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How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Gaussian-process regression and prediction

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Finite Gaussian-process regression observation model

Definition

Let X be a set, let m:X→R, and let k:X×X→R be symmetric and positive semidefinite, meaning that for every finite ordered list t1,…,tr in X and every a∈Rr, ∑i,j=1raiajk(ti,tj)≥0. For finite ordered lists A=(ai)i=1r and B=(bj)j=1s in X, define KAB=(k(ai,bj))i,j; repeated indices are allowed.

Given a real Gaussian process f=(ft)t∈X with mean m(t)=E[ft] and covariance k(s,t)=E[(fs−m(s))(ft−m(t))], and a nonempty finite training list X=(xi)i=1n in X, define fX=(fxi)i=1n and mX=(m(xi))i=1n. For a centered Gaussian noise vector ε∼Nn(0,R) with symmetric positive-semidefinite covariance R that is independent of f, define the observation vector Y=fX+ε and the matrix AX=KXX+R.

For a nonempty finite test list X∗=(xj∗)j=1p in X, define the latent test vector f∗=fX∗. If a centered Gaussian vector η∗∼Np(0,R∗) with symmetric positive-semidefinite covariance R∗ is independent of the pair (f,ε), define the noisy test vector Y∗=f∗+η∗.

Model diagram

diagram failed to render:
\begin{tikzpicture}[>=stealth]
  \node[draw,rounded corners] (prior) at (0,0) {$f\sim\mathrm{GP}(m,k)$};
  \node[draw] (train) at (-2,-1.5) {$f_X$};
  \node[draw] (test) at (3,-1.5) {$f_*$};
  \node[draw] (noise) at (-2,-3) {$\varepsilon$};
  \node[draw,rounded corners] (obs) at (0,-3) {$Y=f_X+\varepsilon$};
  \node[draw] (testnoise) at (3,-3) {$\eta_*$};
  \node[draw,rounded corners] (testobs) at (5,-3) {$Y_*=f_*+\eta_*$};
  \draw[->] (prior) -- (train);
  \draw[->] (prior) -- (test);
  \draw[->] (train) -- (obs);
  \draw[->] (noise) -- (obs);
  \draw[->] (test) -- (testobs);
  \draw[->] (testnoise) -- (testobs);
\end{tikzpicture}

The single prior node represents one process, so fX and f∗ may be correlated. Arrows into Y and Y∗ denote vector addition. The training noise ε is independent of f, and the test noise η∗ is independent of the pair (f,ε); the diagram records no posterior independence claim.

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Projection onto a finite-dimensional subspace of an inner-product space

Statement

Let H be a real inner-product space and V⊆H a finite-dimensional subspace. For each f∈H there is a unique p∈V such that f−p is orthogonal to V, meaning ⟨f−p,v⟩=0 for every v∈V. For every v∈V, ∥f−v∥2=∥f−p∥2+∥p−v∥2. In particular, ∥f∥2=∥p∥2+∥f−p∥2.

Facts & Assumptions

Given: A real inner-product space H, a finite-dimensional subspace V⊆H, and f∈H.

[F1]

Every finite-dimensional real inner-product space has an orthonormal basis; the basis is empty when the dimension is zero. Every finite-dimensional real or complex inner product space has an orthonormal basis

Proof

Given: The hypotheses in the Statement.

1.1F1givenchooseconstructalgebra

Choose an orthonormal basis e1,…,er of V by [F1], allowing r=0, and set p=∑j=1r⟨f,ej⟩ej. For each ℓ, orthonormality gives ⟨f−p,eℓ⟩=⟨f,eℓ⟩−∑j=1r⟨f,ej⟩⟨ej,eℓ⟩=0; since the basis spans V, f−p is orthogonal to V.

2.1givenstep 1.1algebra∎

For any v∈V, p−v∈V and f−p is orthogonal to V. Expanding the squared norm of f−v=(f−p)+(p−v) therefore gives ∥f−v∥2=∥f−p∥2+∥p−v∥2. If q∈V also has f−q orthogonal to V, then p−q=(f−q)−(f−p) lies in V⊥ and in V, hence ∥p−q∥2=0 and p=q. Taking v=0 gives the stated norm identity.

