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FRM Part II · FRM Exam Part II · Credit Scoring and Retail Credit Risk Management

In developing a scorecard, an analyst bins an age variable. In one bin, there are 200 good customers and 20 bad customers. In the total sample there are 4,000 good customers and 800 bad customers. Using WOE = ln(% of goods / % of bads), what is the weight of evidence for this bin (approximately)?

The weight of evidence is about 0.693. The bin holds 5% of all goods (200/4,000) and 2.5% of all bads (20/800); the ratio is 2, and its natural log is 0.693. A positive value indicates the bin is relatively good-heavy.

  1. A0.693Correct
  2. B-0.693
  3. C2.303
  4. D0.000

Explanation

% goods = 200/4,000 = 5%. % bads = 20/800 = 2.5%. Ratio = 2.0, ln(2) = 0.693. The sign-reversed value (-0.693) results from inverting the ratio; 2.303 is ln(10), the result of using raw counts 200/20 = 10 instead of distributions.

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