Theorem. Gluing two Gaussian laws across a separator [fdip-0066]
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  • November 1, 2026

Let \(V=V_1\cup V_2\), \(S=V_1\cap V_2\), \(A=V_1\setminus S\) and \(B=V_2\setminus S\). Suppose positive-definite centered Gaussian laws \(\mu _1,\mu _2\) have the same marginal \(\mu _S\). Write \(p_1,p_2,p_S\) for their positive Lebesgue densities. There is exactly one positive Gaussian extension with \(\phi _A\mathrel {\perp \mkern -9mu\perp }\phi _B\mid \phi _S\). Its density is \[p(a,s,b)=\frac {p_1(a,s)p_2(s,b)}{p_S(s)}.\] If \(S=\varnothing \), take \(p_S=1\); this is the product law.

Let \(\Lambda _1,\Lambda _2,\Lambda _S\) denote the three marginal precision matrices. Padding a matrix on \(T\) with zeros gives the matrix on \(V\) defined by \[\operatorname {pad}_T(M)=\pi _T^TM\pi _T.\] The extension's precision is \[Q=\operatorname {pad}_{V_1}(\Lambda _1) +\operatorname {pad}_{V_2}(\Lambda _2) -\operatorname {pad}_S(\Lambda _S).\] In particular, if a positive global Gaussian law has \(S\) separating \(A\) and \(B\) in its precision graph, this formula reconstructs its precision from the indicated marginals. The matrices \(\Lambda _i\) are marginal precisions, not principal restrictions of \(Q\).

Proof. 

  • November 1, 2026

Factor the proposed density as \(p_S(s)p_1(a\mid s)p_2(b\mid s)\). Successively integrating over \(a,b,s\) gives total mass one. Integrating just \(b\) or just \(a\) recovers \(p_1\) or \(p_2\), respectively; its conditional density is visibly a product. Completing the two local squares writes the quadratic form in its exponent \(-q/2\) as \[\begin {aligned} q(a,s,b)={}&s^T\Lambda _Ss +(a-M_1s)^TD_1(a-M_1s)\\ &+(b-M_2s)^TD_2(b-M_2s), \end {aligned}\] where \(D_1,D_2>0\) are the conditional precisions and \(M_1,M_2\) are the conditional mean maps. This form is positive definite: if it is zero, its three nonnegative terms force first \(s=0\), then \(a=b=0\). The density is therefore a centered positive Gaussian. Subtracting the three density exponents gives the displayed padding formula. These statements include empty coordinate spaces.

For any extension with the asserted conditional independence, its conditional marginals must equal the two prescribed conditional marginals almost everywhere. Its density is consequently the same product almost everywhere, proving uniqueness as a probability law. For an existing global law the Markov theorem supplies this independence. This is the density factorization underlying [speed1986gaussian, Propositions 1–2, pp. 139–141]; the construction here also proves positive-definite existence from the two compatible input laws.

The four-cycle calculation displays the marginal precisions and reconstructs every entry by this formula.