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Econ325ReviewQuestionsforCreditBureausandFICOscores.pdf

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Econ 325— Review Questions For Credit Bureaus, Logodds and FICO Scores

Clarification: Log Odds—Which Base? What Notation?

• In the class, we will always be using the natural log when we are working with “log odds”.

Because many students don’t regularly use logarithms, “log odds” may be written three ways in class, on review problems, or on exams.

That is, log𝑒𝑒(𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1)) = log(𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1)) = ln(𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1)). We never use base 10.

Technically and mathematically, log𝑒𝑒(𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1)) = ln(𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1)). The problem is that the subscript “e” is often hard to read so it gets dropped.

• Mathematicians1 use log to denote the natural log. Physicists and engineers use log to

denote logarithm with base 10. Calculator companies follow the Physicist convention. The class is following the mathematics convention.

• The practical issue is to make sure you use the “ln” button on your calculator.

• It is also often awkward to keep writing (y=1) or (y=0) every time. So, you may see them dropped, such as: logodds or lnodds. Other variations can be log(odds) or ln(odds).

o In those cases, pay attention to whether you have logodds of (y =1) or (y = 0).

FORMULAS PROVIDED ON EXAM 2

Two Point Form of a Line

With two points (𝑥𝑥1,𝑦𝑦1) and (𝑥𝑥2,𝑦𝑦2)

(𝑦𝑦 − 𝑦𝑦1) = � 𝑦𝑦2 − 𝑦𝑦1 𝑥𝑥2 − 𝑥𝑥1

� (𝑥𝑥 − 𝑥𝑥1)

(𝑦𝑦 − 𝑦𝑦2) = � 𝑦𝑦2 − 𝑦𝑦1 𝑥𝑥2 − 𝑥𝑥1

� (𝑥𝑥 − 𝑥𝑥2)

1 Source is http://mathworld.wolfram.com/CommonLogarithm.html.

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LOG ODDS

Concepts

1. Definitions to know and to be able to explain. a. Binary variable (y=1 or y=0) b. Probability, Odds and Log Odds c. Log Odds model or logistic regression model d. Dummy variable (where a variable x = 1 or 0)

2. Given the definition of odds as a function of probability of y = 1 and y = 0, derive the inverse function for probability (y=1) as a function of odds(y=1).

3. Describe in words how odds(y=1) and odds(y=0) are related.

4. Show mathematically how odds(y=1) and odds(y=0) are related.

5. Describe how logodds(y=1) and logodds(y=0) are related.

6. In class, Prof. Sugiyama sometimes writes logodds and sometimes lnodds. Are they different or the same?

7. When trying to figure out log odds of a number, which button do you always press on your calculator?

8. How logodds(default) is changed to become a credit score

Definition of Binary Variable

y is a binary variable that takes on values of 0 or 1.

𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 1) = 𝑝𝑝

𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 0) = 1 − 𝑝𝑝

Since there are only two outcomes

𝑝𝑝 + (1 − 𝑝𝑝) = 1

Or we can write

𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 1) + 𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 0) = 1

To make sure we get 1 or 0, we need to have the following restrictions on p and (1 – p)

0 < 𝑝𝑝 < 1

0 < (1 − 𝑝𝑝) < 1

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Definition of Odds

𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) = 𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 1) 𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 0)

= 𝑝𝑝

1 − 𝑝𝑝

𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 0) = 𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 0) 𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 1)

= 1 − 𝑝𝑝 𝑝𝑝

Because the probability of y =1 and y = 0 are both positive and cannot be zero, then the odds(y=1) and odds(y=0) are always positive.

0 < 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) < ∞

0 < 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 0) < ∞

Relationship Between Odds

If we recall working with reciprocals, then we can find that if we start with

1 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 0)

= 1

�1 − 𝑝𝑝𝑝𝑝 �

Then

�1 − 𝑝𝑝𝑝𝑝 � = 1 ÷ �

1 − 𝑝𝑝 𝑝𝑝

�

= 1 × � 𝑝𝑝

1 − 𝑝𝑝 �

= 𝑝𝑝

1 − 𝑝𝑝

= 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1)

𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) = 1

𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 0)

𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 0) = 1

𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1)

If we do some algebra, we can rewrite the two equations as a product

𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) × 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 0) = 1

