3. We now review k-fold cross-validation.
(a) Explain how k-fold cross-validation is implemented.
The process of K-fold cross validation will divide the set of observations into 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
(b) What are the advantages and disadvantages of k-fold cross- validation relative to:
i. The validation set approach?
Advantages
The advantage is that is much less computationally demanding for common values of k
Disadvantages
The validation MSE can be highly variable . Also only a subset of observation are used to fit the model (training data). statistical methods tend to to perform worse when trained on fewer observations
ii. LOOCV?
Advantages
LOOCV has couple advantages over the validation set approach. The first one being that it has less bias since we fit the statistical learning methods using training data that contains n - 1 observations , almost all of the data set is used. In contrast to the validation set approach which the training set is around half the size of the original data set. The second advantage of LOOCV is that it produces a less variable MSE. In other words while performing LOOCV multiple times will always yield the same results
Disadvantages
The disadvantage that LOOCV has is that its computationally intensive since we are fitting a model n times
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.
library(ISLR)
set.seed(1)
attach(Default)
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.
glm.fit=glm(default ~ income + balance, family='binomial', data = Default, subset = train)
summary(glm.fit)##
## Call:
## glm(formula = default ~ income + balance, family = "binomial",
## data = Default, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.5830 -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.
glm.probs = predict(glm.fit, data = Default[-train, ], type = "response")
glm.pred = rep("No", length(glm.probs))
glm.pred[glm.probs > 0.5] = "Yes"iv. Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified
There is a Test Error rate of 4.76% using the validation approach
mean(glm.pred != Default[-train, 'default'])## [1] 0.0476(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.
1 Test error rate of 4.7 %
train=sample(dim(Default)[1],dim(Default)[1]/2)
glm.fit=glm(default~income+balance, family='binomial', data = Default, subset = train)
glm.probs = predict(glm.fit, data = Default[-train, ], type = "response")
glm.pred = rep("No", length(glm.probs))
glm.pred[glm.probs > 0.5] = "Yes"
mean(glm.pred!= Default[-train, ]$default)## [1] 0.0472 Test error rate of 4.56%
train=sample(dim(Default)[1],dim(Default)[1]/2)
glm.fit=glm(default~income+balance, family='binomial', data = Default, subset = train)
glm.probs = predict(glm.fit, data = Default[-train, ], type = "response")
glm.pred = rep("No", length(glm.probs))
glm.pred[glm.probs > 0.5] = "Yes"
mean(glm.pred!= Default[-train, ]$default)## [1] 0.04563 Test error rate of 4.66%
train=sample(dim(Default)[1],dim(Default)[1]/2)
glm.fit=glm(default~income+balance, family='binomial', data = Default, subset = train)
glm.probs = predict(glm.fit, data = Default[-train, ], type = "response")
glm.pred = rep("No", length(glm.probs))
glm.pred[glm.probs > 0.5] = "Yes"
mean(glm.pred!= Default[-train, ]$default)## [1] 0.0466(d) Now consider a logistic regression model that predicts the probability of default using income, balance, and a dummy variable for student. Estimate the test error for this model using the validation set approach. Comment on whether or not including dummy variable for student leads to a reduction in the test error rate. Including the dummy variable for Student leads to no reduction in test error rate, since it stayed at 4.78%
train=sample(dim(Default)[1],dim(Default)[1]/2)
glm.fit=glm(default~income+balance + student, family='binomial', data = Default, subset = train)
glm.probs = predict(glm.fit, data = Default[-train, ], type = "response")
glm.pred = rep("No", length(glm.probs))
glm.pred[glm.probs > 0.5] = "Yes"
mean(glm.pred!= Default[-train, ]$default)## [1] 0.04786. 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.
set.seed(1)
attach(Default)
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: 8The estimated standard errors for the coefficients associated with income and balance are 4.985e-06 and 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) {
fit = glm(default ~ income + balance, data = data, family = "binomial", subset = index )
return(coef(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.
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-04The estimates of the standard error for the Boot function coefficients are the following
t1* 4.344722e-01
t2* 4.866284e-06
t3* 2.298949e-04
(d) Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function.
The estimated standard error when using the glm() function and the bootstrap function stayed very similar
9. We will now consider the Boston housing data set, from the MASS library.
(a) Based on this data set, provide an estimate for the population mean of medv. Call this estimate μ̂.
library(MASS)
library(boot)
attach(Boston)
mu.hat=mean(medv)
mu.hat## [1] 22.53281(b) Provide an estimate of the standard error of μ̂. 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.
st.er.hat= sd(medv) / sqrt(dim(Boston)[1])
st.er.hat## [1] 0.4088611(c) Now estimate the standard error of μ̂ using the bootstrap. How does this compare to your answer from (b)?
set.seed(1)
boot.fn = function(data, index) {
mu = mean(data[index])
return(mu)
}
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 0.4106622The estimated standard error using the bootstrap method is .4119. which is very similar mu.hat
(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$medv). Hint: You can approximate a 95 % confidence interval using the formula [μ̂ − 2SE(μ̂), μ̂ + 2SE(μ̂)].
t.test(Boston$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
## 22.53281CI.mu.hat=c(22.53281 - 2 * .4106622, 22.53281 + 2 * .4106622)
CI.mu.hat## [1] 21.71149 23.35413(e) Based on this data set, provide an estimate, μ̂med , for the median value of medv in the population.
median.hat= median(medv)
median.hat## [1] 21.2(f) We now would like to estimate the standard error of μ̂med . 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) {
mu=median(data[index])
return(mu)
}
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 0.3770241with this results there is an estimated median value of 21.2 which is the same as the value obtained. Moving on to the standard error we see a decrease being at .3770241 compared to the median value of
(g) Based on this data set, provide an estimate for the tenth percentile of medv in Boston suburbs. Call this quantity μ̂0.1 . (You can use the quantile() function.)
Perc.hat= quantile(medv, c(0.1))
Perc.hat## 10%
## 12.75(h) Use the bootstrap to estimate the standard error of μ̂0.1 . Comment on your findings.
boot.fn = function(data, index) {
mu=quantile(data[index], c(0.1))
return(mu)
}
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 0.4925766we conclude the same estimated .1 percentile value of 12.75 , which is equal to the value on problem G