AS3-3 Predicting Loan Repayment

rm(list=ls(all=T))
options(digits=4, scipen=12)
library(magrittr)

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Introduction

議題：議題：使用貸款人的資料，預測他會不會還款

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1 資料整理 Preparing the Dataset

【1.1 基礎機率】What proportion of the loans in the dataset were not paid in full?

loans = read.csv("loans.csv")
summary(loans)

 credit.policy                 purpose        int.rate      installment    log.annual.inc       dti             fico    
 Min.   :0.000   all_other         :2331   Min.   :0.060   Min.   : 15.7   Min.   : 7.55   Min.   : 0.00   Min.   :612  
 1st Qu.:1.000   credit_card       :1262   1st Qu.:0.104   1st Qu.:163.8   1st Qu.:10.56   1st Qu.: 7.21   1st Qu.:682  
 Median :1.000   debt_consolidation:3957   Median :0.122   Median :268.9   Median :10.93   Median :12.66   Median :707  
 Mean   :0.805   educational       : 343   Mean   :0.123   Mean   :319.1   Mean   :10.93   Mean   :12.61   Mean   :711  
 3rd Qu.:1.000   home_improvement  : 629   3rd Qu.:0.141   3rd Qu.:432.8   3rd Qu.:11.29   3rd Qu.:17.95   3rd Qu.:737  
 Max.   :1.000   major_purchase    : 437   Max.   :0.216   Max.   :940.1   Max.   :14.53   Max.   :29.96   Max.   :827  
                 small_business    : 619                                   NA's   :4                                    
 days.with.cr.line   revol.bal         revol.util    inq.last.6mths   delinq.2yrs        pub.rec      not.fully.paid
 Min.   :  179     Min.   :      0   Min.   :  0.0   Min.   : 0.00   Min.   : 0.000   Min.   :0.000   Min.   :0.00  
 1st Qu.: 2820     1st Qu.:   3187   1st Qu.: 22.7   1st Qu.: 0.00   1st Qu.: 0.000   1st Qu.:0.000   1st Qu.:0.00  
 Median : 4140     Median :   8596   Median : 46.4   Median : 1.00   Median : 0.000   Median :0.000   Median :0.00  
 Mean   : 4562     Mean   :  16914   Mean   : 46.9   Mean   : 1.57   Mean   : 0.164   Mean   :0.062   Mean   :0.16  
 3rd Qu.: 5730     3rd Qu.:  18250   3rd Qu.: 71.0   3rd Qu.: 2.00   3rd Qu.: 0.000   3rd Qu.:0.000   3rd Qu.:0.00  
 Max.   :17640     Max.   :1207359   Max.   :119.0   Max.   :33.00   Max.   :13.000   Max.   :5.000   Max.   :1.00  
 NA's   :29                          NA's   :62      NA's   :29      NA's   :29       NA's   :29                    

table(loans$not.fully.paid)

   0    1 
8045 1533 

1533/nrow(loans)

[1] 0.1601

# 0.1600543

【1.2 檢查缺項】Which of the following variables has at least one missing observation?

# log.annual.inc
# days.with.cr.line
# revol.util
# inq.last.6mths
# delinq.2yrs
# pub.rec

【1.3 決定是否要補缺項】Which of the following is the best reason to fill in the missing values for these variables instead of removing observations with missing data?

missing = subset(loans, is.na(log.annual.inc) | is.na(days.with.cr.line) | is.na(revol.util) | is.na(inq.last.6mths) | is.na(delinq.2yrs) | is.na(pub.rec))
nrow(missing)

[1] 62

table(missing$not.fully.paid)

 0  1 
50 12 

# We want to be able to predict risk for all borrowers, instead of just the ones with all data reported.

【1.4 補缺項工具】What best describes the process we just used to handle missing values?