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Definition of Logodds

𝑙𝑙𝑜𝑜𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) = 𝑙𝑙𝑙𝑙�𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1)� = 𝑙𝑙𝑙𝑙 � 𝑝𝑝

1 − 𝑝𝑝 �

𝑙𝑙𝑜𝑜𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 0) = 𝑙𝑙𝑙𝑙�𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 0)� = 𝑙𝑙𝑙𝑙 � 1 − 𝑝𝑝 𝑝𝑝

�

−∞ < 𝑙𝑙𝑜𝑜𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) < ∞

−∞ < 𝑙𝑙𝑜𝑜𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 0) < ∞

Relationship Between Logodds

One of the logarithm laws is that

𝑙𝑙𝑙𝑙 � 1 𝑥𝑥 � = −𝑙𝑙𝑙𝑙(𝑥𝑥)

So for logodds

𝑙𝑙𝑜𝑜𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) = 𝑙𝑙𝑙𝑙�𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1)�

= 𝑙𝑙𝑙𝑙 � 1

𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 0) �

= −ln �𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 0)�

Thus, we have that

𝑙𝑙𝑜𝑜𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) = −𝑙𝑙𝑜𝑜𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 0) or

𝑙𝑙𝑜𝑜𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 0) = −𝑙𝑙𝑜𝑜𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1)

The sum of the logodds of both outcomes equals to zero, or

𝑙𝑙𝑜𝑜𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) + 𝑙𝑙𝑜𝑜𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 0) = 0

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Inverse Relationships

If we start with the definition of odds,

𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) = 𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 1) 𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 0)

We can do a substation for the prob(y=0) = 1 – prob(y=1), so we get

𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) = 𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 1)

1 − 𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 1)

If we cross multiply, we get

𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) × [1 − 𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 1)] = 𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 1)

Which can be rewritten as

𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) − [𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) × 𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 1)] = 𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 1)

And then rewritten as,

𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) = 𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 1) + [𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) × 𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 1)]

We can factor out prob(y=1) from the right-hand side of the equation to get

𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) = 𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 1) × [1 + 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1)]

If we divide both sides by (1+odds(y=1)), we have prob(y=1) on the right-hand side

𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) 1 + 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1)

= 𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 1)

𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 1) = 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1)

1 + 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1)

Recall that since we are dealing with the natural log, the inverse function is raising to ex

𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) = 𝑒𝑒𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙(𝑦𝑦=1)

Lastly, combining the two equations above, we can solve for the probability based on logodds,

𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 1) = 𝑒𝑒𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙(𝑦𝑦=1)

1 + 𝑒𝑒𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙(𝑦𝑦=1)

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Summary of Relationships

Just to restate what we have seen before:

𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 1) + 𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 0) = 1

𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) × 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 0) = 1

log𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) + 𝑙𝑙𝑜𝑜𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 0) = 0

Key Factoids

If the 𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 1) = 50%, then the 𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 0) = 50%.

Then the 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) = 𝑝𝑝𝑝𝑝𝑙𝑙𝑝𝑝(𝑦𝑦=1) 𝑝𝑝𝑝𝑝𝑙𝑙𝑝𝑝(𝑦𝑦=0)

= 0.50 0.50

= 1 and also then 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 0) = 1

Then the 𝑙𝑙𝑜𝑜𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) = 𝑙𝑙𝑙𝑙(1) = 0 and also 𝑙𝑙𝑜𝑜𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 0) = 𝑙𝑙𝑙𝑙(1) = 0

To summarize, when the probability of an outcome is equal to 50%, then

𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 1) = 𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 0)

𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) = 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 0) = 1

𝑙𝑙𝑜𝑜𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) = 𝑙𝑙𝑜𝑜𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 0) = 0

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Logodds Models

Let the logodds vary linearly with a variable x, for example,

𝑙𝑙𝑜𝑜𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) = 𝑎𝑎 + 𝑝𝑝𝑥𝑥

Then there is a relationship between logodds(y=1) and logodds(y=0), we get

𝑙𝑙𝑜𝑜𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 0) = −𝑎𝑎 − 𝑝𝑝𝑥𝑥

𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) = 𝑒𝑒𝑎𝑎+𝑝𝑝𝑏𝑏

𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 0) = 𝑒𝑒−𝑎𝑎−𝑝𝑝𝑏𝑏

𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 1) = 𝑒𝑒𝑎𝑎+𝑝𝑝𝑏𝑏

1 + 𝑒𝑒𝑎𝑎+𝑝𝑝𝑏𝑏

𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 1) = 𝑒𝑒−𝑎𝑎−𝑝𝑝𝑏𝑏

1 + 𝑒𝑒−𝑎𝑎−𝑝𝑝𝑏𝑏

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Graphs

Let logodds be a linear function of a variable, x. For example,

𝑙𝑙𝑜𝑜𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) = 1 2 𝑥𝑥

For values of x from -20 to 20, we get the following graph of the logodds:

We see that if x increases, then logodds(y=1) decreases.

