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

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

K-fold cross-validation is one of the cross-validation approaches used for Resampling.
This approach involves:
* Divide dataset randomly into K different folds or groups of approximately equal size
* Remove (held out) 1st fold/part, fit model for the remaining K-1 parts. This first part is considered as validation set
* Mean Squared Error(MSE) is computed on the observations in the held out fold (1st fold)
* Repeat this procedure “K” times taking out different parts each time for validation, resulting in K different MSEs.
* The cross validation test error is calculated by averaging the resulting K different MSEs.

In case of classification problems, use Mis-classification Error rate instead of MSEs.

3(b) What are the advantages and disadvantages of k-fold cross-validation relative to:i. The validation set approach? ii. LOOCV?

  1. Validation set approach: The validation set approach has two main drawbacks compared to k-fold cross-validation. Though computation wise, this is less intensive than K-fold, the test error rate is highly variable with this approach. Second, only a subset of the observations are used to fit the model. Since statistical methods tend to perform worse when trained on fewer observations, this suggests that the validation set error rate may tend to overestimate the test error rate for the model fit on the entire data set.

  2. LOOCV is a special case of K-fold in which k=n. LOOCV has two drawbacks compared to K-fold. LOOCV is computationally intensive than K-Fold as it has to be fitted n times where K-fold has to be fitted only K times. LOOCV has high variance compared to K-fold since we are averaging the outputs of n fitted models trained on an almost identical set of observations, these outputs are highly correlated, and the mean of highly correlated quantities has higher variance than less correlated ones.

Here i tried to compare all three approaches.

Category Validation set approach LOOCV K-Fold (K<n)
Computation Relatively less Very High High
Bias Very High Relatively less High
Variance Very High High Relatively less
Randomness Very High Almost none Minimal

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.

# Load the ISLR library
library(ISLR)
# set the seed
set.seed(1)
# See the structure of dataset
str(Default)
## 'data.frame':    10000 obs. of  4 variables:
##  $ default: Factor w/ 2 levels "No","Yes": 1 1 1 1 1 1 1 1 1 1 ...
##  $ student: Factor w/ 2 levels "No","Yes": 1 2 1 1 1 2 1 2 1 1 ...
##  $ balance: num  730 817 1074 529 786 ...
##  $ income : num  44362 12106 31767 35704 38463 ...
5(a) Fit a logistic regression model that uses income and balance to predict default.
# Create model parameters string so that we could reuse it
model_vars <- default~income+balance

# Fit Logistic regression. Using glm instead of lm. Both are same

fit.glm = glm(model_vars,data = Default, family = binomial)
summary(fit.glm)
## 
## Call:
## glm(formula = model_vars, 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

Both income and balance are significant.

5(b) Using the validation set approach, estimate the test error of this model. In order to do this, you must perform the following steps:
5(b)i. Split the sample set into a training set and a validation set.
5(b)ii. Fit a multiple logistic regression model using only the training observations.
5(b)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.
5(b)iv. Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified.

Validation set approach:

# Create a function as we have to repeat this several times and can reuse this code. Set seed as input to use for rerunning

# Initialize the train data set with a sample
train = sample(dim(Default)[1], dim(Default)[1] / 2)

# Create a multiple logistic regression model function to reuse later
mlogit = function(mVars,seed) {

# set the seed
set.seed(seed)
  
# 5(b)i. Splitting the sample into training (train) and validation set. Using dim to get the dimension and dividing it by 2.
train = sample(dim(Default)[1], dim(Default)[1] / 2)

# 5(b)ii. Fit the multiple logistic regression model using only the training observations (train data set)
fit.glm <- glm(mVars, data = Default, subset = train, family = "binomial")

# 5(b)iii. Get the probability for 0.5 as the threshold
probs.glm <- predict(fit.glm, Default[-train, ], type = "response")
pred.glm <- rep("No", length(probs.glm))
pred.glm[probs.glm > 0.5] <- "Yes"

# Return the value
return(mean(pred.glm!=Default[-train, ]$default))
}
mlogit(model_vars,1)
## [1] 0.0254

2.54% of test error rate with this step.

5(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.
# Instantiate a vector to capture the test error rate
cv.error = rep(0,3)
# Iterate over the multiple logistic regression function in previous step 3 times and capture the error rate
for (i in 1:3) { 
  # populate seed with i+1 as i already used seed(1) before. This is just to ensure results are same when i re-run it again.
  cv.error[i] = mlogit(model_vars,i+1) 
}
# Print the vector results
cv.error
## [1] 0.0246 0.0266 0.0252

We got different results each time but the error rates are not too far off.

5(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.
# Create a new model vars variable
model_vars2 <- default~income+balance+student

# Run the model again
# Instantiate a vector to capture the test error rate
cv2.error = rep(0,3)
# Iterate over the multiple logistic regression function in previous step 3 times and capture the error rate
for (i in 1:3) { 
  # populate seed with i+1 as i already used seed(1) before. This is just to ensure results are same when i re-run it again.
  cv2.error[i] = mlogit(model_vars2,i) 
}
# Print the vector results
cv2.error
## [1] 0.0260 0.0246 0.0272

Adding Student predictor didn’t reduce the error rate. However, the error rates were not too significantly different.

