Data Mining Homework 4

Problem 3

We now review k-fold cross-validation.

Explain how k-fold cross-validation is implemented.

We use K-fold cross validation to estimate the test error by averaging the K resulting MSE estimates. Thtis is done by taking the n observations and randomly diving them into k groups of length n/k that don’t overlap. Using these groups as validation sets, and the remeinder act as a training set.

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

The validation set approach?

Using the validation approach has 2 drawbacks comapred to K-fold cross validation. The first drawback is that the validation estimate for the test error has high variability. It depends on what is included in the training set and which are in the observed sets for the validation set. The second drawback involves that only a subset of observations are used to fit the model. When we use fewer observations we can expect an overestimation on the data set.

LOOCV?

Using the LOOCV cross validation approach is a unique case of K-fold cross validation with k=n. This approach also has two drawbacks compared to normal K-fold cross validation. The first issue is that it requires us to fit our model n times compared to only needing to fit the model k times. Secondly, the LOOCV cross validation approach can give unbiased estimates of the test error since each training set contains n-1 observations. The approach his a higher variance though than the original K-fold cross validation. With the tradeoff associated with the choice of K in K-fold cross validaiton, we can see using k = 5 or k = 10 yeilds test error rates that estimate neither high bias or high variance.

Problem 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.

Part A

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

library(ISLR)
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

Part B

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.
Fit a multiple logistic regression model using only the training observations.
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.
Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified.

i

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

ii

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  
## -2.3583  -0.1268  -0.0475  -0.0165   3.8116  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.208e+01  6.658e-01 -18.148   <2e-16 ***
## income       1.858e-05  7.573e-06   2.454   0.0141 *  
## balance      6.053e-03  3.467e-04  17.457   <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: 1457.0  on 4999  degrees of freedom
## Residual deviance:  734.4  on 4997  degrees of freedom
## AIC: 740.4
## 
## Number of Fisher Scoring iterations: 8

iii

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

iv

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

## [1] 0.0286

This means we have a test error of 2.86%

Part 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.

# 1st split
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] 0.0236

# 2nd split
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] 0.028

# 3rd split
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] 0.0268

We can see from adjusting our set a few times that our test error rate varies based on what observations are.

Part 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 <- 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] 0.0264

Even with adding the “dummy variable” for student there does not seem to be a substaintial change to the test error so we determine that there does not seem to be a reduction in the test error due to this change.

Problem 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.

Part 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

From this we obtain B1, B2, B3 to equal .435, 4.99 x 10^(-6), 2.27 x 10^(-4) respectively.

Part 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))
}

Part 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 -8.008379e-03 4.239273e-01
## t2*  2.080898e-05  5.870933e-08 4.582525e-06
## t3*  5.647103e-03  2.299970e-06 2.267955e-04

Using bootstrap we get the estimates of the standard erros with coefficients B0, B1, B2 being .424, 4.58 x 10^(-6), 2.27 x 10^(-4) respectively.

Part D

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

The standard errors between between the glm and our bootstrap function are relatively the same. with the biggest difference being the first error with B0’s.

Problem 9

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

Part A

Based on this data set, provide an estimate for the population mean of medv. Call this estimate $\hat\mu$.

library(MASS)
attach(Boston)
mu.hat = mean(medv)
mu.hat

## [1] 22.53281

Part B

Provide an estimate of the standard error of $\hat\mu$. 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.hat <- sd(medv) / sqrt(dim(Boston)[1])
se.hat

## [1] 0.4088611

Part C

Now estimate the standard error of $\hat\mu$ 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 0.008517589   0.4119374

With our estimated standard error of $\hat\mu$ from (b) of .409 is very close to our bootstrap estimated value of .412

Part 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 [$\hat\mu$ - 2SE($\hat\mu$), $\hat\mu$ + 2SE($\hat\mu$)].

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

Our bootstrap confidence interval is close to the provided C.I. by the t.test() function that we have in R.

Part E

Based on this data set, provide an estimate, $\hat\mu$med, for the median value of medv in the population.

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

## [1] 21.2

Part F

We now would like to estimate the standard error of $\hat\mu_{med}$. 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 -0.0098   0.3874004

Through our function we obtain an estimated median value of 21.2 which is the same value that we obtained through median of the function. With a standard error of .387 we see that it’s small compared to the middle value.

Part G

Based on this data set, provide an estimate for the tenth percentile of medv in Boston suburbs. Call this quantity $$0.1. (You can use the quantile() function.)

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

##   10% 
## 12.75

Part H

Use the bootstrap to estimate the standard error of $$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 0.00515   0.5113487

Similarly with (e) and (f), we get the same value from our boot function versus the value we get from (g), 12.75. With a standard error of .511 we get that our estimation is relatively close to the 10th percentile.

Data Mining Homework 4

John Thomas

March 5, 2019

Problem 3

Problem 5

Part A

Part B

i

ii

iii

iv

Part C

Part D

Problem 6

Part A

Part B

Part C

Part D

Problem 9

Part A

Part B

Part C

Part D

Part E

Part F

Part G

Part H