Assignment 4

Problem 3

We now review k-fold cross-validation. A) Explain how k-fold cross-validation is implemented. - This approach involves randomly dividing the set of observations into k groups (folds) of equal size. The first fold is treated as a validation set. 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,. The k-fold CV estimate is computed by averaging these values

What are the advantages and disadvantages of k-fold crossvalidation relative to:
1. The validation set approach?
  - Advantage: It is simple and easy for implementation.
  - Disadvantage: MSE can be very variable and only a subset of observations can be used.
2. LOOCV?
- Advantage: It is less bias
- Disadvantage: It is computationally intensive

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.

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

library(ISLR)

data("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

logmodel1 = glm(default ~ balance + income, data = Default, family = binomial)

summary(logmodel1)

## 
## Call:
## glm(formula = default ~ balance + income, 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 ***
## balance      5.647e-03  2.274e-04  24.836  < 2e-16 ***
## income       2.081e-05  4.985e-06   4.174 2.99e-05 ***
## ---
## 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.
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. 2.74% Test error rate

#i
trainDefault = sample(dim(Default)[1], dim(Default)[1]*0.50)
testDefault = Default[-trainDefault, ]

#ii
LOGmodel2 = glm(default ~ balance + income, data = Default, family = binomial, subset = trainDefault)

#iii
log.prob_def = predict(LOGmodel2, testDefault, type = "response")
log.pred_def = rep("No", dim(Default)[1]*0.50)
log.pred_def[log.prob_def > 0.5] = "Yes"
table(log.pred_def, testDefault$default)

##             
## log.pred_def   No  Yes
##          No  4800  121
##          Yes   25   54

#iv
mean(log.pred_def !=testDefault$default)

## [1] 0.0292

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.

All three splits gave me a different testing error which means observations differ across each training and validation set.

#i
trainDefault = sample(dim(Default)[1], dim(Default)[1]*0.50)
testDefault = Default[-trainDefault, ]

#ii
LOGmodel2 = glm(default ~ balance + income, data = Default, family = binomial, subset = trainDefault)

#iii
log.prob_def = predict(LOGmodel2, testDefault, type = "response")
log.pred_def = rep("No", dim(Default)[1]*0.50)
log.pred_def[log.prob_def > 0.5] = "Yes"
table(log.pred_def, testDefault$default)

##             
## log.pred_def   No  Yes
##          No  4808  121
##          Yes   17   54

#iv
mean(log.pred_def !=testDefault$default)

## [1] 0.0276

#i
trainDefault = sample(dim(Default)[1], dim(Default)[1]*0.50)
testDefault = Default[-trainDefault, ]

#ii
LOGmodel2 = glm(default ~ balance + income, data = Default, family = binomial, subset = trainDefault)

#iii
log.prob_def = predict(LOGmodel2, testDefault, type = "response")
log.pred_def = rep("No", dim(Default)[1]*0.50)
log.pred_def[log.prob_def > 0.5] = "Yes"
table(log.pred_def, testDefault$default)

##             
## log.pred_def   No  Yes
##          No  4810  119
##          Yes   21   50

#iv
mean(log.pred_def !=testDefault$default)

## [1] 0.028

#i
trainDefault = sample(dim(Default)[1], dim(Default)[1]*0.50)
testDefault = Default[-trainDefault, ]

#ii
LOGmodel2 = glm(default ~ balance + income, data = Default, family = binomial, subset = trainDefault)

#iii
log.prob_def = predict(LOGmodel2, testDefault, type = "response")
log.pred_def = rep("No", dim(Default)[1]*0.50)
log.pred_def[log.prob_def > 0.5] = "Yes"
table(log.pred_def, testDefault$default)

##             
## log.pred_def   No  Yes
##          No  4817  117
##          Yes   15   51

#iv
mean(log.pred_def !=testDefault$default)

## [1] 0.0264

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 errorrate.

2.7% error rate, not much differenct than the other splits. Adding the student variable doesn’t have much impact on the error rate.

#i
trainDefault = sample(dim(Default)[1], dim(Default)[1]*0.50)
testDefault = Default[-trainDefault, ]

#ii
LOGmodel2 = glm(default ~ balance + income + student, data = Default, family = binomial, subset = trainDefault)

#iii
log.prob_def = predict(LOGmodel2, testDefault, type = "response")
log.pred_def = rep("No", dim(Default)[1]*0.50)
log.pred_def[log.prob_def > 0.5] = "Yes"
table(log.pred_def, testDefault$default)

##             
## log.pred_def   No  Yes
##          No  4808  118
##          Yes   19   55

#iv
mean(log.pred_def !=testDefault$default)

## [1] 0.0274

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:

using the bootstrap, and
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.

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.

