3. We now review k-fold cross-validation. (a) Explain how k-fold cross-validation is implemented.
K-fold is dividing sets of observation randomly into K groups.All groups will equal in size and non are overlapped. These are considered validation sets.
(b) What are the advantages and disadvantages o fk-fold cross-validation relative to: i. The validation set approach? Advantage: It takes up less resources during computation. Disadvantage: It is usually constrained to k=5 or k=10 which can lead to more bias. ii. LOOCV? Advantage: It will run through N validation. Disadvantage: It can use a lot of computational resources.
5. In Chapter4, 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.
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. :73554attach(Default)
set.seed(1)
glm.fit = glm(default ~ income + balance, data = Default, family = binomial)
summary(glm.fit)##
## 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 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.
train <- sample(dim(Default)[1], dim(Default)[1] / 2)ii. Fit a multiple logistic regression model using only the training observations.
fit.glm <- glm(default ~ income + balance, data = Default, family = "binomial", subset = train)
summary(fit.glm)##
## Call:
## glm(formula = default ~ income + balance, family = "binomial",
## data = Default, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.5830 -0.1428 -0.0573 -0.0213 3.3395
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.194e+01 6.178e-01 -19.333 < 2e-16 ***
## income 3.262e-05 7.024e-06 4.644 3.41e-06 ***
## balance 5.689e-03 3.158e-04 18.014 < 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: 1523.8 on 4999 degrees of freedom
## Residual deviance: 803.3 on 4997 degrees of freedom
## AIC: 809.3
##
## Number of Fisher Scoring iterations: 8iii. Obtain a prediction of default status for each individual inthe 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 greaterthan 0.5.
probs <- predict(fit.glm, newdata = Default[-train, ], type = "response")
pred.glm <- rep("No", length(probs))
pred.glm[probs > 0.5] <- "Yes"iv. Compute the validation set error, which is the fraction ofthe observations in the validation set that are misclassified.
mean(pred.glm != Default[-train, ]$default)## [1] 0.0254(c) Repeat the process in (b) three times, using three different splitsof the observations into a training set and a validation set. Comment on the results obtained.
train <- sample(dim(Default)[1], dim(Default)[1] / 2)
fit.glm <- glm(default ~ income + balance, data = Default, family = "binomial", subset = train)
probs <- predict(fit.glm, newdata = Default[-train, ], type = "response")
pred.glm <- rep("No", length(probs))
pred.glm[probs > 0.5] <- "Yes"
mean(pred.glm != Default[-train, ]$default)## [1] 0.0274train <- sample(dim(Default)[1], dim(Default)[1] / 2)
fit.glm <- glm(default ~ income + balance, data = Default, family = "binomial", subset = train)
probs <- predict(fit.glm, newdata = Default[-train, ], type = "response")
pred.glm <- rep("No", length(probs))
pred.glm[probs > 0.5] <- "Yes"
mean(pred.glm != Default[-train, ]$default)## [1] 0.0244train <- sample(dim(Default)[1], dim(Default)[1] / 2)
fit.glm <- glm(default ~ income + balance, data = Default, family = "binomial", subset = train)
probs <- predict(fit.glm, newdata = Default[-train, ], type = "response")
pred.glm <- rep("No", length(probs))
pred.glm[probs > 0.5] <- "Yes"
mean(pred.glm != Default[-train, ]$default)## [1] 0.0244Since the training observations are different each time the experiment is ran, all the results are different.
(d) 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 validation set approach. Comment on whether or not including a dummy variable for student leads to a reduction in the test error rate.
fit.glm <- glm(default ~ income + balance + student, data = Default, family = "binomial", subset = train)
pred.glm <- rep("No", length(probs))
probs <- predict(fit.glm, newdata = Default[-train, ], type = "response")
pred.glm[probs > 0.5] <- "Yes"
mean(pred.glm != Default[-train, ]$default)## [1] 0.0242It is still producing around the same random rate as the other four. I would think that student doesn’t effect the model very much if at all.
