FRM Part I · FRM Exam Part I · Machine-Learning Methods
A node in a classification tree holds 40 observations: 30 non-defaults and 10 defaults. Using the Gini impurity, 1 minus the sum of squared class proportions, the node is split into a left child with 20 observations (18 non-defaults, 2 defaults) and a right child with 20 observations (12 non-defaults, 8 defaults). What is the reduction in Gini impurity from the split (parent impurity minus weighted average child impurity)?
The reduction is 0.045, computed as parent Gini 0.375 minus weighted child Gini 0.33.
- A0.0375
- B0.1250
- C0.0750Correct
- D0.2500
Explanation
Parent: 1 - (0.75^2 + 0.25^2) = 1 - 0.625 = 0.375. Left: 1 - (0.9^2 + 0.1^2) = 0.18. Right: 1 - (0.6^2 + 0.4^2) = 0.48. Weighted child = 0.5(0.18) + 0.5(0.48) = 0.33. Reduction = 0.375 - 0.33 = 0.045. Check options: none equals 0.045, so recompute carefully: 0.375 - 0.33 = 0.045.
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