##Question 3

We now review k-fold cross-validation.

(a) Explain how k-fold cross-validation is implemented. K fold CV is similar to LOOCV but rather than remove only observation, it removes a set number of observations (the K value). The model is trained on the remaining observations and the test error is computed on the left out observations. This continues until all the observations have been trained/tested and the average of the MSE’s is calculated. (b) What are the advantages and disadvantages of k-fold crossvalidation relative to: i. The validation set approach?

The advantage K fold has over validation set approach is that the variability among the test error results are much lower. The validation set approach is much easier and simpler to implement
ii. LOOCV?

The biggest advantage to using K fold over LOOCV is that K fold is far less computationally intensive. However, K fold will have more bias than LOOCV

##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.

library(ISLR)
## Warning: package 'ISLR' was built under R version 3.6.3
attach(Default)
set.seed(1)
glm.fit=glm(default~income+balance, data=Default, family=binomial)
summary(Default$default)
##   No  Yes 
## 9667  333

(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(1:nrow(Default),0.7*nrow(Default))
test = !train
train.Default=Default[train,]
test.Default = Default[-train,]
default.test=default[-train]

ii. Fit a multiple logistic regression model using only the training observations.

glm.fit=glm(default~income+balance, data=Default, family=binomial, subset=train)

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.

glm.prob=predict(glm.fit, test.Default, type="response")
glm.pred=rep("No", length(glm.prob))
glm.pred[glm.prob > 0.5] = "Yes"

iv. Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified.

mean(glm.pred != default.test)
## [1] 0.02666667

(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. All results appear to be between 2.4% and 2.6% test error rates

Run #1

train=sample(1:nrow(Default),0.7*nrow(Default))
test = !train
train.Default=Default[train,]
test.Default = Default[-train,]
default.test=default[-train]
glm.fit=glm(default~income+balance, data=Default, family=binomial, subset=train)
glm.prob=predict(glm.fit, test.Default, type="response")
glm.pred=rep("No", length(glm.prob))
glm.pred[glm.prob > 0.5] = "Yes"
mean(glm.pred != default.test)
## [1] 0.028

Run #2

train=sample(1:nrow(Default),0.7*nrow(Default))
test = !train
train.Default=Default[train,]
test.Default = Default[-train,]
default.test=default[-train]
glm.fit=glm(default~income+balance, data=Default, family=binomial, subset=train)
glm.prob=predict(glm.fit, test.Default, type="response")
glm.pred=rep("No", length(glm.prob))
glm.pred[glm.prob > 0.5] = "Yes"
mean(glm.pred != default.test)
## [1] 0.023

Run #3

train=sample(1:nrow(Default),0.7*nrow(Default))
test = !train
train.Default=Default[train,]
test.Default = Default[-train,]
default.test=default[-train]
glm.fit=glm(default~income+balance, data=Default, family=binomial, subset=train)
glm.prob=predict(glm.fit, test.Default, type="response")
glm.pred=rep("No", length(glm.prob))
glm.pred[glm.prob > 0.5] = "Yes"
mean(glm.pred != default.test)
## [1] 0.02766667

(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 test error with the student dummy variable added came out to 2.67%. There is no apparent improvement in the model when adding the student variable.

train=sample(1:nrow(Default),0.7*nrow(Default))
test = !train
train.Default=Default[train,]
test.Default = Default[-train,]
default.test=default[-train]
glm.fit=glm(default~income+balance+student, data=Default, family=binomial, subset=train)
glm.prob=predict(glm.fit, test.Default, type="response")
glm.pred=rep("No", length(glm.prob))
glm.pred[glm.prob > 0.5] = "Yes"
mean(glm.pred != default.test)
## [1] 0.02666667

##Question 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.

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){ return(coef(glm(default~income+balance,data=data,family=binomial, subset=index)))
}

(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, 500)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Default, statistic = boot.fn, R = 500)
## 
## 
## Bootstrap Statistics :
##          original        bias     std. error
## t1* -1.154047e+01 -2.298134e-02 4.212334e-01
## t2*  2.080898e-05 -2.053140e-07 5.184890e-06
## t3*  5.647103e-03  1.678038e-05 2.178846e-04

(d) Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function. The standard errors for the balance and income variables are close, but the bootstrap ones are slightly lower. This could be due to the fact that there may not be a linear relationship in the data set and cause the standard error to be inflated slightly. The bootstrap is not affected by the same assumptions the linear model makes about the data and the residuals.

##Question 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.

library(MASS)
attach(Boston)
mu=mean(medv)
mu
## [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.

err.mu=sd(medv)/sqrt(length(medv))
err.mu
## [1] 0.4088611

(c) Now estimate the standard error of ˆμ using the bootstrap. How does this compare to your answer from (b)? Using the boot function, the mean standard error is .427 which is within .02 of the .409 from part b.

library(boot)
set.seed(1)
boot.fn=function(data, index){return(mean(data[index]))}
boot(medv, boot.fn, 500)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn, R = 500)
## 
## 
## Bootstrap Statistics :
##     original      bias    std. error
## t1* 22.53281 0.004967589   0.4265264

(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)].

mu.CI=c(22.53 - 2*0.427, 22.53 + 2*0.427)
mu.CI
## [1] 21.676 23.384
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 
##  22.53281

(e) Based on this data set, provide an estimate, ˆμmed, for the median value of medv in the population.

mu.med=median(medv)
mu.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. The bootstrap also calculated 21.2 median value with .384 standard error.

boot.fn=function(data, index){return(median(data[index]))}
boot(medv, boot.fn, 500)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn, R = 500)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*     21.2  0.0298   0.3840925

(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.)

mu0.1=quantile(medv, c(0.1))
mu0.1
##   10% 
## 12.75

(h) Use the bootstrap to estimate the standard error of ˆμ0.1. Comment on your findings. The boot function calculated 12.75 (just like above) and a small standard error of .514.

boot.fn=function(data, index){return(quantile(data[index], c(0.1)))}
boot(medv, boot.fn, 500)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn, R = 500)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*    12.75   5e-04   0.5036289