Link-loss identifiability from split and coverage

For each vertex $v$, put $q_v={\prod }_{e\in P(o,v)}{p}_e$, with $q_o=1$. Because $N_d\ge 1$, the repeated-probe law for design $d$ has one-probe marginal $P_{p,d}$; conversely, equality of all one-probe laws gives equality of $\mathsf{L}(p)$. For each covered leaf $r$, the marginal success probability read from any design containing $r$ is $m_r={\mathrm{Pr}}_p(Y_r=1)=q_r$. Thus equality of experiment laws determines every $q_r$ when all receivers are covered.

If a receiver $r$ is uncovered, its terminal edge $e=(u,r)$ occurs in no observed path. Changing $p_e$ to a different value in $(0,1)$ leaves every observed outcome law unchanged, because $r$ is a leaf and no other receiver path uses $e$. The parameter-to-law map is then not injective.

Suppose a nonroot internal vertex $v$ is not split by any design. Let $e_0=(u,v)$ be its incoming edge and let $e_c=(v,c)$ range over its child edges. Since $p_{e_0}<1$ and each of the finitely many $p_{e_c}<1$, an open interval around $1$ keeps $t{p}_{e_0}<1$ and $p_{e_c}\mathrm{/}t<1$ for every child; all changed probabilities remain positive there. Choose $t\mathrm{\ne }1$ in that interval. Define $p_{e_0}^{\mathrm{\prime }}=t{p}_{e_0}$ and $p_{e_c}^{\mathrm{\prime }}=p_{e_c}\mathrm{/}t$, leaving every other probability fixed. Then $p_{e_0}^{\mathrm{\prime }}{p}_{e_c}^{\mathrm{\prime }}=p_{e_0}{p}_{e_c}$ for every child $c$. In each design, receivers below $v$, if any, lie in one child subtree $c$; their outcomes all contain the common factor $B=X_{e_0}{X}_{e_c}$. This factor is Bernoulli with the same success probability under $p$ and $p^{\mathrm{\prime }}$, and is independent of every other link variable. Receivers outside that child subtree use no changed edge; their possible shared ancestor variables remain unchanged and independent of $B$. Thus the joint receiver-outcome law for every design is unchanged, although $p^{\mathrm{\prime }}\mathrm{\ne }p$.

Let $v$ be a nonroot internal vertex. By [F2], some design contains receivers $r,s$ in distinct child subtrees of $v$; since $T$ is a tree their last common vertex on the root paths is $v$. Write $a_r={\prod }_{e\in P(v,r)}{p}_e$ and $a_s={\prod }_{e\in P(v,s)}{p}_e$. Independence in [F1] gives $m_r=q_v{a}_r$, $m_s=q_v{a}_s$, and $m_{rs}:={\mathrm{Pr}}_p(Y_r=Y_s=1)=q_v{a}_r{a}_s$. All factors are positive, so $m_{rs}>0$ and $q_v=m_r{m}_s\mathrm{/}{m}_{rs}$. Hence the law determines $q_v$ for every nonroot internal vertex.

The value $q_o=1$ is fixed; steps 1.1 and 2.1 determine $q_v$ for every other vertex. For each edge $e=(u,v)$, $p_e=q_v\mathrm{/}{q}_u$, with $q_u>0$. Therefore equal experiment laws give equal $p_e$ on every edge, so $\mathsf{L}$ is injective. By [F3], the link-probability identity target is identifiable.

If the split/coverage condition fails, either a receiver is uncovered or a nonroot internal vertex is not split. Steps 1.2 and 1.3 give two distinct parameter vectors with the same full sampling law in either case. Thus injectivity of $\mathsf{L}$ implies the condition; together with step 3.1 this proves the equivalence.