# install.packages("mice")
library(mice)
set.seed(144)
vars.for.imputaton = setdiff(names(loans), "not.fully.paid")
imputed = complete(mice(loans[vars.for.imputaton]))

 iter imp variable
  1   1  log.annual.inc  days.with.cr.line  revol.util  inq.last.6mths  delinq.2yrs  pub.rec
  1   2  log.annual.inc  days.with.cr.line  revol.util  inq.last.6mths  delinq.2yrs  pub.rec
  1   3  log.annual.inc  days.with.cr.line  revol.util  inq.last.6mths  delinq.2yrs  pub.rec
  1   4  log.annual.inc  days.with.cr.line  revol.util  inq.last.6mths  delinq.2yrs  pub.rec
  1   5  log.annual.inc  days.with.cr.line  revol.util  inq.last.6mths  delinq.2yrs  pub.rec
  2   1  log.annual.inc  days.with.cr.line  revol.util  inq.last.6mths  delinq.2yrs  pub.rec
  2   2  log.annual.inc  days.with.cr.line  revol.util  inq.last.6mths  delinq.2yrs  pub.rec
  2   3  log.annual.inc  days.with.cr.line  revol.util  inq.last.6mths  delinq.2yrs  pub.rec
  2   4  log.annual.inc  days.with.cr.line  revol.util  inq.last.6mths  delinq.2yrs  pub.rec
  2   5  log.annual.inc  days.with.cr.line  revol.util  inq.last.6mths  delinq.2yrs  pub.rec
  3   1  log.annual.inc  days.with.cr.line  revol.util  inq.last.6mths  delinq.2yrs  pub.rec
  3   2  log.annual.inc  days.with.cr.line  revol.util  inq.last.6mths  delinq.2yrs  pub.rec
  3   3  log.annual.inc  days.with.cr.line  revol.util  inq.last.6mths  delinq.2yrs  pub.rec
  3   4  log.annual.inc  days.with.cr.line  revol.util  inq.last.6mths  delinq.2yrs  pub.rec
  3   5  log.annual.inc  days.with.cr.line  revol.util  inq.last.6mths  delinq.2yrs  pub.rec
  4   1  log.annual.inc  days.with.cr.line  revol.util  inq.last.6mths  delinq.2yrs  pub.rec
  4   2  log.annual.inc  days.with.cr.line  revol.util  inq.last.6mths  delinq.2yrs  pub.rec
  4   3  log.annual.inc  days.with.cr.line  revol.util  inq.last.6mths  delinq.2yrs  pub.rec
  4   4  log.annual.inc  days.with.cr.line  revol.util  inq.last.6mths  delinq.2yrs  pub.rec
  4   5  log.annual.inc  days.with.cr.line  revol.util  inq.last.6mths  delinq.2yrs  pub.rec
  5   1  log.annual.inc  days.with.cr.line  revol.util  inq.last.6mths  delinq.2yrs  pub.rec
  5   2  log.annual.inc  days.with.cr.line  revol.util  inq.last.6mths  delinq.2yrs  pub.rec
  5   3  log.annual.inc  days.with.cr.line  revol.util  inq.last.6mths  delinq.2yrs  pub.rec
  5   4  log.annual.inc  days.with.cr.line  revol.util  inq.last.6mths  delinq.2yrs  pub.rec
  5   5  log.annual.inc  days.with.cr.line  revol.util  inq.last.6mths  delinq.2yrs  pub.rec

loans[vars.for.imputaton] = imputed
# We predicted missing variable values using the available independent variables for each observation.

