AS3-3 Predicting Loan Repayment

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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?

and 'data.frame':   9578 obs. of  14 variables:
 $ credit.policy    : int  1 1 1 1 1 1 1 1 1 1 ...
 $ purpose          : chr  "debt_consolidation" "credit_card" "debt_consolidation" "debt_consolidation" ...
 $ int.rate         : num  0.119 0.107 0.136 0.101 0.143 ...
 $ installment      : num  829 228 367 162 103 ...
 $ log.annual.inc   : num  11.4 11.1 10.4 11.4 11.3 ...
 $ dti              : num  19.5 14.3 11.6 8.1 15 ...
 $ fico             : int  737 707 682 712 667 727 667 722 682 707 ...
 $ days.with.cr.line: num  5640 2760 4710 2700 4066 ...
 $ revol.bal        : int  28854 33623 3511 33667 4740 50807 3839 24220 69909 5630 ...
 $ revol.util       : num  52.1 76.7 25.6 73.2 39.5 51 76.8 68.6 51.1 23 ...
 $ inq.last.6mths   : int  0 0 1 1 0 0 0 0 1 1 ...
 $ delinq.2yrs      : int  0 0 0 0 1 0 0 0 0 0 ...
 $ pub.rec          : int  0 0 0 0 0 0 1 0 0 0 ...
 $ not.fully.paid   : int  0 0 0 0 0 0 1 1 0 0 ...
 - attr(*, "spec")=List of 2
  ..$ cols   :List of 14
  .. ..$ credit.policy    : list()
  .. .. ..- attr(*, "class")= chr  "collector_integer" "collector"
  .. ..$ purpose          : list()
  .. .. ..- attr(*, "class")= chr  "collector_character" "collector"
  .. ..$ int.rate         : list()
  .. .. ..- attr(*, "class")= chr  "collector_double" "collector"
  .. ..$ installment      : list()
  .. .. ..- attr(*, "class")= chr  "collector_double" "collector"
  .. ..$ log.annual.inc   : list()
  .. .. ..- attr(*, "class")= chr  "collector_double" "collector"
  .. ..$ dti              : list()
  .. .. ..- attr(*, "class")= chr  "collector_double" "collector"
  .. ..$ fico             : list()
  .. .. ..- attr(*, "class")= chr  "collector_integer" "collector"
  .. ..$ days.with.cr.line: list()
  .. .. ..- attr(*, "class")= chr  "collector_double" "collector"
  .. ..$ revol.bal        : list()
  .. .. ..- attr(*, "class")= chr  "collector_integer" "collector"
  .. ..$ revol.util       : list()
  .. .. ..- attr(*, "class")= chr  "collector_double" "collector"
  .. ..$ inq.last.6mths   : list()
  .. .. ..- attr(*, "class")= chr  "collector_integer" "collector"
  .. ..$ delinq.2yrs      : list()
  .. .. ..- attr(*, "class")= chr  "collector_integer" "collector"
  .. ..$ pub.rec          : list()
  .. .. ..- attr(*, "class")= chr  "collector_integer" "collector"
  .. ..$ not.fully.paid   : list()
  .. .. ..- attr(*, "class")= chr  "collector_integer" "collector"
  ..$ default: list()
  .. ..- attr(*, "class")= chr  "collector_guess" "collector"
  ..- attr(*, "class")= chr "col_spec"

summary(loans)

 credit.policy     purpose             int.rate      installment   
 Min.   :0.000   Length:9578        Min.   :0.060   Min.   : 15.7  
 1st Qu.:1.000   Class :character   1st Qu.:0.104   1st Qu.:163.8  
 Median :1.000   Mode  :character   Median :0.122   Median :268.9  
 Mean   :0.805                      Mean   :0.123   Mean   :319.1  
 3rd Qu.:1.000                      3rd Qu.:0.141   3rd Qu.:432.8  
 Max.   :1.000                      Max.   :0.216   Max.   :940.1  
 log.annual.inc       dti             fico     days.with.cr.line
 Min.   : 7.55   Min.   : 0.00   Min.   :612   Min.   :  179    
 1st Qu.:10.56   1st Qu.: 7.21   1st Qu.:682   1st Qu.: 2820    
 Median :10.93   Median :12.66   Median :707   Median : 4140    
 Mean   :10.93   Mean   :12.61   Mean   :711   Mean   : 4562    
 3rd Qu.:11.29   3rd Qu.:17.95   3rd Qu.:737   3rd Qu.: 5730    
 Max.   :14.53   Max.   :29.96   Max.   :827   Max.   :17640    
   revol.bal         revol.util    inq.last.6mths   delinq.2yrs    
 Min.   :      0   Min.   :  0.0   Min.   : 0.00   Min.   : 0.000  
 1st Qu.:   3187   1st Qu.: 22.6   1st Qu.: 0.00   1st Qu.: 0.000  
 Median :   8596   Median : 46.3   Median : 1.00   Median : 0.000  
 Mean   :  16914   Mean   : 46.8   Mean   : 1.58   Mean   : 0.164  
 3rd Qu.:  18250   3rd Qu.: 70.9   3rd Qu.: 2.00   3rd Qu.: 0.000  
 Max.   :1207359   Max.   :119.0   Max.   :33.00   Max.   :13.000  
    pub.rec      not.fully.paid
 Min.   :0.000   Min.   :0.00  
 1st Qu.:0.000   1st Qu.:0.00  
 Median :0.000   Median :0.00  
 Mean   :0.062   Mean   :0.16  
 3rd Qu.:0.000   3rd Qu.:0.00  
 Max.   :5.000   Max.   :1.00  

table(loans$not.fully.paid)

