Homework4

3. We now review k-fold cross-validation.

1. Explain how k-fold cross-validation is implemented.

KFCV is performed by taking the sample space of n observations and partitioning them into k distinct sets of data. Each of theses sets act as validation set, and the remainder as a training set. The test error would be estimated by averaging the MSE estimates of each k.

2. What are the advantages and disadvantages of k-fold crossvalidation relative to:

1. The validation set approach?

Validation set is pretty easy to implement as it’s only partitioning the training set, but you can’t account for the results being skewed as the result of the validation set taking in similar/drastically different observations. With this, error may also be overestimated.

2. LOOCV? LOOCV is computationally intensive and has lower bias, but naturally higher variance than KFCV

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.

1. Fit a logistic regression model that uses income and balance to predict default.

library(ISLR)

## Warning: package 'ISLR' was built under R version 3.6.2

attach(Default)
set.seed(12)
glm.fit = glm(default ~ income + balance, data = Default, family = binomial)

2. Using the validation set approach, estimate the test error of this model. In order to do this, you must perform the following steps:

1. Split the sample set into a training set and a validation set.
2. Fit a multiple logistic regression model using only the training observations.
3. 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.
4. Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified.

FiveB = function() {
    train = sample(dim(Default)[1], dim(Default)[1]/2)
    glm.fit = glm(default ~ income + balance, data = Default, family = binomial, 
        subset = train)
    glm.pred = rep("No", dim(Default)[1]/2)
    glm.probs = predict(glm.fit, Default[-train, ], type = "response")
    glm.pred[glm.probs > 0.5] = "Yes"
    return(mean(glm.pred != Default[-train, ]$default))
}
epic = FiveB() * 100

print("The validation test error % is:")

## [1] "The validation test error % is:"

epic

## [1] 2.7

2.7% test error rate when using the validation approach

3. 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.

trial1=FiveB()
trial2=FiveB()
trial3=FiveB()
trial1

## [1] 0.0266

trial2

## [1] 0.0262

trial3

## [1] 0.0248

print("The average is:")

## [1] "The average is:"

(trial1+trial2+trial3)/3

## [1] 0.02586667

4. 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)
glm.fit = glm(default ~ income + balance + student, data = Default, family = binomial, 
    subset = train)
glm.pred = rep("No", dim(Default)[1]/2)
glm.probs = predict(glm.fit, Default[-train, ], type = "response")
glm.pred[glm.probs > 0.5] = "Yes"
mean(glm.pred != Default[-train, ]$default)

## [1] 0.0292

Our test error rate is 2.6%, which is not significantly better than our previous results. Dummy variable has smol brain

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.

1. 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(69)
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

2. 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)))

3. 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, 50)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Default, statistic = boot.fn, R = 50)
## 
## 
## Bootstrap Statistics :
##          original        bias     std. error
## t1* -1.154047e+01 -7.623099e-02 4.044117e-01
## t2*  2.080898e-05  8.002984e-07 4.872492e-06
## t3*  5.647103e-03  3.210896e-05 2.115457e-04

4. Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function.

similar answers to previous significant digits.

9. We will now consider the Boston housing data set, from the MASS library.

1. Based on this data set, provide an estimate for the population mean of medv. Call this estimate ˆμ.

library(MASS)

## Warning: package 'MASS' was built under R version 3.6.2

set.seed(420)
attach(Boston)
medv.mean = mean(medv)
medv.mean

## [1] 22.53281

2. 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.err = sd(medv)/sqrt(length(medv))
medv.err

## [1] 0.4088611

3. 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]))
library(boot)
bstrap = boot(medv, boot.fn, 1000)
bstrap

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original     bias    std. error
## t1* 22.53281 0.01103281   0.4196645

similar results to exercize b

4. 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), ˆμ + 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 
##  22.53281

c(bstrap$t0 - 2 * 0.4119, bstrap$t0 + 2 * 0.4119)

## [1] 21.70901 23.35661

on the dot with bootstrap estimate

5. 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

6. 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  -0.014   0.3740038

median of 21.2 with a standard error of .37

7. 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

8. 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(0.1)))
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 0.03335   0.5037152

Median value of 12.75 with standard error of .504