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2 建立模型 Prediction Models

【2.1 顯著性】Which independent variables are significant in our model?

set.seed(144)
library(caTools)
split = sample.split(loans$not.fully.paid, SplitRatio = 0.7)
train = subset(loans, split==T)
test = subset(loans, split==F)
loansLog = glm(not.fully.paid ~ . , data = train, family = "binomial")
summary(loansLog)

Call:
glm(formula = not.fully.paid ~ ., family = "binomial", data = train)

Deviance Residuals: 
   Min      1Q  Median      3Q     Max  
-2.187  -0.621  -0.495  -0.361   2.639  

Coefficients:
                             Estimate  Std. Error z value     Pr(>|z|)    
(Intercept)                9.15800765  1.55514067    5.89 0.0000000039 ***
credit.policy             -0.34924658  0.10082598   -3.46      0.00053 ***
purposecredit_card        -0.61439777  0.13440351   -4.57 0.0000048472 ***
purposedebt_consolidation -0.32165576  0.09182845   -3.50      0.00046 ***
purposeeducational         0.13578079  0.17527403    0.77      0.43853    
purposehome_improvement    0.17435349  0.14791842    1.18      0.23851    
purposemajor_purchase     -0.48141854  0.20079306   -2.40      0.01650 *  
purposesmall_business      0.41343357  0.14183237    2.91      0.00356 ** 
int.rate                   0.62211379  2.08490225    0.30      0.76541    
installment                0.00127297  0.00020918    6.09 0.0000000012 ***
log.annual.inc            -0.43129368  0.07145262   -6.04 0.0000000016 ***
dti                        0.00462735  0.00549967    0.84      0.40013    
fico                      -0.00929381  0.00170831   -5.44 0.0000000532 ***
days.with.cr.line          0.00000219  0.00001588    0.14      0.89044    
revol.bal                  0.00000303  0.00000117    2.60      0.00928 ** 
revol.util                 0.00191601  0.00153293    1.25      0.21134    
inq.last.6mths             0.08073987  0.01586849    5.09 0.0000003617 ***
delinq.2yrs               -0.08338381  0.06554209   -1.27      0.20330    
pub.rec                    0.33104300  0.11375811    2.91      0.00361 ** 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 5896.6  on 6704  degrees of freedom
Residual deviance: 5487.4  on 6686  degrees of freedom
AIC: 5525

Number of Fisher Scoring iterations: 5

# credit.policy
# purpose2 (credit card)
# purpose3 (debt consolidation)
# purpose6 (major purchase)
# purpose7 (small business)
# installment
# log.annual.inc
# fico
# revol.bal
# inq.last.6mths
# pub.rec

【2.2 從回歸係數估計邊際效用】Consider two loan applications, which are identical other than the fact that the borrower in Application A has FICO credit score 700 while the borrower in Application B has FICO credit score 710. What is the value of Logit(A) - Logit(B)? What is the value of O(A)/O(B)?

# the difference of logits
-0.009317 * (700-710)

[1] 0.09317

# the ratio of odds
# exp(A + B + C) = exp(A)*exp(B)*exp(C)
exp(0.09317)

[1] 1.098

【2.3 混淆矩陣、正確率 vs 底線機率】What is the accuracy of the logistic regression model? What is the accuracy of the baseline model?

predicted.risk = predict(loansLog, newdata = test, type="response")
test$predicted.risk = predicted.risk
table(test$not.fully.paid, predicted.risk>0.5)

   
    FALSE TRUE
  0  2400   13
  1   457    3

# test accuracy
accuracy = (2400+3)/nrow(test); accuracy

[1] 0.8364

# 0.8364
# baseline accuracy
table(train$not.fully.paid)

   0    1 
5632 1073 

table(test$not.fully.paid)

   0    1 
2413  460 

2413/nrow(test)

[1] 0.8399

# 0.8399

【2.4 ROC & AUC】Use the ROCR package to compute the test set AUC.

library(ROCR)
pred = prediction(test$predicted.risk, test$not.fully.paid)
as.numeric(performance(pred, "auc")@y.values)

[1] 0.6719

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3 提高底線 Smart Baseline

【3.1 高底線模型】The variable int.rate is highly significant in the bivariate model, but it is not significant at the 0.05 level in the model trained with all the independent variables. What is the most likely explanation for this difference?

