Simpson's Paradox, Stratification, and Causal Inference in Statistics

UC Berkeley Admissions Case and Simpson's Paradox

  • Historical Context and Lawsuit:

    • In the 1970s, a high-profile lawsuit was brought against the University of California, Berkeley alleging gender bias in graduate school admissions.

    • Aggregate university data revealed that male applicants were accepted at a significantly higher overall rate than female applicants.

    • The university administration had not set out to create discriminatory policies; admissions decisions were executed independently by departmental chairs.

  • Departmental Analysis and Resolution:

    • Berkeley hired statistician Betty Scott to perform a comprehensive subgroup analysis on the admissions data.

    • When admissions rates were evaluated separately across individual departments, nearly every department accepted female applicants at a equal or higher fractional rate than male applicants.

    • The apparent bias in the aggregate data vanished and reversed when conditioned on individual departments.

  • Definition of Simpson's Paradox:

    • First identified mathematically by Edward H. Simpson in 1951 during subgroup analyses.

    • Simpson's Paradox occurs when statistical trends, associations, or directional relationships observed in aggregate data completely reverse, disappear, or invert when the data is split into underlying subgroups or conditioned on a confounding variable.

Mathematical Framework of Simpson's Paradox

  • Departmental Acceptance Rates and Aggregation:

    • Let WAW_A and WBW_B represent the number of women applying to Department A and Department B, respectively.

    • Let wAw_A and wBw_B represent the number of women accepted to Department A and Department B, where w_A \begin{matrix} \frac{w_A + w_B}{W_A + W_B} \frac{w_A}{W_A} \frac{w_B}{W_B} \frac{m_A + m_B}{M_A + M_B} \frac{m_A}{M_A} \frac{m_B}{M_B} 10,000 21 2,100 100,000 10 10,000 110,000 12,100 \frac{12,100}{110,000} \times 100 \text{ per cent} \rightarrow 11 \text{ per cent} 2,000 20 400 100 10 10 2,100 410 \frac{410}{2,100} \times 100 \text{ per cent} \rightarrow 19.5 \text{ per cent} 80 \text{ per cent} 10 \text{ per cent} 100 50 50 50 \times 0.80 = 40 50 \times 0.10 = 5 45 100 \frac{45}{100} = 0.45 \rightarrow 45 \text{ per cent} 200 100 100 100 \times 0.80 = 80 100 \times 0.10 = 10 90 200 \frac{90}{200} = 0.45 \rightarrow 45 \text{ per cent} 100 90 10 90 \times 0.80 + 10 \times 0.10 = 72 + 1 = 73 100 \frac{73}{100} = 0.73 \rightarrow 73 \text{ per cent} 200 20 180 20 \times 0.80 + 180 \times 0.10 = 16 + 18 = 34 200 \frac{34}{200} = 0.17 \rightarrow 17 \text{ per cent} 10 80 \text{ per cent} 5 20 \text{ per cent} 10 \times 0.80 = 8 5 \times 0.20 = 1 9 15 \frac{9}{15} = \frac{3}{5} = 0.60 \rightarrow 60 \text{ per cent} 5 80 \text{ per cent} 10 20 \text{ per cent} 5 \times 0.80 = 4 10 \times 0.20 = 2 6 15 \frac{6}{15} = \frac{2}{5} = 0.40 \rightarrow 40 \text{ per cent} 0.80 0.80 0.20 0.20 \text{Weighted Average} = \frac{10 \times 0.80 + 5 \times 0.20}{15} = \frac{8 + 1}{15} = 0.60 \text{Weighted Average} = \frac{10}{15} \times 0.80 + \frac{5}{15} \times 0.20 = 0.60 \frac{10}{15} P(\text{Apply } A \text{ } M) \frac{5}{15} P(\text{Apply } B \text{ } M) P(\text{Accepted} \text{ } \text{Man}) = P(A \text{ } \text{Man}) \times P(\text{Accepted} \text{ } \text{Man}, A) + P(B \text{ } \text{Man}) \times P(\text{Accepted} \text{ } \text{Man}, B) p (1-p) M p \times M (1-p) \times M p \times W (1-p) \times W \text{Rate}_{\text{male, CF}} = \frac{p \times M \times P(\text{Accepted} \text{ } \text{Male}, A) + (1-p) \times M \times P(\text{Accepted} \text{ } \text{Male}, B)}{M} \text{Rate}_{\text{male, CF}} = p \times P(\text{Accepted} \text{ } \text{Male}, A) + (1-p) \times P(\text{Accepted} \text{ } \text{Male}, B) \text{Rate}_{\text{female, CF}} = p \times P(\text{Accepted} \text{ } \text{Female}, A) + (1-p) \times P(\text{Accepted} \text{ } \text{Female}, B) P(\text{Accepted} \text{ } \text{Male}, A) = P(\text{Accepted} \text{ } \text{Female}, A) P(\text{Accepted} \text{ } \text{Male}, B) = P(\text{Accepted} \text{ } \text{Female}, B) p \text{Rate}_{\text{male, CF}} = \text{Rate}_{\text{female, CF}} 80 \text{ per cent} \rightarrow 100 \text{ per cent} 60 \text{ per cent} \rightarrow 80 \text{ per cent} 20 \text{ per cent} 80 \text{ per cent} 95 \text{ per cent} 60 \text{ per cent} 90 \text{ per cent} 10 \text{ per cent} 85 \text{ per cent} 40 \text{ per cent} \text{Proficiency}_{\text{School A}} = 0.20 \times 0.95 + 0.80 \times 0.60 = 0.19 + 0.48 = 0.67 \rightarrow 67 \text{ per cent} \text{Proficiency}_{\text{School B}} = 0.90 \times 0.85 + 0.10 \times 0.40 = 0.765 + 0.040 = 0.805 \rightarrow 80.5 \text{ per cent} 95 \text{ per cent} > 85 \text{ per cent} 60 \text{ per cent} > 40 \text{ per cent} 80.5 \text{ per cent} 67 \text{ per cent} 1,000 200 800 500 50 \text{ per cent} 75 \text{ per cent} 200 150 50 \text{ per cent} 800 400 75 \text{ per cent} 50 \text{ per cent} 400 150 \begin{matrix}\text{end}\{cases}\right.