DTM.5:5 - Archetypal Grounding

DTM.5:5.1 - Within-group increase is canceled by changing contribution weights

Take two equally weighted source groups. Their initial shares of variant B are 0.2 and 0.8. Along each group’s transmitted contribution the B share increases by 0.1. The groups supply relative transmitted-instance counts of 2 and 1.

The initial overall B share is 0.5. The receiving share is:

(2 × 0.3 + 1 × 0.9) / 3 = 0.5.

The contribution-weight change is −0.1; the within-contribution change is +0.1. They cancel. An unchanged aggregate would therefore conceal two material changes.

The weights count transmitted instances, not new organizations and not usefulness. The result does not establish why the low-B group supplied twice as much. Better results, broader access or a sampling difference are possible explanations requiring different follow-up.

If the two groups instead supply equal counts, the new B share becomes 0.6. This changed condition shows which contribution the aggregate depended on.

DTM.5:5.2 - A copied method can change without being selected

Suppose every teaching group sends one equally weighted procedure, but recipients reconstruct it with a different timing convention. A change in the receiving mean belongs to the within-contribution term. There is no differential contribution in this stipulated case.

The next action is to study reconstruction with DTM.2 and the relevant learning or notation method. Calling the change selection would direct attention away from how the recipient obtained the new variant.

DTM.5:5.3 - Counting successor collectives answers another question

A project network may form successor teams with different arrangements. If the question is which arrangement becomes common among successor teams, count those teams and recover their formation relations. If the question is which procedure most workers later use, count the corresponding use and contribution relations.

One prolific source team can dominate the latter count without dominating the former. The choice of unit is part of the question, not a defect to remove by finding one universal fitness number.

DTM.5:5.4 - External arrival changes the receiving mean

One procedure instance A continues from the identified source, with B-share z_old = z_cov = 0. One instance B arrives from outside that source set. The receiving population has two instances, so c=1/2 and z_ext=1. Its B-share is (1/2)×0 + (1/2)×1 = 1/2. A source-only calculation would return zero because it answers only for the covered part.

If the external instance’s variant is unknown, the whole-population B-share lies between 0 and 1/2. Additional source tracing is needed only if that uncertainty changes the receiving decision. Neither result attributes the external arrival to selection within the original source.