hw4

Question 3:

a. Explain how k-fold cross-validation is implemented. The k-fold cross-validation is based on the principle of spliting and testing the dataset. For this purpose we first take the random splited data k out of n toal observation. These groups acts as a validation set and the remainder acts as a training set. The test error is then estimated by averaging the k resulting MSE estimates. b.what are the advantages and disadvantages of k fold crossvalidation relative to: i. Validation set approach: The disadvantages of validation set approach is the estimate of the test error rate can be highly variable depending upon selection of testing and validating sets. Since only a subset of the observations are used to fit the model, the validation set error may overestimate the test error rate for the model fit on the entire dataset. ii. LOOCV It has less bias. We repeatedly fit the statistical learning method using training data that contains n-1 observation.The validation approach produces different MSE because of the randomness in the splitting process. whereas in LOOCV, multiple attempt always give the same result because we split based on the 1 observation. Each time LOOCV is computationally intensive. We fit each model n number of times.

Question 5

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

library(ISLR)

## Warning: package 'ISLR' was built under R version 4.0.2

attach(Default)
set.seed(1)
fit.glm <- glm(default ~ income+balance, family='binomial')
summary(fit.glm)

## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial")
## 
## 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. Using the validation set approach, estimate the test error of this mode. In order to do this, you must perform the following steps:
3. Split the sample set into a training set and a validation set.

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

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

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

## 
## 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

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 vategory if the posterior probability is greater than 0.5.

probs <-  predict(fit.glm, newdata =  Default[-train,], type = 'response')
pred.glm <- rep('No', length(probs))
pred.glm[probs >0.5] <- 'Yes'

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

mean(pred.glm != Default [-train,] $default)

## [1] 0.0254

The validation set approach gave us 2.54% test error rate.

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.

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

## [1] 0.0274

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

## [1] 0.0244

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

## [1] 0.0244

We can see that three different splits produce a different resulting error rate which shows that the rate varies by which observation are in the training/validation sets.

4. Now consider a logistic regression model that predicts the probability of default using income, balance, and a dummy variable for students. 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)
fit.glm <- glm(default ~ income + balance + student, data = Default, family = "binomial", subset = train)
pred.glm <- rep("No", length(probs))
probs <-  predict(fit.glm, newdata = Default[-train,], type = "response")
pred.glm[probs > 0.5 ] <- "Yes"
mean(pred.glm != Default[-train, ] $default)

## [1] 0.0278

We can see that including a dummy variable did not result in a reduction in the test error rate.

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.

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(1)
attach(Default)

## The following objects are masked from Default (pos = 3):
## 
##     balance, default, income, student

fit.glm <-  glm(default ~ income + balance, data = Default, family = "binomial")
summary(fit.glm)

## 
## 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

The glm() estimates of the standard errors for the coefficients intercept = 4.348e-01, income =4.985e-06 and balance = 2.274e-04respectively.

2. write a functin, 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){
  fit <- glm(default ~ income + balance, data = data, family = "binomial", subset = index)
  return (coef(fit))
}

3. use the boot() funciton 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

The bootstrap estimates of the standard errors for the coefficients are: B0 =4.340879e-01, B1=4.946329e-06 and B2=2.277448e-04.

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

The estimated standard errors obtained by each of the two methods are very similar.

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 m.hat

library(MASS)

## Warning: package 'MASS' was built under R version 4.0.2

attach(Boston)
m.hat <-  mean(medv)
m.hat

## [1] 22.53281

2. Provide an estimate of the standard error of (m.hat). Interpret this result.

s.hat <-  sd(medv) /sqrt(dim(Boston)[1])
s.hat

## [1] 0.4088611

3. Now estimate the std. error of m.hat using bootstrap. How does it compare to answer in b.

set.seed(1)
boot.fn <- function(data, index){
  m <-  mean(data [index])
  return(m)
}
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 bootstrap estimated standard error of m.hat of 0.410662 is similar to the estimate found in b) of 0.4088611

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 95% confidence interval using the formula [[m.hat] - 2SE[m.hat],[m.hat] + 2SE(m.hat)]

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

CI.m.hat <-  c(22.53 - 2 * 0.4119, 22.53 + 2 * 0.4110)
CI.m.hat

## [1] 21.7062 23.3520

We observed the confidence interval is very close to the one provided by the t.test() function.

5. Based on this data set, provide an estimate, (m.hat)med, for the median value of medv in the population.

med.hat <-  median(medv)
med.hat

## [1] 21.2

6. We now would like to estimate the standard error of (m.hat)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){
  mu <- median(data[index])
  return (mu)
  
}
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

we got estimated median = 21.2 which is equal to the value obtained in (e), with a standard error of 0.377 which is relatively smalle compared to median value.

7. Based on this data set, provide an estimate for the tenth percentile of medv in Boston suburbs. Call this quantity (m.hat)01. (You can use the quantile() function.)

percent.hat <-  quantile(medv,c(0.1))
percent.hat

##   10% 
## 12.75

h.Use the bootstrap to estimate the standard error of (mu-hat)0.1. Comment on your findings.

boot.fn <- function(data, index){
  mu <-  quantile (data[index], c(0.1))
  return(mu)
}
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

We get an estimated 10th percentile value of 12.75 which is equal to the value obtained in (g) with a standard error of 0.5113 which is relatively small compared to percentile value