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Positive definiteness is exactly the inverse condition for the observed GP covariance

Statement

In the finite observation model, let K=KXX and let R be symmetric and positive semidefinite. Then A=K+R is symmetric and positive semidefinite. Its ordinary inverse exists if and only if A is positive definite, equivalently if and only if all singular values of A are positive. Thus the standard inverse formulas are defined exactly under A≻0; no ordinary inverse formula applies when A is singular.

Facts & Assumptions

Given: A finite observation matrix K=KXX from a symmetric positive-semidefinite kernel and a symmetric positive-semidefinite noise covariance R.

[F1]

Every finite Gram matrix of the kernel is positive semidefinite, and the observed matrix is A=KXX+R. Finite Gaussian-process regression observation model

[F3]

The singular values of an endomorphism are the eigenvalues of ∣A∣=A∗A. The singular values of a linear map as the eigenvalues of the positive square root of T^*T

[F4]

A nonnegative finite-dimensional operator has a unique nonnegative square root. A non-negative operator has a unique non-negative square root

[F5]

A finite-dimensional linear map has a singular-value decomposition with exactly its positive singular values in the nonzero diagonal block. Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition

Proof

Given: The matrices K and R in the Statement.

1.1F1givenalgebra

For every z∈Rn, zTAz=zTKz+zTRz≥0 by [F1], and A is symmetric. Hence A is a nonnegative endomorphism of the finite-dimensional real inner-product space Rn.

2.1F2step 1.1algebra

By [F2], choose an orthonormal eigenbasis e1,…,en of A with eigenvalues λj≥0. In this basis, A is invertible exactly when no λj is zero, while zTAz=∑jλjcj2 for z=∑jcjej is positive for every nonzero z exactly when all λj>0. Thus A is invertible if and only if it is positive definite.

3.1F3F4F5step 2.1algebra∎

In the same basis, A∗A=A2 has eigenvalues λj2. By [F4], its unique nonnegative square root has eigenvalues λj, so [F3] says the singular values of A are precisely these λj. By [F5], the SVD has full rank exactly when all n singular values are positive, which is exactly the invertibility condition in step 2.1.

TheoremStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Log marginal likelihood for a finite Gaussian-process regression model

Statement

Assume AC. Let X be a set, let f=(ft)t∈X be a real Gaussian process with mean m and symmetric positive-semidefinite covariance kernel k, and let X=(xi)i=1n∈Xn with n≥1. For independent observation noise ε∼Nn(0,R) with symmetric positive-semidefinite covariance R, set Y=fX+ε, mX=(m(xi)), KXX=(k(xi,xj)), and AX=KXX+R. Suppose AX is positive definite. Define the log marginal likelihood at y∈Rn as the logarithm of the Lebesgue density of Y at y under this prior-predictive model. Then Y∼Nn(mX,AX) and log⁡p(y∣X,m,k,R)=−12(y−mX)TAX−1(y−mX)−12log⁡det⁡(AX)−n2log⁡(2π). This formula is asserted only for the nonsingular observed-covariance case AX≻0.

Facts & Assumptions

Given: AC; the finite Gaussian-process observation model, a nonempty evaluation list X, independent noise ε∼Nn(0,R) with R⪰0, and AX=KXX+R≻0.

[A1]

The Axiom of Choice states that every family of nonempty sets has a choice function. The Axiom of Choice

[F1]

The finite observation model defines Y=fX+ε, mX, KXX, and AX=KXX+R, with ε independent of f. Finite Gaussian-process regression observation model

[F2]

A Gaussian process has multivariate-normal finite evaluation vectors, including singular covariance matrices. Gaussian process

[F3]

Under AC, a multivariate-normal law is characterized by its normal scalar projections and allows symmetric positive-semidefinite covariance. Multivariate normal law, including singular covariance

[F4]

A finite stack of independent Gaussian vectors is jointly multivariate normal with block-diagonal covariance. A finite stack of independent Gaussian vectors is jointly Gaussian

[F5]

Under AC, Nn(a,Σ) with Σ≻0 has Lebesgue density (2π)−n/2det⁡(Σ)−1/2exp⁡ ⁣(−12(x−a)TΣ−1(x−a)). A positive-definite multivariate normal law has its Gaussian density

Proof

Given: The hypotheses in the Statement.

