= Symptom equivalence does not establish mechanistic identity
Let $Z$ index either of two candidate pathways and suppose $P(Y=1\mid Z=1)=P(Y=1\mid Z=2)=p$ for a measured <symptom> $Y$. For any mixing <probability> $\pi=P(Z=1)$, marginalization gives $P(Y=1)=\pi p+(1-\pi)p=p$. An independent sample with $k$ positive observations out of $n$ therefore has <likelihood> $p^k(1-p)^{n-k}$, independent of $\pi$. The <symptom> alone does not identify the mixture proportion or pathway. This is an <identifiability> counterexample, not an empirical claim that real clinical pathways necessarily have identical <symptom> distributions. Additional indicators or <interventions> may distinguish them.
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