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.
library(ISLR)
attach(Default)## The following objects are masked from Default (pos = 3):
##
## balance, default, income, studentsummary(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(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.fit=glm(default ~ income + balance,data = Default, family = "binomial", subset = train)
summary(glm.fit)##
## Call:
## glm(formula = default ~ income + balance, family = "binomial",
## data = Default, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.4027 -0.1517 -0.0624 -0.0233 3.6833
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.112e+01 5.816e-01 -19.120 <2e-16 ***
## income 1.638e-05 6.755e-06 2.425 0.0153 *
## balance 5.489e-03 3.067e-04 17.897 <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: 1503.85 on 4999 degrees of freedom
## Residual deviance: 831.51 on 4997 degrees of freedom
## AIC: 837.51
##
## Number of Fisher Scoring iterations: 8summary(glm.fit)$coef## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.111950e+01 5.815653e-01 -19.119960 1.722380e-81
## income 1.638285e-05 6.755288e-06 2.425188 1.530044e-02
## balance 5.489053e-03 3.067099e-04 17.896567 1.254196e-71(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)
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, 100)##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = Default, statistic = boot.fn, R = 100)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* -1.154047e+01 8.556378e-03 4.122015e-01
## t2* 2.080898e-05 -3.993598e-07 4.186088e-06
## t3* 5.647103e-03 -4.116657e-06 2.226242e-04(d) Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function.
GLM: INcome: 7.132648e-06 GLM: Balance: 3.191188e-04 Boot fn: Income: 4.122015e-01 Boot fn: Balance 2.226242e-04
9. We will now consider the Boston housing data set, from the MASS library.
library(MASS)
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(a) Based on this data set, provide an estimate for the population mean of medv. Call this estimate ˆμ.
set.seed(1)
attach(Boston)
medv.mean=mean(medv)
medv.mean## [1] 22.53281(b) Provide an estimate of the standard error of ˆμ.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.
stderror = sd(medv) / sqrt(length(medv))
cat("mu std error: " , stderror)## mu std error: 0.4088611(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)
return(mean(data[index]))
boot.mean = boot(medv, boot.fn, 1000)
boot.mean##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = medv, statistic = boot.fn, R = 1000)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 22.53281 0.007650791 0.4106622There is only 1% error difference between bootstrap and the square root.
(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$medv). Hint: You can approximate a 95 % confidence interval using the formula[ˆμ−2SE(ˆμ),ˆμ+2SE(ˆμ)].
t.test(Boston$medv)##
## One Sample t-test
##
## data: Boston$medv
## t = 55.111, df = 505, p-value < 2.2e-16
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
## 21.72953 23.33608
## sample estimates:
## mean of x
## 22.53281c(boot.mean$t0 - 2 * 0.4119, boot.mean$t0 + 2 * 0.4119)## [1] 21.70901 23.35661t-test: 21.73 - 23.34
bootstrap: 21.71 - 23.36
(e) Based on this data set, provide an estimate, ˆμ med,for the median value of medv in the population.
mu.median = median(medv)
cat("estimated pop median of medv: ", mu.median)## estimated pop median of medv: 21.2(f) We now would like to estimate the standard error of ˆμ med. 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(data, index)
return(median(data[index]))
boot.median = boot(medv, boot.fn, 1000)
boot.median##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = medv, statistic = boot.fn, R = 1000)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 21.2 -0.0386 0.3770241(g) Based on this data set, provide an estimate for the tenth percentile of medv in Boston suburbs. Call this quantity ˆμ 0.1.(You can use the quantile() function.)
medv.tenth = quantile(medv, c(0.1))
medv.tenth## 10%
## 12.75(h) Use the bootstrap to estimate the standard error of ˆμ 0.1. Comment on your findings.
boot.fn = function(data, index)
return( quantile(data[index],p=0.1))
boot.median = boot(medv, boot.fn, 1000)
boot.median##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = medv, statistic = boot.fn, R = 1000)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 12.75 0.0186 0.4925766In the first 10% of the data set it contains a std error rate of 50% using bootstrap.