-15

-10

-5

-20 -15 -10 -5 0 5 10 15 20 logodds(y=1)

logodds(y=1) = ½ x

Page 9 of 35

The graph of the odds(y=1) when x varies from -20 to 20 is below:

We see that if x increases, then odds(y=1) decreases.

Lastly, we can graph the probability(y=1) as x varies from -20 to 20.

We see that if x increases, then prob(y=1) decreases.

5,000

10,000

15,000

20,000

25,000

-20 -15 -10 -5 0 5 10 15 20

odds(y=1)

odds(y=1) = e1/2x

10%

20%

30%

40%

50%

60%

70%

80%

90%

100%

-20 -15 -10 -5 0 5 10 15 20

prob(y=1)

prob(y=1) = e1/2x/(1+e1/2x)

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Graph 2

Let logodds be another linear function of a variable, x. For example,

𝑙𝑙𝑜𝑜𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) = − 1 2 𝑥𝑥

We see that if x increases, then logodds(y=1) decreases.

We see that if x increases, then logodds(y=1) increases.

-15

-10

-5

-20 -15 -10 -5 0 5 10 15 20 logodds(y=1)

logodds(y=1) = ‒ ½ x

5,000

10,000

15,000

20,000

25,000

-20 -15 -10 -5 0 5 10 15 20

odds(y=1)

odds(y=1) = e‒1/2x

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We see that if x increases, then logodds(y=1) decreases.

10%

20%

30%

40%

50%

60%

70%

80%

90%

100%

-20 -15 -10 -5 0 5 10 15 20

prob(y=1)

prob(y=1) = e‒1/2x/(1+e‒1/2x)

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Graph 3 and 4

Lets compare two similar logodds functions:

𝑙𝑙𝑜𝑜𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) = 1 2 𝑥𝑥 + 5

𝑙𝑙𝑜𝑜𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) = 1 2 𝑥𝑥 − 5

We see that if x increases, then logodds(y=1) increases for both functions.

-15

-10

-5

-30 -25 -20 -15 -10 -5 0 5 10 logodds(y=1)

logodds(y=1) = ½ x + 5

-15

-10

-5

-10 -5 0 5 10 15 20 25 30 logodds(y=1)

logodds(y=1) = ½ x - 5

Page 13 of 35

Then the graphs of the probability for both functions are:

We see that if x increases, then prob(y=1) increases for both functions.

10%

20%

30%

40%

50%

60%

70%

80%

90%

100%

-30 -25 -20 -15 -10 -5 0 5 10

prob(y=1)

prob(y=1) = e1/2x+5/(1+e1/2x+5)

10%

20%

30%

40%

50%

60%

70%

80%

90%

100%

-10 -5 0 5 10 15 20

prob(y=1)

prob(y=1) = e1/2x-5/(1+e1/2x-5)

Page 14 of 35

Graph 5

Let’s compare graphs of logodds(y=1) and logodds(y=0)

𝑙𝑙𝑜𝑜𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) = 1 2 𝑥𝑥

𝑙𝑙𝑜𝑜𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 0) = − 1 2 𝑥𝑥

In the graph, we can see that

𝑙𝑙𝑜𝑜𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) + 𝑙𝑙𝑜𝑜𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 0) = 0

-15

-10

-5

-20 -15 -10 -5 0 5 10 15 20 logodds

logodds(y=1) logodds(y=0)

Page 15 of 35

𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) = 𝑒𝑒 1 2𝑏𝑏

𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 0) = 𝑒𝑒− 1 2𝑏𝑏

While we know that 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 1) × 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑦𝑦 = 0) = 1

It isn’t clear from the graph above this relationship holds.

