#Problem 3 We now review k-fold cross-validation.
##a) Explain how k-fold cross-validation is implemented. In K-fold cross-validation, the dataset is divided into K equal subsets. The model is trained on K-1 of these subsets and tested on the remaining 1 subset. This process is repeated K times, with each of the K subsets used exactly once as the test set. The results are then averaged to provide a more robust estimate of model performance.
##b) What are the advantages and disadvantages of k-fold cross-validation relative to: i. The validation set approach? Advantage of k-fold: Better data utilization, average of results across multiple splits, reduces overfitting Disadvantages: Computationally Expensive, Validation set may be sufficient for big datasets.
#Problem 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(23)
d1 = Default
glm1 = glm(default ~ income + balance, data = d1, family = "binomial")
summary(glm1)##
## Call:
## glm(formula = default ~ income + balance, family = "binomial",
## data = d1)
##
## 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(23)
index <- sample(nrow(d1),0.7*nrow(d1),replace = FALSE)
train_d1 <- d1[index, ]
val_d1 <- d1[-index, ]set.seed(23)
glm2 = glm(default ~ income + balance, data = train_d1, family = "binomial")
summary(glm2)##
## Call:
## glm(formula = default ~ income + balance, family = "binomial",
## data = train_d1)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.189e+01 5.410e-01 -21.975 < 2e-16 ***
## income 2.798e-05 6.086e-06 4.597 4.28e-06 ***
## balance 5.686e-03 2.794e-04 20.351 < 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: 1976.0 on 6999 degrees of freedom
## Residual deviance: 1067.3 on 6997 degrees of freedom
## AIC: 1073.3
##
## Number of Fisher Scoring iterations: 8d1_prob = predict(glm2, val_d1, type = "response")
d1_pred = rep("No", nrow(val_d1))
d1_pred[d1_prob > .5] = "Yes"mean(d1_pred!=val_d1$default)## [1] 0.02733333##(c) Repeat the process in (b) three times, using three different splits of the observations into a training set and a validation set. Comment on the results obtained.
1st time
set.seed(23)
index <- sample(nrow(d1),0.5*nrow(d1),replace = F)
train_d1 <- d1[index, ]
val_d1 <- d1[-index, ]
glm2 = glm(default ~ income + balance, data = train_d1, family = "binomial")
d1_prob = predict(glm2, val_d1, type = "response")
d1_pred = rep("No", nrow(val_d1))
d1_pred[d1_prob > .5] = "Yes"
mean(d1_pred!=val_d1$default)## [1] 0.0292nd time
set.seed(23)
index <- sample(nrow(d1),0.8*nrow(d1),replace = F)
train_d1 <- d1[index, ]
val_d1 <- d1[-index, ]
glm2 = glm(default ~ income + balance, data = train_d1, family = "binomial")
d1_prob = predict(glm2, val_d1, type = "response")
d1_pred = rep("No", nrow(val_d1))
d1_pred[d1_prob > .5] = "Yes"
mean(d1_pred!=val_d1$default)## [1] 0.02453rd time
set.seed(23)
index <- sample(nrow(d1),0.9*nrow(d1),replace = F)
train_d1 <- d1[index, ]
val_d1 <- d1[-index, ]
glm2 = glm(default ~ income + balance, data = train_d1, family = "binomial")
d1_prob = predict(glm2, val_d1, type = "response")
d1_pred = rep("No", nrow(val_d1))
d1_pred[d1_prob > .5] = "Yes"
mean(d1_pred!=val_d1$default)## [1] 0.027##(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(23)
index <- sample(nrow(d1),0.9*nrow(d1),replace = F)
train_d1 <- d1[index, ]
val_d1 <- d1[-index, ]
glm2 = glm(default ~ income + balance + as.factor(student), data = train_d1, family = "binomial")
d1_prob = predict(glm2, val_d1, type = "response")
d1_pred = rep("No", nrow(val_d1))
d1_pred[d1_prob > .5] = "Yes"
mean(d1_pred!=val_d1$default)## [1] 0.027Finally by add the ‘Student’ predictor increased the test error.
