R Notebook

#CHAPTER 5 #Question 3 ##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-fold CV 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.

###(b) What are the advantages and disadvantages of k-fold crossvalidation relative to: ####i. The validation set approach? PROS: Less variance in test error estimates since all observations are used for both training and validation across folds. PROS: Makes better use of data, especially when sample size is small. CONS: High variance — test error estimate depends on how data is split. CONS: Only a subset of the data is used for training, potentially underfitting the model. ####ii. LOOCV? PROS:Much faster to compute, especially for complex models (since only k fits vs n fits). PROS:Can have lower variance than LOOCV when 𝑘<𝑛k<n. PROS: Nearly unbiased estimate of the true test error (uses almost all data for training in each iteration). CONS: But high variance: each fold is similar and thus errors can be highly correlated.

library(ISLR2)

Warning: package ‘ISLR2’ was built under R version 4.4.2

set.seed(1)
train <- sample(392, 196)

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

model = glm(default ~ income + balance, data = Default, family = binomial)
summary(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

###(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.

set.seed(69)
train = sample(x = c(TRUE, FALSE), size = nrow(Default), replace = TRUE)

train_data = Default[train, ]
validation_data = Default[!train, ]

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

model_train = glm(default ~ income + balance, data = train_data, family = binomial)
summary(model_train)

Call:
glm(formula = default ~ income + balance, family = binomial, 
    data = train_data)

Coefficients:
              Estimate Std. Error z value Pr(>|z|)    
(Intercept) -1.118e+01  5.774e-01 -19.361  < 2e-16 ***
income       2.070e-05  6.595e-06   3.139  0.00169 ** 
balance      5.488e-03  3.002e-04  18.283  < 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: 1570.66  on 5010  degrees of freedom
Residual deviance:  853.82  on 5008  degrees of freedom
AIC: 859.82

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.

preds = predict(model_train, newdata = validation_data, type =  "response")
pred_class = ifelse(preds > 0.5, "Yes", "No")
pred_class = factor(pred_class, levels = levels(Default$default))

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

validatino_error = mean(pred_class != validation_data$default)
validatino_error

[1] 0.0244538

###(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.

set.seed(69)
train = sample(x = c(TRUE, FALSE), size = nrow(Default), replace = TRUE, prob = c(0.7, 0.3))

train_data = Default[train, ]
validation_data = Default[!train, ]
model_train = glm(default ~ income + balance, data = train_data, family = binomial)
summary(model_train)

Call:
glm(formula = default ~ income + balance, family = binomial, 
    data = train_data)

Coefficients:
              Estimate Std. Error z value Pr(>|z|)    
(Intercept) -1.164e+01  5.199e-01 -22.391  < 2e-16 ***
income       2.510e-05  5.940e-06   4.225 2.39e-05 ***
balance      5.629e-03  2.695e-04  20.891  < 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: 2047.2  on 7049  degrees of freedom
Residual deviance: 1111.8  on 7047  degrees of freedom
AIC: 1117.8

Number of Fisher Scoring iterations: 8

preds = predict(model_train, newdata = validation_data, type =  "response")
pred_class = ifelse(preds > 0.5, "Yes", "No")
pred_class = factor(pred_class, levels = levels(Default$default))
validatino_error = mean(pred_class != validation_data$default)
validatino_error

[1] 0.02610169

this raises the error to 2.61% up from the 2.44% in the 50/50 split

set.seed(69)
train = sample(x = c(TRUE, FALSE), size = nrow(Default), replace = TRUE, prob = c(0.8, 02))

train_data = Default[train, ]
validation_data = Default[!train, ]

model_train = glm(default ~ income + balance, data = train_data, family = binomial)
summary(model_train)

Call:
glm(formula = default ~ income + balance, family = binomial, 
    data = train_data)

Coefficients:
              Estimate Std. Error z value Pr(>|z|)    
(Intercept) -1.123e+01  7.967e-01  -14.09   <2e-16 ***
income       8.908e-06  9.278e-06    0.96    0.337    
balance      5.709e-03  4.280e-04   13.34   <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: 864.20  on 2818  degrees of freedom
Residual deviance: 458.12  on 2816  degrees of freedom
AIC: 464.12

Number of Fisher Scoring iterations: 8

preds = predict(model_train, newdata = validation_data, type =  "response")
pred_class = ifelse(preds > 0.5, "Yes", "No")
pred_class = factor(pred_class, levels = levels(Default$default))
validatino_error = mean(pred_class != validation_data$default)
validatino_error

[1] 0.02631945

this raises the error to 2.63% up from the 2.44% in the 50/50 and is the highest error rate of the four models

set.seed(69)
train = sample(x = c(TRUE, FALSE), size = nrow(Default), replace = TRUE, prob = c(0.6, 0.4))

train_data = Default[train, ]
validation_data = Default[!train, ]

model_train = glm(default ~ income + balance, data = train_data, family = binomial)
summary(model_train)

Call:
glm(formula = default ~ income + balance, family = binomial, 
    data = train_data)

Coefficients:
              Estimate Std. Error z value Pr(>|z|)    
(Intercept) -1.172e+01  5.568e-01 -21.050  < 2e-16 ***
income       2.540e-05  6.321e-06   4.018 5.86e-05 ***
balance      5.697e-03  2.889e-04  19.723  < 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: 1826.17  on 6078  degrees of freedom
Residual deviance:  969.66  on 6076  degrees of freedom
AIC: 975.66

Number of Fisher Scoring iterations: 8

preds = predict(model_train, newdata = validation_data, type =  "response")
pred_class = ifelse(preds > 0.5, "Yes", "No")
pred_class = factor(pred_class, levels = levels(Default$default))
validatino_error = mean(pred_class != validation_data$default)
validatino_error

