Assignment #4: Resampling Methods

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

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

The K-fold cross-validation technique involves splitting the data set into k subsets, or “folds”, of approximately equal size. Then, it is trained on k-1 of these folds, and the remaining fold is used as a validation set to measure the model’s performance.

The k-fold cross-validation is implemented by:

1. Divide the data set into k subsets or “folds” of approximately equal size.
2. For each fold i, train the model on all the data except for fold i.
3. Use the trained model to make predictions on the data in fold i.
4. Calculate the performance metric (e.g. accuracy, mean squared error) on the predictions for fold i.
5. Repeat steps 2-4 for each fold i, using a different fold as the validation set each time.
6. Average the performance metric over all k folds to obtain an estimate of the model’s performance.

The choice of k will depend on the size of the data set and the computational resources available. A common choice is k=10, which is known as 10-fold cross-validation. However, larger or smaller values of k can also be used.

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

1. The validation set approach?

Advantages:

• K-fold cross-validation uses all available data for training and testing, whereas the validation set approach only uses a portion of the data for training and testing. This makes k-fold cross-validation less prone to high variability in the test error estimate.
• K-fold cross-validation allows for multiple test error estimates, whereas the validation set approach only provides a single test error estimate. This can help to reduce the variance of the test error estimate and provide more reliable performance estimates for the model.

Disadvantages:

• K-fold cross-validation can be computationally expensive, especially for large data sets or complex models, because it requires fitting the model k times.
• The test error estimate from k-fold cross-validation may be biased if the data set is imbalanced or if there are other issues with the data partitioning.

2. LOOCV?

Advantages:

• K-fold cross-validation is less computationally expensive than LOOCV for large data sets, because it requires fitting the model k times instead of n times, where n is the number of observations in the data set.
• K-fold cross-validation can be more reliable than LOOCV in the presence of high variability in the data or outliers, because it averages the test error estimates over multiple subsets of the data.

Disadvantages:

• K-fold cross-validation may provide a less accurate test error estimate than LOOCV for small data sets, because it uses a smaller proportion of the data for testing in each fold.
• K-fold cross-validation may provide a biased test error estimate if the data set is imbalanced or if there are issues with the data partitioning.

 

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.

library(ISLR2)
attach(Default)
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.

set.seed(1)
glm.default = glm(default ~ income + balance, data = Default, family = binomial)

(b) 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.

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

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

glm.default2 = glm(default ~ income + balance, data = Default, subset = train.default, family = "binomial")
summary(glm.default2)

## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default, subset = train.default)
## 
## 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 category if the posterior probability is greater than 0.5.

glm.probs = predict(glm.default2, newdata = Default[-train.default,], type = 'response')
glm.pred = rep("No", length(glm.probs))
glm.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(glm.pred != Default[-train.default, ]$default)

## [1] 0.0254

• 2.5% is the fraction of obs. that are missclassified.

(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.default = sample(dim(Default)[1], dim(Default)[1] / 2)
glm.default2 = glm(default ~ income + balance, data = Default, family = "binomial", subset = train.default)
probs = predict(glm.default2, newdata = Default[-train.default, ], type = "response")
glm.pred = rep("No", length(probs))
glm.pred[probs > 0.5] <- "Yes"
mean(glm.pred != Default[-train.default, ]$default)

## [1] 0.0274

• The validation set error is 2.82%

train.default <- sample(dim(Default)[1], dim(Default)[1] / 2)
glm.default2 <- glm(default ~ income + balance, data = Default, family = "binomial", subset = train.default)
probs <- predict(glm.default2, newdata = Default[-train.default, ], type = "response")
glm.pred <- rep("No", length(probs))
glm.pred[probs > 0.5] <- "Yes"
mean(glm.pred != Default[-train.default, ]$default)

## [1] 0.0244

• The validation set error is 2.72%

train.default <- sample(dim(Default)[1], dim(Default)[1] / 2)
glm.default2 <- glm(default ~ income + balance, data = Default, family = "binomial", subset = train.default)
probs <- predict(glm.default2, newdata = Default[-train.default, ], type = "response")
glm.pred <- rep("No", length(probs))
glm.pred[probs > 0.5] <- "Yes"
mean(glm.pred != Default[-train.default, ]$default)

## [1] 0.0244

• The validation set error is 2.68%

(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.default = sample(dim(Default)[1], dim(Default)[1] / 2)
glm.default2 = glm(default ~ income + balance + student, data = Default, family = "binomial", subset = train.default)
glm.pred = rep("No", length(probs))
probs = predict(glm.default2, newdata = Default[-train.default, ], type = "response")
glm.pred[probs > 0.5] <- "Yes"
mean(glm.pred != Default[-train.default, ]$default)

## [1] 0.0278

• The validation set error is 3.18 after the adition of the student dummy variable. The rror rate did not go down.

 

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.

set.seed(1)
attach(Default)

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

(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.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

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

(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)
set.seed(1)
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 estimated st.errors obtaining from the glm() and boot() function are both pretty close.

 

Question #9. We will now consider the Boston housing data set, from the ISLR2 library.

detach(Default)
attach(Boston)

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

mu.hat = mean(medv)
mu.hat

## [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.

st.error = sd(medv)/sqrt(nrow(Boston))
st.error

## [1] 0.4088611

• the estimate of st.error is 40.89%

(c) Now estimate the standard error of ˆµ using the bootstrap. How does this compare to your answer from (b)?

set.seed(1)
boot.boston = function(data, index)
  return(mean(data[index]))

boot(medv, boot.boston, 1000)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.boston, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original      bias    std. error
## t1* 22.53281 0.007650791   0.4106622

• the st.error for (c) is 41.07%, which is close to the (b) st.error. Still, (c) is higher.

(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)].

CI = c(mu.hat - 2*0.4106622, mu.hat + 2*0.4106622)
CI

## [1] 21.71148 23.35413

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 mean from the medv one sample t-test falls in 95% confidence interval from mean by bootstrap.

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

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

## [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.

set.seed(1)
boot.boston = function(data, index)
  return(median(data[index]))

boot(medv, boot.boston, 1000)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.boston, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*     21.2 0.02295   0.3778075

• the st. esrror estimate of median is 0.3778075. It is equal to that calculated from the data.

(g) Based on this data set, provide an estimate for the tenth percentile of medv in Boston census tracts. Call this quantity ˆµ0.1. (You can use the quantile() function.)

mu_0.1 = quantile(medv, 0.1)
mu_0.1

##   10% 
## 12.75

(h) Use the bootstrap to estimate the standard error of ˆµ0.1. Comment on your findings.9. We will now consider the Boston housing data set, from the ISLR2 library.

set.seed(1)
boot.boston = function(data, index)
  return(quantile(data[index], 0.1))
boot(medv, boot.boston, 1000)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.boston, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*    12.75  0.0339   0.4767526

• standard error of 0.47675 which is pretty small compared to the percentile value.