1.1A1F1F2F3F4givenalgebra

By [F1] and [F2], fX∼Nn(mX,KXX). The independent noise has law Nn(0,R), so [F4] makes U=(fX,ε) jointly Gaussian with mean (mX,0) and block-diagonal covariance diag⁡(KXX,R). For any a∈Rn, aTY=(aT,aT)U is a normal scalar projection by [F3]; its mean is aTmX and its variance is aT(KXX+R)a. Thus the projection definition [F3] gives Y∼Nn(mX,AX).

2.1A1F5givenstep 1.1algebra∎

Apply [F5] to the law from step 1.1, using the hypothesis AX≻0. Its density at y is (2π)−n/2det⁡(AX)−1/2exp⁡ ⁣(−12(y−mX)TAX−1(y−mX)). This is positive, so taking its logarithm and expanding products gives the displayed formula.

CorollaryStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Finite Gaussian-process regression posterior and predictive law

Statement

Assume AC. Let X be a set, m:X→R, and k:X×X→R a symmetric kernel whose every finite Gram quadratic form on points of X is nonnegative. Let f=(ft)t∈X be the canonical Gaussian process with m(t)=E[ft] and k(s,t)=E[(fs−m(s))(ft−m(t))]. For finite ordered lists A=(ai)i=1r and B=(bj)j=1s in X, set KAB=(k(ai,bj)); repetitions are allowed. Take X=(xi)i=1n∈Xn and X∗=(xj∗)j=1p∈Xp with n,p≥1, and write mX=(m(xi)), m∗=(m(xj∗)), fX=(fxi), and f∗=(fxj∗). Let ε∼Nn(0,R) be independent of f, where R is symmetric positive semidefinite, and set Y=fX+ε and AX=KXX+R. Suppose AX is positive definite. Then for every y∈Rn the conditional law of f∗ given Y=y has mean m∗+K∗XAX−1(y−mX) and covariance K∗∗−K∗XAX−1KX∗. This covariance is positive semidefinite and may be singular. If η∗∼Np(0,R∗) is independent of the pair (f,ε), where R∗ is symmetric positive semidefinite, then the conditional law of f∗+η∗ has the same mean and covariance with R∗ added. The displayed kernel is defined for every y, although regular conditional versions are determined only PY-almost surely.

Facts & Assumptions

Given: AC; a set X, mean m, symmetric positive-semidefinite kernel k; finite ordered lists X and X∗ of positive lengths; observation noise ε∼Nn(0,R) independent of the process with R⪰0; optional test noise η∗∼Np(0,R∗) independent of the pair (f,ε) with R∗⪰0; and AX=KXX+R≻0.

[A1]

The Axiom of Choice states that every family of nonempty sets has a choice function. The Axiom of Choice

[F1]

Under AC, a mean function and symmetric positive-semidefinite kernel on an arbitrary index set determine a unique canonical process on the cylinder sigma-algebra with the stated finite Gaussian marginals; the construction supplies no sample-path continuity. A mean and positive-semidefinite kernel define a Gaussian process

[F2]

A real Gaussian process has multivariate-normal finite evaluation vectors, including repeated indices and singular covariance matrices. Gaussian process

[F3]

For finite lists, independent Gaussian observation noise with PSD covariance and positive-definite observed covariance gives the stated all-y conditional Gaussian kernel; its predictive covariance may be singular, and Gaussian test noise independent of the pair (f,ε) adds its covariance. Finite Gaussian-process posterior and predictive laws

[F4]

In this model, AX=KXX+R is invertible exactly when it is positive definite. Positive definiteness is exactly the inverse condition for the observed GP covariance

Proof

Given: The assumptions in the Statement.

