LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)

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Cross-covariance of finite Gaussian Wick powers

Statement

Assume the Axiom of Choice. Let $(X, Y)$ have a centered, possibly degenerate, jointly Gaussian law with covariance matrix $(\begin{array}{cc} σ^{2} & c \\ c & τ^{2} \end{array})$ , where $σ^{2}, τ^{2} \geq 0$ and the matrix is positive semidefinite. Define $: X^{n} :_{σ^{2}}$ and $: Y^{m} :_{τ^{2}}$ by the finite covariance Wick-power generating identity. For all integers $n, m \geq 0$ ,

$E [: X^{n} :_{σ^{2}} : Y^{m} :_{τ^{2}}] = {\begin{cases} n! c^{n}, & n = m, \\ 0, & n \neq m, \end{cases}$

with $c^{0} : = 1$ .

Statement references

Finite covariance Wick normal powers

Finite Gaussian normalization and Wick moments

Facts & Assumptions

Given: Assume the Axiom of Choice. Let $(X, Y)$ have the centered Gaussian law and covariance matrix in the Statement, and fix $n, m \geq 0$ .

[A1]

The Axiom of Choice is the stated hypothesis of the finite Gaussian suppliers. The Axiom of Choice

[F1]

For every finite list of centered jointly Gaussian linear coordinates, each moment is the finite sum over pairings of covariance factors; this includes singular covariance matrices. Finite Gaussian normalization and Wick moments

[F2]

The formal generating identities are $\exp (s X - σ^{2} s^{2} / 2) = \sum_{j \geq 0} : X^{j} :_{σ^{2}} s^{j} / j!$ and $\exp (t Y - τ^{2} t^{2} / 2) = \sum_{k \geq 0} : Y^{k} :_{τ^{2}} t^{k} / k!$ . Finite covariance Wick normal powers

Proof

technique · formal coefficient extraction from the finite Gaussian pairing formula

Given: [A1], [F1], [F2], and the finite centered Gaussian vector $(X, Y)$ .

1.1A1F1givenalgebra

In the formal power-series ring $R ⟦ s, t ⟧$ , set $M (s, t) = \sum_{a, b \geq 0} E [X^{a} Y^{b}] s^{a} t^{b} / (a! b!)$ . By [F1], a pairing with $j$ $X$ -self-pairs, $k$ $Y$ -self-pairs, and $ℓ$ cross-pairs can occur only when $a = 2 j + ℓ$ and $b = 2 k + ℓ$ . The number of such pairings is $a! b! / (2^{j} j! 2^{k} k! ℓ!)$ : choose the $ℓ$ cross-indices on each side, match them in $ℓ!$ ways, then pair the remaining indices on each side. Thus coefficient-by-coefficient $M (s, t) = \exp (σ^{2} s^{2} / 2) \exp (c s t) \exp (τ^{2} t^{2} / 2)$ . This is an identity of formal series, so no moment-generating-function convergence is assumed.

2.1F2step 1.1algebra∎

Multiplying the series in step 1.1 by $\exp (- σ^{2} s^{2} / 2) \exp (- τ^{2} t^{2} / 2)$ leaves $\exp (c s t) = \sum_{r \geq 0} c^{r} s^{r} t^{r} / r!$ . By [F2], the coefficient of $s^{n} t^{m}$ on the multiplied left side is $E [: X^{n} :_{σ^{2}} : Y^{m} :_{τ^{2}}] / (n! m!)$ . The right side has zero coefficient when $n \neq m$ and coefficient $c^{n} / n!$ when $n = m$ , giving the claimed formula. For $n = m = 0$ the constant coefficient is $1$ , including when $c = 0$ .

Cross-covariance of finite Gaussian Wick powers

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources