##Problem 3
We now review k-fold cross-validation.
(a) Explain how k-fold cross-validation is implemented.
K-fold cross-validation divides the dataset into K number of groups or folds that have the same or close to the same number of observations. The fist fold acts as our testing data and we train with the remaining groups. This process is repeated for the remaining number of folds each time a different fold being withheld as the testing data. The validation error is calculated for each run and then averaged together to represent the mean square error of the model being tested.
(b) What are the advantages and disadvantages of k-fold crossvalidation relative to: i. The validation set approach?
The validation set approach is very simple to implement, but it the MSE can be higly variable. Since you are only using a subset of the training data to fit the model, you are likely to get a poorer performing model that K-fold cross-validation due to the smaller size of observations and only one sample to test with.
ii. LOOCV?
LOOCV is less bias tha K-fold cross-validation, but it has a higher variance, so there is a trade off. LOOCV is also much more computaionally expensive than K-fold cross-validation.
##Problem 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)## Warning: package 'ISLR' was built under R version 3.6.3library(MASS)
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.
set.seed(1)
train=sample(nrow(Default),nrow(Default)*0.5)ii. Fit a multiple logistic regression model using only the training observations.
glm.fit2=glm(default~income+balance, data=Default, subset=train, family=binomial)
summary(glm.fit2)##
## 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.fit2, Default[-train,], type='response')
glm.pred=rep("No",length(glm.probs)) # creates a vector with No for all entries
glm.pred[glm.probs>0.5]="Yes" #if glm.prob is greater then .5, the obs gets reassigned as Yes
table(glm.pred, Default[-train,]$default)##
## glm.pred No Yes
## No 4824 108
## Yes 19 49iv. Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified.
mean(glm.pred!=Default[-train,]$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.
I repeated the same steps as above, but changed the set.seed value on the next 3 runs in order to obtain a different sample set. The original run above resulted in a 2.54% validation error. The second, third, and fourth runs had the following validation errors respectivley, 2.76%, 2.48%, and 2.8%. These are all very close values in comparison. The average validation error for these four runs is 2.645%.
set.seed(123)
train=sample(nrow(Default),nrow(Default)*0.5)
glm.fit2=glm(default~income+balance, data=Default, subset=train, family=binomial)
summary(glm.fit2)##
## Call:
## glm(formula = default ~ income + balance, family = binomial,
## data = Default, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.2810 -0.1348 -0.0529 -0.0185 3.7767
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.194e+01 6.451e-01 -18.504 < 2e-16 ***
## income 2.210e-05 7.381e-06 2.995 0.00275 **
## balance 5.874e-03 3.362e-04 17.474 < 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: 1429.88 on 4999 degrees of freedom
## Residual deviance: 752.69 on 4997 degrees of freedom
## AIC: 758.69
##
## Number of Fisher Scoring iterations: 8glm.probs = predict(glm.fit2, Default[-train,], type='response')
glm.pred=rep("No",length(glm.probs)) # creates a vector with No for all entries
glm.pred[glm.probs>0.5]="Yes" #if glm.prob is greater then .5, the obs gets reassigned as Yes
table(glm.pred, Default[-train,]$default)##
## glm.pred No Yes
## No 4808 117
## Yes 21 54mean(glm.pred!=Default[-train,]$default)## [1] 0.0276set.seed(999)
train=sample(nrow(Default),nrow(Default)*0.5)
glm.fit2=glm(default~income+balance, data=Default, subset=train, family=binomial)
summary(glm.fit2)##
## Call:
## glm(formula = default ~ income + balance, family = binomial,
## data = Default, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.2382 -0.1486 -0.0591 -0.0216 3.7000
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.129e+01 5.840e-01 -19.329 <2e-16 ***
## income 1.590e-05 6.846e-06 2.322 0.0202 *
## balance 5.688e-03 3.144e-04 18.092 <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: 1550.17 on 4999 degrees of freedom
## Residual deviance: 829.71 on 4997 degrees of freedom
## AIC: 835.71
##
## Number of Fisher Scoring iterations: 8glm.probs = predict(glm.fit2, Default[-train,], type='response')
glm.pred=rep("No",length(glm.probs)) # creates a vector with No for all entries
glm.pred[glm.probs>0.5]="Yes" #if glm.prob is greater then .5, the obs gets reassigned as Yes
table(glm.pred, Default[-train,]$default)##
## glm.pred No Yes
## No 4819 96
## Yes 28 57mean(glm.pred!=Default[-train,]$default)## [1] 0.0248set.seed(76)
train=sample(nrow(Default),nrow(Default)*0.5)
glm.fit2=glm(default~income+balance, data=Default, subset=train, family=binomial)
summary(glm.fit2)##
## Call:
## glm(formula = default ~ income + balance, family = binomial,
## data = Default, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.4528 -0.1353 -0.0552 -0.0202 3.3635
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.176e+01 6.497e-01 -18.101 < 2e-16 ***
## income 2.515e-05 7.437e-06 3.381 0.000722 ***
## balance 5.626e-03 3.300e-04 17.048 < 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: 1368.2 on 4999 degrees of freedom
## Residual deviance: 740.4 on 4997 degrees of freedom
## AIC: 746.4
##
## Number of Fisher Scoring iterations: 8glm.probs = predict(glm.fit2, Default[-train,], type='response')
glm.pred=rep("No",length(glm.probs)) # creates a vector with No for all entries
glm.pred[glm.probs>0.5]="Yes" #if glm.prob is greater then .5, the obs gets reassigned as Yes
