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.