Lab 6

  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.

    library(ISLR2)
    ## Warning: package 'ISLR2' was built under R version 4.3.3
    library(MASS)
    ## 
## Attaching package: 'MASS'
    ## The following object is masked from 'package:ISLR2':
## 
##     Boston
    attach(Default)
set.seed(1)
    1. Fit a logistic regression model that uses income and balance to predict default.
    lr <- glm(default ~ income + balance, family = binomial, data = Default)
summary(lr)
    ## 
## 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
    pred <- predict(lr, Default$default, type = "response")
pred.class <- ifelse(pred > 0.5, "Yes", "No")
round(mean(Default$default != pred.class), 4)
    ## [1] 0.0263
    1. Using the validation set approach, estimate the rest error of this model. In order to do this, you must perform the following steps:

    • Split the sample set into a training set and a validation set.

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

    • 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.

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

    subset <- sample(nrow(Default), nrow(Default) * 0.9)
default.train <- Default[subset, ]
default.test <- Default[-subset, ]

lr2 <- glm(default ~ income + balance, family = binomial, data = default.train)
pred2 <- predict(lr2, default.test, type = "response")
class2 <- ifelse(pred2 > 0.5, "Yes", "No")

table(default.test$default, class2, dnn = c("Actual", "Predicted"))
    ##       Predicted
## Actual  No Yes
##    No  962   1
##    Yes  28   9
    mr <- round(mean(class2 != default.test$default), 4)
print(mr)
    ## [1] 0.029
    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.
    subset<-sample(nrow(Default),nrow(Default)*0.8)
default.train<-Default[subset,]
default.test<-Default[-subset,]
lr3<-glm(default~income+balance,family = binomial,data=default.train)
pred3<-predict(lr3,default.test,type="response")
class3<-ifelse(pred3>0.5,"Yes","No")

mr3<-round(mean(class3!=default.test$default),4)


subset<-sample(nrow(Default),nrow(Default)*0.7)
default.train<-Default[subset,]
default.test<-Default[-subset,]
lr4<-glm(default~income+balance,family = binomial,data=default.train)
pred4<-predict(lr4,default.test,type="response")
class4<-ifelse(pred4>0.5,"Yes","No")

mr4<-round(mean(class4!=default.test$default),4)

subset<-sample(nrow(Default),nrow(Default)*0.5)
default.train<-Default[subset,]
default.test<-Default[-subset,]
lr5<-glm(default~income+balance,family = binomial,data=default.train)
pred5<-predict(lr5,default.test,type="response")
class5<-ifelse(pred5>0.5,"Yes","No")

mr5<-round(mean(class5!=default.test$default),4)

print(mr3)
    ## [1] 0.0235
    print(mr4)
    ## [1] 0.0283
    print(mr5)
    ## [1] 0.0292
    • I split the data into three different ratios of training vs test data. The misclassification rates are 0.0295, 0.027, and 0.03 for the ratios 80:20, 70:30, and 50:50 respectively.
    1. Now consider a logistic regression model that predicts the probability of default using income, balance, and a dummy variable for students. 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)
subset <- sample(nrow(Default), nrow(Default) * 0.7)
default.train <- Default[subset, ]
default.test <- Default[-subset, ]

lr6 <- glm(default ~ income + balance + student, family = binomial, data = default.train)
summary(lr6)
    ## 
## Call:
## glm(formula = default ~ income + balance + student, family = binomial, 
##     data = default.train)
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.095e+01  5.880e-01 -18.616   <2e-16 ***
## income       6.273e-06  9.845e-06   0.637    0.524    
## balance      5.678e-03  2.741e-04  20.715   <2e-16 ***
## studentYes  -7.167e-01  2.886e-01  -2.483    0.013 *  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 2030.3  on 6999  degrees of freedom
## Residual deviance: 1073.5  on 6996  degrees of freedom
## AIC: 1081.5
## 
## Number of Fisher Scoring iterations: 8
    pred6 <- predict(lr6, default.test, type = "response")
class6 <- ifelse(pred6 > 0.7, "Yes", "No")
table(default.test$default, class6, dnn = c("Actual", "Predicted"))
    ##       Predicted
## Actual   No  Yes
##    No  2898    0
##    Yes   94    8
    mr6 <- round(mean(class6 != default.test$default), 4)
print(mr6)
    ## [1] 0.0313
    • The inclusion of the student variable does not lead to a reduction in the test error rate.

  2. 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)
lr <- glm(default ~ income + balance, family = binomial, data = Default)
summary(lr)
    ## 
## 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
    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.
    boot.fn <- function(data, index) {
  fit <- glm(default ~ income + balance, data = data, family = binomial, subset = index)
  return(coef(fit))
}
## boot.fn takes data and gives the coefficient estimate for the given index
    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.
    library(boot)
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  8.556378e-03 4.122015e-01
## t2*  2.080898e-05 -3.993598e-07 4.186088e-06
## t3*  5.647103e-03 -4.116657e-06 2.226242e-04
    1. Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function.
    • The standard errors obtained from the two methods are very similar.

