Why Fairness Criteria Can Conflict
A predictive decision can be assessed in several legitimate ways. For a hypothetical program deciding which applications receive an additional review, one stakeholder may focus on whether applicants with the same predicted risk receive comparable scores. Another may focus on whether error rates are similar across groups. A third may ask whether a positive decision is equally reliable in each group. These are different questions, captured by calibration, equalized odds, and predictive parity, respectively.
When groups have different outcome prevalence, a single decision rule generally cannot satisfy all three criteria at once. This is not a contradiction in the definitions: the criteria compare different conditional probabilities. The practical lesson is to report which criterion is being assessed, how the decision is defined, and what population and outcome the analysis concerns. This tutorial is aligned with the SOA Exam PA (Predictive Analytics) syllabus.
Equalized odds: the true-positive rates and false-positive rates are equal across groups. Predictive parity: the positive predictive values are equal across groups. Calibration within groups: for a risk score \(R\), the observed event frequency among members of group \(g\) assigned score \(r\) equals \(r\), so \(P(Y=1\mid R=r,G=g)=r\) wherever that score occurs.
Calibration is a property of a score, whereas equalized odds and predictive parity are often evaluated for a thresholded decision. A calibrated score does not by itself guarantee equal predictive values above every possible threshold: the score distributions above the threshold may differ. A useful, precise special case is a binary decision represented as a two-level risk score: every positive decision receives the same score \(r_+\) in both groups, and every negative decision receives the same score \(r_-\) in both groups. Calibration of that score then requires the positive predictive value to equal \(r_+\) in each group. In this special case, calibration implies predictive parity for positive decisions. The examples below distinguish this special case from group-wise calibration more generally.
The Algebra Behind the Tradeoff
Write the prevalence in group \(g\) as \(\pi_g=P(Y=1\mid G=g)\), its true-positive rate as \(t_g\), and its false-positive rate as \(f_g\). Among the whole group, the proportion who are true positives is \(\pi_g t_g\). The proportion who are false positives is \((1-\pi_g)f_g\). Therefore, provided the group has positive decisions, its positive predictive value is
This is Bayes' rule expressed using rates per group member. It makes the source of the conflict visible: predictive value depends not only on the error rates but also on prevalence. Under equalized odds, set \(t_g=t\) and \(f_g=f\) in every group. If \(t>0\) and \(f>0\), the positive predictive value is an increasing function of prevalence:
Thus, if two groups have different prevalences and share the same nonzero true-positive and false-positive rates, their positive predictive values differ. Equalized odds and predictive parity cannot both hold in this non-degenerate setting. If the positive decision is also represented by one common calibrated score level, calibration cannot join both criteria: calibration would require that common positive score to equal each group's different positive predictive value.
Worked Examples
Worked Example: Equalized Odds With Different Positive Predictive Values
A hypothetical review program evaluates 1,000 applicants in each of two groups. The outcome is whether an applicant would experience a specified adverse event. The event prevalence is 20% in Group A and 50% in Group B. At a chosen decision threshold, both groups have a true-positive rate of 0.80 and a false-positive rate of 0.20. Calculate the confusion-matrix counts and positive predictive values. Determine which criteria are met.
Method. For each group, multiply the number with the event by the true-positive rate to obtain true positives, and multiply the number without the event by the false-positive rate to obtain false positives.
Group A. There are \(1{,}000(0.20)=200\) applicants with the event and \(1{,}000-200=800\) without it. Thus \(TP=200(0.80)=160\), \(FN=200-160=40\), \(FP=800(0.20)=160\), and \(TN=800-160=640\). The positive predictive value is \(160/(160+160)=0.5000\).
Group B. There are \(1{,}000(0.50)=500\) applicants with the event and 500 without it. Thus \(TP=500(0.80)=400\), \(FN=500-400=100\), \(FP=500(0.20)=100\), and \(TN=500-100=400\). The positive predictive value is \(400/(400+100)=0.8000\).
Check. Both groups have \(TPR=0.80\) and \(FPR=0.20\), so equalized odds holds. Predictive parity does not hold: the positive predictive values are 0.5000 and 0.8000. To avoid a common mistake, do not compare the numbers of false positives directly; the relevant false-positive rates are calculated among those without the event. The calculations confirm the algebraic result: equal error rates do not equalize the reliability of positive decisions when prevalence differs.
Worked Example: Predictive Parity and Calibration, but Not Equalized Odds
Use the same hypothetical group sizes and prevalences as in the preceding example. A revised decision rule has a true-positive rate of 0.80 in each group. The program chooses group-specific false-positive rates so that both groups have a positive predictive value of 0.75. Calculate the necessary rates and show how a two-level score can be calibrated for positive decisions.
Method. If the desired positive predictive value is \(q\), then for each group the true-positive proportion is \(\pi t\) and the false-positive proportion is \((1-\pi)f\). Solving the predictive-value equation for \(f\) gives
Group A. With \(\pi=0.20\), \(t=0.80\), and \(q=0.75\), the false-positive proportion per group member must be \(0.20(0.80)(0.25)/0.75=0.053333\). Since 80% of Group A does not have the event, its false-positive rate must be \(0.053333/0.80=0.066667\). Among 1,000 applicants, \(TP=160\), \(FN=40\), \(FP=800(0.066667)\), which is approximately 53.333, and \(TN\) is approximately 746.667. The positive predictive value is \(160/(160+53.333)=0.7500\).
