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:
| Response group | Population share | (A(0),A(1)) | Mean (Y(0)) | Mean (Y(1)) |
|---|---|---|---|---|
| Always-takers | 0.2 | (1,1) | 0.10 | 0.60 |
| Compliers | 0.4 | (0,1) | 0.40 | 0.65 |
| Never-takers | 0.4 | (0,0) | 0.30 | 0.30 |
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.