table(glm.pred, Default[-train,]$default)##
## glm.pred No Yes
## No 4809 129
## Yes 11 51mean(glm.pred!=Default[-train,]$default)## [1] 0.028(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.
The validation error with the student variable added was 2.4%, so it doesn’t appear that adding that variable made much of a difference in the error rate. However, I did notice that with the student variable added, the income variable is no longer significant.
set.seed(125)
train=sample(nrow(Default),nrow(Default)*0.5)
glm.fit3=glm(default~income+balance+student, data=Default, subset=train, family=binomial)
summary(glm.fit3)##
## Call:
## glm(formula = default ~ income + balance + student, family = binomial,
## data = Default, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.4632 -0.1537 -0.0604 -0.0219 3.5029
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.061e+01 6.696e-01 -15.846 <2e-16 ***
## income 1.187e-06 1.146e-05 0.104 0.9175
## balance 5.661e-03 3.158e-04 17.928 <2e-16 ***
## studentYes -6.886e-01 3.231e-01 -2.131 0.0331 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1530.39 on 4999 degrees of freedom
## Residual deviance: 826.38 on 4996 degrees of freedom
## AIC: 834.38
##
## Number of Fisher Scoring iterations: 8glm.probs = predict(glm.fit3, Default[-train,], type='response')
glm.pred=rep("No",length(glm.probs)) # creates a vector with No for all entries
glm.pred[glm.probs>0.5]="Yes" #if glm.prob is greater then .5, the obs gets reassigned as Yes
table(glm.pred, Default[-train,]$default)##
## glm.pred No Yes
## No 4827 103
## Yes 17 53mean(glm.pred!=Default[-train,]$default)## [1] 0.024##Problem 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.
(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.
As seen below, the glm() estimates of the standard errors for the coefficents for the intercept, income, and balance are respectivley 0.4348, 4.985x106(-6), and 2.274x10^(-4).
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: 8coef(glm.fit)## (Intercept) income balance
## -1.154047e+01 2.080898e-05 5.647103e-03(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) {
glm.fit=glm(default~income+balance, data=data, subset=index, family="binomial")
return(coef(glm.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.
Using the bootstrap method, the estimates of the standard errors for the coefficents for the intercept, income, and balance are respectivley 0.4163, 4.261x106(-6), and 2.228x10^(-4).
library(boot)
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.
The estimated standard errors obtained from both methods are very close.
##Problem 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 ˆμ.
The population mean of medv which I called mu.hat is 22.53281.
library(ISLR)
summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08204 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. :50.00attach(Boston)mu.hat = mean(medv)
mu.hat## [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.
The estimate for the standard error of mu.hat, which I called semu.hat, is 0.4088611.
semu.hat = sd(medv)/sqrt(length(medv))
semu.hat## [1] 0.4088611(c) Now estimate the standard error of ˆμ using the bootstrap. How does this compare to your answer from (b)?
The standard errror using the bootstrap method is 0.4106622, which very close the the answer we got in (b).
set.seed(1)
boot.fn <- function(data, index) {
bootstrap <- mean(data[index])
return (bootstrap)
}
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$medv).
Hint: You can approximate a 95 % confidence interval using the formula [ˆμ − 2SE(ˆμ), μˆ + 2SE(ˆμ)].
As seen below, there is not much difference in the 95% confidence levels when calculated both ways.
ci.mu.hat = c(mu.hat - 2*semu.hat, mu.hat + 2*semu.hat)
ci.mu.hat## [1] 21.71508 23.35053t.test(Boston$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(e) Based on this data set, provide an estimate, ˆμmed, for the median value of medv in the population.
The median value of medv in the population which I called med.hat is 21.2.
med.hat = median(medv)
med.hat## [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.
The estimated standard error of the median is 0.3778.
set.seed(1)
boot.fn <- function(data, index) {
bootstrap <- median(data[index])
return (bootstrap)
}
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 0.02295 0.3778075(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.)
mu.tenth <- quantile(medv, c(0.1))
mu.tenth## 10%
## 12.75(h) Use the bootstrap to estimate the standard error of ˆμ0.1. Comment on your findings.
The estimated 10th percentile value is 12.75 which is the same as calulated above in (g). The standard error is 0.506 which is higher than the standard errors we have been getting in the other problems, but still relatively small compared to the percentile value.
boot.fn <- function(data, index) {
bootstrap <- quantile(data[index],c(0.1))
return (bootstrap)
}
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 0.01455 0.4823468