homework4

library(tidyverse)

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library(ISLR)

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library(caret)

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library(MASS)

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library('fastDummies')

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library('boot')

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

1. Explain how k-fold cross-validation is implemented.

In this method data is put into k equal groups. model is trained on k-1 groups, remaining group is used for testing. this is repeated k times, so every group acts as a test data atleast once. Each occurence is recorded and averaged

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

1. The validation set approach? Advantages : lower variablity, all the data is used, over-estimate the test error Disadvantages: computational advantage for validation

2. LOOCV? advantages : k-fold is less computionally demanding, k fold has a better bias-variance tradeoff Disadvantage : k-fold has variation in the out of sample error compared to LOOCV

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.

set.seed(123)

data("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

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

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

summary(glm_default)

## 
## 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. 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.

index <- createDataPartition(y=Default$default,p= 0.6999, list= F)

train <- Default[index,]

test<- Default[-index,]

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

glm_default_train <- glm(default ~ income+balance, data= train, family = "binomial" )

summary(glm_default_train)

## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = train)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.5089  -0.1466  -0.0595  -0.0219   3.7097  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.165e+01  5.235e-01  -22.25  < 2e-16 ***
## income       2.683e-05  5.988e-06    4.48 7.45e-06 ***
## balance      5.605e-03  2.710e-04   20.68  < 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: 2050.5  on 6999  degrees of freedom
## Residual deviance: 1120.8  on 6997  degrees of freedom
## AIC: 1126.8
## 
## Number of Fisher Scoring iterations: 8

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.

pred_test <- factor(ifelse(predict(glm_default_train,newdata = test, type= 'response')>0.5,"Yes", "No"))
pred_test[0:9]

##  6 12 19 23 24 26 29 33 34 
## No No No No No No No No No 
## Levels: No Yes

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

mean(test$default != pred_test)

## [1] 0.02633333

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.

first split

set.seed(12)
index <- createDataPartition(y=Default$default,p= 0.75, list= F)

train <- Default[index,]

test<- Default[-index,]

glm_default_train <- glm(default ~ income+balance, data= train, family = "binomial" )

pred_test <- factor(ifelse(predict(glm_default_train,newdata = test, type= 'response')>0.5,"Yes", "No"))

mean(test$default != pred_test)

## [1] 0.02721088

second split

set.seed(12)
index <- createDataPartition(y=Default$default,p= 0.80, list= F)

train <- Default[index,]

test<- Default[-index,]

glm_default_train <- glm(default ~ income+balance, data= train, family = "binomial" )

pred_test <- factor(ifelse(predict(glm_default_train,newdata = test, type= 'response')>0.5,"Yes", "No"))

mean(test$default != pred_test)

## [1] 0.02951476

third split

set.seed(12)
index <- createDataPartition(y=Default$default,p= 0.85, list= F)

train <- Default[index,]

test<- Default[-index,]

glm_default_train <- glm(default ~ income+balance, data= train, family = "binomial" )

pred_test <- factor(ifelse(predict(glm_default_train,newdata = test, type= 'response')>0.5,"Yes", "No"))

mean(test$default != pred_test)

## [1] 0.02535023

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.

set.seed(12)

index <- createDataPartition(y = Default$default, p = 0.7, list = F)
train <- Default[index, ]
test <- Default[-index, ]

log_def <- glm(default ~ ., data = train, family = "binomial")
summary(log_def)

## 
## Call:
## glm(formula = default ~ ., family = "binomial", data = train)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.4433  -0.1473  -0.0583  -0.0218   3.7086  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.059e+01  5.749e-01 -18.420   <2e-16 ***
## studentYes  -7.118e-01  2.788e-01  -2.553   0.0107 *  
## balance      5.640e-03  2.726e-04  20.691   <2e-16 ***
## income       7.322e-07  9.730e-06   0.075   0.9400    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 2050.6  on 7000  degrees of freedom
## Residual deviance: 1120.9  on 6997  degrees of freedom
## AIC: 1128.9
## 
## Number of Fisher Scoring iterations: 8

test_pred <- factor(ifelse(predict(log_def, newdata = test, type = "response") > 0.5, "Yes", "No"))

mean(test$default != test_pred)

