3a) Explain how k-fold cross-validation is implemented
Randomly dividing the set of observation into k groups. The first is treated as a validation set, and the method is fit on the remaining k - 1 folds. Mean Squared Error is then computed on the observations in the field the held-out fold.
3b) Explain advantages and disadvantages of k-fold cross-validation relative to The validation set approach? and LOOCV?
LOOCV requires fitting the statistical learning method n times which has the potential to be computationally expensive. While cross-validation is very general and can be applied to almost any statistical learning method.
5)n 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.
5a)Fit a logistic regression model that uses income and balance to predict default.
library(ISLR)## Warning: package 'ISLR' was built under R version 4.0.5summary(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. :73554glm.fit = glm(default ~ income + balance, data = Default, family = binomial)5b)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 sample set into training set and validation set. II. Fit multiple logistic regression model using only the training observations. 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. IV. Compute validation set error, which is the fraction of the observations in the validation set that are misclassified.
set.seed(1)
#I
train = sample(dim(Default)[1], dim(Default)[1]/2)
#II
glm.fit = glm(default ~ income + balance, data = Default, family = binomial)
#III
glm.pred = rep("No", dim(Default)[1]/2)
glm.probs = predict(glm.fit, Default[-train, ], type = "response")
glm.pred[glm.probs > 0.5] = "Yes"
#IV
mean(glm.pred != Default[-train, ]$default)## [1] 0.02485c)Repeat the process in (b) three times using different splits of the observations into a training set and a validation set. What are the results
Average results were about 2.6% or .026 for error rate
5d)Now consider a logistic regression model that predicts the probability of default using income, balance, and a dummy variable for student. Estimate the test error for this model using the validation set approach. Comment on whether or not including a dummy variable for student leads to a reduction in the test error rate.
attach(Default)
train2 = sample(dim(Default)[1], dim(Default)[1]/2)
glm.fit2 = glm(default ~ income + balance + student, family = binomial, subset = train)
glm.pred2 = rep("No", dim(Default)[1]/2)
glm.probs2 = predict(glm.fit2, Default[-train, ], type = "response")
glm.pred[glm.probs2 > .5] = "Yes"
mean(glm.pred2 != Default[-train, ]$default)## [1] 0.0314detach(Default)With the validation approach the test error rate is 3.14% with the student variable, it appears to make the error rate larger.
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.
6a)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.
attach(Default)
set.seed(1)
glm.fit3 = glm(default ~ income + balance, family = binomial)
summary(glm.fit3)##
## Call:
## glm(formula = default ~ income + balance, family = binomial)
##
## 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: 82.08e and 5.64e
6b)Write boot.fn(), that takes the input 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)))6c)Use boot() function together with your boot.fn() to estimate standard errors of the logistic regression coefficients for income and balance
library(boot)
boot(Default, boot.fn, 50)##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = Default, statistic = boot.fn, R = 50)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* -1.154047e+01 -5.661486e-02 4.847786e-01
## t2* 2.080898e-05 -7.436578e-08 4.456965e-06
## t3* 5.647103e-03 1.854126e-05 2.639029e-04detach(Default)6d)Comment on the estimated standard errors obtained using the glm() function & using your bootstrap function. The glm and bootstrap are both the same or very similar 2.08 and 5.64 respectively on both.
9)Now we consider the Boston housing data set, from the MASS library.
library(MASS)
library(boot)
attach(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. :50.009a)Based on this data set, provide an estimate for the population mean of medv. Call this estimate μ
medv.mean = mean(medv)
medv.mean## [1] 22.53281medv mean is 22.53
9b)Provide an estimate of the standard error of μ. Interpret this result.
medv.err = sd(medv)/sqrt(length(medv))22.53 standard error
9c)Now estimate standard error of μ using bootstrap. How does this compare from answer b?
boot.fn2 = function(data, index) return(mean(data[index]))
bstrap = boot(medv, boot.fn2, 1000)
bstrap##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = medv, statistic = boot.fn2, R = 1000)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 22.53281 0.01485415 0.3983795It appears to have grown in standard error significantly.
9d)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$medv)
t.test(medv)##
## One Sample t-test
##
## data: medv
## t = 55.111, df = 505, p-value < 2.2e-16
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
## 21.72953 23.33608
## sample estimates:
## mean of x
## 22.53281c(bstrap$t0 - 2 * 0.3944303, bstrap$t0 + 2 * 0.3944303)## [1] 21.74395 23.32167About 3 points different from the t test and bootstrap.
9e)Based on this data set, provide an estimate, ˆμmed, for the median value of medv in the population.
medv.med = median(medv)
medv.med## [1] 21.29f)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.fn3 = function(data, index) return(median(data[index]))
boot(medv, boot.fn3, 1000)##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = medv, statistic = boot.fn3, R = 1000)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 21.2 -0.0071 0.3765852The median is 21.2 with the stdnard error being .386
9g)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.)
medv.10th = quantile(medv, c(.1))
medv.10th## 10%
## 12.759h)Use the bootstrap to estimate the std error of μ. Comment on findings.
boot.fn4 = function(data, index) return(quantile(data[index], c(0.1)))
boot(medv, boot.fn4, 1000)##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = medv, statistic = boot.fn4, R = 1000)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 12.75 0.00355 0.511695