Assignment 4

*#3. We now review k-fold cross-validation. (a) Explain how k-fold cross-validation is implemented. K-fold cross validation is where you break up the entire population of observations into “k” number of sample group observations. Then you leave one group of observations out of the calculation as your “test/validation set” and then use the other sample group observations as your “training set”. You calculate the MSE on your “test/validation set” based upon the model developed on your “training set”. A common example in Design of Experiment statistical classes is the perfect, normal, binomial distribution with n=100 observations, k=5, and each k-sample has 20 observations each. i.e. You take all the observations in the first sample set, k=1, and set aside as your test/validation set (your first fold). You build your model from the other 80 observations in the combined k=2, 3, 4, & 5 sets. You test that model against your k = 1 set, then repeat the same process for each of the sample groups (k = 2, 3, 4, & 5) of 20 observations, for a total of 5 folds. Then the MSE of each k-sample group is averaged for an overall population MSE.

2. What are the advantages and disadvantages of k-fold cross validation relative to:

1. The validation set approach? The k-fold CV has advantages in that instead of simply dividing the population in half (half training, half test/validation), it’s divided into a k-number of computationally manageable groups. With the validation set, the variability of the model is higher than with k-fold; but it can also give you a good starting point before performing the k-fold CV approach. If your population is extremely large, than dividing in half maybe all you can do in the interim while you wait on your k-fold CV or LOOCV to compute.

2. LOOCV? Advantages would be presumably a very tightly fitted model that has very low MSE. Disadvantages this could result in 1) over-fitting 2) burning out your server from computing all those MSEs on such a large population (n) size. As much as people are complaining about crypto-mining data farms straining resources, I can only imagine what would happen if all organizations started using LOOCV on their data sets. *

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.

1. Fit a logistic regression model that uses income and balance to predict default. 5.4 Exercises 221
2. 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.
2. Fit a multiple logistic regression model using only the training observations.

#load ISLR and MASS library to access Default data

library(ISLR)

library(MASS)

attach(Default)


head(Default)

##   default student   balance    income
## 1      No      No  729.5265 44361.625
## 2      No     Yes  817.1804 12106.135
## 3      No      No 1073.5492 31767.139
## 4      No      No  529.2506 35704.494
## 5      No      No  785.6559 38463.496
## 6      No     Yes  919.5885  7491.559

#data structure/characteristics

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 ...

# set seed
set.seed(997)

Default$default <- as.factor(Default$default)

#create logistic regression model using income and balance as factors to predict default

log.reg = glm(default ~ income + balance, family = binomial, data = Default)

log.reg

## 
## Call:  glm(formula = default ~ income + balance, family = binomial, 
##     data = Default)
## 
## Coefficients:
## (Intercept)       income      balance  
##  -1.154e+01    2.081e-05    5.647e-03  
## 
## Degrees of Freedom: 9999 Total (i.e. Null);  9997 Residual
## Null Deviance:       2921 
## Residual Deviance: 1579  AIC: 1585

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.

# divide population in half

set.seed(997)

train = sample(1000,500)

test <- Default[-train,]

pred.logreg <- predict(log.reg, test, type = "response")

class.logreg <- ifelse(pred.logreg > 0.5, "Yes", "No")

table(test$default, class.logreg, dnn = c("Actual", "Predicted"))

##       Predicted
## Actual   No  Yes
##    No  9153   38
##    Yes  207  102

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

#MSE with the 50/50 split

round(mean(class.logreg != test$default), 4)

## [1] 0.0258

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.

# divide population in half

set.seed(997)

train2 = sample(1000,333)

test2 <- Default[-train2,]

pred.logreg <- predict(log.reg, test2, type = "response")

class.logreg <- ifelse(pred.logreg > 0.5, "Yes", "No")

table(test2$default, class.logreg, dnn = c("Actual", "Predicted"))

##       Predicted
## Actual   No  Yes
##    No  9309   38
##    Yes  216  104

#MSE with the 1/3

round(mean(class.logreg != test2$default), 4)

## [1] 0.0263

# divide population 

set.seed(997)

train3 = sample(1000,800)

test3 <- Default[-train3,]

pred.logreg <- predict(log.reg, test3, type = "response")

class.logreg <- ifelse(pred.logreg > 0.5, "Yes", "No")

table(test3$default, class.logreg, dnn = c("Actual", "Predicted"))

##       Predicted
## Actual   No  Yes
##    No  8858   36
##    Yes  205  101

#MSE with the 900

round(mean(class.logreg != test3$default), 4)

## [1] 0.0262

4. 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.

