Assignment 04

Exercise 3

We now review k-fold cross-validation.

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

It divides the data set into k groups (folds) of approximately equal size, just like LOOCV, each fold (group) will be treated as a validation (test) set, and the model gets fit on the remaining k-1 folds (groups). LOOCV and K-Fold Cross Validation result in the same if k = n (number of observations), however we usually use k=5 or k=10.

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

i. The validation set approach?

Advantages: The validation set approach is a very simple and easy to implement approach, since our model gets only to be fitted once.

Disadvantages: However, since only one portion of the data set is used for validation and the other for training, the test error rate (MSE) can be highly variable, since it depends on which observations are used for training and which ones for validation. Following the same line, the validation MSE might even tend to overestimate the test error rate for the model fit on the entire data set.

ii. LOOCV?

Advantages: The first and most known advantage of k-fold CV over LOOCV is that this last one way more computationally intense, since we have to fit each model, and with big data sets that could almost be impossible. When it comes to bias reduction, LOOCV is preferred over k-fold CV, LOOCV will give approximately unbiased estimates of the test error, since each training set contains n − 1 observations, which is almost as many as the number of observations in the full data set. And performing k-fold CV for, k = 5 or k = 10 will lead to an intermediate level of bias, since each training set contains (k − 1)n/k observations—fewer than in the LOOCV approach, but substantially more than in the validation set approach.

Disadvantages: On the other hand, when it comes to variance, LOOCV has higher variance than K-fold CV when k>n, this is because by averaging the outputs of n fitted models, those will be highly correlated with each other. In result the test error estimate resulting from LOOCV tends to have higher variance than does the test error estimate resulting from k-fold CV.

Exercise 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.

library(ISLR)

## Warning: package 'ISLR' was built under R version 4.1.2

library(MASS)
library(caTools)

## Warning: package 'caTools' was built under R version 4.1.2

library(boot)

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

attach(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

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

summary(glm.fit)

## 
## 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.

sample_data = sample.split(Default, SplitRatio = 0.50)
training.set = subset(Default, sample_data==TRUE)
test.set=subset(Default,sample_data==FALSE)
test.default = test.set$default

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

glm.train = glm(default~income+balance, data=training.set, family = binomial)
summary(glm.train)

## 
## Call:
## glm(formula = default ~ income + balance, family = binomial, 
##     data = training.set)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.5287  -0.1436  -0.0570  -0.0204   3.3287  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.162e+01  6.066e-01 -19.164  < 2e-16 ***
## income       1.944e-05  6.931e-06   2.805  0.00504 ** 
## balance      5.772e-03  3.178e-04  18.161  < 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: 1569.84  on 4999  degrees of freedom
## Residual deviance:  821.43  on 4997  degrees of freedom
## AIC: 827.43
## 
## 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.

glm.probs = predict(glm.train, test.set, type = "response")
glm.preds = rep("No",length(test.set$default))
glm.preds[glm.probs>0.5]="Yes"

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

table(glm.preds,test.default)

##          test.default
## glm.preds   No  Yes
##       No  4827  100
##       Yes   23   50

Test error rate is: $(100+23)/(5000) = 2.5%$

(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.

Another random split

set.seed(12)
sample_data = sample.split(Default, SplitRatio = 0.50)
training.set = subset(Default, sample_data==TRUE)
test.set=subset(Default,sample_data==FALSE)
test.default = test.set$default

Fitting the model

glm.train = glm(default~income+balance, data=training.set, family = binomial)
glm.probs = predict(glm.train, test.set, type = "response")
glm.preds = rep("No",length(test.set$default))
glm.preds[glm.probs>0.5]="Yes"
table(glm.preds,test.default)

##          test.default
## glm.preds   No  Yes
##       No  4798  122
##       Yes   27   53

Test error rate is: $(122+27)/(5000) = 2.98%$

Validation set size of 70% (original seed = 1)

set.seed(1)
index = sample(1:nrow(Default), 0.7*nrow(Default))
training.set = Default[index,]
test.set=Default[-index,]
test.default = test.set$default

Fitting the model

glm.train = glm(default~income+balance, data=training.set, family = binomial)
glm.probs = predict(glm.train, test.set, type = "response")
glm.preds = rep("No",length(test.set$default))
glm.preds[glm.probs>0.5]="Yes"
table(glm.preds,test.default)

##          test.default
## glm.preds   No  Yes
##       No  2896   78
##       Yes    2   24

Test error rate is: $(2+78)/(3000) = 2.67%$

Validation set size of 80% (original seed = 1)

set.seed(1)
index = sample(1:nrow(Default), 0.8*nrow(Default))
training.set = Default[index,]
test.set=Default[-index,]
test.default = test.set$default

