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Finite-dimensional heavy Gaussian source shift

Physical scope

Pure mathematics; no physical premises or physical claim.

Statement

Assume the Axiom of Choice. Let $d \geq 1$ , let $K \in R^{d \times d}$ be symmetric positive definite, and let $j \in R^{d}$ . Let $γ_{K^{- 1}} = N_{d} (0, K^{- 1})$ , where covariance is the matrix $E [X X^{T}]$ . Then $\frac{\int_{R^{d}} \exp (- \frac{1}{2} x^{T} K x + j^{T} x) d x}{\int_{R^{d}} \exp (- \frac{1}{2} x^{T} K x) d x} = \int_{R^{d}} e^{j^{T} x} d γ_{K^{- 1}} (x) = \exp ⁣ (\frac{1}{2} j^{T} K^{- 1} j) .$ Equivalently, in the finite real Gaussian regulator with normalized weight $e^{- S}$ and $ℏ = 1$ , completing the source shift $χ = K^{- 1} j + η$ gives $\frac{1}{2} χ^{T} K χ - j^{T} χ = \frac{1}{2} η^{T} K η - \frac{1}{2} j^{T} K^{- 1} j .$ Use $ℏ = 1$ and coordinate/source units for which $x^{T} K x$ and $j^{T} x$ are dimensionless. Both integrals are finite; their common Gaussian normalization cancels in the ratio. No infinite-dimensional Lebesgue measure or determinant is asserted.

Facts & Assumptions

Given: AC; a finite dimension $d \geq 1$ ; symmetric positive-definite $K$ ; $j \in R^{d}$ ; and the covariance convention in the Statement.

[A1]

The Axiom of Choice: AC is the assumption used by the finite-dimensional normal-law and density suppliers.

[F1]

Multivariate normal law, including singular covariance: $N_{d} (m, Σ)$ has covariance $Σ$ for every real symmetric positive-semidefinite $Σ$ .

[F2]

A positive-definite multivariate normal law has its Gaussian density: if $Σ$ is positive definite, $N_{d} (m, Σ)$ has density $(2 π)^{- d / 2} \det (Σ)^{- 1 / 2} \exp (- \frac{1}{2} (x - m)^{T} Σ^{- 1} (x - m))$ .

[F3]

Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation: Lebesgue measure is invariant under translations.

[F4]

Measure-preserving transformations and systems: a measurable self-map is measure preserving when inverse images preserve measure.

[F5]

Integral invariance under measure-preserving maps: a measure-preserving map preserves integrals of nonnegative measurable functions, including extended-valued integrals.

Proof

technique · direct Gaussian density and translation

Given: The objects and conventions in the Statement.

1.1A1F1F2given

By [F1]–[F2], $γ_{K^{- 1}}$ has Lebesgue density $c_{K} \exp (- \frac{1}{2} x^{T} K x)$ , where $c_{K} = (2 π)^{- d / 2} \det (K^{- 1})^{- 1 / 2} > 0$ . The expectation in the Statement is therefore the ratio of the displayed Lebesgue integrals, provided the numerator is finite.

2.1F2F3F4F5step 1.1algebra∎

Since $K$ is invertible, direct expansion gives $\frac{1}{2} x^{T} K x - j^{T} x = \frac{1}{2} (x - K^{- 1} j)^{T} K (x - K^{- 1} j) - \frac{1}{2} j^{T} K^{- 1} j .$ The translation $x \mapsto x - K^{- 1} j$ preserves Lebesgue measure by [F3]–[F5], so the numerator is the denominator multiplied by $\exp (\frac{1}{2} j^{T} K^{- 1} j)$ . The denominator is finite and positive because it is the normalizing integral of the probability density in [F2]; hence the numerator is also finite, and division proves both equalities. Expanding the same square proves the action identity.

Finite-dimensional heavy Gaussian source shift

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources