We now review k-fold cross validation.
(a) Explain how k-fold cross validation is implemented.
K-Fold cross validation is implemented by taking a set of observations and splitting them randomly into k groups/folds that are of equal size. The first is treated as a validation set and the method is then fit onto the remaining k - 1 folds. The MSE is then calculated on the observations from the held out fold. This is repeated k times, with each time having a different group of observations being 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 of k-fold cross validation relative to the validation set approach is that the test error rate is not as highly variable. Reduces the likeliness that the test error rate is overestimated compared to the validation set approach. Disadvantage is that it is more computationally intensive compared to the relatively simple method of the validation set approach.
ii. LOOCV?
The LOOCV method is more computationally intense, and has a higher variance, but lower bias than k-fold cross validation.
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.
library(ISLR)
summary(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) Fit a logistic regression model that uses income and balance to predict 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.
ii. Fit a 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 the validation set error, which is the fraction of the observations in the validation set that are misclassified.
i.
train = sample(dim(Default)[1], 0.75*dim(Default)[1])ii.
glm.fit = glm(default ~ income + balance, data = Default, family = binomial, subset = train)iii.
glm.probs = predict(glm.fit, Default[-train,], type = "response")
glm.preds = rep('No', dim(Default)[1])
glm.preds[glm.probs>0.5] = 'Yes'iv.
mean(glm.preds != Default[-train, 'default'])## [1] 0.026From this result, we can see that the validation set error was 2.6%
(c) Repeat the process in (b) three times, using different splits of the observations into a training set and a validation set. Comment on the results obtained.
train = sample(dim(Default)[1], 0.80*dim(Default)[1])
glm.fit = glm(default ~ income + balance, data = Default, family = binomial, subset = train)
glm.probs = predict(glm.fit, Default[-train,], type = "response")
glm.preds = rep('No', dim(Default)[1])
glm.preds[glm.probs>0.5] = 'Yes'
mean(glm.preds != Default[-train, 'default'])## [1] 0.026train = sample(dim(Default)[1], 0.90*dim(Default)[1])
glm.fit = glm(default ~ income + balance, data = Default, family = binomial, subset = train)
glm.probs = predict(glm.fit, Default[-train,], type = "response")
glm.preds = rep('No', dim(Default)[1])
glm.preds[glm.probs>0.5] = 'Yes'
mean(glm.preds != Default[-train, 'default'])## [1] 0.027train = sample(dim(Default)[1], 0.95*dim(Default)[1])
glm.fit = glm(default ~ income + balance, data = Default, family = binomial, subset = train)
glm.probs = predict(glm.fit, Default[-train,], type = "response")
glm.preds = rep('No', dim(Default)[1])
glm.preds[glm.probs>0.5] = 'Yes'
mean(glm.preds != Default[-train, 'default'])## [1] 0.022Using different splits, the test error rate did not dramatically increase or decrease compared to the first version of the split. Two of the test error stayed very vlose to the original 2.6% and the other was only a few tenths above that at 3.2%.
(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.
train = sample(dim(Default)[1], dim(Default)[1]/2)
fit.glm = glm(default ~ income + balance + student, data = Default, family = binomial, subset = train)
glm.probs = predict(glm.fit, Default[-train,], type = 'response')
glm.preds = rep('No', dim(Default)[1])
glm.preds[glm.probs > 0.5] = 'Yes'
mean(glm.preds != Default[-train, 'default'])## [1] 0.0258It does not appear as though the addition of the student dummy variable leads to a reduction in the test error rate since it increased by 0.01, which is not a significant change.
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.
library(boot)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) 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.
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-04(d) Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function.
The standard errors from the glm() functin and the bootstrap function were actually quite similar.
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. :50.00set.seed(1)
attach(Boston)(a) Based on this data set, provide an estimate for the population mean of medv. Call this estimate ˆμ.
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.
medv.error = sd(medv)/sqrt(length(medv))
medv.error## [1] 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]))
bootstrap = boot(medv, boot.fn, 1000)
bootstrap##
## 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$medv). Hint: You can approximate a 95 % confidence interval using the formula [ˆμ − 2SE(ˆμ), μˆ + 2SE(ˆμ)].
t.test(medv)##
## One Sample t-test
##
## data: medv
## t = 55.111, df = 505, p-value < 2.2e-16
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
## 21.72953 23.33608
## sample estimates:
## mean of x
## 22.53281c(bootstrap$t0 - 2 *.4106622, bootstrap$t0 + 2 * .4106622)## [1] 21.71148 23.35413It is very close to the mean that was obtained from the t test.
(e) 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] 21.2(f)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.fn = function(data, index) return(median(data[index]))
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 0.3770241Standard Error of about .38 ***
(g) 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.tenth = quantile(medv, c(0.1))
medv.tenth ## 10%
## 12.75(h) Use the bootstrap to estimate the standard error of ˆμ0.1. Comment on your findings.
boot.fn = function(data, index) return(quantile(data[index], c(.01)))
boot(medv, boot.fn, 1000)##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = medv, statistic = boot.fn, R = 1000)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 7.01 -0.07244 0.5335053Standard Error of .58 which compared to the value of the tenth percentile, it is actually quite small.