[1] 0.02499362

this raises the error to 2.49% up from the 2.44% in the 50/50 and is the second lowest error rate. What is notable is the 50/50 split had the lowest error rate and as we change the train and validation sets, the more lopsided they get the worse the error rate gets

###(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.

set.seed(69)
train = sample(x = c(TRUE, FALSE), size = nrow(Default), replace = TRUE)

train_data = Default[train, ]
validation_data = Default[!train, ]

model_train = glm(default ~ income + balance + as.factor(student), data = train_data, family = binomial)
summary(model_train)

Call:
glm(formula = default ~ income + balance + as.factor(student), 
    family = binomial, data = train_data)

Coefficients:
                        Estimate Std. Error z value Pr(>|z|)    
(Intercept)           -1.064e+01  6.549e-01 -16.248   <2e-16 ***
income                 6.487e-06  1.085e-05   0.598   0.5501    
balance                5.569e-03  3.069e-04  18.145   <2e-16 ***
as.factor(student)Yes -5.326e-01  3.224e-01  -1.652   0.0985 .  
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 1570.66  on 5010  degrees of freedom
Residual deviance:  851.11  on 5007  degrees of freedom
AIC: 859.11

Number of Fisher Scoring iterations: 8

preds = predict(model_train, newdata = validation_data, type =  "response")
pred_class = ifelse(preds > 0.5, "Yes", "No")
pred_class = factor(pred_class, levels = levels(Default$default))
validatino_error = mean(pred_class != validation_data$default)
validatino_error

[1] 0.02485468

including the student variable gave me an error rate inbetween the 50/50 split and the 60/40 split. But also the income p value changed wildly.

#Question 6 ##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.

model = glm(default ~ income + balance, data = Default, family = binomial)
summary(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

the SE for income is .4348 and the SE for balance is .0002274

###(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.

library(boot)

boot_fn = function(data, index) {
  fit = glm(default ~ income + balance, data = data[index, ], family = binomial())
  return(coef(fit)[c("income", "balance")])
}

####(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.

set.seed(420)
boot_results = boot(data = Default, statistic = boot_fn, R = 1000)

boot_results

ORDINARY NONPARAMETRIC BOOTSTRAP


Call:
boot(data = Default, statistic = boot_fn, R = 1000)


Bootstrap Statistics :
        original       bias     std. error
t1* 2.080898e-05 2.327731e-07 4.790983e-06
t2* 5.647103e-03 1.849025e-05 2.244517e-04

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

the bootstrap standard errors are very close to the glm() standard errors. which means the glm() is a good fit model. the low bias means it’s not overly sensitive to sampling variability. and using the bootstrap method is a good data=driven check and is hlepful in small samples. Basically, the bootstrap method confirms the reliability of the SE’s obtained from the glm().

#Question 9 ##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.

mu_hat = mean(Boston$medv)
mu_hat

[1] 22.53281

the meadian home value is $22,532

###(b) Provide an estimate of the standard error of ˆμ. Interpret thisresult. 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.

s = sd(Boston$medv)
n = nrow(Boston)
se_mu = s / sqrt(n)
se_mu

[1] 0.4088611

The standard error (SE) of 0.408 means that if you repeatedly took random samples of the same size from the population and computed the mean each time, the standard deviation of those sample means would be about 0.41.

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

boot_fn = function(data, index) {
  mean(data[index])
}

set.seed(420)
boot_mu = boot(
  data = Boston$medv, statistic = boot_fn, R = 1000
)

boot_mu

ORDINARY NONPARAMETRIC BOOTSTRAP


Call:
boot(data = Boston$medv, statistic = boot_fn, R = 1000)


Bootstrap Statistics :
    original     bias    std. error
t1* 22.53281 0.01103281   0.4196645

the SE went up slightly from .409 to .419, which suggests the distribution of the mean is close to normal, and the bootstrap confirms our previous result

###(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)].

boot_ci

[1] 21.69348 23.37214

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

mu_med_hat = median(Boston$medv)
mu_med_hat

[1] 21.2

the median home value is $21,200

###(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_median_fn = function(data, index) {
  median(data[index])
}

set.seed(420)
boot_median = boot(
  data = Boston$medv,
  statistic = boot_median_fn,
  R = 1000
)

boot_median

ORDINARY NONPARAMETRIC BOOTSTRAP


Call:
boot(data = Boston$medv, statistic = boot_median_fn, R = 1000)


Bootstrap Statistics :
    original  bias    std. error
t1*     21.2 0.00235   0.3802059

se_median = sd(boot_median$t)
se_median

[1] 0.3802059

the estimated SE is .3802059 or a sample medians would vary about $380

###(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.)

quantile(Boston$medv, probs = 0.1)

  10% 
12.75 

the 10th percential of the median value is about $12,750 and shows us what the lower end of the housing market looks like

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

boot_quantile_fn = function(data, index) {
  quantile(data[index], probs = 0.1)
}

set.seed(420)
boot_q10 <- boot(
  data = Boston$medv,
  statistic = boot_quantile_fn,
  R = 1000
)

boot_q10

ORDINARY NONPARAMETRIC BOOTSTRAP


Call:
boot(data = Boston$medv, statistic = boot_quantile_fn, R = 1000)


Bootstrap Statistics :
    original  bias    std. error
t1*    12.75 0.00985   0.5323616

the SE for the 10th percential is .5323616 or taking various samples, the cost would change by about $532 and the small SE means the lower quantile is fairly stable.