5,000

10,000

15,000

20,000

25,000

-20 -15 -10 -5 0 5 10 15 20

odds

odds(y=1) odds(y=0)

Page 16 of 35

We know that

𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 1) = 𝑒𝑒 1 2𝑏𝑏

1 + 𝑒𝑒 1 2𝑏𝑏

𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 0) = 𝑒𝑒−

1 2𝑏𝑏

1 + 𝑒𝑒− 1 2𝑏𝑏

In the graph above, it is somewhat clear that we have

𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 1) + 𝑝𝑝𝑝𝑝𝑜𝑜𝑝𝑝(𝑦𝑦 = 0) = 1

10%

20%

30%

40%

50%

60%

70%

80%

90%

100%

-20 -15 -10 -5 0 5 10 15 20

prob

prob(y=1) prob(y=0)

Page 17 of 35

Credit Bureau and Credit Score Concepts

Be familiar with these concepts or definitions. Be able to explain these terms or concepts. Be able to provide examples to demonstrate understanding.

1. Credit Bureau 2. Credit Report 3. Credit Score

4. Fair Isaac Co. 5. FICO Score 6. FICO Score Range of 300 to 850 7. Negative Performance Category

a. Charge-off2 b. 30, 60, 90 days late c. Bankruptcy

8. What information is used in a FICO score

9. What is a “safer” borrower—someone with a high FICO score or someone with a low FICO score?

10. Points to double odds—how it is used to define a credit score 11. Credit bureaus reduce asymmetric information. 12. Credit bureaus have existed in the US since the late 19th century.

13. There are three major credit bureaus in US. 14. Special Credit Report Information

a. 30 day, 60 day, 90 day, 120 day, 150 day, 180 day b. Payment activity, account balances, credit limits, initial loan amount

c. Open account; closed account, delinquent account, “good standing” d. Repossession, Foreclosure, and Voluntary Surrender e. Derogatory, Charge off, and collections amount f. Bankruptcy

g. Length of credit history

h. Revolving credit and installment credit

2 A charge-off is when a lender believes that a debt is not collectible. For Federal tax purposes, this may vary by the type of the loan, but commonly 120 or 180 days after a payment is due. This allows for the debt to be deducted for tax purposes.

Page 18 of 35

Part A: Simple Logodds, Odds, and Probability Problems

1. Fill in the following table

Probability of (y=1) Prob of (y=0)

15%

25%

50%

95%

2. Fill in the following table

Probability of (y=1) Odds of (y=1) Logodds of (y=1)

15%

25%

50%

95%

3. Fill in the following table

Odds(y=1) Odds(y=0)

4. Fill in the following table

Logodds(y=1) Logodds(y=0)

-2

-1

Page 19 of 35

5. Fill in the following table

Logodds(y=1) Logodds(y=0)

ln(1/3)

ln(1/2)

ln(1)

ln(3)

ln(9)

6. Fill in the following table

Odds(y=1) Prob(y=1)

0.25

0.75

7. Fill in the following table

Odds(y=1) Prob(y=1)

e-2

e-1

Page 20 of 35

8. Fill in the following table

Odds(y=1) Odds(y=0)

e-2

e-1

9. Fill in the following table

Odds(y=1) Prob(y=1) Prob(y=0)

0.25

0.75

10. Fill in the following table

Log Odds of y=1 Odds of y = 1 Prob of y=1

-1.25

-0.75

-0.25

0.25

0.75

1.25

Page 21 of 35

Part B: Log Odds Models—Simple

1. Suppose you are given the following logistic model:

log odds( 1) 3 4e y x= = − +

(a) What is the odds of success, i.e. y =1 , when x = 1? (b) What is the probability of success when x = 1? (c) For what value of x do we have a 50% probability of success? (d) What is the probability of success when x = ½? (e) Does the probability of success increase or decrease as x increases? 2. Suppose you are given the following logistic model:

= = − + +1 2 1

log odds( 1) 1 2 3e

y x x

where the two variables, 1x and 2x are binary variables, i.e., 

=  

1 0

x and 

=  

1 0

1x 2x Logodds(y=1) Odds(y=1) Prob(y=1)

0 0

0 1

1 0

1 1

Page 22 of 35

Part C: Working with Logodds Models

1. Suppose you have log(odds(y=1)) = 2 – x. (a) What is the formula for odds(y=1)? (b) What is the formula for prob(y=1)?

2. Suppose you have log(odds(y=1)) = 2 – x

(a) What is the formula for odds(y=0)? (b) What is the formula for prob(y=0)?

