library(ISLR2)assignment 4
Question 3
We now review k-fold cross-validation.
(a) Explain how k-fold cross-validation is implemented.
This approach involves randomly dividing 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. The mean squared error, MSE1, is then computed on the observations in the held-out fold. This procedure is repeated k times; each time, a different group of observations is treated as a validation set. This process results in k estimates of the test error,. The k-fold CV estimate is computed by averaging these values
(b) What are the advantages and disadvantages of k-fold cross-validation relative to:
- The validation set approach?
Advantages:The validation set approach is conceptually simple and is easy to implement
Disadvantages: The validation MSE can be highly variable and Only a subset of observations are used to fit the model (training data).
- LOOCV?
Advantages: LOOCV have less bias. The validation approach produces different MSE when applied repeatedly due to randomness in the splitting process, while performing LOOCV multiple times will always yield the same results, because we split based on 1 obs. each time.
Disadvantage: LOOCV is computationally intensive.
Question 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.
data("Default")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 logmodel1 = glm(default ~ balance + income, data = Default, family = binomial)
summary(logmodel1)
Call:
glm(formula = default ~ balance + income, family = binomial,
data = Default)
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -1.154e+01 4.348e-01 -26.545 < 2e-16 ***
balance 5.647e-03 2.274e-04 24.836 < 2e-16 ***
income 2.081e-05 4.985e-06 4.174 2.99e-05 ***
---
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:
- Split the sample set into a training set and a validation set.
trainDefault = sample(dim(Default)[1], dim(Default)[1]*0.50)
testDefault = Default[-trainDefault, ]- Fit a multiple logistic regression model using only the training observations.
LOGmodel2 = glm(default ~ balance + income, data = Default, family = binomial, subset = trainDefault)- 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.
log.prob_def = predict(LOGmodel2, testDefault, type = "response")
log.pred_def = rep("No", dim(Default)[1]*0.50)
log.pred_def[log.prob_def > 0.5] = "Yes"
table(log.pred_def, testDefault$default)
log.pred_def No Yes
No 4812 117
Yes 18 53- Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified.
mean(log.pred_def !=testDefault$default)[1] 0.027(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.
#i
trainDefault = sample(dim(Default)[1], dim(Default)[1]*0.50)
testDefault = Default[-trainDefault, ]
#ii
LOGmodel2 = glm(default ~ balance + income, data = Default, family = binomial, subset = trainDefault)
#iii
log.prob_def = predict(LOGmodel2, testDefault, type = "response")
log.pred_def = rep("No", dim(Default)[1]*0.50)
log.pred_def[log.prob_def > 0.5] = "Yes"
table(log.pred_def, testDefault$default)
log.pred_def No Yes
No 4821 106
Yes 19 54#iv
mean(log.pred_def !=testDefault$default)[1] 0.025#i
trainDefault = sample(dim(Default)[1], dim(Default)[1]*0.50)
testDefault = Default[-trainDefault, ]
#ii
LOGmodel2 = glm(default ~ balance + income, data = Default, family = binomial, subset = trainDefault)
#iii
log.prob_def = predict(LOGmodel2, testDefault, type = "response")
log.pred_def = rep("No", dim(Default)[1]*0.50)
log.pred_def[log.prob_def > 0.5] = "Yes"
table(log.pred_def, testDefault$default)
log.pred_def No Yes
No 4819 101
Yes 25 55#iv
mean(log.pred_def !=testDefault$default)[1] 0.0252#i
trainDefault = sample(dim(Default)[1], dim(Default)[1]*0.50)
testDefault = Default[-trainDefault, ]
#ii
LOGmodel2 = glm(default ~ balance + income, data = Default, family = binomial, subset = trainDefault)
#iii
log.prob_def = predict(LOGmodel2, testDefault, type = "response")
log.pred_def = rep("No", dim(Default)[1]*0.50)
log.pred_def[log.prob_def > 0.5] = "Yes"
table(log.pred_def, testDefault$default)
log.pred_def No Yes
No 4828 104
Yes 24 44#iv
mean(log.pred_def !=testDefault$default)[1] 0.0256(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.