bivariateModel = glm(not.fully.paid~ int.rate, data = train, family = "binomial")
summary(bivariateModel)

Call:
glm(formula = not.fully.paid ~ int.rate, family = "binomial", 
    data = train)

Deviance Residuals: 
   Min      1Q  Median      3Q     Max  
-1.055  -0.627  -0.544  -0.436   2.291  

Coefficients:
            Estimate Std. Error z value Pr(>|z|)    
(Intercept)   -3.673      0.169   -21.8   <2e-16 ***
int.rate      15.921      1.270    12.5   <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: 5896.6  on 6704  degrees of freedom
Residual deviance: 5734.8  on 6703  degrees of freedom
AIC: 5739

Number of Fisher Scoring iterations: 4

# int.rate is correlated with other risk-related variables, and therefore does not incrementally improve the model when those other variables are included.

【3.2 高底線模型的預測值】What is the highest predicted probability of a loan not being paid in full on the testing set? With a logistic regression cutoff of 0.5, how many loans would be predicted as not being paid in full on the testing set?

biPred = predict(bivariateModel, newdata = test, type = 'response')
max(biPred)

[1] 0.4266

# 0.4266
table(biPred>0.5)

FALSE 
 2873 

# 0

【3.3 高底線模型的辨識率】What is the test set AUC of the bivariate model?

library(ROCR)
pred = prediction(biPred, test$not.fully.paid)
as.numeric(performance(pred, "auc")@y.values)

[1] 0.6239

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4 預估投資獲利 Computing the Profitability of an Investment

【4.1 投資價值的算法】How much does a $10 investment with an annual interest rate of 6% pay back after 3 years, using continuous compounding of interest?

10*exp(0.06*3)

[1] 11.97

# 11.97

【4.2 投資獲利的算法，合約完成】While the investment has value c * exp(rt) dollars after collecting interest, the investor had to pay $c for the investment. What is the profit to the investor if the investment is paid back in full?

# c * exp(rt) - c

【4.3 投資獲利的算法，違約】Now, consider the case where the investor made a $c investment, but it was not paid back in full. Assume, conservatively, that no money was received from the borrower (often a lender will receive some but not all of the value of the loan, making this a pessimistic assumption of how much is received). What is the profit to the investor in this scenario?

# c * exp(rt) - c correct
# -c

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5 簡單投資策略 A Simple Investment Strategy

【5.1 計算測試資料的實際投報率】What is the maximum profit of a $10 investment in any loan in the testing set?

test$profit = exp(test$int.rate*3) - 1
test$profit[test$not.fully.paid == 1] = -1
max(test$profit)*10

[1] 8.895

# 8.895

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6 面對不確定性的投資策略 An Investment Strategy Based on Risk

A simple investment strategy of equally investing in all the loans would yield profit $20.94 for a $100 investment. But this simple investment strategy does not leverage the prediction model we built earlier in this problem.

【6.1 高利率、高風險】What is the average profit of a $1 investment in one of these high-interest loans (do not include the $ sign in your answer)? What proportion of the high-interest loans were not paid back in full?

highInterest = subset(test, int.rate>=0.15)
mean(highInterest$profit)

[1] 0.2251

# 0.2251
table(highInterest$not.fully.paid)

  0   1 
327 110 

110/nrow(highInterest)

[1] 0.2517

# 0.2517

【6.2 高利率之中的低風險】What is the profit of the investor, who invested $1 in each of these 100 loans? How many of 100 selected loans were not paid back in full?

cutoff = sort(highInterest$predicted.risk, decreasing=FALSE)[100]
selectedLoans = subset(highInterest, highInterest$predicted.risk <= cutoff)
sum(selectedLoans$profit)

[1] 31.28

# 31.28
table(selectedLoans$not.fully.paid)

 0  1 
81 19 

# 19

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【Q】利用我們建好的模型，你可以設計出比上述的方法獲利更高的投資方法嗎？請詳述你的作法？

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