3. Suppose you have log(odds(y=1)) = –0.25 + 0.325x.

(a) What is the formula for odds(y=1)? (b) What is the formula for prob(y=1)? (c) What value of x gives us a probability of 50%? Show your work.

Part D: Graphing

1. Suppose you have log(odds(y=1)) = 1 – 0.5x. (a) What does the graph of logodds(y=1) look like, as x varies? (b) What does the graph of odds(y=1) look like, as x varies? (c) What does the graph of the probability of y=1 look like, as x varies?

2. Suppose you have log(odds(y=1)) = 1 – 0.5x.

(a) What does the graph of log(odds(y=0)) look like, as x varies? (b) What does the graph of odds(y=0) look like, as x varies? (c) What does the graph of the probability of y=0 look like, as x varies?

Page 23 of 35

Part E: Creating Credit Score Scales

1. Suppose that the probability of not defaulting is 50% at a credit score of 500 and that odds of not defaulting double for every increase of 100 points. Find the equation of the line that defines the relationship between logodds and credit scores. Let x = logodds and y = credit score. Hint:

Odds(y=1) Credit Score 1 500 2 600

2. Suppose that you wish to build a credit model where the odds of not-defaulting doubles every 75 points and the odds of defaulting is equal to 2 at a score of 300. Determine the formula for this credit model.

3. Suppose you have the following two points, find the equation of the line that defines the relationship between logodds and credit scores. Let x = logodds and y = credit score.

Credit Score LogOdds(y=1) 500 0 600 1

4. Suppose you have the following two points, find the equation of the line that defines the relationship between logodds and credit scores.

Logodds(y=1) Credit Score 0 400

Ln(2) 450

5. Suppose you have the following two points where you know the odds and the credit score. Find the equation of the line that defines the relationship between logodds and credit scores.

Odds(y=1) Credit Score 1 600 3 640

6. Suppose you have the following two points where you know the odds and the credit score. Find the equation of the line that defines the relationship between logodds and credit scores.

Odds(y=1) Credit Score 3 550 6 600

Page 24 of 35

Part F: FICO Scores

1. Suppose that a credit card company uses a Fico model to predict new customer defaults. The Fico model predicts the logodds of a charge-off2(or loan default). Thus, y takes on two values: y = “charge-off” or “no charge-off” or we can write y=“default” or “no-default”.

The model is given by: ( ) = − 19 50

logodds default Fico

(a) Fill out the following table:

Fico Score Log odds(default) Odds(default) Prob(default)

800

720

640

530

(b) Now fill out the table but we want to look at no-defaults as the outcome:

Fico Score Logodds(no-default) Odds(no-default) Prob(no-default)

800

720

640

530

2. Suppose that a credit card company uses a FICO scoes to predict new customer defaults. The FICO model predicts the logodds of a default.

The model is given by: ( ) = − 19 50

logodds default Fico

(a) What would be the logodds equation for no-default?

(b) The bank wants to approve only applicants (or new customers) if the probability of a default is 4% or less. For what FICO scores would the bank decline a new customer?

Page 25 of 35

3. Suppose you wish to build a credit model. You want a score of 400 to be an odds of no default = 1. Then you want the odds to double every 50 points. Calculate a linear equation that describes the relationship of logodds of no default and your credit score.

4. Suppose you wish to build a credit model. You want a score of 300 to be an odds of no default = 4. Then you want the odds to double every 60 points. Calculate a linear equation that describes the relationship of logodds of no default and your credit score.

Page 26 of 35

Selected Solutions

Part A: Simple Logodds, Odds and Probability Problems--Solutions

1. Fill in the following table

Probability of (y=1) Prob of (y=0)

5% 95%

15% 85%

25% 75%

50% 50%

95% 5% 2. Fill in the following table

Probability of (y=1) Odds of (y=1) LogOdds of (y=1)

5% 0.5/0.95 ln(0.5/0.95)

15% 0.15/0.85 ln(0.15/0.85)

25% 0.25/0.75 ln(0.25/0.75)

50% 0.50/0.50 ln(0.50/0.50)

95% 0.95/0.05 ln(0.95/0.05)

3. Fill in the following table

Odds(y=1) Odds(y=0)

1 1

3 1/3

9 1/9

27 1/27

81 1/81

4. Fill in the following table

Logodds(y=1) logodds(y=0)

