library(ISLR2)
## Warning: package 'ISLR2' was built under R version 4.3.1
library(tidyr)
## Warning: package 'tidyr' was built under R version 4.3.1
library(moments)
library(ggplot2)
## Warning: package 'ggplot2' was built under R version 4.3.1
library(MASS)
## 
## Attaching package: 'MASS'
## The following object is masked from 'package:ISLR2':
## 
##     Boston
library(class)
library("corrplot")
## Warning: package 'corrplot' was built under R version 4.3.1
## corrplot 0.92 loaded
library(boot)
  1. Fit a logistic regression model that uses income and balance to predict default.
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
def.fit<-glm(default ~ income + balance, data = Default, family = binomial)
summary(def.fit)
## 
## Call:
## glm(formula = default ~ income + balance, family = binomial, 
##     data = Default)
## 
## 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
  1. 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.
# splitting data
set.seed(42)
ind<-sample(1:nrow(Default), size = 0.5*nrow(Default))
train<-Default[ind,]
test<-Default[-ind,]
  1. Fit a multiple logistic regression model using only the training observations.
cross.fit<-glm(default ~ income + balance, data = train, family = "binomial")
summary(cross.fit)
## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = train)
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.123e+01  5.898e-01 -19.033  < 2e-16 ***
## income       1.808e-05  6.975e-06   2.592  0.00954 ** 
## balance      5.528e-03  3.115e-04  17.746  < 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:  813.39  on 4997  degrees of freedom
## AIC: 819.39
## 
## Number of Fisher Scoring iterations: 8
  1. 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.
  2. Compute the validation set error, which is the fraction of the observations in the validation set that are mis-classified.
def_pred<-predict(cross.fit, test)
def_prob<-rep("No", length(def_pred))
def_prob[def_pred > 0.5] = "Yes"
table(def_prob, test$default)
##         
## def_prob   No  Yes
##      No  4831  124
##      Yes    6   39
mean(def_prob != test$default)
## [1] 0.026
  1. 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.
#siplt data 60/40
set.seed(42)
ind<-sample(1:nrow(Default), size = 0.6*nrow(Default))
train6<-Default[ind,]
test6<-Default[-ind,]
cross.fit6<-glm(default ~ income + balance, data = train6, family = "binomial")
summary(cross.fit6)
## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = train6)
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.151e+01  5.602e-01 -20.540  < 2e-16 ***
## income       1.681e-05  6.415e-06   2.621  0.00876 ** 
## balance      5.718e-03  2.982e-04  19.177  < 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: 1753.7  on 5999  degrees of freedom
## Residual deviance:  941.4  on 5997  degrees of freedom
## AIC: 947.4
## 
## Number of Fisher Scoring iterations: 8
def_pred6<-predict(cross.fit6, test6)
def_prob6<-rep("No", length(def_pred6))
def_prob[def_pred6 > 0.5] = "Yes"
table(def_prob6, test6$default)
##          
## def_prob6   No  Yes
##        No 3867  133
mean(def_prob6 != test6$default)
## [1] 0.03325
# split data 70/30
set.seed(42)
ind<-sample(1:nrow(Default), size = 0.7*nrow(Default))
train7<-Default[ind,]
test7<-Default[-ind,]
cross.fit7<-glm(default ~ income + balance, data = train7, family = "binomial")
summary(cross.fit)
## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = train)
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.123e+01  5.898e-01 -19.033  < 2e-16 ***
## income       1.808e-05  6.975e-06   2.592  0.00954 ** 
## balance      5.528e-03  3.115e-04  17.746  < 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:  813.39  on 4997  degrees of freedom
## AIC: 819.39
## 
## Number of Fisher Scoring iterations: 8
def_pred7<-predict(cross.fit7, test7)
def_prob7<-rep("No", length(def_pred7))
def_prob[def_pred7 > 0.5] = "Yes"
table(def_prob7, test7$default)
##          
## def_prob7   No  Yes
##        No 2904   96
mean(def_prob7 != test7$default)
## [1] 0.032
# Split data 80/20
set.seed(42)
ind<-sample(1:nrow(Default), size = 0.8*nrow(Default))
train8<-Default[ind,]
test8<-Default[-ind,]
cross.fit8<-glm(default ~ income + balance, data = train, family = "binomial")
summary(cross.fit)
## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = train)
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.123e+01  5.898e-01 -19.033  < 2e-16 ***
## income       1.808e-05  6.975e-06   2.592  0.00954 ** 
## balance      5.528e-03  3.115e-04  17.746  < 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:  813.39  on 4997  degrees of freedom
## AIC: 819.39
## 
## Number of Fisher Scoring iterations: 8
def_pred8<-predict(cross.fit8, test8)
def_prob8<-rep("No", length(def_pred8))
def_prob[def_pred8 > 0.5] = "Yes"
table(def_prob8, test8$default)
##          
## def_prob8   No  Yes
##        No 1939   61
mean(def_prob8 != test8$default)
## [1] 0.0305
  1. 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.
dummy.fit<-glm(default ~ income + balance + student, data = train, family = "binomial")
summary(dummy.fit)
## 
## Call:
## glm(formula = default ~ income + balance + student, family = "binomial", 
##     data = train)
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.052e+01  6.677e-01 -15.763   <2e-16 ***
## income      -4.771e-07  1.126e-05  -0.042   0.9662    
## balance      5.616e-03  3.171e-04  17.713   <2e-16 ***
## studentYes  -6.790e-01  3.223e-01  -2.107   0.0352 *  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 1483.8  on 4999  degrees of freedom
## Residual deviance:  809.0  on 4996  degrees of freedom
## AIC: 817
## 
## Number of Fisher Scoring iterations: 8
dummy_pred<-predict(dummy.fit, test)
dummy_prob<-rep("No", length(dummy_pred))
dummy_prob[dummy_pred > 0.5] = "Yes"
table(dummy_prob, test$default)
##           
## dummy_prob   No  Yes
##        No  4829  121
##        Yes    8   42
mean(dummy_prob != test$default)
## [1] 0.0258

