信用违约

1. 定义

＊ 银行与借款人之间的协议 — 贷款＋按揭偿还本金和利息

2. 预期损失 Expected loss (EL)

  • 组成：
    • Probability of default (PD)
    • Exposure at default (EAD)
    • Loss given default (LGD): as % of EAD
  • 公式：EL = PD * EAD * LGD

3. 银行用来分析信用风险的信息

  • 申请表信息
    • 收入
    • 婚姻状况
  • 申请人信息
    • 目前账户余额
    • 应付欠款历史记录

4. 实战数据介绍

head(loan.data,10)
##    age                    education year_emp income debt_income cred_debt
## 1   47 Did not complete high school       22     81         5.5  1.505790
## 2   40 Did not complete high school       22     95         3.6  0.632700
## 3   35 Did not complete high school       16     36         3.4  0.178704
## 4   43 Did not complete high school       16     89         0.4  0.159488
## 5   47 Did not complete high school       26    100        12.8  4.582400
## 6   52 Did not complete high school       24     64        10.0  3.929600
## 7   35 Did not complete high school       13     35         4.5  0.431550
## 8   36 Did not complete high school       16     32        10.9  0.544128
## 9   49           High school degree       14     63        15.8  0.935676
## 10  35           High school degree       14     82         0.8  0.468384
##    other_debt Loan      Logistc PGR_1 Dis_1    Dis1_1     Discrim
## 1    2.949210   No 0.0006254570    No    No 0.9884384 0.011561555
## 2    2.787300   No 0.0016333330    No    No 0.9733337 0.026666274
## 3    1.045296   No 0.0009656821    No    No 0.9888641 0.011135941
## 4    0.196512   No 0.0013862700    No    No 0.9741002 0.025899755
## 5    8.217600   No 0.0123452723    No    No 0.9319657 0.068034281
## 6    2.470400   No 0.0002818009    No    No 0.9916055 0.008394544
## 7    1.143450   No 0.0029283941    No    No 0.9764911 0.023508945
## 8    2.943872 <NA> 0.0036198182    No    No 0.9703037 0.029696254
## 9    9.018324 <NA> 0.0429850328    No    No 0.9000222 0.099977832
## 10   0.187616   No 0.0060533168    No    No 0.9011298 0.098870237
CrossTable(loan.data$education)
## 
##  
##    Cell Contents
## |-------------------------|
## |                       N |
## |         N / Table Total |
## |-------------------------|
## 
##  
## Total Observations in Table:  100 
## 
##  
##                              | Did not complete high school |           High school degree |                 Some college |               College degree | 
##                              |------------------------------|------------------------------|------------------------------|------------------------------|
##                              |                           53 |                           25 |                           17 |                            5 | 
##                              |                        0.530 |                        0.250 |                        0.170 |                        0.050 | 
##                              |------------------------------|------------------------------|------------------------------|------------------------------|
## 
## 
## 
## 
CrossTable(loan.data$education, loan.data$Loan, prop.r = T, prop.c = F, prop.t = F, prop.chisq = F)
## 
##  
##    Cell Contents
## |-------------------------|
## |                       N |
## |           N / Row Total |
## |-------------------------|
## 
##  
## Total Observations in Table:  84 
## 
##  
##                              | loan.data$Loan 
##          loan.data$education |        No |       Yes | Row Total | 
## -----------------------------|-----------|-----------|-----------|
## Did not complete high school |        35 |         7 |        42 | 
##                              |     0.833 |     0.167 |     0.500 | 
## -----------------------------|-----------|-----------|-----------|
##           High school degree |        15 |         6 |        21 | 
##                              |     0.714 |     0.286 |     0.250 | 
## -----------------------------|-----------|-----------|-----------|
##                 Some college |        10 |         6 |        16 | 
##                              |     0.625 |     0.375 |     0.190 | 
## -----------------------------|-----------|-----------|-----------|
##               College degree |         4 |         1 |         5 | 
##                              |     0.800 |     0.200 |     0.060 | 
## -----------------------------|-----------|-----------|-----------|
##                 Column Total |        64 |        20 |        84 | 
## -----------------------------|-----------|-----------|-----------|
## 
##
  • There are 100 observations.
  • Row-wise proportion of default by education category （注意反常识的结果，推断可能原因）

