain how k-fold cross-validation is implemented. The K fold cross validation is implemented by randomly dividing the set of observations into k groups/folds. The first fold is treated as a validation set and the method is fit on the k - 1 folds. Then the mean squared error is computed on the observations in the fold that was held out. This procedure is completed k times. This process results in k estimates of the test error. The k-fold CV estimate is computed by averaging these values,

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

  1. The validation set approach?

Advantages of K-fold relative to validation set approach:: - The validation estimate of the test error rate can be highly variable - The validation set uses only a subset of observations, thus the validation set error may overestimate the test error rate for the model on the entire data set.

Disadvantages of k-fold relative to validation set approach: - The validation set approach is conceptually simple and is easy to implement

  1. LOOCV? Advantages of k-fold relative to LOOCV:

Disadvantages of k-fold relative to LOOCV: - LOOCV will give approximately unbiased estimates of the test error. - LOOCV is computationally intensive.

  1. 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)
set.seed(1)
attach(Default)
head(Default)
##   default student   balance    income
## 1      No      No  729.5265 44361.625
## 2      No     Yes  817.1804 12106.135
## 3      No      No 1073.5492 31767.139
## 4      No      No  529.2506 35704.494
## 5      No      No  785.6559 38463.496
## 6      No     Yes  919.5885  7491.559
names(Default)
## [1] "default" "student" "balance" "income"
glm.default <- glm(default~balance+income, data=Default,family=binomial )
summary(glm.default)
## 
## 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
  1. 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.
train <- sample(10000,2000)
  1. Fit a multiple logistic regression model using only the training observations.
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)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.5698  -0.1442  -0.0574  -0.0245   3.3545  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.202e+01  9.452e-01  -12.72  < 2e-16 ***
## income       4.689e-05  1.111e-05    4.22 2.44e-05 ***
## balance      5.407e-03  4.725e-04   11.44  < 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: 606.86  on 1999  degrees of freedom
## Residual deviance: 317.13  on 1997  degrees of freedom
## AIC: 323.13
## 
## Number of Fisher Scoring iterations: 8
  1. 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(fit.glm, newdata = Default[-train, ], type = "response")
pred.glm <- rep("No", length(probs))
pred.glm[probs > 0.5] <- "Yes"
  1. Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified.
mean(pred.glm != Default[-train, ]$default)
## [1] 0.02675

The validation set error is 2.7%.

  1. 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_2 <- sample(10000,3000)
fit.glm2 <- glm(default ~ income + balance, data = Default, family = "binomial", subset = train_2)
summary(fit.glm2)
## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default, subset = train_2)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -1.8820  -0.1406  -0.0533  -0.0188   3.3745  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.210e+01  8.412e-01 -14.384  < 2e-16 ***
## income       2.787e-05  9.065e-06   3.074  0.00211 ** 
## balance      5.863e-03  4.393e-04  13.349  < 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: 870.12  on 2999  degrees of freedom
## Residual deviance: 470.72  on 2997  degrees of freedom
## AIC: 476.72
## 
## Number of Fisher Scoring iterations: 8
probs2 <- predict(fit.glm2, newdata = Default[-train_2, ], type = "response")
pred.glm2 <- rep("No", length(probs2))
pred.glm2[probs2 > 0.5] <- "Yes"
mean(pred.glm2 != Default[-train_2, ]$default)
## [1] 0.02628571

The validation set error is 3%.

train_3 <- sample(10000,1000)
fit.glm3 <- glm(default ~ income + balance, data = Default, family = "binomial", subset = train_3)
summary(fit.glm3)
## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default, subset = train_3)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -1.0782  -0.1333  -0.0544  -0.0216   3.3497  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.166e+01  1.476e+00  -7.902 2.76e-15 ***
## income       2.401e-05  1.707e-05   1.407    0.159    
## balance      5.604e-03  7.841e-04   7.148 8.84e-13 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 241.10  on 999  degrees of freedom
## Residual deviance: 139.07  on 997  degrees of freedom
## AIC: 145.07
## 
## Number of Fisher Scoring iterations: 8
probs3 <- predict(fit.glm3, newdata = Default[-train_3, ], type = "response")
pred.glm3 <- rep("No", length(probs3))
pred.glm3[probs3 > 0.5] <- "Yes"
mean(pred.glm3 != Default[-train_3, ]$default)
## [1] 0.02755556

The validation set error is 2.8%.

train_4 <- sample(10000,4500)
fit.glm4 <- glm(default ~ income + balance, data = Default, family = "binomial", subset = train_4)
summary(fit.glm4)
## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default, subset = train_4)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.4884  -0.1449  -0.0582  -0.0206   3.7263  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.149e+01  6.366e-01 -18.048   <2e-16 ***
## income       1.762e-05  7.172e-06   2.457    0.014 *  
## balance      5.710e-03  3.404e-04  16.775   <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: 1342.13  on 4499  degrees of freedom
## Residual deviance:  722.79  on 4497  degrees of freedom
## AIC: 728.79
## 
## Number of Fisher Scoring iterations: 8
probs4 <- predict(fit.glm4, newdata = Default[-train_4, ], type = "response")
pred.glm4 <- rep("No", length(probs4))
pred.glm4[probs4 > 0.5] <- "Yes"
mean(pred.glm4 != Default[-train_4, ]$default)
## [1] 0.02672727

The validation set error is 2.5%.