1.1A1F1F2givenconstruct

By [A1] and [F1], the specified mean and kernel give the canonical cylinder-space Gaussian process. By [F2], its finite evaluations on X and X∗ are jointly multivariate normal, including when either list has repetitions or a covariance block is singular.

2.1A1F3F4givenstep 1.1∎

In the notation of the finite model, the observed block in [F3] is C=KXX+R=AX. By the hypothesis and [F4], C≻0, so every inverse required by [F3] exists. Under the optional noisy-prediction hypothesis, η∗ is independent of the pair (f,ε), exactly the joint-independence premise for [F3]'s Gaussian stack. Applying [F3] with these lists, means, noise covariance, and observed block gives the displayed conditional mean and covariance; its statement also gives positive semidefiniteness, possible singularity, the all-y version, and the test-noise covariance addition.

RemarkRemark: AI-adaptedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Computational costs and limits of finite Gaussian-process regression

Remark

A real Gaussian process is a family whose finite evaluation vectors are multivariate normal. Let T be a set and let f=(ft)t∈T be a real Gaussian process with mean m(t)=Eft and symmetric positive-semidefinite covariance kernel k(s,t)=E[(fs−m(s))(ft−m(t))]; positive semidefinite means every finite Gram quadratic form is nonnegative. For n≥1 and a finite training list X=(xi)i=1n in T, set fX=(fxi), KXX=(k(xi,xj)), and let ε be centered Gaussian observation noise independent of f, with symmetric positive-semidefinite covariance R. Set Y=fX+ε and AX=KXX+R≻0. Write the dense Cholesky factorization as AX=LLT, where L is lower triangular with positive diagonal. In the scalar-operation model, where each addition, subtraction, multiplication, division, or square root counts as one operation, factorization costs n3/3+O(n2) operations. For a test input x∗∈T, define kx∗X=(k(x∗,xi))i=1n and kXx∗=kx∗XT. After factorization, the latent predictive variance k(x∗,x∗)−kx∗XAX−1kXx∗ requires O(n2) operations, where O(n2) denotes an asymptotic operation count as n grows. These are arithmetic counts, not accuracy guarantees. For an invertible real n×n matrix A, define its spectral condition number by κ2(A)=∥A∥2∥A−1∥2, where ∥A∥2=sup⁡∥x∥2=1∥Ax∥2 for x∈Rn. For 0<ϵ<1, Aϵ=diag⁡(1,ϵ) has condition number 1/ϵ, while E=diag⁡(0,−ϵ) has relative spectral-norm size ∥E∥2/∥Aϵ∥2=ϵ even though Aϵ+E is singular.

The negative log marginal likelihood need not be convex in a covariance hyperparameter. On the one-point set T={t}, take the centered Gaussian prior ft∼N(0,θ) with θ≥0, covariance kernel kθ(t,t)=θ, and independent observation noise ε∼N(0,1). For observation Y=ft+ε at y=0, the observed covariance is Aθ=1+θ>0 and the negative log marginal likelihood is L(θ)=12log⁡(1+θ)+12log⁡(2π). For every θ>0, its second derivative is L′′(θ)=−1/(2(1+θ)2)<0, so L is not convex on its parameter domain.

The finite Gaussian-process regression posterior formula uses Gaussian observation noise. Write ϕ(z)=(2π)−1/2e−z2/2 for the standard normal density and Φ(z)=∫−∞zϕ(t) dt for its cumulative distribution function. With one latent value z∼N(0,1) and one positive label, the logistic likelihood P(y=1∣z)=σ(z)=(1+e−z)−1 or probit likelihood P(y=1∣z)=Φ(z) gives a non-Gaussian posterior: the posterior log density has nonzero third derivative at z=log⁡3 for logistic and z=0 for probit. Both likelihoods lie in (0,1), so multiplying by the standard normal density gives a finite, positive normalizing integral. Finally, a Gaussian process specifies finite-dimensional laws but need not have continuous paths. A continuous modification means a process on the same probability space that agrees almost surely at each fixed input and has almost surely continuous sample paths. Under AC, on T=[0,1] the equality kernel k(s,t)=1 when s=t and 0 otherwise has no continuous modification.