5. 数据分析 — 直方图 ＋ 异常值

1. 直方图

hist(loan.data$debt_income, main = 'Histogram of debt-to-income ratio (*100)', xlab = 'Debt-to-income ratio')

hist(loan.data$income, main = 'Histogram of household income in thousands', xlab = 'Income')
hist(loan.data$income, main = 'Histogram of household income in thousands', xlab = 'Income')$breaks

##  [1]   0  20  40  60  80 100 120 140 160 180
hist(loan.data$income, breaks = sqrt(nrow(loan.data)), main = 'Histogram of income with breaks argument', xlab = 'Income')

2. 异常值

plot(loan.data$income,col=ifelse(loan.data$income>=150,"orange","black"), pch=ifelse(loan.data$income>=150,19,1), ylab = 'Income')

  • 异常值判断
    • 专业领域判断
    • 常规判断：Q1-1.5*IQR — Q3+1.5*IQR
    • 两者兼具
outlier1 = which(loan.data$income>150)
data.nooutlier1 = loan.data[-outlier1,]
outlier.cutoff = quantile(loan.data$income, 0.75) + 1.5 * IQR(loan.data$income)
outlier2 = which(loan.data$income>outlier.cutoff)
data.nooutlier2 = loan.data[-outlier2,]
hist(data.nooutlier1$income, breaks = sqrt(nrow(data.nooutlier1)), main = 'Histogram of income without outliers', xlab = 'Income')

plot(loan.data$year_emp, loan.data$income, col=ifelse(loan.data$income>=150,"orange","black"), pch=ifelse(loan.data$income>=150,19,1), main = 'Bivariate plot', xlab = 'Years with current employer', ylab = 'Income')

  • Bivariate scatterplot helps to check outliers on two-dimentional variables

6. 数据分析 — 缺失值

sapply(loan.data, function(x) all(!is.na(x)))
##         age   education    year_emp      income debt_income   cred_debt 
##        TRUE        TRUE        TRUE        TRUE        TRUE        TRUE 
##  other_debt        Loan     Logistc       PGR_1       Dis_1      Dis1_1 
##        TRUE       FALSE        TRUE        TRUE        TRUE        TRUE 
##     Discrim 
##        TRUE
temp = loan.data
set.seed(10)
temp$year_emp[sample(1:nrow(loan.data),10)] = NA
summary(temp$year_emp)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max.    NA's 
##   0.000   3.000   6.500   8.378  13.750  26.000      10
na.index = which(is.na(temp$year_emp))
data.nona1 = temp[-na.index,]
data.nona2 = temp
data.nona2$year_emp[na.index] = median(temp$year_emp, na.rm=T)
temp$year_emp_cat <- rep(NA, length(temp$year_emp))
temp$year_emp_cat[which(temp$year_emp <= 5)] <- "0-5"
temp$year_emp_cat[which(temp$year_emp > 5 & temp$year_emp <= 10)] <- "5-10"
temp$year_emp_cat[which(temp$year_emp > 10 & temp$year_emp <= 15)] <- "10-15"
temp$year_emp_cat[which(temp$year_emp > 15 & temp$year_emp <= 20)] <- "15-20"
temp$year_emp_cat[which(temp$year_emp > 20)] <- "20+"
temp$year_emp_cat[which(is.na(temp$year_emp))] <- "Missing"
temp$year_emp_cat <- factor(temp$year_emp_cat, levels = c("0-5","5-10","10-15","15-20","20+","Missing"))
plot(temp$year_emp_cat)

  • 处理方法（也可用于处理异常值）:
    • 删除行／列（适用于连续型和分类型数据）
    • 替换 (4个主要计算方法，如连续型数据用median，分类型数据用频率最高的类别，待以后文章详解)
    • 保留 (连续型数据用bin分类；分类型数据创建NA类别)

7. 数据模型检测 — training/test set + confusion matrix

Cross-validation

  • Accuracy = (TN+TP)/(TN+TP+FP+FN)
  • Sensitivity = TP/(TP+FN)
  • Specificity = TN/(TN+FP)
set.seed(1)
train.index <- sample(1:nrow(loan.data), 2/3 * nrow(loan.data))
training <- loan.data[train.index, ]
test <- loan.data[-train.index, ]
#table(test$Loan, model_pred)