Scientific model contracts and applicability — Examples

Consider eight enrolled participants indexed by $i\in \{1,\dots ,8\}$. Each participant is both the biological unit and the assignment unit; all eight participants are included, so this example does not posit probability sampling from a wider population. Exactly four participants are assigned to treatment ($Z_i=1$), uniformly over $\mathcal{Z}=\{z\in \{0,1{\}}^8:{\sum }_{i=1}^8{z}_i=4\},$ which has $\left(\genfrac{}{}{0px}{}{8}{4}\right)=70$ assignments. The response is recorded at baseline $t=0$ and four weeks after assignment $t=1$. At each participant-visit combination one specimen is read in three technical assay replicates $r\in \{1,2,3\}$.

For each $i,t,a$, let $u_{it}(a)\in \mathbb{R}$ be a fixed response value under arm $a\in \{0,1\}$. Let $d_{itr}\in \mathbb{R}$ be a fixed reading offset for technical replicate $r$. The stipulated deterministic observation mechanism is $R_{itr}=u_{it}(Z_i)+d_{itr}\mathrm{.}$ The full data object has one row per participant and six response cells indexed by $(t,r)$; for this example the participant is the analysis unit as well as the biological and assignment unit. The assignment vector $Z$ is random under $\pi$, so the response vector $R$ is random under $P_{\theta }$. Both $Z$ and the 48 response coordinates are recorded. All $u_{it}(a)$ and $d_{itr}$ are fixed latent model quantities; in particular, $u_{it}(1-Z_i)$ is unobserved. The target population is these eight participants, and the estimand is the finite-schedule mean contrast at the four-week visit, $\psi (u,d)=\frac{1}{8}{\sum }_{i=1}^8(u_{i1}(1)-u_{i1}(0))\mathrm{.}$ The experiment law averages over the stated assignment distribution. Its support is the finite set of records generated by the 70 assignments. The assignment law is stipulated, with no implementation evidence supplied; the deterministic measurement rule is an unvalidated modeling assumption. No independence between the 48 readings or generalization beyond the eight participants is asserted.

The construction is finite and choice-free. Here $\pi$ is a stipulated finite design law; no physical protocol or implementation is asserted. In an empirical contract, this assignment is design-controlled only if a protocol imposes it and implementation records confirm it. The deterministic measurement rule is empirically assessable with calibration or validation measurements, but none are supplied here, so the construction establishes no physical adequacy.

Suppose one instrument reading $R$ is modeled by $R=q+b,$ where $q\in \mathbb{R}$ is the true quantity and $b\in \mathbb{R}$ is an unvalidated calibration offset. Take sample space $(\mathbb{R},\mathcal{P}(\mathbb{R}))$, parameter set $\mathrm{\Theta }={\mathbb{R}}^2$, and sampling law $P_{(q,b)}={\delta }_{q+b}$. The target is $\psi (q,b)=q$. The recorded reading determines the sum $q+b$, but the experiment does not separately determine $q$ and $b$. Thus the unvalidated assumption $b=0$ is not established by this reading model; an external calibration standard or additional model structure would be needed to support it.

The calibration offset is substantively untestable from this one-reading law alone. If calibration data identify $b$ at a known value $b_0$, or the model fixes $b=b_0$, then the observation $R=q+b_0$ determines $q=R-b_0$ in the resulting model. Additional data or nontrivial restrictions on $b$ may still leave $q$ unidentified. The proof uses no axiom of choice.