Adding the student variable doesnt seem to change the error, with the student variable not indicating rejection of the null.

  1. 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(1)
def_log<-glm(default ~ income + balance, data = Default, family = 'binomial')
summary(def_log)
## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default)
## 
## 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
  1. 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_fun<- function(data = Default, index) 
  {fit <- glm(default ~ income + balance, data = Default, family = "binomial", subset = index)
    return (coef(fit))}
  1. Use the boot() function together with your boot.fn() function to estimate the standard errors of the logistic regression coefficients for income and balance.
boot(data = Default, boot_fun, R = 1000)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Default, statistic = boot_fun, 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
  1. Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function.

The bootstrap function appears to produce a marginal increase in Sd as compared to GLM, but close enough to validate its use.

  1. In Sections 5.3.2 and 5.3.3, we saw that the cv.glm() function can be used in order to compute the LOOCV test error estimate. Alternatively, one could compute those quantities using just the glm() and predict.glm() functions, and a for loop. You will now take this approach in order to compute the LOOCV error for a simple logistic regression model on the Weekly data set. Recall that in the context of classification problems, the LOOCV error is given in (5.4).
  1. Fit a logistic regression model that predicts Direction using Lag1 and Lag2.
attach(Weekly)
week_log<-glm(Direction ~ Lag1 + Lag2, data = Weekly, family = "binomial")

  1. Fit a logistic regression model that predicts Direction using Lag1 and Lag2 using all but the first observation.

  2. Use the model from (b) to predict the direction of the first observation.

You can do this by predicting that the first observation will go up if P(Direction = “Up”|Lag1, Lag2) > 0.5. Was this observation correctly classified?

set.seed(1)
sub_log<-glm(Direction ~ Lag1 + Lag2, data = Weekly[-1,], family = "binomial")
pred_sub<-predict(sub_log, Weekly[1,], type = "response") > 0.5
print(pred_sub)
##    1 
## TRUE
print(Weekly[1,])
##   Year  Lag1  Lag2   Lag3   Lag4   Lag5   Volume Today Direction
## 1 1990 0.816 1.572 -3.936 -0.229 -3.484 0.154976 -0.27      Down

Looks like the prediction was that the value would go up, the actual value was down.

  1. Write a for loop from i = 1 to i = n, where n is the number of observations in the data set, that performs each of the following steps:
  1. Fit a logistic regression model using all but the ith observation to predict Direction using Lag1 and Lag2.
  2. Compute the posterior probability of the market moving up for the ith observation.
  3. Use the posterior probability for the ith observation in order to predict whether or not the market moves up.
  4. Determine whether or not an error was made in predicting the direction for the ith observation. If an error was made, then indicate this as a 1, and otherwise indicate it as a 0.
  1. Take the average of the n numbers obtained in (d)iv in order to obtain the LOOCV estimate for the test error. Comment on the results.
err_total = rep(0, dim(Weekly)[1])
for (i in 1:(dim(Weekly)[1])) {
    ith_fit= glm(Direction ~ Lag1 + Lag2, data = Weekly[-i, ], family = binomial)
    pred_ith = predict.glm(ith_fit, Weekly[i, ], type = "response") > 0.5
    tru_ith = Weekly[i, ]$Direction == "Up"
    if (pred_ith != tru_ith) 
        err_total[i] = 1
}
sum(err_total)
## [1] 490
mean(err_total)
## [1] 0.4499541
  1. We will now perform cross-validation on a simulated data set.
  1. Generate a simulated data set as follows:
set.seed (1)
x <- rnorm(100)
y <- x - 2 * x^2 + rnorm(100)
data<-data.frame(x,y)

In this data set, what is n and what is p? Write out the model used to generate the data in equation form. n = 100, p (number of predictors) = 1, formula is y = x - 2x^2+e

  1. Create a scatterplot of X against Y . Comment on what you find.
plot(x, y)

Looks like the relationship is curvlinear, indicating a quadratic relationship.