   0    1 
8045 1533 

1533/(8045+1533) #0.1600543

[1] 0.1601

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

summary(loans)

 credit.policy     purpose             int.rate      installment   
 Min.   :0.000   Length:9578        Min.   :0.060   Min.   : 15.7  
 1st Qu.:1.000   Class :character   1st Qu.:0.104   1st Qu.:163.8  
 Median :1.000   Mode  :character   Median :0.122   Median :268.9  
 Mean   :0.805                      Mean   :0.123   Mean   :319.1  
 3rd Qu.:1.000                      3rd Qu.:0.141   3rd Qu.:432.8  
 Max.   :1.000                      Max.   :0.216   Max.   :940.1  
 log.annual.inc       dti             fico     days.with.cr.line
 Min.   : 7.55   Min.   : 0.00   Min.   :612   Min.   :  179    
 1st Qu.:10.56   1st Qu.: 7.21   1st Qu.:682   1st Qu.: 2820    
 Median :10.93   Median :12.66   Median :707   Median : 4140    
 Mean   :10.93   Mean   :12.61   Mean   :711   Mean   : 4562    
 3rd Qu.:11.29   3rd Qu.:17.95   3rd Qu.:737   3rd Qu.: 5730    
 Max.   :14.53   Max.   :29.96   Max.   :827   Max.   :17640    
   revol.bal         revol.util    inq.last.6mths   delinq.2yrs    
 Min.   :      0   Min.   :  0.0   Min.   : 0.00   Min.   : 0.000  
 1st Qu.:   3187   1st Qu.: 22.6   1st Qu.: 0.00   1st Qu.: 0.000  
 Median :   8596   Median : 46.3   Median : 1.00   Median : 0.000  
 Mean   :  16914   Mean   : 46.8   Mean   : 1.58   Mean   : 0.164  
 3rd Qu.:  18250   3rd Qu.: 70.9   3rd Qu.: 2.00   3rd Qu.: 0.000  
 Max.   :1207359   Max.   :119.0   Max.   :33.00   Max.   :13.000  
    pub.rec      not.fully.paid
 Min.   :0.000   Min.   :0.00  
 1st Qu.:0.000   1st Qu.:0.00  
 Median :0.000   Median :0.00  
 Mean   :0.062   Mean   :0.16  
 3rd Qu.:0.000   3rd Qu.:0.00  
 Max.   :5.000   Max.   :1.00  

#log.annual.inc,days.with.cr.line,revol.until,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] 0

table(missing$not.fully.paid) 

< table of extent 0 >

12/62 #0.19

[1] 0.1935

#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?

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

 iter imp variable
  1   1
  1   2
  1   3
  1   4
  1   5
  2   1
  2   2
  2   3
  2   4
  2   5
  3   1
  3   2
  3   3
  3   4
  3   5
  4   1
  4   2
  4   3
  4   4
  4   5
  5   1
  5   2
  5   3
  5   4
  5   5

Number of logged events: 1

loans[vars.for.imputation] = 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?

library(caTools)
set.seed(144)
split=sample.split(loans$not.fully.paid,SplitRatio=0.7)
train=subset(loans,split==TRUE)
test=subset(loans,split==FALSE)
model1=glm(not.fully.paid~.,data=train,family=binomial)
summary(model1)

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

Deviance Residuals: 
   Min      1Q  Median      3Q     Max  
-2.172  -0.620  -0.495  -0.361   2.635  

Coefficients:
                             Estimate  Std. Error z value     Pr(>|z|)    
(Intercept)                9.23868531  1.55340639    5.95 0.0000000027 ***
credit.policy             -0.35962512  0.10049091   -3.58      0.00035 ***
purposecredit_card        -0.61018465  0.13434286   -4.54 0.0000055724 ***
purposedebt_consolidation -0.31801924  0.09181777   -3.46      0.00053 ***
purposeeducational         0.13933851  0.17520994    0.80      0.42646    
purposehome_improvement    0.17762530  0.14786550    1.20      0.22965    
purposemajor_purchase     -0.47785935  0.20067564   -2.38      0.01725 *  
purposesmall_business      0.41640011  0.14181047    2.94      0.00332 ** 
int.rate                   0.69224856  2.08421219    0.33      0.73978    
installment                0.00126980  0.00020911    6.07 0.0000000013 ***
log.annual.inc            -0.43103508  0.07138624   -6.04 0.0000000016 ***
dti                        0.00449540  0.00549797    0.82      0.41356    
fico                      -0.00940183  0.00170766   -5.51 0.0000000368 ***
days.with.cr.line          0.00000413  0.00001582    0.26      0.79406    
revol.bal                  0.00000300  0.00000116    2.58      0.01001 *  
revol.util                 0.00176946  0.00152968    1.16      0.24737    
inq.last.6mths             0.07630349  0.01552192    4.92 0.0000008840 ***
delinq.2yrs               -0.08756530  0.06557839   -1.34      0.18179    
pub.rec                    0.31888334  0.11367874    2.81      0.00503 ** 
---
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: 5489.5  on 6686  degrees of freedom
AIC: 5527

Number of Fisher Scoring iterations: 5

【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
-0.00929*(-10) #0.0929

[1] 0.0929

# the ratio of odds
exp(0.0929) #1.097352

[1] 1.097

【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?

