Matthew Westley Ch5 pg198: 3, 5, 6, 9
library(ISLR)(a) Explain how k-fold cross-validation is implemented.
Observations are randomly divided into k groups of approximately equal size. The first fold is treated as a validation set and the method is fit on the remaining k-1 groups. The mean squared error (MSE) is then computed on the observations of the held out groups. The procedure is repeated k times, and each time a different group of observations is treated as the validation set.
(b) What are the advantages and disadvantages of k-fold cross-validation relative to:
i. The validation set approach?
An advantage to k-fold CV over the validation set approach is that k-fold CV is less likely to overestimate the test error rate than the validation set approach. A disadvantage to k-fold CV is that it is more complicated to implement than the validation set approach.
ii. LOOCV?
The advantages of k-fold cross-validation relative to LOOCV is that LOOCV can be much more computationally expensive if n is very large. Another advantage is that k-fold CV often gives more accurate estimates of the test error rate. A disadvantage is that k-fold CV has more bias than LOOCV.
(a) Fit a logistic regression model that uses income and balance to predict default.
attach(Default)
set.seed(1)
dfault.glm <- glm(default~income + balance, data = Default, family = "binomial")
summary(dfault.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 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.
Val.GLM <- glm(default~income + balance, data = Default, family = "binomial", subset = train)
summary(Val.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 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.
prob.def <- predict(Val.GLM, newdata = Default[-train, ], type = "response")
pred.glm <- rep("No", length(prob.def))
pred.glm[prob.def > 0.5] <- "Yes"iv. Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified.
The validation set error is 0.0254
mean(pred.glm != Default[-train, ]$default)## [1] 0.0254(c) Repeat the process in (b) three times, using three different splits of the observations into a training set and a validation set. Comment on the results obtained.
The three different splits of the observations result in three different error rates, which indicates that the error rate changes depending on which observations are included in the two sets.
train <- sample(dim(Default)[1], dim(Default)[1] / 2)
Val.GLM <- glm(default~income + balance, data = Default, family = "binomial", subset = train)
prob.def <- predict(Val.GLM, newdata = Default[-train, ], type = "response")
pred.glm <- rep("No", length(prob.def))
pred.glm[prob.def > 0.5] <- "Yes"
mean(pred.glm != Default[-train, ]$default)## [1] 0.0274train <- sample(dim(Default)[1], dim(Default)[1] / 2)
Val.GLM <- glm(default~income + balance, data = Default, family = "binomial", subset = train)
prob.def <- predict(Val.GLM, newdata = Default[-train, ], type = "response")
pred.glm <- rep("No", length(prob.def))
pred.glm[prob.def > 0.5] <- "Yes"
mean(pred.glm != Default[-train, ]$default)## [1] 0.0244train <- sample(dim(Default)[1], dim(Default)[1] / 2)
Val.GLM <- glm(default~income + balance, data = Default, family = "binomial", subset = train)
prob.def <- predict(Val.GLM, newdata = Default[-train, ], type = "response")
pred.glm <- rep("No", length(prob.def))
pred.glm[prob.def > 0.5] <- "Yes"
mean(pred.glm != Default[-train, ]$default)## [1] 0.0244(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. The test error using the dummy variable for student improved a little.
train <- sample(dim(Default)[1], dim(Default)[1] / 2)
Val.GLM <- glm(default~income + balance + student, data = Default, family = "binomial", subset = train)
prob.def <- predict(Val.GLM, newdata = Default[-train, ], type = "response")
pred.glm <- rep("No", length(prob.def))
pred.glm[prob.def > 0.5] <- "Yes"
mean(pred.glm != Default[-train, ]$default)## [1] 0.0278(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. The estimated standar errors for the coefficients associated with income is 4.985e-06 and 2.274e-04 for balance.
set.seed(1)
attach(Default)## The following objects are masked from Default (pos = 3):
##
## balance, default, income, studentsix.GLM <- glm(default~income + balance, data = Default, family = "binomial")
summary(six.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) 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) {
bt.fit <- glm(default~income + balance, data = data, family = "binomial", subset = index)
return(coef(bt.fit))
}(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.
The bootstrap function estimates of the standard errors of the logistic regression coefficients for income is 4.866284e-06 and 2.298949e-04 for balance.
library(boot)
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 obtained using the glm function are pretty close the estimated standard errors obtained from the bootsrap function.
(a) Based on this data set, provide an estimate for the population mean of medv. Call this estimate μˆ.
μˆ= 22.53281
library(MASS)
attach(Boston)
μˆ <- mean(medv)
μˆ## [1] 22.53281(b) Provide an estimate of the standard error of μˆ. 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.
Standard error of μˆ is 0.4088611
sterr <- sd(medv) / sqrt(dim(Boston)[1])
sterr## [1] 0.4088611(c) Now estimate the standard error of μˆ using the bootstrap. How does this compare to your answer from (b)?
The standard errors from the two methods are fairly close.
set.seed(1)
boot.fn <- function(data, index) {
μ <- mean(data[index])
return(μ)
}
boot(medv, boot.fn, 1000)##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = medv, statistic = boot.fn, R = 1000)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 22.53281 0.007650791 0.4106622(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 [μˆ − 2SE(μˆ), μˆ + 2SE(μˆ)]. The confidence interval from the t test (21.72953-23.33608) and from the bootstrap estimate (21.71149-23.35413) are pretty close.
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
## 22.53281conf.int <- c(22.53281 - 2 * 0.4106622, 22.53281 + 2 * 0.4106622)
conf.int## [1] 21.71149 23.35413(e) Based on this data set, provide an estimate, μˆmed, for the median value of medv in the population.
21.2 is the median value of medv
med <- median(medv)
med## [1] 21.2(f) We now would like to estimate the standard error of μˆmed.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.
The standard error of the median is 0.3770241, which is lower than the standard error 0.4088611 of the mean.
boot.fn <- function(data, index) {
μ <- median(data[index])
return(μ)
}
boot(medv, boot.fn, 1000)##
## 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 per- centile of medv in Boston suburbs. Call this quantity μˆ0.1. (You can use the quantile() function.)
12.75
ten.per <- quantile(medv, c(0.1))
ten.per## 10%
## 12.75(h) Use the bootstrap to estimate the standard error of μˆ0.1. Comment on your findings.
The standard error for μˆ0.1 (0.4925766) is much higher than standard error or the mean (0.4088611) or median (0.3770241).
boot.fn <- function(data, index) {
μ <- quantile(data[index], c(0.1))
return(μ)
}
boot(medv, boot.fn, 1000)##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = medv, statistic = boot.fn, R = 1000)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 12.75 0.0186 0.4925766