ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)

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Prior whitening turns an anisotropic Gaussian posterior into Tikhonov form

Example

Assume AC. Let $X \sim N_{2} (0, diag (4, 1))$ , $η \sim N_{2} (0, diag (1, 4))$ independently, $A = I_{2}$ , and observe $y = (2, 4)^{T}$ in $Y = A X + η$ . Put $L = diag (2, 1)$ and $W = diag (1, 1 / 2)$ . The coordinate change $x = L z$ makes the prior on $z$ standard Gaussian, and $W$ whitens the observation noise in that $W Γ W^{T} = I_{2}$ . The preconditioned Tikhonov objective is $∥ W A L z - W y ∥_{2}^{2} + ∥ z ∥_{2}^{2} = ∥ diag (2, 1 / 2) z - (2, 2)^{T} ∥_{2}^{2} + ∥ z ∥_{2}^{2} .$ The posterior mean, MAP point, and minimizer in the original parameter coordinates are $(8 / 5, 4 / 5)^{T}$ , and the posterior covariance is $(4 / 5) I_{2}$ .

Example references

The posterior formula handles general positive-definite observation covariance.

Prior whitening gives the support-coordinate Tikhonov objective whose unique minimizer is the MAP and posterior mean.

Facts & Assumptions

Given: AC, the model and datum in the Example.

[A1]

AC is inherited from the Gaussian-law and finite Gaussian-conditioning interfaces. The Axiom of Choice

[F1]

The finite linear Gaussian posterior formula is valid for singular or positive-definite prior covariance and positive-definite noise covariance. Finite linear Gaussian inverse posterior with singular prior covariance

[F2]

The finite Gaussian MAP and posterior mean equal the support-restricted Tikhonov minimizer. Finite Gaussian MAP, posterior mean, and support-restricted Tikhonov minimizer coincide

[F3]

For $λ > 0$ , the Tikhonov objective is $∥ D z - b ∥_{2}^{2} + λ ∥ z ∥_{2}^{2}$ , with unique minimizer $(D^{T} D + λ I)^{- 1} D^{T} b$ . The Tikhonov regularised least-squares objective $∥ A x - b ∥_{2}^{2} + λ ∥ x ∥_{2}^{2}$ for $λ > 0$ , For every $λ > 0$ , the Tikhonov objective is strictly convex and has the unique minimiser $(A^{*} A + λ I)^{- 1} A^{*} b$

Verification

Given: The hypotheses and facts in the Example and Facts & Assumptions.

1.1A1F1givenalgebra

Here $S = C_{0} + Γ = 5 I_{2}$ , so [F1] gives $m (y) = C_{0} S^{- 1} y = (8 / 5, 4 / 5)^{T}$ and $C (y) = C_{0} - C_{0} S^{- 1} C_{0} = (4 / 5) I_{2}$ .

2.1F2F3step 1.1algebra∎

The whitening matrices satisfy $W Γ W^{T} = I_{2}$ and $D = W A L = diag (2, 1 / 2)$ , while $b = W y = (2, 2)^{T}$ . Applying [F3] with $λ = 1$ gives $\hat{z} = (D^{T} D + I_{2})^{- 1} D^{T} b = (4 / 5, 4 / 5)^{T}$ . Therefore $x = L \hat{z} = (8 / 5, 4 / 5)^{T}$ . By [F2] this is the unique MAP, posterior mean, and support-restricted minimizer.

Depends on

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Sources

Christian Clason, Regularization of Inverse Problems: Lecture Notes, arXiv:2001.00617v2 (2021) (standard reference, not scraped)