Default$student <- as.factor(Default$student)

set.seed(997)

log.reg2 = glm(default ~ income + balance + student, family = binomial, data = Default, subset = train)

log.reg2

## 
## Call:  glm(formula = default ~ income + balance + student, family = binomial, 
##     data = Default, subset = train)
## 
## Coefficients:
## (Intercept)       income      balance   studentYes  
##  -1.099e+01    6.314e-06    6.187e-03   -4.810e-01  
## 
## Degrees of Freedom: 499 Total (i.e. Null);  496 Residual
## Null Deviance:       192.6 
## Residual Deviance: 97.69     AIC: 105.7

set.seed(997)

train4 = sample(1000,500)

test4 <- Default[-train4,]

pred.logreg <- predict(log.reg, test4, type = "response")

class.logreg <- ifelse(pred.logreg > 0.5, "Yes", "No")

table(test4$default, class.logreg, dnn = c("Actual", "Predicted"))

##       Predicted
## Actual   No  Yes
##    No  9153   38
##    Yes  207  102

#MSE with student factor included

round(mean(class.logreg!=test4$default),4)

## [1] 0.0258

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(997)

train5 <- sample(dim(Default)[1], dim(Default)[1]/2)

fit.glm <- glm(default ~ income + balance, data = Default, family = "binomial", subset = train5)

summary(fit.glm)

## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default, subset = train5)
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.185e+01  6.323e-01 -18.744  < 2e-16 ***
## income       2.382e-05  7.063e-06   3.373 0.000743 ***
## balance      5.807e-03  3.317e-04  17.507  < 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: 1483.83  on 4999  degrees of freedom
## Residual deviance:  798.72  on 4997  degrees of freedom
## AIC: 804.72
## 
## Number of Fisher Scoring iterations: 8

2. 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 <- glm(default ~ income + balance, data = data[index, ], family = "binomial") 
  return(coef(fit.glm)[2:3])
  }

fit.glm

## 
## Call:  glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default, subset = train5)
## 
## Coefficients:
## (Intercept)       income      balance  
##  -1.185e+01    2.382e-05    5.807e-03  
## 
## Degrees of Freedom: 4999 Total (i.e. Null);  4997 Residual
## Null Deviance:       1484 
## Residual Deviance: 798.7     AIC: 804.7

3. 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(998)

results <- boot(Default, boot.fn, R = 1000)

results

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Default, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##         original       bias     std. error
## t1* 2.080898e-05 6.044926e-08 4.884814e-06
## t2* 5.647103e-03 1.079285e-05 2.273975e-04

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

glm_se <- summary(fit.glm)$coefficients[, "Std. Error"]

bootstrap_se <- apply(results$t, 2, sd)

cat("GLM Standard Errors:\n", glm_se, "\n")

## GLM Standard Errors:
##  0.6322513 7.062876e-06 0.0003317074

cat("Bootstrap Standard Errors:\n", bootstrap_se, "\n")

## Bootstrap Standard Errors:
##  4.884814e-06 0.0002273975

The standard errors are slightly better with the bootstrap methods.

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

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

library(ISLR2)

## 
## Attaching package: 'ISLR2'

## The following object is masked from 'package:MASS':
## 
##     Boston

## The following objects are masked from 'package:ISLR':
## 
##     Auto, Credit

data(Boston)

mean_medv <- mean(Boston$medv)
mean_medv

## [1] 22.53281

2. 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.

n <- length(Boston$medv)
s <- sd(Boston$medv)
se_sample <- s/sqrt(n)
se_sample

## [1] 0.4088611

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

set.seed(997) 
B <- 1000
boot_means <- replicate(B, {
  sample_data <- sample(Boston$medv, n, replace = TRUE)
  mean(sample_data)
})
boot_se1 <- sd(boot_means)

boot_se1

## [1] 0.413037

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 a 95 % confidence interval using the formula [ˆμ − 2SE(ˆμ), ˆμ + 2SE(ˆμ)].

CI_lower <- mean_medv - 2 * boot_se1

CI_upper <- mean_medv + 2 * boot_se1

c(CI_lower, CI_upper)

## [1] 21.70673 23.35888

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

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

med_medv <- median(Boston$medv)
med_medv

## [1] 21.2

6. 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.

boot_medians <- replicate(B, {
  sample_data <- sample(Boston$medv, n, replace = TRUE)
  median(sample_data)
})
boot_se_median <- sd(boot_medians)
boot_se_median

## [1] 0.3849359

7. 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(Boston$medv, 0.1)
mu_0_1

##   10% 
## 12.75

8. Use the bootstrap to estimate the standard error of ˆμ0.1. Comment on your findings.#

boot_tenth <- replicate(B, {
  sample_data <- sample(Boston$medv, n, replace = TRUE)
  quantile(sample_data, 0.1)
})
boot_se_tenth <- sd(boot_tenth)
boot_se_tenth

## [1] 0.4820805