# test accuracy
test$predicted.risk=predict(model1,type="response",newdata=test)
table(test$not.fully.paid,test$predicted.risk>0.5)

   
    FALSE TRUE
  0  2401   12
  1   457    3

(2400+3)/(2400+13+457+3)  #0.8364

[1] 0.8364

# baseline accuracy
table(test$not.fully.paid)

   0    1 
2413  460 

(2413)/(2413+460) #0.8398886

[1] 0.8399

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

# test 
# baseline accuracy
library(ROCR)
ROCRpredTest=prediction(test$predicted.risk,test$not.fully.paid)
auc=as.numeric(performance(ROCRpredTest,"auc")@y.values);auc #0.6718878

[1] 0.6718

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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?

summary(model1)

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

Deviance Residuals: 
   Min      1Q  Median      3Q     Max  
-2.172  -0.620  -0.495  -0.361   2.635  

Coefficients:
                             Estimate  Std. Error z value     Pr(>|z|)    
(Intercept)                9.23868531  1.55340639    5.95 0.0000000027 ***
credit.policy             -0.35962512  0.10049091   -3.58      0.00035 ***
purposecredit_card        -0.61018465  0.13434286   -4.54 0.0000055724 ***
purposedebt_consolidation -0.31801924  0.09181777   -3.46      0.00053 ***
purposeeducational         0.13933851  0.17520994    0.80      0.42646    
purposehome_improvement    0.17762530  0.14786550    1.20      0.22965    
purposemajor_purchase     -0.47785935  0.20067564   -2.38      0.01725 *  
purposesmall_business      0.41640011  0.14181047    2.94      0.00332 ** 
int.rate                   0.69224856  2.08421219    0.33      0.73978    
installment                0.00126980  0.00020911    6.07 0.0000000013 ***
log.annual.inc            -0.43103508  0.07138624   -6.04 0.0000000016 ***
dti                        0.00449540  0.00549797    0.82      0.41356    
fico                      -0.00940183  0.00170766   -5.51 0.0000000368 ***
days.with.cr.line          0.00000413  0.00001582    0.26      0.79406    
revol.bal                  0.00000300  0.00000116    2.58      0.01001 *  
revol.util                 0.00176946  0.00152968    1.16      0.24737    
inq.last.6mths             0.07630349  0.01552192    4.92 0.0000008840 ***
delinq.2yrs               -0.08756530  0.06557839   -1.34      0.18179    
pub.rec                    0.31888334  0.11367874    2.81      0.00503 ** 
---
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: 5489.5  on 6686  degrees of freedom
AIC: 5527

Number of Fisher Scoring iterations: 5

model2=glm(not.fully.paid~int.rate,data=train,family=binomial)
summary(model2)

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?

predictTest=predict(model2,type="response",newdata=test)
summary(predictTest) #0.42662  #0

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  0.062   0.116   0.151   0.160   0.189   0.427 

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

ROCRpredTest2=prediction(predictTest,test$not.fully.paid)
as.numeric(performance(ROCRpredTest2,"auc")@y.values)

[1] 0.6239

#0.6239081

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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?

#c=10,r=0.06,t=3
10*exp(0.06*3) #11.97

[1] 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

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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
summary(test$profit)

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 -1.000   0.286   0.411   0.209   0.498   0.889 

#0.8895*10=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)
summary(highinterest$profit) #0.2251

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 -1.000  -1.000   0.599   0.225   0.638   0.889 

table(highinterest$not.fully.paid) 

  0   1 
327 110 

110/(327+110) #0.2517162

[1] 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$predicted.risk, decreasing=FALSE)[100]
selectedLoans=subset(highinterest,predicted.risk<=cutoff)
sum(selectedLoans$profit) #31.27825

[1] 31.23

table(selectedLoans$not.fully.paid) #19

 0  1 
81 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？

highinterest1=subset(test,int.rate>=0.14)
summary(highinterest1)
cutoff1 = sort(highinterest1$predicted.risk, decreasing=FALSE)[100]
selectedLoans1=subset(highinterest1,predicted.risk<=cutoff1)
sum(selectedLoans1$profit) #34.67
table(selectedLoans1$not.fully.paid) #14
#   一開始,  我們認為選擇高的期望利率,會有高的報酬,但發現,提高利率之後,報酬不增反減,因為沒有履約償還的人數變多了,於是,我們嘗試將利率降低,雖然期望利率變低了,但發現不履約的人減少,因而獲利增加

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