#i
trainDefault = sample(dim(Default)[1], dim(Default)[1]*0.50)
testDefault = Default[-trainDefault, ]
#ii
LOGmodel2 = glm(default ~ balance + income + student, data = Default, family = binomial, subset = trainDefault)
#iii
log.prob_def = predict(LOGmodel2, testDefault, type = "response")
log.pred_def = rep("No", dim(Default)[1]*0.50)
log.pred_def[log.prob_def > 0.5] = "Yes"
table(log.pred_def, testDefault$default)
log.pred_def No Yes
No 4812 126
Yes 14 48#iv
mean(log.pred_def !=testDefault$default)[1] 0.028Question 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 sm.GLM() function. Do not forget to set a random seed before beginning your analysis.
(a) Using the summarize() and sm.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.
LOGmodel3 = glm(default ~ balance + income, data = Default, family = binomial)
summary(LOGmodel3)
Call:
glm(formula = default ~ balance + income, family = binomial,
data = Default)
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -1.154e+01 4.348e-01 -26.545 < 2e-16 ***
balance 5.647e-03 2.274e-04 24.836 < 2e-16 ***
income 2.081e-05 4.985e-06 4.174 2.99e-05 ***
---
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) return(coef(glm(default ~ balance + income, data = data, family = binomial, subset = index)))(c) Following the bootstrap example in the lab, use your boot_fn() function to estimate the standard errors of the logistic regression coefficients for income and balance.
library(boot)(d) Comment on the estimated standard errors obtained using the sm.GLM() function and using the bootstrap.
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 -3.348419e-02 4.578754e-01
t2* 5.647103e-03 2.626665e-05 2.507091e-04
t3* 2.080898e-05 -4.251706e-07 4.051833e-06Question 9
We will now consider the Boston housing data set, from the ISLP library.
library(MASS)
Attaching package: 'MASS'The following object is masked from 'package:ISLR2':
Bostondata("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. :50.00 (a) Based on this data set, provide an estimate for the population mean of medv. Call this estimateˆµ.
attach(Boston)
mean.medv = mean(medv)
mean.medv[1] 22.53281(b) Provide an estimate of the standard error ofˆµ. 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.
stderr.mean = sd(medv)/sqrt(length(medv))
stderr.mean[1] 0.4088611(c) Now estimate the standard error ofˆµ using the bootstrap. How does this compare to your answer from (b)?
boot.fn2 = function(data, index) return(mean(data[index]))
boot2 = boot(medv, boot.fn2, 100)
boot2
ORDINARY NONPARAMETRIC BOOTSTRAP
Call:
boot(data = medv, statistic = boot.fn2, R = 100)
Bootstrap Statistics :
original bias std. error
t1* 22.53281 0.0008003953 0.3785282(d) Based on your bootstrap estimate from (c), provide a 95 % confidence interval for the mean of medv. Compare it to the results obtained by using Boston[‘medv’].std() and the two standard error rule (3.9).
Hint: You can approximate a 95 % confidence interval using the formula [ˆ µ− 2SE(ˆ µ), ˆµ + 2SE(ˆ µ)].
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
22.53281 CI.bos = c(22.53 - 2 * 0.4174872, 22.53 + 2 * 0.4174872)
CI.bos[1] 21.69503 23.36497(e) Based on this data set, provide an estimate,ˆµmed, for the median value of medv in the population.
median.medv = median(medv)
median.medv[1] 21.2(f) We now would like to estimate the standard error ofˆµmed. 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]))
boot3 = boot(medv, boot.fn3, 1000)
boot3
ORDINARY NONPARAMETRIC BOOTSTRAP
Call:
boot(data = medv, statistic = boot.fn3, R = 1000)
Bootstrap Statistics :
original bias std. error
t1* 21.2 -0.02835 0.37681(g) Based on this data set, provide an estimate for the tenth percentile of medv in Boston census tracts. Call this quantityˆµ0.1. (You can use the np.percentile() function.)
tenth.medv = quantile(medv, c(0.1))
tenth.medv 10%
12.75 (h) Use the bootstrap to estimate the standard error ofˆµ0.1. Comment on your findings.
boot.fn4 = function(data, index) return(quantile(data[index], c(0.1)))
boot4 = boot(medv, boot.fn4, 1000)
boot4
ORDINARY NONPARAMETRIC BOOTSTRAP
Call:
boot(data = medv, statistic = boot.fn4, R = 1000)
Bootstrap Statistics :
original bias std. error
t1* 12.75 0.00865 0.5121083