ANALYSIS: Effect of Credit Limit and Education on Credit Card Default (in Taiwan) -- MODEL 2 (with INTERACTIONS)

Sameer Mathur

YOUR BOSS WANTS TO KNOW:

How does the PROBABILITY of credit card DEFAULT depend on the level of EDUCATION of the customer and the CREDIT LIMIT offered to the customer?

YOUR BOSS WANTS TO KNOW:

(A) How much does the PROBABILITY of credit card default change, if the CREDIT LIMIT of customers without college education (i.e. Education = '1'), is raised by $10,000?

YOUR BOSS WANTS TO KNOW:

(B) How much does the PROBABILITY of credit card default change, if the CREDIT LIMIT of customers having a bachelor's degree (i.e. Education = '2'), is raised by $10,000?

Model 2 -- Framework

\( Default = \beta_0 + \beta_1*CreditLimit + \beta_2*Education + \beta_3*Education*CreditLimit \)

Model 2 -- Logistic Regression

\[ log\frac{p}{1-p} = \beta_0 + \beta_1*CreditLimit + \beta_2*Education2 + \beta_3*Education3 + \beta_4*Education4 \]

\[ + \beta_5*Education2*CreditLimit + \beta_6*Education3*CreditLimit \]

\[ + \beta_7*Education4*CreditLimit \]

….(2)

Model 2 – Logistic Regression Model With Interaction

# fitting logistic regression model with interaction
Model2 <- glm(Default ~ CreditLimit
                      + Education
                      + Education*CreditLimit,
                       data = CCdefault.dt, 
                       family = binomial())
# summary of the model
summary(Model2)


Call:
glm(formula = Default ~ CreditLimit + Education + Education * 
    CreditLimit, family = binomial(), data = CCdefault.dt)

Deviance Residuals: 
    Min       1Q   Median       3Q      Max  
-2.8191   0.4536   0.6529   0.7733   0.8940  

Coefficients:
                         Estimate Std. Error z value Pr(>|z|)    
(Intercept)             9.262e-01  4.470e-02  20.719  < 2e-16 ***
CreditLimit             2.555e-06  2.005e-07  12.747  < 2e-16 ***
Education2             -2.493e-01  5.453e-02  -4.572 4.83e-06 ***
Education3             -2.507e-01  6.689e-02  -3.748 0.000178 ***
Education4             -1.318e-01  9.503e-01  -0.139 0.889700    
CreditLimit:Education2  1.125e-06  2.824e-07   3.983 6.80e-05 ***
CreditLimit:Education3  9.762e-07  4.067e-07   2.400 0.016390 *  
CreditLimit:Education4  8.732e-06  5.802e-06   1.505 0.132360    
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 31427  on 29600  degrees of freedom
Residual deviance: 30610  on 29593  degrees of freedom
AIC: 30626

Number of Fisher Scoring iterations: 6

Comparing Model 1 & Model 2

AIC of Model1 = 30639

AIC of Model2 = 30626

AIC of Model2 < AIC of Model1

Model 2 is better than the Model 1, since it fits the data better.

Predicting the Probabilities Using Model 2

Probability of Default among Consumers having Credit Limit = $100,000 & Education = “2”

newdata <- data.frame(CreditLimit = 100000, Education = "2")
probability <- predict(Model2, newdata, type= "response")
probability

        1 
0.7397817

Model 2 gives different answers than Model 1.

The predicted probability based on Model 2 are more trustworthy.

Questions from the BOSS

(A) How much does the PROBABILITY of credit card default change, if the CREDIT LIMIT of customers without college education (i.e. Education = '1'), is raised by $10,000?

Model 2

\[ log\frac{p}{1-p} = \beta_0 + \beta_1*CreditLimit + \beta_2*Education2 + \beta_3*Education3 \] \[ + \beta_4*Education4 + \beta_5*Education2*CreditLimit + \beta_6*Education3*CreditLimit \] \[ + \beta_7*Education4*CreditLimit \]

Model 2 for Education = 1

\[ log\frac{p}{1-p} = \beta_0 + \beta_1*CreditLimit \]

Change in log odds ratio for Education = 1

\[ \frac{\partial}{\partial (CreditLimit)}log\frac{p}{1-p} = \beta_1 \]

\[ \frac{\partial}{\partial (CreditLimit)}log\frac{p}{1-p} = 0.000002555 \]

\[ \partial log\frac{p}{1-p} = 0.000002555* \partial (CreditLimit) \]

\[ \partial log\frac{p}{1-p} = 0.000002555* \partial (CreditLimit) \]

\[ \partial log\frac{p}{1-p} = 0.000002555*10000 \]

\[ \partial log\frac{p}{1-p} = 0.02555 \]

\[ \partial (p) = \frac{exp(0.02555)}{1 + exp(0.02555)} \]

\[ \partial (p) = 0.5063872 \]

(B) How much does the PROBABILITY of credit card default change, if the CREDIT LIMIT of customers having a bachelor's degree (i.e. Education = '2'), is raised by $10,000?

Model 2

\[ log\frac{p}{1-p} = \beta_0 + \beta_1*CreditLimit + \beta_2*Education2 + \beta_3*Education3 \] \[ + \beta_4*Education4 + \beta_5*Education2*CreditLimit + \beta_6*Education3*CreditLimit \] \[ + \beta_7*Education4*CreditLimit \]

Model 2 for Education = 2

\[ log\frac{p}{1-p} = (\beta_0 + \beta_2) + (\beta_1+\beta_5)*CreditLimit \]

Change in log odds ratio for Education = 2

\[ \frac{\partial}{\partial (CreditLimit)}log\frac{p}{1-p} = \beta_1 + \beta_5*Education2 \]

\[ \frac{\partial}{\partial (CreditLimit)}log\frac{p}{1-p} = 0.000002555 + 0.000001125*1 \]

\[ \frac{\partial}{\partial (CreditLimit)}log\frac{p}{1-p} = 0.00000368 \]

\[ \partial log\frac{p}{1-p} = 0.00000368* \partial (CreditLimit) \]

\[ \partial log\frac{p}{1-p} = 0.00000368* \partial (CreditLimit) \]

\[ \partial log\frac{p}{1-p} = 0.00000368*10000 \]

\[ \partial log\frac{p}{1-p} = 0.0368 \]

\[ \partial (p) = \frac{exp(0.0368)}{1 + exp(0.0368)} \]

\[ \partial (p) = 0.509199 \]

Education = 1

The probability of default increases by 50.64%, if the credit limit rises by $10,000

Education = 2

The probability of default increases by 50.92%, if the credit limit rises by $10,000