MMP.15:5 - Archetypal Grounding

These constructed population probabilities make the calculations inspectable; they are not empirical results. Each case assumes fixed intervention meanings and no interference between its units. Finite records add estimation uncertainty to the assumption dependence shown here.

MMP.15:5.1 - Separate a production action from the batches receiving it

A production team asks whether enabling a stabilization mode (A=1), rather than (A=0), increases the probability of a conforming item (Y=1). Incoming batch condition (L) is recorded before mode selection. Half the target batches have (L=0), half (L=1). Mode 1 is used on 80% of (L=0) batches and 20% of (L=1) batches.

The subject account asserts consistency and mean exchangeability given (L); both modes occur in both strata. Adjustment therefore gives (\mu_0=0.5(0.10)+0.5(0.70)=0.40), (\mu_1=0.5(0.20)+0.5(0.80)=0.50), and (\Delta=0.10).

The selected groups instead give (E[Y\mid A=1]=0.8(0.20)+0.2(0.80)=0.32) and (E[Y\mid A=0]=0.2(0.10)+0.8(0.70)=0.58): an association of (-0.26). Its sign differs because the two modes receive different batch mixtures.

This positive average effect is conditional on the causal account; the table does not prove exchangeability. An already supported account and sufficient estimate need no new experiment.

MMP.15:5.2 - Use a mediator, then withdraw a pathway exclusion

A service’s command (A) can activate a retry mechanism (M), which affects successful completion (Y). Unrecorded load can affect both command selection and completion. The initial causal account permits (A\to M\to Y) and an unobserved common cause of (A,Y), but no other arrows: in particular, no direct effect of (A) on (Y) and no hidden common cause of (A,M) or (M,Y).

All three variables are binary. The available joint law has (P(A=1)=0.5), (P(M=1\mid A=0)=0.25), (P(M=1\mid A=1)=0.75), and:

There is no observed pretreatment adjustment variable for the shared load. The front-door conditions nevertheless hold. First adjust the mediator’s effect over (A): the inner means are (h(0)=0.5(0.10)+0.5(0.30)=0.20) and (h(1)=0.5(0.70)+0.5(0.90)=0.80). Then average over the mediator law induced by each command:

[ \mu_0=0.75(0.20)+0.25(0.80)=0.35,\qquad \mu_1=0.25(0.20)+0.75(0.80)=0.65. ]

Thus (\Delta=0.30). Directly comparing observed commands instead gives (0.75-0.25=0.50).

Changed condition. Inspection of the service reveals that the command can also change completion through a path bypassing retry. The revised account permits a direct (A\to Y) arrow. Retaining the previous formula is no longer justified.

The same full observational law now admits at least two answers. To demonstrate this, let unobserved (U) be a fair binary variable, let observational command selection be (A=U), and generate (M) with probability (0.25+0.50A) using independent random variation. In two alternative models, generate completion with respective probabilities

[ \text{Model I: }0.10+0.60M+0.20U,\qquad \text{Model II: }0.10+0.60M+0.20A, ]

using a further independent random draw. Every probability lies between zero and one. Because (A=U) observationally, both models give every cell of the supplied joint law.

Under (do(A=a)), (U) remains fair. Model I gives (\mu_0=0.35,\mu_1=0.65); Model II gives (\mu_0=0.25,\mu_1=0.75). Their effects are 0.30 and 0.50. Both satisfy the revised account, which permits the listed influences without requiring every permitted influence to be nonzero. Model II was excluded by the original no-direct-effect premise.

This is a proof that the revised assumptions and available law do not identify the average effect. These are ambiguity witnesses, not discovered service mechanisms. More precise estimation of the same law cannot distinguish them. A substantive restriction might; further evidence is chosen by its value for the use.

MMP.15:5.3 - Transfer an experiment to another mixture of sites

An ecological model asks for the effect of a specified irrigation change (A) on establishment of a seedling (Y), averaged across a target collection of sites. Soil stratum (L) is known from an existing inventory. A selected experiment randomized both irrigation levels within each stratum, recorded every outcome, and used the same irrigation implementations as the target question.

Suppose the subject account supports equality of conditional potential-outcome means between experimental and target sites. Experimental sites are 80% (L=0) and 20% (L=1); the target inventory is 25% (L=0) and 75% (L=1).

The experimental average effect is (0.8(0.30)+0.2(0.10)=0.26). The target means are (\mu_0=0.25(0.20)+0.75(0.60)=0.50) and (\mu_1=0.25(0.50)+0.75(0.70)=0.65). The requested effect is 0.15. Randomization identifies the comparisons inside the experiment; the transport assumption and target inventory justify the different outer average.

Now suppose the source is found to contain experimental outcomes only for (L=0); the supplied (L=1) values were extrapolations, with no justified response relation supporting them. Retain the target inventory and transport within the covered stratum. With no outcome restriction for (L=1) beyond binary outcomes, its mean effect (\delta_1) lies in ([-1,1]). Therefore

[ \Delta=0.25(0.30)+0.75\delta_1\in[-0.675,0.825]. ]

The endpoints are attainable by making every uncovered site respectively harmed or helped: ((Y(0),Y(1))=(1,0)) or ((0,1)). Those choices do not alter any available experimental outcome. The bound is sharp under these assumptions, and the sign of the target effect is not identified. The effect 0.30 remains available for the covered stratum if that is the agreed receiving question; it must not silently replace the original population target.

MMP.15:5.4 - Distinguish an offer effect, a local use effect and an overall effect

A team considers offering help with a work procedure. Let (Z=1) mean an independently randomized offer, (A=1) actual use, and (Y=1) successful completion. Assume fixed versions, no interference, exclusion, monotonicity and complete outcome recording. The following constructed population supplies one possible basis:

Randomization is independent of response group and potential outcomes. The action rates are 0.2 without the offer and 0.6 with it. The outcome means are

[ E[Y\mid Z=0]=0.2(0.60)+0.4(0.40)+0.4(0.30)=0.40, ] [ E[Y\mid Z=1]=0.2(0.60)+0.4(0.65)+0.4(0.30)=0.50. ]

The offer effect is 0.10. The first stage is 0.40, so the ratio gives (0.10/0.40=0.25), the complier mean effect. Under the stipulated full table, the population effect is instead (0.2(0.50)+0.4(0.25)+0.4(0)=0.20).

The observational law does not reveal the full table. Change the never-takers’ mean (Y(1)) from 0.30 to 0.80 while retaining all other quantities. Their action remains zero under both assignments, so this change leaves the full observed law of (Z,A,Y) unchanged and preserves the assumptions. The overall effect becomes 0.40; the identified complier effect remains 0.25. This pair of models shows why the available law does not identify the overall effect under these premises.

Now change the offer itself: it teaches a technique that can improve success without use of the help. Exclusion no longer holds. Retain the randomized offer-effect question, but withdraw the former interpretation of the ratio as the complier use effect until a revised model supplies a valid argument. Observing the same four means would not restore the missing exclusion premise.