Group B. With \(\pi=0.50\), the false-positive proportion per group member must be \(0.50(0.80)(0.25)/0.75=0.133333\). Because half the group does not have the event, its false-positive rate must be \(0.133333/0.50=0.266667\). Among 1,000 applicants, \(TP=400\), \(FN=100\), \(FP=500(0.266667)\), approximately 133.333, and \(TN\) is approximately 366.667. The positive predictive value is \(400/(400+133.333)=0.7500\).
Calibration check and conclusion. If every positive decision receives the common two-level score \(R=0.75\), the observed event proportion among positive decisions is 0.75 in each group, so that positive score level is calibrated. Predictive parity also holds. But equalized odds fails: the true-positive rates are both 0.80, while the false-positive rates, approximately 0.0667 and 0.2667, differ. The non-integer expected counts are rates applied to hypothetical groups of 1,000; rounding those counts before calculating the rates could create small artificial discrepancies.
Worked Example: A Perfect-Prediction Exception
Consider two hypothetical groups with prevalences 10% and 40%. A decision rule identifies every person with the event and no person without it. Check whether equalized odds, predictive parity, and calibration can all hold, despite the different base rates.
Group A. In 1,000 people, 100 have the event and 900 do not. Perfect decisions give \(TP=100\), \(FN=0\), \(FP=0\), and \(TN=900\). The true-positive rate is \(100/100=1\), the false-positive rate is \(0/900=0\), and the positive predictive value is \(100/100=1\).
Group B. In 1,000 people, 400 have the event and 600 do not. The counts are \(TP=400\), \(FN=0\), \(FP=0\), and \(TN=600\). Its true-positive rate is \(400/400=1\), false-positive rate is \(0/600=0\), and positive predictive value is \(400/400=1\).
Check and interpretation. Both groups have the same true-positive and false-positive rates, and both have positive predictive value 1. A score assigning 1 to every positive decision and 0 to every negative decision is calibrated: each score category has exactly its stated event frequency. Thus all three criteria hold in this perfect-prediction exception. This does not contradict the tradeoff: the derivation assumed a nonzero false-positive rate, while this example has \(f=0\).
What the Criteria Do—and Do Not—Establish
The calculations show why a fairness conclusion should not be reduced to a single metric. Equalized odds asks whether the decision misses the same fraction of people with the outcome and flags the same fraction of people without it. Predictive parity asks whether positive decisions have the same observed reliability. Calibration asks whether score values correspond to observed event frequencies within groups. These measures can point in different directions because they condition on different information.
A score can be calibrated within each group even when positive predictive values differ, if the score assigned to positive cases differs by group or if positive cases span a range of scores. In the first worked example, for instance, a two-level score assigning the positive category its group's observed positive predictive value would assign 0.50 in Group A and 0.80 in Group B. The positive category would then be calibrated separately within each group, but predictive parity would fail. This illustrates why a statement such as “the model is calibrated” should specify the score, the groups, and how calibration was assessed.
The two-level common-score case has a sharper implication. If positive decisions in both groups share one calibrated score \(r_+\), calibration requires each group's positive predictive value to equal \(r_+\). That is predictive parity. With different prevalences and nonzero common error rates, however, equalized odds requires different positive predictive values. Therefore all three requirements cannot be met under those conditions.
Specify what \(Y=1\) means and what model output or operational rule makes \(D=1\). The rates are not meaningful without these definitions.
Calculate or report \(P(Y=1\mid G=g)\). Different prevalence is central to interpreting predictive-value differences.
For equalized odds, compare both conditional error rates; for predictive parity, compare positive predictive values; for calibration, compare observed outcome frequencies with score values within groups.
Report which quantities align and which do not, then explain the practical consequences for affected stakeholders rather than labelling the model simply “fair” or “unfair.”
A common exam trap is to claim that calibration always means equal positive predictive value. It does not: calibration relates each score value to its outcome frequency, and groups may have different score distributions or different score values among positive decisions. Another trap is to say equalized odds alone forces different predictive values in every case. The result requires differing prevalence and non-degenerate rates; equal prevalence or perfect prediction can be exceptions. Always state the assumptions before applying the incompatibility argument.
Check Your Understanding
For each problem, identify the relevant rates and check the stated fairness criteria.
- Two groups have prevalences 0.25 and 0.50. Both have \(TPR=0.60\) and \(FPR=0.10\). Calculate each group's positive predictive value to four decimals. (Answer: 0.6667 and 0.8571; equalized odds holds, predictive parity does not.)
- Groups have prevalences 0.20 and 0.40 and share \(TPR=0.75\). What false-positive rate is required in each group for both positive predictive values to equal 0.60? (Answer: Group 1 \(FPR=0.1250\); Group 2 \(FPR=0.3333\).)
- Two groups have the same prevalence of 0.30 and share \(TPR=0.70\) and \(FPR=0.20\). Calculate the common positive predictive value and state whether the three criteria can coexist if positive decisions receive that common calibrated score. (Answer: \(0.21/(0.21+0.14)=0.6000\); equalized odds, predictive parity, and positive-category calibration can coexist.)
- Group A and Group B have different prevalences, but both have \(TPR=1\) and \(FPR=0\). What are their positive predictive values, assuming each group has members with the outcome? (Answer: Both are 1; the positive decisions are perfectly predictive, so differing prevalence does not prevent predictive parity.)
- A report states that a risk score is calibrated within each group and that equalized odds holds at a chosen threshold. Does that statement alone prove that positive predictive parity also holds? (Answer: No. Calibration across a range of score values does not by itself require equal positive predictive values among all cases above the threshold; the score distributions and thresholded decision must also be considered.)