Assignment #4

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

Explain how k-fold cross-validation is implemented.

the K-fold cross-validation is done by putting sets of of observations and splitting them in a random manner into k non-overlapping data sets. each data set acts as a validatior of the other data set the rest are known as training sets.

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

The validation set approach? -the approach is done by separating the training sets into two different sets. This method could cause problems depending on which observations are included in each set because it can end up giving a variable test error rate which is not a good thing. The validation set error rate could create issues by overestimating the test error rate for the whole data set.
LOOCV? -LOOCV is a special type of k-fold cross validation. This type of cross-validation is the most complicated because the model must fit n times. The method also has a higher variance with a lower bias than the k-fold CV.

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

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

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

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

iii. 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.

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  4812  109
##          Yes   25   54

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

mean(log.pred_def !=testDefault$default)

## [1] 0.0268

the test error rate comes out to 2.74%

Repeat the process in (b) thre times, using three different splits of the observations into a training set and a validation set. Comment on the results obtained.

trainDefault = sample(dim(Default)[1], dim(Default)[1]*0.50)
testDefault = Default[-trainDefault, ]
LOGmodel2 = glm(default ~ balance + income, data = Default, family = binomial, subset = trainDefault)
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  4821  124
##          Yes    9   46

mean(log.pred_def !=testDefault$default)

## [1] 0.0266

trainDefault = sample(dim(Default)[1], dim(Default)[1]*0.50)
testDefault = Default[-trainDefault, ]
LOGmodel2 = glm(default ~ balance + income, data = Default, family = binomial, subset = trainDefault)
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  4820  102
##          Yes   28   50

mean(log.pred_def !=testDefault$default)

## [1] 0.026

trainDefault = sample(dim(Default)[1], dim(Default)[1]*0.50)
testDefault = Default[-trainDefault, ]
LOGmodel2 = glm(default ~ balance + income, data = Default, family = binomial, subset = trainDefault)
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  4822  105
##          Yes   15   58

mean(log.pred_def !=testDefault$default)

## [1] 0.024

the above functions are the three different splits to validate info

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.

trainDefault = sample(dim(Default)[1], dim(Default)[1]*0.50)
testDefault = Default[-trainDefault, ]
LOGmodel2 = glm(default ~ balance + income + student, data = Default, family = binomial, subset = trainDefault)
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  4809  122
##          Yes   20   49

mean(log.pred_def !=testDefault$default)

## [1] 0.0284

the test error comes out to 2.82% which is not too different from the other splits. the variable added did not cause a big impacct to the error rate

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.

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.

library(ISLR)
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

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

## 
## 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 standard error coefficients for incime and balance are 2.274e-04 and 4.985e-06

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 toestimate 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  8.556378e-03 4.122015e-01
## t2*  5.647103e-03 -4.116657e-06 2.226242e-04
## t3*  2.080898e-05 -3.993598e-07 4.186088e-06

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

the standard errors found come out to be 4.122015e-01,2.226242e-04,4.186088e-06 using the bootstrap function

9.9.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 ˆμ.

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

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

## [1] 22.53281

the output comes out to 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.

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)?

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 0.02923123   0.3997709

there is a standard error of 39.97% which makes it an apporiate comparison based on the formulas used

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)].

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.

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 -0.0146   0.3804952

we find a median of 21.2 with a standard error of 38.05%

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.

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 0.00575   0.5188339

we see that the quantile value same as above of 12.75 but with a diffrent standard error of 51.88%

Assignment #4

oscar cancino

3/9/2022