Theorem. Precision zeros and Gaussian conditional independence [fdip-0065]
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  • November 1, 2026

Let \(\phi \sim N(0,Q^{-1})\) on \(\mathbb R^V\), where \(Q\) is any real symmetric positive-definite matrix. Its precision graph has an edge \(\{a,b\}\) for distinct vertices exactly when \(Q_{ab}\ne 0\). For pairwise disjoint subsets \(A,B,S\), say that \(S\) separates \(A\) from \(B\) if every graph path from \(A\) to \(B\) meets \(S\). Then \[S\text { separates }A\text { from }B \quad \Longrightarrow \quad \phi _A\mathrel {\perp \mkern -9mu\perp }\phi _B\mid \phi _S.\] Here conditional independence means that the conditional joint law factors into its two conditional marginals almost everywhere in \(\phi _S\). The canonical Gaussian density kernels used below factor at every conditioning value. For distinct vertices there is also the pairwise equivalence \[Q_{ab}=0\quad \Longleftrightarrow \quad \phi _a\mathrel {\perp \mkern -9mu\perp }\phi _b \mid \phi _{V\setminus \{a,b\}}.\] No Laplacian sign condition on \(Q\) is required.

Proof. 

  • November 1, 2026

First suppose \(V=A\sqcup S\sqcup B\). Separation implies \(Q_{AB}=0\). With \(\phi _S=s\) fixed, the quadratic energy in \(a,b\) is the sum \[\begin {aligned} &\frac 12a^TQ_{AA}a+a^TQ_{AS}s\\ &\qquad +\frac 12b^TQ_{BB}b+b^TQ_{BS}s +\frac 12s^TQ_{SS}s. \end {aligned}\] The last term is constant. The two remaining positive Gaussian factors normalize separately, proving independence at every \(s\). Equivalently, the conditional precision is \(\operatorname {diag}(Q_{AA},Q_{BB})\); its inverse has zero cross block. The square completion is the coordinate case of Proposition [fdip-0060].

For smaller disjoint \(A,B,S\), no connected component of the graph on \(V\setminus S\) meets both \(A\) and \(B\). Assign its components to two unions \(A',B'\) with \(A\subseteq A'\), \(B\subseteq B'\) and \(A'\sqcup B'=V\setminus S\); components meeting neither may be assigned arbitrarily. The preceding argument gives conditional independence of \(\phi _{A'}\) and \(\phi _{B'}\) given \(s\). Taking coordinate marginals gives the asserted independence of \(\phi _A\) and \(\phi _B\).

For the pairwise statement, fixing all other coordinates leaves conditional precision and cross covariance \[\begin {pmatrix}Q_{aa}&Q_{ab}\\Q_{ab}&Q_{bb}\end {pmatrix}, \qquad \operatorname {Cov}(\phi _a,\phi _b\mid \phi _{V\setminus \{a,b\}}) =\frac {-Q_{ab}}{Q_{aa}Q_{bb}-Q_{ab}^2}.\] The denominator is positive. Independence implies zero covariance, and zero \(Q_{ab}\) makes the conditional density a product. This proves both directions.

A positive Gaussian law is Markov with respect to a graph \(\Gamma \) if its precision is zero on all distinct nonedges of \(\Gamma \); extra zeros on edges are allowed. The same proof gives conditional independence across every separator of \(\Gamma \). Conversely its separator condition implies these zeros by taking \(S=V\setminus \{a,b\}\) for each nonedge. This is the finite Gaussian global Markov equivalence of [speed1986gaussian, Proposition 1 and Corollary 1, p. 139; Proposition 2, pp. 140–141]. The lattice harmonic-field statements of [friedli2017statistical, Theorems 8.17 and 8.26, pp. 391 and 399] concern their specified lattice precisions, not this arbitrary-precision theorem.

Mathlib's ProbabilityTheory.HasGaussianLaw.indepFun_of_covariance_eval supports the independence step for two finite coordinate families with a jointly Gaussian law and zero cross covariances. Apply it separately to the conditional Gaussian at each fixed \(s\); the block inverse above verifies its covariance hypothesis. The declaration asserts ordinary independence under that probability law. Graph separation, the passage to conditional kernels and the pairwise converse are supplied by the argument above.