  3. In sections 5.3.2 and 5.3.3, we saw that the cv.glm() function can be used in order to compute the LOOCV test error estimate. Alternatively, one could compute those quantities using just the glm() and predict.glm() functions, and a for loop. You will now take this approach in order to compute the LOOCV error for a simple logistic regression model on the weekly data set. Recall that in the context of classification problems, the LOOCV error is given in (5.4).

    1. Fit a logistic regression model that predicts direction using lag1 and lag2.
    attach(Weekly)
set.seed(1)
weekly <- glm(Direction ~ Lag1 + Lag2, family = binomial, data = Weekly)
summary(weekly)
    ## 
## Call:
## glm(formula = Direction ~ Lag1 + Lag2, family = binomial, data = Weekly)
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)    
## (Intercept)  0.22122    0.06147   3.599 0.000319 ***
## Lag1        -0.03872    0.02622  -1.477 0.139672    
## Lag2         0.06025    0.02655   2.270 0.023232 *  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 1496.2  on 1088  degrees of freedom
## Residual deviance: 1488.2  on 1086  degrees of freedom
## AIC: 1494.2
## 
## Number of Fisher Scoring iterations: 4
    1. Fit a logistic regression model that predicts direction using lag1 and lag2 using all but the first observation.
    weekly1 <- glm(Direction ~ Lag1 + Lag2, family = binomial, data = Weekly[-1, ])
    1. Use the model from (b) to predict the direction of the first observation. You can do this by predicting that the first observation will go up if P(direction = ‘Up’ | lag1, lag2) > 0.5. Was this observation correctly classified?
    predweekly <- predict(weekly, Weekly[1, ], type = "response")
predweekly.class <- ifelse(predweekly > 0.5, "Up", "Down")
predweekly.class
    ##    1 
## "Up"
    1. Write a for loop from i = 1 to i = n, where n is the number of observations in th data set, that performs each of the following steps:

    • Fit a logistic regression model using all but the ith observation to predict direction using lag1 and lag2.

    • Compute the posterior probability of the market moving up for the ith observation.

    • Use the posterior probability for the ith observation in order to predict whether of not the market moves up.

    • Determine whether or not an error was made in predicting the direction for the ith observation. If an error was made, then idicate this as a 1, and otherwise indicate it as a 0.

    error<-rep(0,dim(Weekly)[1])
for (i in 1:dim(Weekly)[1]){
  fit.glm<-fit.glm <- glm(Direction ~ Lag1 + Lag2, data = Weekly[-i, ],  family = "binomial")
  pred.up <- predict.glm(fit.glm, Weekly[i, ], type = "response") > 0.5
  true.up <- Weekly[i, ]$Direction == "Up"
  if (pred.up != true.up)
    error[i] <- 1
}
    1. Take the average of the n numbers obtained in (d) in order to obtain the LOOCV estimates for the test error. Comment on the results.
    mean(error)
    ## [1] 0.4499541
    ## The LOOCV estimate is about 45%.

  1. We will now consider the Boston housing data set, from the islr2 library.

    1. Based on this data set, provide an estimate for the population mean of medv. Call this estimate mu.
    library(MASS)
attach(Boston)
sample_mu <- mean(medv)
sample_mu
    ## [1] 22.53281
    1. Provide an estimate of the standard error of mu. Interpret this result.
    standarderror <- sd(medv) / sqrt(nrow(Boston))
standarderror
    ## [1] 0.4088611
    ## The estimated standard error is .4088611. This gives us the accuracy of the estimate or how much the mean will vary based on the sample chosen.
    1. Now estimate the standard error of mu using the bootstrap. How does this compare to your answer from (b)?
    library(boot)
set.seed(1)
boot.fn <- function(data, index) {
  mu <- mean(data[index])
  return(mu)
}
set.seed(1)
boot(medv, boot.fn, 100)
    ## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn, R = 100)
## 
## 
## Bootstrap Statistics :
##     original      bias    std. error
## t1* 22.53281 0.009027668   0.3482331
    ## The standard error obtained here is very similar to that obtained in part (b).
    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).
    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
    confint <- c(sample_mu - 2 * standarderror, sample_mu + 2 * standarderror)
confint
    ## [1] 21.71508 23.35053
    ## The confidence intervals obtained from the two methods are very very similar.
    1. Based on this data set, provide an estimate, mu-med, for the median value of medv in the population.
    mu_median <- median(medv)
mu_median
    ## [1] 21.2
    1. We now would like to estimate the standard error of 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.
    library(boot)
set.seed(1)
boot.fn <- function(data, index) {
  median_mu <- median(data[index])
  return(median_mu)
}
set.seed(1)
boot(medv, boot.fn, 100)
    ## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn, R = 100)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*     21.2  -0.029   0.3461316
    ## The standard error of the median is found here using bootstrapping.
    1. Based on this data set, provide an estimate for the tenth percentile of medv in Boston census tracts. Call this quantity mu-.1.
    quantile_mu <- quantile(medv, c(0.1))
quantile_mu
    ##   10% 
## 12.75
    1. Use the bootstrap to estimate the standard error of mu-.1. Comment on your findings.
    boot.fn <- function(data, index) {
  mu_quantile <- quantile(data[index], c(0.1))
  return(mu_quantile)
}
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  0.0285   0.4861472
    ## This standard error is small compared to the tenth percentile so we know that we have a good estimate of the population.