Assignment #4

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

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

In this approach, the dataset is randomly divided into k groups, or folds, each of similar size. One fold is designated as the validation set, while the remaining k - 1 folds are used for training the model. The validation set is then evaluated to compute the mean squared error (MSE1). This process is repeated k times, with each fold taking turns as the validation set. Consequently, k estimates of the test error are generated. Finally, the k-fold cross-validation estimate is obtained by averaging these computed values.

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

1. The validation set approach?

The validation set approach offers simplicity and straightforward implementation, making it accessible for many applications. However, it comes with drawbacks. The validation mean squared error (MSE) can exhibit significant variability, potentially affecting the reliability of model evaluation. Additionally, this method utilizes only a portion of the available observations for model training, which may limit its effectiveness, particularly with smaller datasets.

2. LOOCV?

LOOCV offers the advantage of lower bias compared to the validation approach. Unlike the validation method, which produces varying MSE values upon repeated application due to the random splitting process, LOOCV consistently yields the same results when performed multiple times, as each iteration involves splitting based on one observation at a time. However, a notable disadvantage of LOOCV is its high computational intensity, as it requires fitting the model repeatedly for each individual observation, which can be resource-intensive, particularly with large datasets.

5. 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)

data("Default")

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

## 
## 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

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]*0.50)
test <- Default[-train, ]

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

logit_model2 <- glm(default ~ income+ balance, data = Default, family = 'binomial', subset = train)

summary(logit_model2)

## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default, subset = train)
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.147e+01  6.147e-01 -18.653  < 2e-16 ***
## income       2.229e-05  7.085e-06   3.146  0.00166 ** 
## balance      5.594e-03  3.194e-04  17.512  < 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: 1477.13  on 4999  degrees of freedom
## Residual deviance:  808.97  on 4997  degrees of freedom
## AIC: 814.97
## 
## 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.

log_prob = predict(logit_model2, test, type = "response")
log_pred = rep("No", dim(Default)[1]*0.50)
log_pred[log_prob > 0.5] = "Yes"
table(log_pred, test$default)

##         
## log_pred   No  Yes
##      No  4814  112
##      Yes   22   52

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

mean(log_pred !=test$default)

## [1] 0.0268

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.

train <- sample(dim(Default)[1], dim(Default)[1]*0.50)
test <- Default[-train, ]

logit_model2 <- glm(default ~ income+ balance, data = Default, family = 'binomial', subset = train)

log_prob = predict(logit_model2, test, type = "response")
log_pred = rep("No", dim(Default)[1]*0.50)
log_pred[log_prob > 0.5] = "Yes"
table(log_pred, test$default)

##         
## log_pred   No  Yes
##      No  4816  116
##      Yes   15   53

mean(log_pred !=test$default)

## [1] 0.0262

train <- sample(dim(Default)[1], dim(Default)[1]*0.50)
test <- Default[-train, ]

logit_model2 <- glm(default ~ income+ balance, data = Default, family = 'binomial', subset = train)

log_prob = predict(logit_model2, test, type = "response")
log_pred = rep("No", dim(Default)[1]*0.50)
log_pred[log_prob > 0.5] = "Yes"
table(log_pred, test$default)

##         
## log_pred   No  Yes
##      No  4813  123
##      Yes   12   52

mean(log_pred !=test$default)

## [1] 0.027

train <- sample(dim(Default)[1], dim(Default)[1]*0.50)
test <- Default[-train, ]

logit_model2 <- glm(default ~ income+ balance, data = Default, family = 'binomial', subset = train)

log_prob = predict(logit_model2, test, type = "response")
log_pred = rep("No", dim(Default)[1]*0.50)
log_pred[log_prob > 0.5] = "Yes"
table(log_pred, test$default)

##         
## log_pred   No  Yes
##      No  4807  117
##      Yes   27   49

mean(log_pred !=test$default)

## [1] 0.0288

It looks like the validation set error that is calculated is different every single time.

4. Now consider a logistic regression model that predicts the prob- ability of default using income, balance, and a dummy variable for student. Estimate the test error for this model using the val- idation set approach. Comment on whether or not including a dummy variable for student leads to a reduction in the test error rate.

train <- sample(dim(Default)[1], dim(Default)[1]*0.50)
test <- Default[-train, ]

logit_model3 <- glm(default ~ income+ balance+student, data = Default, family = 'binomial', subset = train)

log_prob = predict(logit_model2, test, type = "response")
log_pred = rep("No", dim(Default)[1]*0.50)
log_pred[log_prob > 0.5] = "Yes"
table(log_pred, test$default)

##         
## log_pred   No  Yes
##      No  4802  112
##      Yes   19   67

mean(log_pred !=test$default)

## [1] 0.0262

It looks like adding the dummy variable of student to the logistic model reduces the test error rate very slighlty to 0.0252.