Facts & Assumptions

Given: The finite regression model, dense arithmetic-operation model, scalar covariance-parameter example, logistic and probit one-coordinate likelihoods, and arbitrary-index Gaussian-process construction described in the Remark.

[A1]

The Axiom of Choice is used for arbitrary-index finite Gaussian laws and Kolmogorov extension. The Axiom of Choice

[F1]

For a positive-definite observed covariance, the Gaussian-process log marginal likelihood is −12(y−mX)TAX−1(y−mX)−12log⁡det⁡AX−n2log⁡(2π). Log marginal likelihood for a finite Gaussian-process regression model

[F2]

A mean and symmetric positive-semidefinite kernel under AC define a Gaussian process with the stated finite Gaussian marginals, but the construction asserts no sample-path continuity. A mean and positive-semidefinite kernel define a Gaussian process

[F3]

Every finite evaluation vector of a Gaussian process is multivariate normal with the corresponding mean and covariance. Gaussian process

[F4]

A multivariate-normal law with mean a and covariance Σ has characteristic function u↦exp⁡(iuTa−12uTΣu), which uniquely determines the law. Characteristic function of a multivariate normal law

[F5]

Dense Cholesky factorization costs n3/3+O(n2) scalar operations, and the two triangular solves cost 2n2+O(n). Cholesky solves Hermitian positive-definite systems and has about half the factorisation cost of LU

[F6]

The finite Gaussian-process posterior formula has covariance K∗∗−K∗XAX−1KX∗ under independent Gaussian observation noise and permits that covariance to be singular. Finite Gaussian-process regression posterior and predictive law

[F7]

For decreasing measurable sets under a finite measure, the measure of their intersection is the limit of their measures. Continuity from above when one set has finite measure

Proof

Given: The hypotheses and models in the Remark.

1.1F5F6givenalgebra

Cholesky factorization costs n3/3+O(n2) by [F5]. For one test point, the posterior variance formula requires solving AXu=kX∗ and forming k∗∗−k∗Xu. Factoring AX=LLT reduces this to two triangular solves, each with O(n2) cost by [F5], followed by an O(n) inner product. Therefore the variance calculation after factorization is O(n2).

1.2givenalgebra

The spectral norm of Aϵ=diag⁡(1,ϵ) is 1, its smallest singular value is ϵ, and therefore its spectral condition number is 1/ϵ. The perturbation E=diag⁡(0,−ϵ) has norm ϵ, so ∥E∥2/∥Aϵ∥2=ϵ, while Aϵ+E=diag⁡(1,0) is singular. Thus operation counts alone do not control sensitivity to finite-precision perturbations near singularity.

1.3F1givenalgebra

The scalar kernel kθ(t,t)=θ is positive semidefinite for θ≥0, and the observed covariance is Aθ=1+θ>0. Apply [F1] with m=0 and y=0 to get L(θ)=12log⁡(1+θ)+12log⁡(2π). Differentiating twice gives L′′(θ)=−1/(2(1+θ)2)<0 for every θ>0, so this negative log marginal likelihood is not convex on its parameter domain.

1.4givenalgebra

Write σ(z)=(1+e−z)−1; then 0<σ(z)<1 for all z, so ∫Rϕ(z)σ(z) dz is finite and positive. The logistic posterior is therefore a positive density whose log, up to an additive constant, is q(z)=−z2/2−log⁡(1+e−z). Differentiation gives q′′′(z)=−σ(z)(1−σ(z))(1−2σ(z)). At z=log⁡3 this equals 3/32≠0. A positive density on all of R cannot be a degenerate Gaussian; every nondegenerate univariate Gaussian log density is quadratic and has third derivative zero, so this posterior is not Gaussian.

