Exercise 3

We now review k-fold cross-validation.

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

This approach involves randomly dividing the set of observations into k groups, or folds, of approximately equal size. The first fold is treated as a validation set, and the method is fit on the remaining k − 1 folds. The mean squared error, MSE1, is then computed on the observations in the held-out fold. This procedure is repeated k times; each time, a different group of observations is treated as a validation set. This process results in k estimates of the test error,MSE1, MSE2,…, MSEk. The k-fold CV estimate is computed by averaging these values.

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

The validation set approach?

Advantages:The validation set approach is conceptually simple and is easy to implement

Disadvantages: The validation MSE can be highly variable and Only a subset of observations are used to fit the model (training data).

LOOCV?

Advantages: LOOCV have less bias. The validation approach produces different MSE when applied repeatedly due to randomness in the splitting process, while performing LOOCV multiple times will always yield the same results, because we split based on 1 obs. each time.

Disadvantage: LOOCV is computationally intensive.

library(ISLR2)
library(MASS)

## 
## Attaching package: 'MASS'

## The following object is masked from 'package:ISLR2':
## 
##     Boston

library(boot)

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.

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

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

## 
## 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 thismodel. 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.

#i
train=sample(dim(Default)[1], dim(Default)[1]*0.50)
test=Default[-train, ]

#ii
glm.fit=glm(default ~ income + balance, data = Default, family = binomial, subset = train)

#iii
glm.probs = predict(glm.fit, test, type = "response")
glm.pred = rep("No", dim(Default)[1]*0.50)
glm.pred[glm.probs > 0.5] = "Yes"
table(glm.pred, test$default)

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

#iv
mean(glm.pred != test$default)

## [1] 0.0254

(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.

#i
train=sample(dim(Default)[1], dim(Default)[1]*0.50)
test=Default[-train, ]

#ii
glm.fit1=glm(default ~ income + balance, data = Default, family = binomial, subset = train)

#iii
glm.probs1 = predict(glm.fit1, test, type = "response")
glm.pred1 = rep("No", dim(Default)[1]*0.50)
glm.pred1[glm.probs1 > 0.5] = "Yes"
table(glm.pred1, test$default)

##          
## glm.pred1   No  Yes
##       No  4813  112
##       Yes   25   50

#iv
mean(glm.pred1 != test$default)

## [1] 0.0274

#i
train=sample(dim(Default)[1], dim(Default)[1]*0.50)
test=Default[-train, ]

#ii
glm.fit2=glm(default ~ income + balance, data = Default, family = binomial, subset = train)

#iii
glm.probs2 = predict(glm.fit2, test, type = "response")
glm.pred2 = rep("No", dim(Default)[1]*0.50)
glm.pred2[glm.probs2 > 0.5] = "Yes"
table(glm.pred2, test$default)

##          
## glm.pred2   No  Yes
##       No  4827  103
##       Yes   19   51

#iv
mean(glm.pred2 != test$default)

## [1] 0.0244

#i
train=sample(dim(Default)[1], dim(Default)[1]*0.50)
test=Default[-train, ]

#ii
glm.fit3=glm(default ~ income + balance, data = Default, family = binomial, subset = train)

#iii
glm.probs3 = predict(glm.fit3, test, type = "response")
glm.pred3 = rep("No", dim(Default)[1]*0.50)
glm.pred3[glm.probs3 > 0.5] = "Yes"
table(glm.pred3, test$default)

##          
## glm.pred3   No  Yes
##       No  4824  106
##       Yes   16   54

#iv
mean(glm.pred3 != test$default)

## [1] 0.0244

All three different splits have the different testing error that means the observations are different for each training set and validation set.

(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.

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

glm.fit4=glm(default ~ income + balance + student, data = Default, family = binomial, subset = train)

glm.probs4 = predict(glm.fit4, test, type = "response")
glm.pred4 = rep("No", dim(Default)[1]*0.50)
glm.pred4[glm.probs4 > 0.5] = "Yes"
table(glm.pred4, test$default)

##          
## glm.pred4   No  Yes
##       No  4808  121
##       Yes   18   53

mean(glm.pred4 != test$default)

## [1] 0.0278

Including a student dummy variable does not lead to a reduction in the test error rate.

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.fit5=glm(default ~ income + balance, data = Default, family = "binomial")
summary(glm.fit5)

## 
## 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) 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)))
}

(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

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

The estimated standard errors for the coefficients obtained using the glm() for income and balance are 4.985e-06 and 2.274e-04, respectively.

The estimated standard errors for the coefficients obtained using the bootstrap function for income and balance are 4.988746e-06 and 2.347490e-04, respectively.

Exercise 9

We will now consider the `Boston` housing data set, from the `ISLR2` library.

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)

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

mean.medv = mean(medv)
mean.medv

## [1] 22.53281

(b) Provide an estimate of the standard error of ˆµ. Interpret this result.

stderr.medv = sd(medv) / sqrt(dim(Boston)[1])
stderr.medv

## [1] 0.4088611

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

boot.fn1 <- function(data, index){
    return (mean(data[index]))
}

boot(medv, boot.fn1, 1000)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn1, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original    bias    std. error
## t1* 22.53281 0.0109415    0.414028

It is very closed to the estimation 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)`.

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 = c(22.53281 - 2 * 0.4064926, 22.53281 + 2 * 0.4064926)
CI

## [1] 21.71982 23.34580

t.test confidence interval is simmilar to the bootstap confidence interval.

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

median.medv = median(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.

boot.fn2 <- function(data, index) {
    return (median(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*     21.2 -0.02095   0.3707169

The median value is 21.2 which is similar from part(e), and the standard error is 0.3705591.

(g) Based on this data set, provide an estimate for the tenth percentile of `medv` in Boston census tracts. Call this quantity ˆµ0.1. (You can use the `quantile()` function.)

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

##   10% 
## 12.75

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

boot.fn3 <- function(data, index) {
    return (quantile(data[index], c(0.1)))
}
boot(medv, boot.fn3, 1000)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn3, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*    12.75 0.02795   0.5077737

The estimate for the tenth percentile of medv is 12.75, which is the same with part(g), and the standard error is 0.4842804.

Assignment #4 - Chapter 5

Tien Vo

2023-03-21

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:

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 thismodel. In order to do this, you must perform the following steps:

(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`.

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

Exercise 9

We will now consider the `Boston` housing data set, from the `ISLR2` 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.

(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)`.

(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 census tracts. 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.

Assignment #4 - Chapter 5

Tien Vo

2023-03-21

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:

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 thismodel. In order to do this, you must perform the following steps:

(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.

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

Exercise 9

We will now consider the Boston housing data set, from the ISLR2 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.

(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).

(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 census tracts. 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.

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

(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`.

We will now consider the `Boston` housing data set, from the `ISLR2` library.

(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)`.

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

(g) Based on this data set, provide an estimate for the tenth percentile of `medv` in Boston census tracts. Call this quantity ˆµ0.1. (You can use the `quantile()` function.)