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.

6(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.
# Fit Logistic regression.

fit2.glm = glm(model_vars,data = Default, family = binomial)
summary(fit2.glm)
## 
## Call:
## glm(formula = model_vars, 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 standard errors estimated by glm are 4.348e-01, 4.985e-06 and 2.274e-04 for each of the coefficients.

6(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.
library(boot)
## Warning: package 'boot' was built under R version 3.6.2
set.seed(1)
boot.fn <- function(data,index) {
  fit3.glm <- glm(default~income+balance, data=data, subset = index , family = binomial)
  return(coef(fit3.glm))
}
6(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.
boot(Default, boot.fn, R=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 standard errors estimated by bootstrap function are 4.344722e-01, 4.8662845e-06 and 2.298949e-04 for each of the coefficients.

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

The estimated standard errors for both glm() and the bootstrap function are very close. Bootstrap function helped with smoothing.

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

# Load the MASS library
library(MASS)
## Warning: package 'MASS' was built under R version 3.6.2
9(a) Based on this data set, provide an estimate for the population mean of medv. Call this estimate μˆ.
# Get the mean of medv 
mu.hat <- mean(Boston$medv)
mu.hat
## [1] 22.53281
9(b) Provide an estimate of the standard error of μˆ. Interpret this result. Hint: We can compute the standard error of the sample mean by dividing the sample standard deviation by the square root of the number of observations.
SE = sd(Boston$medv)/sqrt(nrow(Boston))
SE
## [1] 0.4088611

The Standard Error is 0.4088611. This is small and hence we can interpret that the sample mean mu.hat is closer to the true population mean.

9(c) Now estimate the standard error of μˆ using the bootstrap. How does this compare to your answer from (b)?
# set the seed for repeatability
set.seed(1)

# Define boot function to return the mean
boot.fn = function(data, index) {
  mu <- mean(data[index])
  return (mu)
}

# call the boot function with R = 1000
SE_boot = boot(Boston$medv, boot.fn, R=1000)

# Print the statistics
SE_boot
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original      bias    std. error
## t1* 22.53281 0.007650791   0.4106622

Bootstrap estimated the standard error of \(\mu\hat{}\) as 0.4106622, where as standard error of sample mean is 0.4088611. The values are very close.

9(d) Based on your bootstrap estimate from (c), provide a 95 % con- fidence 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 [μˆ − 2SE(μˆ), μˆ + 2SE(μˆ)].

Lets calculate the confidence intervals for the boostrap estimate

# Get the μˆ and SE(μˆ) from above bootstrap 
mu.hat_b = SE_boot$t0
SE_b = sd(SE_boot$t)
# calculate the 95% confidence interval using the formula [μˆ − 2SE(μˆ), μˆ + 2SE(μˆ)]
CI <- c(mu.hat_b - 2 * SE_b,mu.hat_b + 2 * SE_b)
CI
## [1] 21.71148 23.35413

Lets do the t-test for Boston$medv.

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

The 95% confidence intervals from t-test of Boston$medv are 21.72953 and 23.33608 where as the 95% confidence intervals calculated from the bootstrap estimate are 21.71148 and 23.35413. They are very close.

9(e) Based on this dataset, provide an estimate, μˆmed, for the median value of medv in the population.
# use the median function and get the μˆmed
med.hat <- median(Boston$medv)
med.hat
## [1] 21.2
9(f) We now would like to estimate the standard error of μˆmed. 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.
# Estimate the standard error of μˆmed using the bootstrap
# set the seed for repeatability
set.seed(1)

# Define boot function to return the median
boot.fn = function(data, index) {
  med <- median(data[index])
  return (med)
}

# call the boot function with R = 1000
SE_boot_med = boot(Boston$medv, boot.fn, R=1000)

# Print the statistics
SE_boot_med
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*     21.2 0.02295   0.3778075

The standard error of 0.3778075 for the median of bootstrap estimate is relatively small.

9(g) Based on this data set, provide an estimate for the tenth percentile of medv in Boston suburbs. Call this quantity μˆ0.1. (You can use the quantile() function.)
# Get the tenth percentile of medv in Boston suburbs using Quantile function with .1
mu.hat.0.1 = quantile(Boston$medv,.1)
mu.hat.0.1
##   10% 
## 12.75
9(h) Use the bootstrap to estimate the standard error of μˆ0.1. Comment on your findings.
# Use bootstrap and estimate the standard error of μˆ0.1
# set the seed for repeatability
set.seed(1)

# Define boot function to return the quantile at 0.1 i.e. tenth percentile
boot.fn = function(data, index) {
  mu.hat.0.1.b <- quantile(data[index],c(0.1))
  return (mu.hat.0.1.b)
}

# call the boot function with R = 1000
SE_boot_quant = boot(Boston$medv, boot.fn, R=1000)

# Print the statistics
SE_boot_quant
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*    12.75  0.0339   0.4767526

The standard error of the 10th percentile is 0.4767526 which is relatively small.