-2 2

-1 1

0 0

2 -2

4 -4

Page 27 of 35

5. Fill in the following table

Logodds(y=1) logodds(y=0)

ln(1/3) −ln(1/3)

ln(1/2) −ln(1/2)

ln(1) −ln(1)

ln(3) −ln(3)

ln(9) −ln(9)

6. Fill in the following table

Odds(y=1) Prob(y=1)

3 3/(1+3)

10 10/(1+10)

1 1/(1+1)

0.25 0.25/(1+0.25)

0.75 0.75/(1+0.75)

7. Fill in the following table

Odds(y=1) Prob(y=1)

e-2 e-2/(1+ e-2 )

e-1 e-1/(1+ e-1 )

e0 e0/(1+ e0)

e1 e1/(1+ e1)

e2 e2/(1+ e2)

Page 28 of 35

8. Fill in the following table

Odds(y=1) Odds(y=0)

e-2 1/e-2 = e2

e-1 1/e-1 = e1

e0 1/e0 = 1

e1 1/e1= e-1

e2 1/e2= e-2

9. Fill in the following table

Odds(y=1) Prob(y=1) Prob(y=0)

3 3/(1+3) = 3/4 1 – (3/(1+3))

10 10/(1+10) = 1/11 1 – (10/(1+10))

1 1/(1+1) = ½ 1 – (1/(1+1))

0.25 0.25/(1.25) = 1/5 1 – (0.25/(1.25))

0.75 0.75/1.75 = 3/7 1 – (0.75/1.75)

10. Fill in the following table

Log Odds of y=1 Odds of y = 1 Prob of y=1

-1.25 e-1.25 e-1.25/(1 + e-1.25 )

-0.75 e-0.75 e-0.75/(1 + e-0.75)

-0.25 e-0.25 e-0.25/(1 + e-0.25 )

0.25 e0.25 e0.25/(1 + e0.25 )

0.75 e0.75 e0.75/(1 + e0.75 )

1.25 e1.25 e1.25/(1 + e1.25)

Page 29 of 35

Part B—Partial Solutions

2. Suppose you are given the following logistic model:

= = − + +1 2 1

log odds( 1) 1 2 3e

y x x

where the two variables, 1x and 2x are binary variables, i.e., 

=  

1 0

x and 

=  

1 0

1x 2x Logodds(y=1) Odds(y=1) Prob(y=1)

0 0 -1 e-1 e-1/(1 + e-1)

0 1 1 e1 e1/(1 + e1)

1 0 -2/3 e-2/3 e-2/3/(1 + e-2/3 )

1 1 4/3 e4/3 e4/3/( 1+ e4/3)

Part C: Working with Logodds Models—Partial Solutions

1. Suppose you have ln(odds(y=1)) = 2– x. (a) What is the formula for odds(y=1)?

odds(y=1) = e2-x

(b) What is the formula for prob(y=1)? prob(y=1) = e2-x / (1 + e2-x)

2. Suppose you have ln(odds(y=1)) = 2 - x

(a) What is the formula for odds(y=0)?

ln(odds(y=0)) = – 2 + x

odds(y=0) = e-2 + x

(b) What is the formula for prob(y=0)? prob(y=0) = e– 2 + x / (1 + e–2+x)

Page 30 of 35

Part D—Partial Solutions

1. Suppose you have log(odds(y=1)) = 1 – 0.5x.

As x increases, the logodds falls, so the graph of logodds looks something like:

As x increases, the odds falls, so the graph of odds looks something like:

-15

-10

-5

-20 -15 -10 -5 0 5 10 15 20 logodds

logodds(y=1)

-1,000

4,000

9,000

14,000

19,000

24,000

-24 -19 -14 -9 -4 1 6 11 16

odds

odds(y=1)

Page 31 of 35

As x increases, the prob falls, so the graph of prob looks something like:

2. Suppose you have log(odds(y=1)) = 1 – 0.5x, then we have log(odds(y=0)) = 0.5x – 1.

The graphs will be similar to the ones from problem 1, except they will be flipped on the y-axis. Here is what the graph of logodds(y=0) will look like.