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 coef- ficients 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 esti- mated standard errors for the coefficients associated with income and balance in a multiple logistic regression model that uses both predictors.

logit_model4 <- glm(default ~ income + balance, data = Default, family = binomial)

summary(logit_model4)

## 
## 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

The estimated standard error for Income is 4.985e-06 , adn the estimated standard error for Balance is 2.274e-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) return(coef(glm(default ~ balance + income, 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)

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 -3.411727e-02 4.589800e-01
## t2*  5.647103e-03  1.203954e-05 2.459425e-04
## t3*  2.080898e-05  7.179969e-07 4.670885e-06

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

The estimated standard errors for the coefficients for income and balance are 2.294044e-04 and 3.875525e-06.

9. 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 μˆ.

library(MASS)
data("Boston")
summary(Boston)

##       crim                zn             indus            chas        
##  Min.   : 0.00632   Min.   :  0.00   Min.   : 0.46   Min.   :0.00000  
##  1st Qu.: 0.08205   1st Qu.:  0.00   1st Qu.: 5.19   1st Qu.:0.00000  
##  Median : 0.25651   Median :  0.00   Median : 9.69   Median :0.00000  
##  Mean   : 3.61352   Mean   : 11.36   Mean   :11.14   Mean   :0.06917  
##  3rd Qu.: 3.67708   3rd Qu.: 12.50   3rd Qu.:18.10   3rd Qu.:0.00000  
##  Max.   :88.97620   Max.   :100.00   Max.   :27.74   Max.   :1.00000  
##       nox               rm             age              dis        
##  Min.   :0.3850   Min.   :3.561   Min.   :  2.90   Min.   : 1.130  
##  1st Qu.:0.4490   1st Qu.:5.886   1st Qu.: 45.02   1st Qu.: 2.100  
##  Median :0.5380   Median :6.208   Median : 77.50   Median : 3.207  
##  Mean   :0.5547   Mean   :6.285   Mean   : 68.57   Mean   : 3.795  
##  3rd Qu.:0.6240   3rd Qu.:6.623   3rd Qu.: 94.08   3rd Qu.: 5.188  
##  Max.   :0.8710   Max.   :8.780   Max.   :100.00   Max.   :12.127  
##       rad              tax           ptratio          black       
##  Min.   : 1.000   Min.   :187.0   Min.   :12.60   Min.   :  0.32  
##  1st Qu.: 4.000   1st Qu.:279.0   1st Qu.:17.40   1st Qu.:375.38  
##  Median : 5.000   Median :330.0   Median :19.05   Median :391.44  
##  Mean   : 9.549   Mean   :408.2   Mean   :18.46   Mean   :356.67  
##  3rd Qu.:24.000   3rd Qu.:666.0   3rd Qu.:20.20   3rd Qu.:396.23  
##  Max.   :24.000   Max.   :711.0   Max.   :22.00   Max.   :396.90  
##      lstat            medv      
##  Min.   : 1.73   Min.   : 5.00  
##  1st Qu.: 6.95   1st Qu.:17.02  
##  Median :11.36   Median :21.20  
##  Mean   :12.65   Mean   :22.53  
##  3rd Qu.:16.95   3rd Qu.:25.00  
##  Max.   :37.97   Max.   :50.00

attach(Boston)
mean_medv = mean(medv)
mean_medv

## [1] 22.53281

2. Provide an estimate of the standard error of μˆ. 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.

st_mean = sd(medv)/sqrt(length(medv))
st_mean

## [1] 0.4088611

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

boot.fn2 = function(data, index) return(mean(data[index]))
boot2 = boot(medv, boot.fn2, 100)
boot2

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn2, R = 100)
## 
## 
## Bootstrap Statistics :
##     original    bias    std. error
## t1* 22.53281 0.0261581   0.3808429

4. Based on your bootstrap estimate from (c), provide a 95 % con- fidence 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 [μˆ − 2SE(μˆ), μˆ + 2SE(μˆ)].

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

CI_bos = c(22.53 - 2 * 0.4174872, 22.53 + 2 * 0.4174872)
CI_bos

## [1] 21.69503 23.36497

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

median_medv = median(medv)
median_medv

## [1] 21.2

6. We now would like to estimate the standard error of μˆmed.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.fn3 = function(data, index) return(median(data[index]))
boot3 = boot(medv, boot.fn3, 1000)
boot3

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn3, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*     21.2 -0.0082   0.3744569

7. Based on this data set, provide an estimate for the tenth per- centile of medv in Boston census tracts. Call this quantity μˆ0.1. (You can use the quantile() function.)

ten_medv = quantile(medv, c(0.1))
ten_medv

##   10% 
## 12.75

8. Use the bootstrap to estimate the standard error of μˆ0.1. Com- ment on your findings.

boot.fn4 = function(data, index) return(quantile(data[index], c(0.1)))
boot4 = boot(medv, boot.fn4, 1000)
boot4

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn4, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*    12.75  0.0045   0.4976871