The Positive Test

Question

A disease affects 1% of a population. A test detects it with probability 95% when present, and gives a false positive with probability 5% when absent. A randomly chosen person tests positive. What is the probability they actually have the disease?

Accepted Answer

Use natural frequencies on 10,000 people: - Sick: 100, of whom 95 test positive (true positives). - Healthy: 9,900, of whom 5% = 495 test positive (false positives). A positive person is one of 95 + 495 = 590, so P(textsick mid +) = frac95590 = frac19118 approx 0.161. A "95% accurate" test leaves you at 16%: the false positives from the enormous healthy pool swamp the true positives from the tiny sick pool. Approach 2, odds form Prior odds 1:99; the likelihood ratio of a positive is frac0.950.05 = 19. Posterior odds 19:99, hence frac19118. One multiplication, no denominators. Why traders care This is base-rate neglect, the most common quantitative reasoning failure. The identical arithmetic governs trading signals: a backtest pattern that "fires correctly 95% of the time" but hunts for a regime present 1% of the time is mostly noise when it fires. Always ask: out of all the times the signal fires, what fraction are real?

The Positive Test

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