MMP.18:5.2 - Join two analyses without counting their prior twice

Two analyses concern the same binary condition z. Both start from P(z=1)=0.2 and P(z=0)=0.8. Each has one positive observation with the supplied law:

P(positive | z=1) = 0.75
P(positive | z=0) = 0.25.

The two observations are distinct and conditionally independent given z. Each separate posterior gives:

P(z=1 | one positive) = (0.2*0.75)/(0.2*0.75 + 0.8*0.25) = 3/7.

Multiplying the two posterior mass functions and normalizing gives 9/(9+16)=9/25=0.36. This has counted the shared prior twice.

Construct the joint model from one prior and the two likelihood factors:

P(z=1 | two positives)
 = (0.2*0.75^2)/(0.2*0.75^2 + 0.8*0.25^2)
 = 9/13, approximately 0.692.

The same result is recoverable from the separate posteriors by dividing their product by the common prior before normalizing. That operation preserves their intended contributions under the supplied conditional independence.

Now discover that the two reports contain the same observation, copied into two analyses. There is only one likelihood factor. The correct result under the original observation model is again 3/7; the 9/13 calculation is no longer supported. If there are two dependent observations instead, their joint conditional law is needed.

The exchange therefore includes the identity and dependence of the contributing information, not just two numbers labeled “probability.” In a method assessment, the same problem appears when two models of performance use overlapping case records.