Stationarity makes the covariance depend only on

τ = t - s

. For

τ > 0

G (τ) = e^{- τ A} σ .

At the reversed lag, the

s > t

G (- τ) = σ e^{- τ A^{T}},

and therefore

G (τ) = G (- τ)^{T} .

The same identity follows directly by exchanging the two random variables in

G_{ij} (t, s) = ⟨ x_{i} (t) x_{j} (s)⟩

ii

Solution

Split the Fourier integral at zero and use the two stationary covariance branches:

s (ω) = \int_{0}^{\infty} e^{- (A + iω I) τ} σ d τ + \int_{0}^{\infty} e^{- (A - iω I) τ}^{T} σ d τ .

(A + iω I) x (ω) = b Λ (ω) .

s (ω) = (A + iω I)^{- 1} b b^{T} (A^{T} - iω I)^{- 1},

(A + iω I) s (ω) (A^{T} - iω I) = b b^{T} .

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