DTM.3:5 - Archetypal Grounding

DTM.3:5.1 - Equal average readiness can imply different spread

Consider an illustrative readiness x between 0 and 1. Suppose a relevant demonstration occurs once per period and produces the receiving event with probability q(x)=x². This law is a declared hypothesis for the example, not a general law of learning.

In population A every source has readiness 0.5, giving q=0.25. In population B half have readiness 0.1 and half 0.9. Both have mean readiness 0.5, but population B has mean contribution (0.01+0.81)/2=0.41.

A model using only mean readiness would treat both populations identically. If the comparison is about transmitted successful practice, that reduction loses the effect. Preserve the readiness distribution or sufficient classes. If the actual event law were linear and contact opportunities equal, the same particular loss would disappear.

This case also separates evidence obligations. Measurements of readiness do not establish q(x)=x²; that link requires observations of the chosen receiving event.

DTM.3:5.2 - Growth can reduce the support available to each learner

Suppose local readiness follows the conditional model:

x' = α(f)(1−x) − δx
α(f) = α₀/(1+kf).

Here f is the fraction adopting, α is the effective support rate per learner, δ a loss rate, and k specifies how adoption loads the support arrangement. The signs and functional form express the assumed sharing mechanism; they would have to change if adoption added support faster than demand.

For α₀=1, δ=0.25 and k=3 in the chosen time unit, the stationary readiness is α/(α+δ). It is about 0.714 at f=0.2 and 0.541 at f=0.8. If q=x² remains the receiving-event law, the corresponding contributions are about 0.510 and 0.292.

The loop is now visible: adoption changes support per participant, readiness changes the receiving-event rate, and that rate changes adoption. These stationary substitutions are useful only when readiness adjusts fast enough. Immediately after a large influx of beginners, replacing their states with these equilibria can be wrong.

A trial to increase adoption might therefore need additional teaching support. The equation does not prove that such support exists or how to teach the missing operation; those are provider questions.

DTM.3:5.3 - Spread changes compatibility without changing a learner

Consider a group comparing two exchange standards. Let f be the share of its relevant partners on the new standard. Over the same horizon, the group compares switching with continuing. Suppose it pays conversion costs while its old-standard partners keep their process unchanged. Its gain from switching is b−c−l(1−f): improvement b, switching cost c and conversion cost l for the remaining old-standard partner share.

Here the feedback is through compatible exchanges. No readiness state or training equation is needed. DTM.2 constructs the actual adoption event; the resulting adoption changes f and thus the gain available to later groups. A broadcast announcement is not the same as an authorized, affordable transition.

Use ECO.8 to retain the coordination and cost-bearing relations. Use the engineering compatibility method to establish that an adapter actually works. The spread model couples those supplied results rather than replacing them.

DTM.3:5.4 - Repeated opportunities do not create repeated first uptake

For one initially eligible recipient, suppose two independent comparable opportunities each succeed with probability 0.8. The probability of first uptake during those opportunities is 1−0.2²=0.96. It can also be obtained as 0.8+0.2×0.8: the second term includes only the recipients still eligible after the first opportunity.

The product 2×0.8=1.6 instead gives the expected number of successes when both attempts are actually performed and success is repeatable. It is not a first-uptake probability. Either quantity can be useful, but they feed different receiving models. The participant-state coupling must preserve that distinction even when the same internal readiness supplies both probabilities.