1.5givenalgebra

Since 0<Φ(z)<1 for all z, the probit posterior normalizer ∫Rϕ(z)Φ(z) dz is finite and positive. Its positive log density, up to an additive constant, is q(z)=−z2/2+log⁡Φ(z). Put h=ϕ/Φ and c=h(0)=2/π. Since (log⁡Φ)′=h and h′=−zh−h2, differentiation yields (log⁡Φ)′′′(0)=h′′(0)=−c+2c3=c(4/π−1)≠0; the last inequality uses π<4. A positive density on all of R cannot be a degenerate Gaussian; every nondegenerate univariate Gaussian log density is quadratic and has third derivative zero, so this posterior is not Gaussian.

1.6A1F2givenconstructalgebra

Let D={ti:1≤i≤r} for any finite list t1,…,tr∈[0,1]. The equality kernel has Gram form ∑t∈D(∑i:ti=tai)2≥0. Therefore it is positive semidefinite, and by [A1, F2] it defines a cylinder-space Gaussian process with standard normal coordinates and covariance 1s=t.

2.1F3F4step 1.6algebra

For any finite list of distinct inputs in [0,1], [F3] gives a centered Gaussian vector with covariance identity. By [F4] its characteristic function is exp⁡(−12∑iui2), which factors into the univariate standard-normal characteristic functions; uniqueness therefore gives independent standard-normal coordinates.

3.1F2F7step 2.1givenalgebra

Let qj=1/(j+2). If a continuous modification G existed, the equalities Gqj=fqj for all j and G0=f0 would hold simultaneously almost surely, since there are countably many coordinates. Let E0={∣f0∣<1}, which has positive probability, and Ej={∣fqj∣>2}, each of probability p>0 by [F2]. The finite-coordinate independence from step 2.1 gives Pr⁡(E0∩⋂j=NMEjc)=Pr⁡(E0)(1−p)M−N+1. By [F7], letting M→∞ gives Pr⁡(E0∩⋂j=N∞Ejc)=0 for each N; the countable union over N then shows E0 and infinitely many Ej occur together with probability Pr⁡(E0)>0. On that event ∣G0∣<1 while ∣Gqj∣>2 infinitely often, contradicting continuity at 0. No continuous modification exists.

4.1F6step 1.4step 1.5∎

The scope distinction in [F6] concerns a Gaussian observation likelihood; the explicit logistic and probit posterior calculations in steps 1.4 and 1.5 have nonzero third derivatives and are not Gaussian. Thus the exact finite Gaussian regression formula does not give an exact Gaussian posterior for those classification models.

TheoremStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Kernel ridge regression equals the Gaussian-process posterior mean

Statement

Assume AC. Let T be a set, let k:T×T→R be a symmetric kernel whose every finite Gram quadratic form is nonnegative, and let m:T→R. Let H be a real Hilbert space of functions on T such that, for every x∈T, the section kx:T→R, kx(t)=k(t,x), belongs to H and satisfies h(x)=⟨h,kx⟩H for every h∈H. Assume m∈H. For a nonempty finite ordered list X=(xi)i=1n (repetitions allowed), let K=KXX=(k(xi,xj)), mX=(m(xi)), and take y∈Rn and σ>0. On the affine space m+H, define J(g)=∑i=1n(yi−g(xi))2+σ2∥g−m∥H2. Then J has a unique minimizer g∗=m+∑i=1nαikxi,α=(K+σ2In)−1(y−mX). Its evaluation vector is (g∗(xi))i=1n=mX+K(K+σ2In)−1(y−mX). Under AC, the mean and kernel determine a canonical cylinder-space Gaussian-process prior. If it is observed at X with independent Nn(0,σ2In) noise, this vector is the conditional mean of the latent evaluation vector given the observation Y=y. The Hilbert-space existence and reproducing-section properties are hypotheses here; the theorem makes no general kernel-to-Hilbert-space existence claim.

