HW4_iwl322

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

(a) Explain how k-fold cross-validation is implemented.

In this method, the set of observations are randomly divided into k equally sized groups, or folds. The first fold is the validation set and the remaining k – 1 folds are the training set. The model is fit on the remaining folds and the MSE is computed on the validation set. After being repeated k times there are k estimates of the MSE. Averaging the K different MSEs results in the k-fold CV estimate.

(b) What are the advantages and disadvantages of k-fold cross validation relative to:

The validation set approach?

Although it is simpler and easier to implement, the variability in test error estimates from the validation set approach tends to be much higher than the variability in k-fold CV estimates. The validation set method can also lead to overestimates of the test error rate.

LOOCV?

The LOOCV approach is potentially computationally expensive, whereas the k-fold CV is typically performed with k = 5 or k = 10, making it more feasible. The k-fold approach also gives more accurate estimates of test error rates. However if our main goal was to decrease bias, then LOOCV would be preferred over k-fold.

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.

library(ISLR)

## Warning: package 'ISLR' was built under R version 4.1.2

set.seed(1)
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.

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:

Split the sample set into a training set and a validation set.

train=sample(dim(Default)[1], dim(Default)[1]/2)

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

glm.fit2=glm(default~income+balance, data=Default, family="binomial", subset=train)
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: 8

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.fit2, Default[-train, ], type="response")
glm.pred=rep("No", dim(Default)[1])
glm.pred[glm.prob > 0.5]="Yes"

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.

train=sample(dim(Default)[1], dim(Default)[1]/2)
glm.fit3=glm(default~income+balance, data=Default, family="binomial", subset=train)
glm.prob=predict(glm.fit3, Default[-train, ], type="response")
glm.pred=rep("No", dim(Default)[1])
glm.pred[glm.prob > 0.5]="Yes"
mean(glm.pred != Default[-train, "default"])

## [1] 0.0274

train=sample(dim(Default)[1], dim(Default)[1]/2)
glm.fit4=glm(default~income+balance, data=Default, family="binomial", subset=train)
glm.prob=predict(glm.fit4, Default[-train, ], type="response")
glm.pred=rep("No", dim(Default)[1])
glm.pred[glm.prob > 0.5]="Yes"
mean(glm.pred != Default[-train, "default"])

## [1] 0.0244

train=sample(dim(Default)[1], dim(Default)[1]/2)
glm.fit5=glm(default~income+balance, data=Default, family="binomial", subset=train)
glm.prob=predict(glm.fit5, Default[-train, ], type="response")
glm.pred=rep("No", dim(Default)[1])
glm.pred[glm.prob > 0.5]="Yes"
mean(glm.pred != Default[-train, "default"])

## [1] 0.0244

The results of the validation set errors are variable, but fairly close in range.

(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)
glm.fit6=glm(default~income+balance+student, data=Default, family="binomial", subset=train)
glm.prob=predict(glm.fit6, Default[-train, ], type="response")
glm.pred=rep("No", dim(Default)[1])
glm.pred[glm.prob > 0.5]="Yes"
mean(glm.pred != Default[-train, "default"])

## [1] 0.0278

Adding the dummy variable does not seem to affect the test error rate, as it is still within the same range as the other rates.

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.

set.seed(1)

(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.

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, subset=index, family="binomial")))

(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, 1000)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Default, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##          original        bias     std. error
## t1* -1.154047e+01 -3.945460e-02 4.344722e-01
## t2*  2.080898e-05  1.680317e-07 4.866284e-06
## t3*  5.647103e-03  1.855765e-05 2.298949e-04

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

The standard error estimates from the glm() and bootstrap functions are quite close to each other.

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

library(MASS)
attach(Boston)

(a) Based on this data set, provide an estimate for the population mean of medv. Call this estimate $\hat{\mu}$

mu.hat=mean(medv)
mu.hat

## [1] 22.53281

(b) Provide an estimate of the standard error of $\hat{\mu}$. 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.

sd(medv)/sqrt(dim(Boston)[1])

## [1] 0.4088611

(c) Now estimate the standard error of $\hat{\mu}$ using the bootstrap. How does this compare to your answer from (b)?

set.seed(1)

boot.fn=function(data, index)
  return(mean(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* 22.53281 0.007650791   0.4106622

The standard errors computed in (b) and (c) are very close in range.

(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$medv). Hint: You can approximate a 95% confidence interval using the formula [$\hat{\mu}$ − 2SE($\hat{\mu}$), $\hat{\mu}$ + 2SE($\hat{\mu}$)].

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

[22.53281 - 2(0.4106622), 22.53281 + 2(0.4106622)] = [21.71149, 23.35413]

The confidence interval results are very similar.

(e) Based on this data set, provide an estimate, $\hat{\mu}_{med}$, for the median value of medv in the population.

median(medv)

## [1] 21.2

(f) We now would like to estimate the standard error of $\hat{\mu}_{med}$. 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   0.3770241

The median estimate is the same and the standard error is relatively small.

(g) Based on this data set, provide an estimate for the tenth percentile of medv in Boston suburbs. Call this quantity $\hat{\mu}_{0.1}$. (You can use the quantile() function.)

mu.percentile=quantile(medv, c(0.1))
mu.percentile

##   10% 
## 12.75

(h) Use the bootstrap to estimate the standard error of $\hat{\mu}_{0.1}$. 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  0.0186   0.4925766

The percentile estimate is the same and the standard error is relatively small.

HW4_iwl322

Hayley Miller

2022-03-10

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

(a) Explain how k-fold cross-validation is implemented.

(b) What are the advantages and disadvantages of k-fold cross validation relative to:

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

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

(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.

(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.

(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.

(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.

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

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 \(\hat{\mu}\)

(b) Provide an estimate of the standard error of \(\hat{\mu}\). 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.

(c) Now estimate the standard error of \(\hat{\mu}\) using the bootstrap. How does this compare to your answer from (b)?

(e) Based on this data set, provide an estimate, \(\hat{\mu}_{med}\), for the median value of medv in the population.

(f) We now would like to estimate the standard error of \(\hat{\mu}_{med}\). 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.

(g) Based on this data set, provide an estimate for the tenth percentile of medv in Boston suburbs. Call this quantity \(\hat{\mu}_{0.1}\). (You can use the quantile() function.)

(h) Use the bootstrap to estimate the standard error of \(\hat{\mu}_{0.1}\). Comment on your findings.