MMP.18:5 - Archetypal Grounding

These constructed cases keep the model assumptions visible so that the reader can change the exchange and recompute its consequence.

MMP.18:5.1 - Preserve a transfer across different time resolutions

Two components model stores A and B. Material flows from A to B with the supplied rate q(t)=k*t, where t is time since the interval’s start and k=1 unit per minute squared. The stores have enough material and capacity for the stipulated transfer over T=1 minute. Initially A=10 units and B=0.

The donor computes the interval amount:

Q = integral from 0 to T of k*t dt = k*T^2/2 = 0.5 units.
A(T) = 10 - 0.5 = 9.5 units.

The receiving simulator takes the initial rate q(0)=0 and holds it throughout the minute. It obtains B(T)=0. The combined stores now total 9.5 units, although the model contains no external removal.

The error is in the exchange. Both components must use the same transferred amount over the same interval. Sending Q=0.5 units and applying A(T)=10-Q, B(T)=Q gives a total of 10 units. C.29.BB supplies the balance; this construction makes the exchange between the two component representations satisfy it.

Now change the requested result: when does B first reach 0.125 units? Under the supplied continuous rate, B(t)=k*t^2/2, so it reaches the threshold at t=0.5 minutes. A receiver that inserts the entire amount only at T=1 minute reports a different crossing time despite preserving the final balance.

For that question, send or reconstruct the cumulative transfer Q(t)=k*t^2/2 over the interval, or use a computation with adequate intermediate and event resolution. Exchanging only the interval amount is sufficient for the final stores but insufficient for the crossing. The changed question reopens the temporal representation, not the already correct conservation argument.

A nonlinear receiver can likewise need more than an averaged input. Suppose two equally weighted fine cells supply x values (0,2), and the receiver requires the average of x^2. The mean input is 1, but the required response is (0+4)/2=2. Sending only the mean and squaring it gives 1. Supplying variance 1 restores 1^2+1=2; a justified closure could supply the same missing contribution in a larger model.

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.