The estimated standard errors for the coefficients for balance and income are 2.274e-04. and 4.985e-06

LOGmodel3 = glm(default ~ balance + income, data = Default, family = binomial)

summary(LOGmodel3)

## 
## Call:
## glm(formula = default ~ balance + income, 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 ***
## balance      5.647e-03  2.274e-04  24.836  < 2e-16 ***
## income       2.081e-05  4.985e-06   4.174 2.99e-05 ***
## ---
## 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

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

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, 100)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Default, statistic = boot.fn, R = 100)
## 
## 
## Bootstrap Statistics :
##          original        bias     std. error
## t1* -1.154047e+01  7.361967e-03 3.471039e-01
## t2*  5.647103e-03 -4.198188e-06 1.900556e-04
## t3*  2.080898e-05 -1.969407e-07 4.499751e-06

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

The estimated standard errors for the coefficients for income and balance are 2.294044e-04 and 3.875525e-06.

Problem 9

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

library(MASS)
data("Boston")
summary(Boston)

##       crim                zn             indus            chas        
##  Min.   : 0.00632   Min.   :  0.00   Min.   : 0.46   Min.   :0.00000  
##  1st Qu.: 0.08205   1st Qu.:  0.00   1st Qu.: 5.19   1st Qu.:0.00000  
##  Median : 0.25651   Median :  0.00   Median : 9.69   Median :0.00000  
##  Mean   : 3.61352   Mean   : 11.36   Mean   :11.14   Mean   :0.06917  
##  3rd Qu.: 3.67708   3rd Qu.: 12.50   3rd Qu.:18.10   3rd Qu.:0.00000  
##  Max.   :88.97620   Max.   :100.00   Max.   :27.74   Max.   :1.00000  
##       nox               rm             age              dis        
##  Min.   :0.3850   Min.   :3.561   Min.   :  2.90   Min.   : 1.130  
##  1st Qu.:0.4490   1st Qu.:5.886   1st Qu.: 45.02   1st Qu.: 2.100  
##  Median :0.5380   Median :6.208   Median : 77.50   Median : 3.207  
##  Mean   :0.5547   Mean   :6.285   Mean   : 68.57   Mean   : 3.795  
##  3rd Qu.:0.6240   3rd Qu.:6.623   3rd Qu.: 94.08   3rd Qu.: 5.188  
##  Max.   :0.8710   Max.   :8.780   Max.   :100.00   Max.   :12.127  
##       rad              tax           ptratio          black       
##  Min.   : 1.000   Min.   :187.0   Min.   :12.60   Min.   :  0.32  
##  1st Qu.: 4.000   1st Qu.:279.0   1st Qu.:17.40   1st Qu.:375.38  
##  Median : 5.000   Median :330.0   Median :19.05   Median :391.44  
##  Mean   : 9.549   Mean   :408.2   Mean   :18.46   Mean   :356.67  
##  3rd Qu.:24.000   3rd Qu.:666.0   3rd Qu.:20.20   3rd Qu.:396.23  
##  Max.   :24.000   Max.   :711.0   Max.   :22.00   Max.   :396.90  
##      lstat            medv      
##  Min.   : 1.73   Min.   : 5.00  
##  1st Qu.: 6.95   1st Qu.:17.02  
##  Median :11.36   Median :21.20  
##  Mean   :12.65   Mean   :22.53  
##  3rd Qu.:16.95   3rd Qu.:25.00  
##  Max.   :37.97   Max.   :50.00

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

attach(Boston)
mean.medv = mean(medv)
mean.medv

## [1] 22.53281

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.

stderr.mean = sd(medv)/sqrt(length(medv))
stderr.mean

## [1] 0.4088611

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

standard error estimate in (b) is simmilar to the bootstap standard error

boot.fn2 = function(data, index) return(mean(data[index]))
boot2 = boot(medv, boot.fn2, 100)
boot2

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn2, R = 100)
## 
## 
## Bootstrap Statistics :
##     original     bias    std. error
## t1* 22.53281 0.07417391   0.4009085

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)].

t.test confidence interval is simmilar to the bootstap confidence interval

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

CI.bos = c(22.53 - 2 * 0.4174872, 22.53 + 2 * 0.4174872)
CI.bos

## [1] 21.69503 23.36497

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

median.medv = median(medv)
median.medv

## [1] 21.2

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.

estimated median value of 21.2 fromm bootstrap which is sames as the value in (e), and the standard error of 0.377

boot.fn3 = function(data, index) return(median(data[index]))
boot3 = boot(medv, boot.fn3, 1000)
boot3

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn3, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*     21.2 -0.0027   0.3856247

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

tenth.medv = quantile(medv, c(0.1))
tenth.medv

##   10% 
## 12.75

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

estimated quantile value of 12.75 from bootstap which is same as the value in (g), and the standard error of 0.487

boot.fn4 = function(data, index) return(quantile(data[index], c(0.1)))
boot4 = boot(medv, boot.fn4, 1000)
boot4

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn4, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original   bias    std. error
## t1*    12.75 -0.00135   0.5012304

Assignment 4

Nicole Hayek

2022-12-03

Problem 3

Problem 5

Problem 6

Problem 9