Fitting the model

glm.train = glm(default~income+balance, data=training.set, family = binomial)
glm.probs = predict(glm.train, test.set, type = "response")
glm.preds = rep("No",length(test.set$default))
glm.preds[glm.probs>0.5]="Yes"
table(glm.preds,test.default)

##          test.default
## glm.preds   No  Yes
##       No  1928   50
##       Yes    2   20

Test error rate is: $(2+50)/(2000) = 2.6%$

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

Fitting the model (with the 80/20 split)

glm.train = glm(default~income+balance+student, data=training.set, family = binomial)
glm.probs = predict(glm.train, test.set, type = "response")
glm.preds = rep("No",length(test.set$default))
glm.preds[glm.probs>0.5]="Yes"
table(glm.preds,test.default)

##          test.default
## glm.preds   No  Yes
##       No  1928   53
##       Yes    2   17

Test error rate is: $(2+53)/(2000) = 2.75%$

Exercise 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)
glm.fit2 = glm(default~income+balance, data=Default, family=binomial)
summary(glm.fit2)$coefficients[2:3,2]

##       income      balance 
## 4.985167e-06 2.273731e-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)
  return(coef(glm(default~income+balance, data=data, subset=index, family = binomial)))

(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.

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

summary(glm.fit2)$coefficients

##                  Estimate   Std. Error    z value      Pr(>|z|)
## (Intercept) -1.154047e+01 4.347564e-01 -26.544680 2.958355e-155
## income       2.080898e-05 4.985167e-06   4.174178  2.990638e-05
## balance      5.647103e-03 2.273731e-04  24.836280 3.638120e-136

Comments: We can see that the coefficients in the original glm model of the Std. Errors for Intercept, Income and balance are:4.347564e-01, 4.985167e-06, 2.273731e-04 respectively, and the bootstrap Std. Error coefficients are not very different at all: 4.344722e-01, 4.866284e-06, 2.298949e-04 (same order: Intercept, Income and Balance)

Exercise 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.

detach(Default)
attach(Boston)

summary(medv)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##    5.00   17.02   21.20   22.53   25.00   50.00

mu = mean(medv)

(b) 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.

length=length(medv)
sd=sqrt(sum((medv-mu)^2)/(length-1))
se=sd/sqrt(length)
se

## [1] 0.4088611

Comments:

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

se=sd/sqrt(length)
se

## [1] 0.4088611

set.seed(1)
boot.fn2 = function(data,index)
  return(mean(data[index]))

boot(medv,boot.fn2,1000)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn2, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original      bias    std. error
## t1* 22.53281 0.007650791   0.4106622

Comments: The standard error using bootstrapping with R=1000 is 0,41, almost the same as the one from the original mean function 0.408.

(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).

Hint: You can approximate a 95% confidence interval using the formula [ˆμ − 2SE(ˆμ), ˆμ + 2SE(ˆμ)].

mu-2*se #lower bound

## [1] 21.71508

mu+2*se #higher bound

## [1] 23.35053

T-Test

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

Comments: The 95 percent confidence interval from the T-test and the interval we calculated is almost the same.

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

median.medv = median(Boston$medv)
median.medv

## [1] 21.2

(f) 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)
  return(median(data[index]))

boot(medv,boot.fn3,1000)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn3, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*     21.2 0.02295   0.3778075

Comments: The median of our bootstrapping method is 21.2, which is the same as the calculated one, the std. error is 0.37, which is very small compared to the 21.2 median.

(g) 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.)

tenth.perc = quantile(medv, 0.1)
tenth.perc

##   10% 
## 12.75

(h) Use the bootstrap to estimate the standard error of ˆμ0.1. Comment on your findings.

set.seed(1)
boot.fn4 = function(data,index){
  X=medv[index]
  Y=quantile(X,0.1)
  return(Y)
}

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  0.0339   0.4767526

Comments: Our calculated 10th quantile was 12.75 and the bootstrapping one was 12.75, exactly the same, our Std. Error was 0.4767526.

Assignment 04

Aranzazu Chaparro Molina - wvg029

3/9/2022

Exercise 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?

ii. LOOCV?

Exercise 5

(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.

(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.

Exercise 6

(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.

(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.

Exercise 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.

(b) 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.

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

(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).

Hint: You can approximate a 95% confidence interval using the formula [ˆμ − 2SE(ˆμ), ˆμ + 2SE(ˆμ)].

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

(f) 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.

(g) 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.)

(h) Use the bootstrap to estimate the standard error of ˆμ0.1. Comment on your findings.