Why does improving precision hurt recall, and vice versa?

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AcadiFi · Accepted Answer

Precision = TP/(TP+FP); Recall = TP/(TP+FN). They trade off because as you tighten the classification threshold, fewer positives are flagged (both TP and FP decrease, but FN rises), so precision climbs while recall falls. ```mermaid flowchart LR A[Raise
threshold] --> B[Fewer
positives flagged] B --> C[Precision
UP] B --> D[Recall
DOWN] E[Lower
threshold] --> F[More
positives flagged] F --> G[Precision
DOWN] F --> H[Recall
UP] ``` Example: Heronshaw Insurance operates a fraudulent claims classifier. Base rate 2% fraud in 60,000 annual claims = 1,200 fraud cases. At threshold 0.30: flags 4,800 claims. TP=960, FP=3,840. Precision=20%, Recall=80%. Adjusters overwhelmed. At threshold 0.55: flags 1,400 claims. TP=780, FP=620. Precision=56%, Recall=65%. Manageable volume but 420 frauds slip through. Choice depends on cost structure: if investigation costs $220/claim and missed fraud averages $9,000 loss, optimum threshold minimizes total cost. Heronshaw's analysis shows threshold 0.42 (precision 38%, recall 74%) minimizes expected cost at ~$6.8M/yr vs. $8.4M at naive threshold. Tricks to improve both simultaneously: (1) collect more features, (2) use ensemble methods (random forest, gradient boosting typically beat single tree Pareto frontier), (3) cost-sensitive training with class weights, (4) SMOTE oversampling for rare positive class. But along any single model's ROC curve, the tradeoff is mathematically forced.

Why does improving precision hurt recall, and vice versa?

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