Chapter 05 (page 197): 3, 5, 6, 9

3.) k-fold cross-validation.

Explain how k-fold cross-validation is implemented.

K-fold CV is an alternative to LOOCV. This method randomly divides a set of observations into K groups, or folds, of equal-ish size. The first fold is a validation set, and the method is fit on the rest of the k − 1 folds. The mean squared error, MSE1, is then computed on the observations in the held-out fold. This procedure is repeated k times; each time, a different group of observations is treated as a validation set. This process results in k estimates of the test error, MSE1, MSE2,…, MSEk

What are the advantages and disadvantages of k-fold crossvalidation relative to:
1. The validation set approach? Cons- -The validation estimate of the test error can be highly variable, depending on how precise the included observations are in the training set and which observations are in the validation set. -Since a subset of observations are used to fit the model (training), there are fewer observations to train on, decreasing the models performance. Pros- simple and easy to implement

LOOCV ? Similar to Validation Set Approach but tries to address the disadvantages listed above..

This approach has two drawbacks compared to k-fold cross-validation.

it requires fitting the potentially computationally expensive model n times compared to k-fold cross-validation which requires the model to be fitted only k times. 2. the LOOCV cross-validation approach may give approximately unbiased estimates of the test error, since each training set contains n−1 observations; however, this approach has higher variance than k-fold cross-validation (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). So, there is a bias-variance trade-off associated with the choice of k in k-fold cross-validation; typically using k=5 or k=10 yield test error rate estimates that suffer neither from excessively high bias nor from very high variance. Advantages inculde: 1. Less bias, b/c tends not to overestimate the test error rate as much as the validation set approach does 2. Yields the same result when ran multiple times b/c there is no randomness in the training/validation set splits.

5.) Applied

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 4.0.3

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

attach(Default)
set.seed(1)
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

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. 5.4 Exercises 199

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

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.

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

Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified. Validation Set Approach yields a Error rate = 2.86% (below)

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

## [1] 0.0254

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) #splitting
fit.glm=glm(default ~ income + balance, data = Default, family = "binomial", subset = train) #multiple regression w/ training subset
probs=predict(fit.glm, newdata = Default[-train, ], type = "response") #predicting
pred.glm <- rep("No", length(probs))
pred.glm[probs > 0.5] <- "Yes"
mean(pred.glm != Default[-train, ]$default) #compute error

## [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

After running the three, you can see the test error can be variable depending on which observations are in the training set and which observations are in the validation set.

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)
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

No improvement in the Validation Set Estimate of the test error seen when adding the dummy “student” variable in the model.

6.) Applied - glm() & Boot Strap

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 computes 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.

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.

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 β0, β1 and β2 are respectively 0.4347564, 4.985167210^{-6} and 2.273731410^{-4}.

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)

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 standard error is shown in the last column for the coefficients of β0, β1 and β2

D.) Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function. The standard error from both the glm() method and the bootstrap method are close.

9.) Applied -Boston Data Set

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 μ^.

library(MASS)
attach(Boston)

mu.hat=mean(medv)
mu.hat

## [1] 22.53281

se.hat=sd(medv) / sqrt(dim(Boston)[1])
se.hat

## [1] 0.4088611

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

set.seed(1)
boot.fn <- function(data, index) {
    mu <- mean(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* 22.53281 0.007650791   0.4106622

The bootstrap estimated standard error of μ^ of 0.4119 is very close to the estimate found in (b) of 0.4089.

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).

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

## [1] 21.7062 23.3538

The bootstrap confidence interval is similar to the one provided by the t.test() function.

E.) Based on this data set, provide an estimate, μ^med, 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 μ^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

an estimated median value of 21.2 which is equal to the value obtained in (e), with a standard error of 0.3874 which is relatively small compared to median value.

G. ) Based on this data set, provide an estimate for the tenth percentile of “medv” in Boston suburbs. Call this quantity μ^0.1.

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

##   10% 
## 12.75

H. Use the bootstrap to estimate the standard error of μ^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

Estimate of 10th percentile value is 12.75, equal to the value previously seen in G. w/ a standard error of 0.5113 which is small compared to percentile value

Assignment 4 Ch. 5

Brittany Parffrey

3/23/2021

Chapter 05 (page 197): 3, 5, 6, 9

3.) k-fold cross-validation.

5.) Applied

6.) Applied - glm() & Boot Strap

9.) Applied -Boston Data Set