FRM Part I · FRM Exam Part I · Machine-Learning Methods
A node in a classification tree contains 40 loans, of which 20 defaulted. A candidate split sends 20 loans to the left child with 2 defaults and 20 loans to the right child with 18 defaults. Using the Gini impurity measure, 1 - sum of squared class proportions, what is the reduction in impurity from the split (parent minus weighted average of children)?
The reduction is 0.32. The parent node has Gini impurity 0.50, while each child has proportions 10/90 or 90/10, giving impurity 0.18. The weighted child impurity is 0.18, so the split lowers impurity by 0.50 minus 0.18, which equals 0.32.
- A0.18
- B0.32Correct
- C0.50
- D0.68
Explanation
Parent Gini = 1 - (0.5² + 0.5²) = 0.50. Left: p = 0.1, Gini = 1 - (0.01 + 0.81) = 0.18. Right: p = 0.9, Gini = 0.18. Weighted average = 0.18. Reduction = 0.50 - 0.18 = 0.32.
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