#Problem 6 We continue to consider the use of a logistic regression model to predict the probability of default using income and balance on the Default data set. In particular, we will now compute estimates for the standard errors of the income and balance logistic regression coefficients in two different ways: (1) using the bootstrap, and (2) using the standard formula for computing the standard errors in the glm() function. Do not forget to set a random seed before beginning your analysis.
##(a) Using the summary() and glm() functions, determine the estimated standard errors for the coefficients associated with income and balance in a multiple logistic regression model that uses both predictors.
glm3 = glm(default ~ income + balance, data = d1, family = "binomial")
summary(glm3)$coef## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.154047e+01 4.347564e-01 -26.544680 2.958355e-155
## income 2.080898e-05 4.985167e-06 4.174178 2.990638e-05
## balance 5.647103e-03 2.273731e-04 24.836280 3.638120e-136The estimated std error for coeff is for income is 4.985e-06 and for balance is 2.274e-04.
##(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) + coef(glm(default ~ income + balance, data = d1, subset = index, family = "binomial"))set.seed(23)
boot.fn(d1, 1:10000)[-1]## income balance
## 2.080898e-05 5.647103e-03##(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(23)
boot(data = d1, statistic = boot.fn, R = 1000)##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = d1, statistic = boot.fn, R = 1000)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* -1.154047e+01 -6.425518e-03 4.252997e-01
## t2* 2.080898e-05 -4.567628e-08 4.571768e-06
## t3* 5.647103e-03 -3.014214e-07 2.228104e-04The Estimated std error is for income is 4.572e-06 and for balance is 2.228e-04.
##(d) Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function. Using the glm() function, the ESE for income is 4.985e-06 and for balance is 2.274e-04. Using the bootstrap function function, the ESE for income is income is 4.572e-06 and for balance is 2.228e-04.
Standard error was lower when using bootstrap.
#Problem 9 We will now consider the Boston housing data set, from the ISLR2 library.
bos = Boston##(a) Based on this data set, provide an estimate for the population mean of medv.
mean(bos$medv)## [1] 22.53281##(b) Provide an estimate of the standard error of the population mean of medv. Interpret this result. Hint: We can compute the standard error of the sample mean by dividing the sample standard deviation by the square root of the number of observations.
sd(bos$medv)/sqrt(length(bos$medv))## [1] 0.4088611##(c) Now estimate the standard error of the population mean of medv using the bootstrap. How does this compare to your answer from (b)?
boot.fn = function(vector, index) {mean(vector[index])}set.seed(23)
boot_res = boot(bos$medv, boot.fn, 1000)
boot_res##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = bos$medv, statistic = boot.fn, R = 1000)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 22.53281 -0.01904763 0.4000822The results have similar error but negligibly lower (0.400 vs 0.409).
##(d) Based on your bootstrap estimate from (c), provide a 95 % confidence interval for the mean of medv. Compare it to the results obtained using t.test(Boston$medv).
boot_res_sd = sd(boot_res$t)
c(mean(bos$medv) - 2*boot_res_sd, mean(bos$medv) + 2*boot_res_sd)## [1] 21.73264 23.33297t.test(bos$medv)##
## One Sample t-test
##
## data: bos$medv
## t = 55.111, df = 505, p-value < 2.2e-16
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
## 21.72953 23.33608
## sample estimates:
## mean of x
## 22.53281##(e) Based on this data set, provide an estimate, for the median value of medv in the population.
median(bos$medv)## [1] 21.2##(f) We now would like to estimate the standard error of the population median of medv. 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(23)
boot_res = boot(bos$medv, boot.fn, 1000)
boot_res##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = bos$medv, statistic = boot.fn, R = 1000)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 21.2 -0.02375 0.3647828Standard error is small
##(g) Based on this data set, provide an estimate for the tenth percentile of medv in Boston census tracts. Call this quantity mew^0.1. (You can use the quantile() function.)
quantile(bos$medv, 0.1)## 10%
## 12.75##(h) Use the bootstrap to estimate the standard error of mewˆ0.1. Comment on your findings.
boot.fn = function(vector, index) {quantile(vector[index], 0.1)}set.seed(23)
boot_res = boot(bos$medv, boot.fn, 1000)
boot_res##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = bos$medv, statistic = boot.fn, R = 1000)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 12.75 0.028 0.4964549Standard error is actually large than other bootstraps.