Parent dependence persists under nondescendant conditioning

Use the supplied Borel noise densities, or their positive finite Borel versions from [F2] under Countable Choice. Work with their Borel shifted-noise product and its integrated marginals; changing density versions does not change an almost-everywhere conditional-independence factorization. Every variable in $\{i\}\cup P\cup Q$ is a nondescendant of $j$. By [F3], $N_j$ is independent of the vector $(X_i,X_P,X_Q)$. Conditional on $(X_i=x,X_P=p,X_Q=q)$, the structural equation therefore gives the density $p_{X_j\mid X_i,X_P,X_Q}(y\mid x,p,q)=q_j(y-f_j(p,x))$ for almost every $(x,p,q,y)$. This density does not depend on $q$, so the joint conditional density of $(X_j,X_Q)$ given $(X_i,X_P)$ factors; hence $X_j{\perp }_d{X}_Q\mid (X_i,X_P)$.

Suppose, for contradiction, that $X_j{\perp }_d{X}_i\mid X_P$. Since $N_j$ is independent of the parent vector $(X_i,X_P)$, its conditional density is $q_j(y-f_j(p,x))$. The assumed factorization says this density is independent of $x$ for almost every $p$ and almost every $x$. The conditional density of $X_i$ given $X_P=p$ is positive for almost every $x$ at almost every $p$ with $0<p_P(p)<\mathrm{\infty }$, because the full joint density is strictly positive. Therefore, for almost every such $p$, for almost every pair $x,x^{\mathrm{\prime }}$, the two translates $q_j(y-f_j(p,x))$ and $q_j(y-f_j(p,x^{\mathrm{\prime }}))$ agree for almost every $y$.

A positive integrable density cannot have a nonzero period: if $q_j(y-a)=q_j(y-b)$ almost everywhere with $a\mathrm{\ne }b$, then $q_j$ repeats on every translate of an interval of length $\mathrm{\mid }a-b\mathrm{\mid }$; that interval has positive integral, so the integral over the disjoint translates is infinite, contradicting $\int q_j=1$. Thus step 1.2 forces $f_j(p,x)=f_j(p,x^{\mathrm{\prime }})$ for almost every pair $x,x^{\mathrm{\prime }}$ and almost every $p$. Continuity of $f_j$ extends this constancy to every setting of the other parent coordinates: any two unequal values would persist on open neighborhoods of positive Lebesgue measure. This contradicts [F1], so $X_j\mathrm{̸}{\perp }_d{X}_i\mid X_P$.

If $Q$ is empty, $S=P$ and step 2.1 is the required conclusion. Otherwise, suppose $X_j{\perp }_d{X}_i\mid X_{P,Q}$. By the conditional-density equality in step 1.1 and this supposition, for almost every $(p,y)$ one has $p_{X_j\mid X_i=x,X_P=p}(y)=p_{X_j\mid X_P=p,X_Q=q}(y)$ for almost every pair $(x,q)$. The left side depends only on $x$ and the right side only on $q$; by Fubini, each is therefore almost everywhere constant in its varying coordinate. Hence $X_j{\perp }_d{X}_i\mid X_P$, contradicting step 2.1. Thus the factorization given $X_S$ fails.