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Exact elimination of a Gaussian star generates pair couplings

Statement

Assume ${A C}_{ω}$ and let integer $n \geq 2$ . Let $1_{n} \in R^{n}$ have every coordinate equal to $1$ , and define the precision matrix on $n$ leaf variables and one central variable by

$K_{n} : = (\begin{array}{cc} 2 n I_{n} & 1_{n} \\ 1_{n}^{T} & 1 \end{array}) .$

This matrix is positive definite, so it defines a finite centered Gaussian law with precision $K_{n}$ . Its leaf-to-leaf precision entries are initially zero for distinct leaves. Eliminating the central coordinate exactly gives leaf precision

$K_{l e a f} = 2 n I_{n} - 1_{n} 1_{n}^{T},$

whose off-diagonal entry is $- 1$ for every pair of distinct leaves. Thus one exact Gaussian block step creates a nonzero pair interaction between every two leaves, even though those pairs had no direct precision entry in the original star.

Statement references

Correctness of finite Gaussian block elimination

Finite Gaussian probability with a positive precision matrix

Facts & Assumptions

Given: ${A C}_{ω}$ , integer $n \geq 2$ , and the matrix $K_{n}$ in the Statement.

[F1]

Under ${A C}_{ω}$ , the finite exact block algorithm returns the retained Gaussian precision and normalization after any finite sequence of eliminations (Correctness of finite Gaussian block elimination).

[F2]

A real symmetric positive-definite precision defines the centered finite Gaussian law with covariance its inverse (Finite Gaussian probability with a positive precision matrix).

Proof

1.1givenF2algebra

For $x \in R^{n}$ and $y \in R$ , expansion and completion of the scalar square give ${(\begin{array}{c} x \\ y \end{array})}^{T} K_{n} (\begin{array}{c} x \\ y \end{array}) = (y + 1_{n}^{T} x)^{2} + 2 n \sum_{i = 1}^{n} x_{i}^{2} - {(\sum_{i = 1}^{n} x_{i})}^{2} .$ Expanding the finite sum shows $n \sum_{i = 1}^{n} x_{i}^{2} - {(\sum_{i = 1}^{n} x_{i})}^{2} = \sum_{1 \leq i < j \leq n} (x_{i} - x_{j})^{2} \geq 0.$ Consequently the remaining part is $n \sum_{i} x_{i}^{2} + \sum_{i < j} (x_{i} - x_{j})^{2}$ , which is positive when $x \neq 0$ ; when $x = 0$ and $(x, y) \neq 0$ , the first square is $y^{2} > 0$ . Hence $K_{n}$ is symmetric positive definite and the Gaussian law is defined by [F2].

2.1F1step 1.1algebra∎

Use one retained set consisting of the $n$ leaves and eliminate the final central coordinate. The eliminated block is $C = [1]$ , and the cross block is $B = 1_{n}$ . The algorithm and its correctness theorem [F1] therefore return $2 n I_{n} - 1_{n} [1]^{- 1} 1_{n}^{T} = 2 n I_{n} - 1_{n} 1_{n}^{T} .$ For $i \neq j$ , the identity matrix contributes zero and $(1_{n} 1_{n}^{T})_{i j} = 1$ , so the resulting precision entry is $- 1$ . In the original $K_{n}$ , the leaf block is $2 n I_{n}$ and its distinct-leaf entries are zero. This verifies the stated generation of all pair couplings.

Exact elimination of a Gaussian star generates pair couplings

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources