Assignment #4

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

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

The K-fold cross-validation is performed by randomly dividing the obervations into fold or k groups of equal sizes. Our first fold is the “validation set”. The cross-validation is used in order to estimate our model and how it is expected to perform. It consists of a summary of the mean of our model skill scores, the measure of the variance: standard deviation/standard error.

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

1. The validation set approach?

One of the disadvantages of validation set approach is that depending how many observations are within our model the test error rate estimare might be extremely varied. Another disadvantage is that the test error rate for our model fit on the full data set may tend to overstate the validation set error rate. For advantage, the validation set technique is both theoretically and practically straighforward.

2. LOOCV

One of the advantage of LOOCV is that it is completely random, because the model is trained on the full dataset, the bias will be minimized and the test error rate will not be overestimated. For disadvantage, in order to perfomr LOOCV requires much computing time.

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.

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

library(ISLR)

## Warning: package 'ISLR' was built under R version 4.0.5

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

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)

fit.glm.1 <- glm(default ~ income + balance, data = Default, family = binomial)

summary(fit.glm.1)

## 
## 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

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

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

fit.glm.2 <- glm(default ~ balance + income, data = Default, family = binomial, subset = train)

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.

predict.five <-predict(fit.glm.2, test, type= "response")

pred.glm.five <- rep("No", dim(Default)[1]*0.50)

pred.glm.five[predict.five > 0.5] <- "Yes"

table(pred.glm.five, test$default)

##              
## pred.glm.five   No  Yes
##           No  4824  108
##           Yes   19   49

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

mean(pred.glm.five !=test$default)

## [1] 0.0254

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.

1:

train1 <- sample(dim(Default)[1], dim(Default)[1]/2)
test1 <- Default[-train1, ]

fit.glm.c1 <- glm(default ~ balance + income, data = Default, family = binomial, subset = train1)

predict.c1 <-predict(fit.glm.c1, test1, type= "response")

pred.glm.c1 <- rep("No", dim(Default)[1]*0.50)

pred.glm.c1[predict.c1 > 0.5] <- "Yes"

table(pred.glm.c1, test1$default)

##            
## pred.glm.c1   No  Yes
##         No  4813  112
##         Yes   25   50

mean(pred.glm.c1 !=test1$default)

## [1] 0.0274

train2 <- sample(dim(Default)[1], dim(Default)[1]/2)
test2 <- Default[-train2, ]

fit.glm.c2 <- glm(default ~ balance + income, data = Default, family = binomial, subset = train2)

predict.c2 <-predict(fit.glm.c2, test2, type= "response")

pred.glm.c2 <- rep("No", dim(Default)[1]*0.50)

pred.glm.c2[predict.c2 > 0.5] <- "Yes"

table(pred.glm.c2, test2$default)

##            
## pred.glm.c2   No  Yes
##         No  4827  103
##         Yes   19   51

mean(pred.glm.c2 !=test2$default)

## [1] 0.0244

3:

train3 <- sample(dim(Default)[1], dim(Default)[1]/2)
test3 <- Default[-train3, ]

fit.glm.c3 <- glm(default ~ balance + income, data = Default, family = binomial, subset = train3)

predict.c3 <-predict(fit.glm.c3, test3, type= "response")

pred.glm.c3 <- rep("No", dim(Default)[1]*0.50)

pred.glm.c3[predict.c3 > 0.5] <- "Yes"

table(pred.glm.c3, test3$default)

##            
## pred.glm.c3   No  Yes
##         No  4824  106
##         Yes   16   54

mean(pred.glm.c3 !=test3$default)

## [1] 0.0244

First: 0.0258 Second:0.0270 Third: 0.0252

After performing model three times, we can see how our error rate change every time, for .c1 it was 0.0258, .c2 it was 0.0270 and .c3 it was 0.0252. The test error rate is variable, so it depends on what observations are used in training and validation set.

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.

traind <- sample(dim(Default)[1], dim(Default)[1]/2)
testd <- Default[-traind, ]

fit.glm.d <- glm(default ~ balance + income + student, data = Default, family = binomial, subset = traind)

predict.d <-predict(fit.glm.d, testd, type= "response")

pred.glm.d <- rep("No", dim(Default)[1]*0.50)

pred.glm.d[predict.d > 0.5] <- "Yes"

table(pred.glm.d, testd$default)

##           
## pred.glm.d   No  Yes
##        No  4808  121
##        Yes   18   53

mean(pred.glm.d !=testd$default)

## [1] 0.0278

After including student, it gave us a error rate of 0.026 which is close to the other error rates we ran.The addition of the student variable has little effect on the 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.

set.seed(1)

attach(Default)

## The following objects are masked from Default (pos = 3):
## 
##     balance, default, income, student

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.

fit.glm6 <- glm(default ~ income + balance, data = Default, family = "binomial")

summary(fit.glm6)

## 
## 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

The intercept: 4.348e-01 with p-value: < 2e-16 Income: 4.985e-06 with p-value: 2.99e-05 Balance: 2.274e-04 with p-value: 2e-16

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) return(coef(glm(default ~ income + balance, 
    data = data, family = binomial, 
    subset = index)))

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.

library(boot)

boot6 <- boot(Default, boot.fn, 1000)

boot6

## 
## 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 standard errors obtained using the glm function compared to bootstrap are very very close to each other,

9. 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.

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.

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

staerr.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(1)

boot.fn9 <- function(data, index) return(mean(data[index]))

boot9 <- boot(medv, boot.fn9, 1000)

boot9

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn9, 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)].

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

conf.int.mu.hat <- c(22.53 - 2 * 0.414028, 22.53 + 2 * 0.414028)

conf.int.mu.hat

## [1] 21.70194 23.35806

The t.test method provides a confidence interval that is extremely close to the bootstrap confidence interval.

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

umed.hat <- median(medv)

umed.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.fn9f <- function(data, index) return(median(data[index]))

boot9f <- boot(medv, boot.fn9f, 1000)

boot9f

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn9f, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*     21.2 -0.0386   0.3770241

Our estimated median value is 21.2 which is the same result as our umed.hat on e).

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.)

tenth.perc <- quantile(medv, c(0.1))

tenth.perc

##   10% 
## 12.75

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

boot.fn9h <- function(data, index) return(quantile(data[index], c(0.1)))

boot9h <- boot(medv, boot.fn9h, 1000)

boot9h

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn9h, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*    12.75  0.0186   0.4925766

Our estimated median value is 12.75 which is the same result as our tenth.perc on g).And our standard error is 0.5141291 which is smaller.