Free SOA Exam SRM (Statistics for Risk Modeling) Decision Trees Practice Questions
Decision tree methods on SOA Exam SRM include classification and regression trees (CART), random forests, gradient boosting, and bagging. Questions test both algorithm mechanics and practical interpretation of results.
Sample Questions
Question 1
Easy
Random forests improve upon bagging by:
Solution
D is correct.
The key innovation of random forests over bagging is the random selection of a subset of predictors at each split point. This prevents dominant predictors from being used in every tree, thereby reducing the correlation between trees. Since the variance of an average of correlated quantities depends on the correlation, decorrelation leads to greater variance reduction.
The key innovation of random forests over bagging is the random selection of a subset of predictors at each split point. This prevents dominant predictors from being used in every tree, thereby reducing the correlation between trees. Since the variance of an average of correlated quantities depends on the correlation, decorrelation leads to greater variance reduction.
Question 2
Medium
Which statement about decision trees and outliers is correct?
Solution
E is correct.
Decision trees split by comparing predictor values to thresholds. The split only depends on whether an observation falls above or below the cutpoint, not on how far it is from the cutpoint. This makes trees inherently robust to outliers in the predictor space. An extreme value of a predictor simply falls to one side of a split just like any other value on that side.
Decision trees split by comparing predictor values to thresholds. The split only depends on whether an observation falls above or below the cutpoint, not on how far it is from the cutpoint. This makes trees inherently robust to outliers in the predictor space. An extreme value of a predictor simply falls to one side of a split just like any other value on that side.
Question 3
Hard
A gradient boosting model for regression uses absolute error loss . For a terminal node containing five observations with residuals , the optimal leaf value (the value that minimizes the total absolute error within the node) is:
Solution
D is correct.
When the loss function is absolute error , the value that minimizes over the constant is the median of the residuals.
Sorting the residuals: . With 5 observations, the median is the 3rd value: .
When the loss function is absolute error , the value that minimizes over the constant is the median of the residuals.
Sorting the residuals: . With 5 observations, the median is the 3rd value: .