Facts & Assumptions

Given: AC; the kernel, real Hilbert space, reproducing-section identities, mean function m∈H, finite list X, response y, and scalar σ>0 in the Statement; and the Gaussian-process observation model for the final comparison.

[A1]

AC is the axiom used by the Gaussian-process and finite conditional-normal interfaces. The Axiom of Choice

[F1]

The observation model defines KXX, mX, and Y=fX+ε; an independent Gaussian noise vector with covariance R is added to the latent evaluations. Finite Gaussian-process regression observation model

[F2]

Every finite-dimensional subspace of a real inner-product space has an orthonormal basis, and projection onto it gives a unique orthogonal decomposition with the Pythagorean identity. Projection onto a finite-dimensional subspace of an inner-product space

[F3]

A finite-dimensional positive-semidefinite operator has a positive-semidefinite square root whose square is the operator. A non-negative operator has a unique non-negative square root

[F5]

Under AC, the finite Gaussian-process posterior at test list X∗ has mean m∗+K∗X(KXX+R)−1(y−mX) when the observed covariance is positive definite. Finite Gaussian-process regression posterior and predictive law

[F6]

In the finite observation model, KXX+R is positive definite exactly when its ordinary inverse formulas are defined. Positive definiteness is exactly the inverse condition for the observed GP covariance

Proof

Given: The hypotheses in the Statement.

1.1F1givenconstructalgebra

Define the linear evaluation map S:H→Rn by (Sh)i=h(xi). It is bounded because ∥Sh∥22=∑i∣⟨h,kxi⟩H∣2≤∥h∥H2∑i∥kxi∥H2. The reproducing identities then give its adjoint S∗β=∑iβikxi, since ⟨Sh,β⟩=∑iβi⟨h,kxi⟩H=⟨h,S∗β⟩H. Consequently SS∗=K and V=ran⁡(S∗)=span⁡{kxi:1≤i≤n}.

1.2F1F6givenalgebra

For every c∈Rn, cT(K+σ2In)c=cTKc+σ2∥c∥2≥σ2∥c∥2, because K is a Gram matrix of the positive-semidefinite kernel. Hence A=K+σ2In is positive definite and, by [F6], its ordinary inverse exists. Put r=y−mX, α=A−1r, and h0=S∗α.

2.1F2step 1.1givenalgebra

Write g=m+h, so J(m+h)=∥r−Sh∥2+σ2∥h∥H2. By [F2], each h decomposes uniquely as p+q with p∈V and q⊥V. Then Sq=0 by the reproducing identities, and ∥p+q∥H2=∥p∥H2+∥q∥H2. Thus J(m+p+q)=J(m+p)+σ2∥q∥H2, so minimization reduces to V.

2.2F3F4step 1.1step 1.2algebra

Let B=K1/2 be supplied by [F3], and define C:Rn→Rn⊕Rn by Cβ=(Kβ,σBβ). For h=S∗β, its objective is ∥Cβ−(r,0)∥2, because Sh=Kβ and ∥S∗β∥H2=βTKβ=∥Bβ∥2. The normal equation is C∗Cβ=C∗(r,0), that is (K2+σ2K)β=Kr. Our α satisfies it because (K2+σ2K)α=K(K+σ2In)α=Kr. Therefore [F4] makes h0 a minimizer on V.

2.3step 1.1step 1.2algebra

For any d∈H, Sh0−r=Kα−r=−σ2α and S∗(Sh0−r)=−σ2h0. Expanding both squared norms and using the adjoint identity gives J(m+h0+d)−J(m+h0)=∥Sd∥2+σ2∥d∥H2. This is nonnegative and is zero only for d=0, so the minimizer is unique on all of H, not just on V.

3.1A1F1F5step 1.2step 2.2∎

The minimizing function is g∗=m+h0=m+∑iαikxi, and Sh0=Kα, giving the displayed evaluation vector. For Gaussian observations with R=σ2In, [F5] at test list X∗=X gives the same conditional-mean vector; A=K+R≻0 by step 1.2, so its inverse hypothesis holds.

4 · Examples, counterexamples and false statements

None yet.

Sources