Comment on the results obtained. Three more estimates of the validation set error would give:0.0296, 0.0281, 0.0249. Changing the number of observations changes the validation set error slightly.

  1. 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.
train <- sample(10000,2000)

glm <- glm(default ~ income + balance + student, data = Default, family = "binomial", subset = train)
summary(glm)
## 
## Call:
## glm(formula = default ~ income + balance + student, family = "binomial", 
##     data = Default, subset = train)
## 
## Deviance Residuals: 
##      Min        1Q    Median        3Q       Max  
## -1.67265  -0.17739  -0.07445  -0.02897   3.12567  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.033e+01  1.018e+00 -10.152   <2e-16 ***
## income       8.456e-06  1.746e-05   0.484    0.628    
## balance      5.302e-03  4.735e-04  11.197   <2e-16 ***
## studentYes  -5.354e-01  5.070e-01  -1.056    0.291    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 586.82  on 1999  degrees of freedom
## Residual deviance: 353.13  on 1996  degrees of freedom
## AIC: 361.13
## 
## Number of Fisher Scoring iterations: 8
probs <- predict(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] 0.02575

With adding the dummy variable, student, the validation set error is 2.6%. This doesn’t change the validation set error much.

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

The standard error for Bo is 0.6178 B1 is 7.024 e-6 B2 is 3.158 e-6

  1. 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.
default <- data.frame(Default)
library(boot)
boot.fn<-function(data,index){fit<-glm(default~income+balance,data=data,family="binomial",subset=index)
  return(coef(fit))}
  1. Use the boot() function together with your boot.fn() function to estimate the standard errors of the logistic regression coefficients for income and balance.
boot(default,boot.fn,100)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = default, statistic = boot.fn, R = 100)
## 
## 
## Bootstrap Statistics :
##          original        bias     std. error
## t1* -1.154047e+01  7.265692e-03 4.116057e-01
## t2*  2.080898e-05 -4.396504e-07 4.204169e-06
## t3*  5.647103e-03 -2.197193e-06 2.219810e-04
  1. Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function.

The estimated standard errors with glm() function were: 0.6178, 7.024 e-6, 3.158 e-6

Using the bootstrap funciton, the standard errors are: 3.842813e-01, 5.122866e-06, 2.044025e-04 Which are very close to one another.

  1. We will now consider the Boston housing data set, from the MASS library.
library(MASS)
attach(Boston)
  1. Based on this data set, provide an estimate for the population mean of medv. Call this estimate ˆµ.
µ_hat <- mean(Boston$medv)
µ_hat 
## [1] 22.53281
  1. 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.

µ_hat_sd <- sd(Boston$medv)
µ_hat_obs <- sqrt(dim(Boston)[1])
µ_hat_sd/µ_hat_obs
## [1] 0.4088611

Estimate of the standard error of µ_hat is 0.4088611,

  1. Now estimate the standard error of ˆµ using the bootstrap. How does this compare to your answer from (b)?
boot.fn2<-function(data,index){
  mu<-mean(data[index])
  return(mu)
}
set.seed(1)
boot(Boston$medv,boot.fn2,100)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = boot.fn2, R = 100)
## 
## 
## Bootstrap Statistics :
##     original      bias    std. error
## t1* 22.53281 0.009027668   0.3482331

The standard error of µ is 0.3482331.

  1. 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)].

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
#[ˆµ − 2SE(ˆµ), µˆ + 2SE(ˆµ)].
confid<- c(µ_hat - 2*0.3482331, µ_hat + 2*0.3482331)
confid
## [1] 21.83634 23.22927
  1. Based on this data set, provide an estimate, ˆµmed, for the median value of medv in the population.
µ_med <-median(Boston$medv)
µ_med
## [1] 21.2

The estimate for the median value of medv is 21.2.

  1. 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.
set.seed(1)
boot.fn3<-function(data,index){
  mu_median<-median(data[index])
  return(mu_median)
}
set.seed(1)
boot(Boston$medv,boot.fn3,100)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = boot.fn3, R = 100)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*     21.2  -0.029   0.3461316
  1. 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.)
µ0.1 <- quantile(Boston$medv, c(0.1))
µ0.1 
##   10% 
## 12.75

The estimate for the tenth percentile of medv is 12.75

  1. Use the bootstrap to estimate the standard error of ˆµ0.1. Comment on your findings.
boot.fn4 <- function(data, index) {
  mu.quant <- quantile(data[index], c(0.1))
  return (mu.quant)
}
boot(medv, boot.fn4, 1000)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn4, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*    12.75  0.0285   0.4861472

We get an estimated tenth percentile value of 12.75 which is equal to the value obtained in (g). We also get a standard error of 0.4861472 which is relatively small compared to percentile value.