  1. Set a random seed, and then compute the LOOCV errors that result from fitting the following four models using least squares:
  1. Y = β0 + β1X + ϵ
  2. Y = β0 + β1X + β2X2 + ϵ
  3. Y = β0 + β1X + β2X2 + β3X3 + ϵ
  4. Y = β0 + β1X + β2X2 + β3X3 + β4X4 + ϵ. Note you may find it helpful to use the data.frame() function to create a single data set containing both X and Y .
set.seed(1)
# reg fit
fit1 <- glm(y ~ x)
cv.glm(data, fit1)$delta[1]
## [1] 7.288162
# ^2 fit
fit2 <- glm(y ~ poly(x, 2))
cv.glm(data, fit2)$delta[1]
## [1] 0.9374236
# ^3 fit
fit3 <- glm(y ~ poly(x, 3))
cv.glm(data, fit3)$delta[1]
## [1] 0.9566218
# ^4 fit
fit4 <- glm(y ~ poly(x, 4))
cv.glm(data, fit4)$delta[1]
## [1] 0.9539049
  1. Repeat (c) using another random seed, and report your results. Are your results the same as what you got in (c)? Why?
set.seed(42)
cv.error <- rep(0, 4)
for (i in 1:4) {
new1.fit <- glm(y ~ poly(x, i), data = data)
cv.error[i] <- cv.glm(data , new1.fit)$delta[1]
}

cv.error
## [1] 7.2881616 0.9374236 0.9566218 0.9539049
  1. Which of the models in (c) had the smallest LOOCV error? Is this what you expected? Explain your answer.

Poly fit^2 had the least error, which was demonstrated in the quadratic relationship between x and y.

  1. Comment on the statistical significance of the coefficient estimates that results from fitting each of the models in (c) using least squares. Do these results agree with the conclusions drawn based on the cross-validation results?
summary(new1.fit)
## 
## Call:
## glm(formula = y ~ poly(x, i), data = data)
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  -1.55002    0.09591 -16.162  < 2e-16 ***
## poly(x, i)1   6.18883    0.95905   6.453 4.59e-09 ***
## poly(x, i)2 -23.94830    0.95905 -24.971  < 2e-16 ***
## poly(x, i)3   0.26411    0.95905   0.275    0.784    
## poly(x, i)4   1.25710    0.95905   1.311    0.193    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for gaussian family taken to be 0.9197797)
## 
##     Null deviance: 700.852  on 99  degrees of freedom
## Residual deviance:  87.379  on 95  degrees of freedom
## AIC: 282.3
## 
## Number of Fisher Scoring iterations: 2
  1. 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.
mu_hat<-mean(Boston$medv)
print(mu_hat)
## [1] 22.53281
  1. Provide an estimate of the standard error of ˆμ. Interpret this result. Hint: We can compute the standard error of the sample mean by dividing the sample standard deviation by the square root of the number of observations.
stan_err <- sqrt(var(Boston$medv)/nrow(Boston))
print(stan_err)
## [1] 0.4088611
  1. Now estimate the standard error of ˆμ using the bootstrap. How does this compare to your answer from (b)?
set.seed(42)
boot.fn<-function(data, indices){
  resampled<-data[indices]
  result<-mean(resampled)
  return(result)
}
result<-boot(Boston$medv, boot.fn, 1000)
print(result)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original     bias    std. error
## t1* 22.53281 0.02671186   0.4009216

Using bootstrap, the estimated std is .4, which is the approximate estimate of the original output (i.e. .408)

  1. 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$medv). Hint: You can approximate a 95% confidence interval using the formula [ˆμ − 2SE(ˆμ), ˆμ + 2SE(ˆμ)].
t.test(Boston$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
low <- mu_hat - (2* stan_err)
high<- mu_hat + (2*stan_err)
boot_con<-c(low, high)            
boot_con
## [1] 21.71508 23.35053

again, the output is very close.

  1. Based on this data set, provide an estimate, ˆμmed, for the median value of medv in the population.
med<-median(Boston$medv)
print(med)
## [1] 21.2
  1. We now would like to estimate the standard error of ˆμmed. Unfortunately, there is no simple formula for computing the standard error of the median. Instead, estimate the standard error of the median using the bootstrap. Comment on your findings.
set.seed(42)
boot.fn<-function(data, indices){
  resampled<-data[indices]
  result<-median(resampled)
  return(result)
}
result<-boot(Boston$medv, boot.fn, 1000)
print(result)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*     21.2  0.0106   0.3661785

The results are identical. WOW!

  1. 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_hat1<-quantile(Boston$medv, c(0.1))
print(mu_hat1)
##   10% 
## 12.75
  1. Use the bootstrap to estimate the standard error of ˆμ0.1. Comment on your findings.
set.seed(42)
boot.fn<-function(data, indices){
  resampled<-data[indices]
  result<-quantile(Boston$medv, c(0.1))
  return(result)
}
result<-boot(Boston$medv, boot.fn, 1000)
print(result)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*    12.75       0           0