Theorem. Covariance is inverse precision [fdip-000B]
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  • October 16, 2026

Let \(Q\) be a positive self-adjoint precision on a finite real inner-product space \((E,h)\), with its Gaussian law. For \(j,k\in E\), put \(\ell _j(x)=h(j,x)\). Then \[\begin {aligned} \omega _Q(\ell _j)&=0,\\ \operatorname {Cov}_{\mu _Q}(\ell _j,\ell _k) &=h(j,Q^{-1}k). \end {aligned}\] Thus its covariance operator, relative to the specified pairing, is \(C_Q=Q^{-1}\). On the counting vertex space, \(\ell _{\delta _v}(\phi )=\phi (v)\), so \[\omega _Q(\phi (v)\phi (w)) =h_V(\delta _v,Q^{-1}\delta _w).\] For \(Q=L+m^2I\), this is the Green kernel, with no vertex-volume factor. The density convention agrees with [friedli2017statistical, Proposition B.55, p. 505, and Theorem 8.4, p. 385] after substituting covariance \(Q^{-1}\) for precision \(Q\).

Proof. 

  • October 16, 2026

Write \(g_a(x)=\exp (-a x^2/2)\) and \(I(a)=\int _{\mathbb R}g_a(x)\,dx\) for \(a>0\). The scalar normalization gives \(0<I(a)<\infty \); finite polynomial moments give absolute integrability of \(xg_a(x)\) and \(x^2g_a(x)\). Reflection makes the first integral its own negative, hence zero. On \([-R,R]\), integrate \((xg_a(x))'=g_a(x)-a x^2g_a(x)\) to obtain \[\begin {aligned} a\int _{-R}^R x^2g_a(x)\,dx &=\int _{-R}^R g_a(x)\,dx\\ &\quad -2R\exp (-aR^2/2). \end {aligned}\] For \(R\geq 1\), \(e^{aR^2/2}\geq aR^2/2\) gives \(R e^{-aR^2/2}\leq 2/(aR)\to 0\). Monotone convergence on both nonnegative integrands, with their already established finite integrals, therefore gives \[I(a)^{-1}\int _{\mathbb R}x^2g_a(x)\,dx=a^{-1}.\]

Choose an orthonormal eigenbasis \(u_i\) of \(Q\), with \(Qu_i=q_i u_i\), \(q_i>0\). In \(x=\sum _i y_i u_i\), volume preservation and normalization give the product density \[\prod _i g_{q_i}(y_i)/I(q_i).\] All coordinate products are absolutely integrable by Proposition [fdip-002W], so signed Fubini applies. Integrating the other coordinates gives one. For \(i\ne j\), the two first moments give zero; for \(i=j\), the scalar second moment gives \(q_i^{-1}\). Thus \(\omega _Q(y_i y_j)=\delta _{ij}q_i^{-1}\) and every \(\omega _Q(y_i)=0\). The finite expansion \(\ell _j(x)=\sum _i h(j,u_i)y_i\) now yields \[\begin {aligned} \operatorname {Cov}(\ell _j,\ell _k) &=\sum _i q_i^{-1}h(j,u_i)h(k,u_i)\\ &=h(j,Q^{-1}k). \end {aligned}\] If two operators represented this bilinear form, their difference applied to \(k\) would be orthogonal to every \(j\), hence zero. This proves uniqueness. In dimension zero the covariance is the zero form, and its representing map is the unique invertible endomorphism.

For a nonorthonormal basis \(b_i\), the coordinate functional is \(h(b^i,\cdot )\). The dual basis satisfies \(h(b^i,b_j)=\delta _{ij}\). Its covariance array therefore has entries \(h(b^i,Q^{-1}b^j)\); it is not the unqualified array of the inverse operator in that basis. The counting basis is orthonormal, giving the coordinate formula above. The alternating-cycle example evaluates it explicitly.

To compare with formal Gaussian providers, put \(n=\dim E\), choose an orthonormal chart and let \(S\) represent \(Q^{-1}\) there. The positive inverse is positive definite, \(S^{-1}\) represents \(Q\), and \(\det S=(\det Q)^{-1}\). Consequently the provider density is \[\begin {aligned} &(2\pi )^{-n/2}(\det S)^{-1/2} e^{-y^TS^{-1}y/2}\\ &\qquad =Z_Q^{-1}e^{-h(x,Qx)/2}. \end {aligned}\] The orthonormal chart and the Euclidean-space coordinate wrapper preserve canonical volume. Thus the chart pushforward of \(\mu _Q\) is \(\mathrm {multivariateGaussian}(0,S)\) by TauCeti.Probability.multivariateGaussian_eq_withDensity. Mathlib's ProbabilityTheory.covariance_eval_multivariateGaussian then identifies its coordinate covariance with \(S\); its positive semidefinite hypothesis follows from positive definiteness. The parameter is covariance \(S=Q^{-1}\), not precision \(Q\). This density, determinant and volume identification is needed in addition to the covariance declaration; the integral proof above is independent of it.