50%

100%

-20 -10 0 10 20

prob

prob(y=1)

-15

-10

-5

-20 -15 -10 -5 0 5 10 15 20 logodds

logodds(y=0)

Page 32 of 35

Part E —Partial Solutions

Odds(y=1) Logodds(y=1) Credit Score 1 0 500 2 Ln(2) 600

500 600 0

Lo go

dd s

Credit Score

Ln(2)

With two points (𝑥𝑥1,𝑦𝑦1) and (𝑥𝑥2,𝑦𝑦2)

(𝑦𝑦 − 𝑦𝑦1) = � 𝑦𝑦2 − 𝑦𝑦1 𝑥𝑥2 − 𝑥𝑥1

� (𝑥𝑥 − 𝑥𝑥1)

(𝑦𝑦 − 𝑦𝑦2) = � 𝑦𝑦2 − 𝑦𝑦1 𝑥𝑥2 − 𝑥𝑥1

� (𝑥𝑥 − 𝑥𝑥2)

Page 33 of 35

Pick (x1=500, y1=0) and (x2=600, y2=ln(2)) Then substitute into the first two-point equation of a line:

(𝑦𝑦 − 0) = � ln (2) − 0 600− 500

� (𝑥𝑥 − 500)

This simplifies to

𝑦𝑦 = ln (2) 100

(𝑥𝑥 − 500)

Then recall that y = logodds and x=credit score

𝑙𝑙𝑜𝑜𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 = ln (2) 100

𝑐𝑐𝑝𝑝𝑒𝑒𝑜𝑜𝑐𝑐𝑐𝑐 𝑜𝑜𝑐𝑐𝑜𝑜𝑝𝑝𝑒𝑒 − (5 × ln (2))

2. Suppose that the odds of not defaulting is 2 at a credit score of 300 and that odds of not defaulting double for every increase of 75 points. Find the equation of the line that defines the relationship between logodds and credit scores. Solution If we reduce the credit score to 300 – 75 = 225, then the odds of not defaulting = 1.

Odds(y=1) Logodds(y=1) Credit Score 1 0 225 2 Ln(2) 300

So the points to use to define the line are:

Logodds(y=1)

Credit Score 0 225

Ln(2) 300

Then use the same strategy as in the previous problem to find the equation of the line for those two points.

Page 34 of 35

Part F: FICO Scores—Partial Solutions

1. The model is given by: ( ) = − 19 50

logodds default Fico

Fill out the following table:

Fico Score Log odds(default) Odds(default) Prob(default)

800 9 − 800 50

= −7 𝑒𝑒−7 𝑒𝑒−7

1 + 𝑒𝑒−7

720

640

530

Now fill out the table but we want to look at no-defaults as the outcome:

Fico Score Logodds(no-default) Odds(no-default) Prob(no-default)

800 −9 + 800 50

= +7 𝑒𝑒−7 𝑒𝑒+7

1 + 𝑒𝑒∓7

720 −9 + 720 50

640

530

2. Suppose that a credit card company uses a FICO scoes to predict new customer defaults. The FICO model predicts the logodds of a default.

The model is given by: ( ) = − 19 50

logodds default Fico

(a) What would be the logodds equation for no-default?

𝑙𝑙𝑜𝑜𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(𝑙𝑙𝑜𝑜 𝑜𝑜𝑒𝑒𝑑𝑑𝑎𝑎𝑑𝑑𝑙𝑙𝑐𝑐) = −9 + 1

50 𝐹𝐹𝑐𝑐𝑐𝑐𝑜𝑜

(b) The bank wants to approve only applicants (or new customers) if the probability of a default is 4% or less.

If p = 4%, the odds = 4/96 = 1/24. The logodds = ln(1/24) = -ln(24)

Page 35 of 35

Solve for

−ln (24) = −9 + 1

50 𝐹𝐹𝑐𝑐𝑐𝑐𝑜𝑜

−50 ln (24) = −450 + 𝐹𝐹𝑐𝑐𝑐𝑐𝑜𝑜

450 − 50 ln (24) = 𝐹𝐹𝑐𝑐𝑐𝑐𝑜𝑜

𝐹𝐹𝑐𝑐𝑐𝑐𝑜𝑜 = 450 − 50 ln (24)

𝐹𝐹𝑐𝑐𝑐𝑐𝑜𝑜 = 608.903

Since FICO scores are integers, we need FICO > 609

3. See problem 1 and 2 from Part E as a hint. The problems are similar.

4. See problems 1 and 2 from Part E as a hint. The problems are similar.