## [1] 0.02834278

dummy variable

New_default <- dummy_cols(Default,select_columns = 'student')
drop <- c('student')
New_default<- subset(New_default, select = -c(student))
head(New_default)

##   default   balance    income student_No student_Yes
## 1      No  729.5265 44361.625          1           0
## 2      No  817.1804 12106.135          0           1
## 3      No 1073.5492 31767.139          1           0
## 4      No  529.2506 35704.494          1           0
## 5      No  785.6559 38463.496          1           0
## 6      No  919.5885  7491.559          0           1

set.seed(12)

index <- createDataPartition(y = New_default$default, p = 0.7, list = F)
train <- New_default[index, ]
test <- New_default[-index, ]

log_def <- glm(default ~ ., data = train, family = "binomial")
summary(log_def)

## 
## Call:
## glm(formula = default ~ ., family = "binomial", data = train)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.4433  -0.1473  -0.0583  -0.0218   3.7086  
## 
## Coefficients: (1 not defined because of singularities)
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.130e+01  5.156e-01 -21.920   <2e-16 ***
## balance      5.640e-03  2.726e-04  20.691   <2e-16 ***
## income       7.322e-07  9.730e-06   0.075   0.9400    
## student_No   7.118e-01  2.788e-01   2.553   0.0107 *  
## student_Yes         NA         NA      NA       NA    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 2050.6  on 7000  degrees of freedom
## Residual deviance: 1120.9  on 6997  degrees of freedom
## AIC: 1128.9
## 
## Number of Fisher Scoring iterations: 8

test_pred <- factor(ifelse(predict(log_def, newdata = test, type = "response") > 0.5, "Yes", "No"))

## Warning in predict.lm(object, newdata, se.fit, scale = 1, type = if (type == :
## prediction from a rank-deficient fit may be misleading

mean(test$default != test_pred)

## [1] 0.02834278

so including a dummy variable for student does not reduce the test error rate.

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.

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.

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

summary(log_def)$coefficients[, 2]

##  (Intercept)       income      balance 
## 4.347564e-01 4.985167e-06 2.273731e-04

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 = 1:nrow(data)) {
  coef(glm(default ~ income + balance, data = data, subset = index, family = "binomial"))[-1]
}

boot.fn(Default)

##       income      balance 
## 2.080898e-05 5.647103e-03

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.

set.seed(123)

boot(data = Default, statistic = boot.fn, R = 100)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Default, statistic = boot.fn, R = 100)
## 
## 
## Bootstrap Statistics :
##         original       bias     std. error
## t1* 2.080898e-05 1.483646e-07 5.124683e-06
## t2* 5.647103e-03 2.836407e-06 2.150082e-04

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

The estimated standard errors obtained by the two mehods are similiar

9. We will now consider the Boston housing data set, from the MASS library.

1. Based on this data set, provide an estimate for the population mean of medv. Call this estimate ˆμ.

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)?

set.seed(123)

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 -0.01607372   0.4045557

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), ˆμ + 2SE(ˆμ)].

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 * 0.4045557, 22.53 + 2 * 0.4045557)
CI_mu_hat

## [1] 21.72089 23.33911

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.

boot.fn <- function(vector, index) {
  median(vector[index])
}

set.seed(77)

(boot_results <- boot(data = Boston$medv, statistic = boot.fn, R = 1000))

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*     21.2 -0.0119    0.371108

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.)

quantile(Boston$medv, 0.1)

##   10% 
## 12.75

8. Use the bootstrap to estimate the standard error of ˆμ0.1. Comment on your findings.

boot.fn <- function(vector, index) {
  quantile(vector[index], 0.1)
}

set.seed(77)

(boot_results <- boot(data = Boston$medv, statistic = boot.fn, R = 1000))

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*    12.75 0.01305   0.4850335