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Gaussian-process regression and prediction — Examples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

Noisy interpolation on two independent coordinates

Example

Assume AC and take $T = {a, b}$ with distinct points, mean $m = 0$ , and equality kernel $k (s, t) = 1$ if $s = t$ and $0$ otherwise. Use the canonical cylinder-space Gaussian-process prior with this mean and kernel. For $X = (a, b)$ , set $K_{X X} = (k (x_{i}, x_{j})) = I_{2}$ and observe with centered Gaussian noise $ε \sim N_{2} (0, I_{2})$ independent of $f$ , so $A_{X} = K_{X X} + I_{2} = 2 I_{2}$ , with data $y = (2, - 1)^{T}$ . Predict at $X_{*} = X$ . The latent posterior mean is $(1, - \frac{1}{2})^{T}$ and its covariance is $\frac{1}{2} I_{2}$ . If a centered Gaussian test-noise vector $η_{*} \sim N_{2} (0, R_{*})$ with $R_{*} = I_{2}$ is independent of the pair $(f, ε)$ , then the noisy predictive mean is unchanged and its covariance is $\frac{3}{2} I_{2}$ .

Example references

Rasmussen and Williams, GPML, Chapter 2 §2.2, printed pp. 13–18

Bishop, PRML, §6.4.2, printed pp. 306–309

Facts & Assumptions

Given: AC, the two-point set, equality kernel, zero mean, ordered training and test lists, centered Gaussian training noise $ε \sim N_{2} (0, I_{2})$ independent of $f$ , optional test noise $η_{*} \sim N_{2} (0, I_{2})$ independent of $(f, ε)$ , and observation $y$ as stated.

[A1]

AC is the axiom used by the canonical GP construction and finite conditioning interface. The Axiom of Choice

[F0]

Under AC, a symmetric positive-semidefinite kernel and mean define a canonical Gaussian process with the specified finite Gaussian laws. A mean and positive-semidefinite kernel define a Gaussian process

[F1]

The observation model permits finite ordered lists and defines $K_{A B}$ , $m_{X}$ , $Y = f_{X} + ε$ , and test noise independent of the pair $(f, ε)$ . Finite Gaussian-process regression observation model

[F2]

For $A_{X} ≻ 0$ , the posterior latent mean and covariance are $m_{*} + K_{* X} A_{X}^{- 1} (y - m_{X})$ and $K_{* *} - K_{* X} A_{X}^{- 1} K_{X *}$ ; test noise independent of the pair $(f, ε)$ adds its covariance. Finite Gaussian-process regression posterior and predictive law

[F3]

For PSD Gram and noise matrices, the observed covariance is invertible exactly when it is positive definite. Positive definiteness is exactly the inverse condition for the observed GP covariance

Verification

Given: The finite model in the Example.

1.1A1F0F1F2F3givenconstructalgebra

For any finite list $t_{1}, \dots, t_{r} \in T$ and scalars $c_{i}$ , the equality-kernel Gram form is $\sum_{t \in T} (\sum_{i : t_{i} = t} c_{i})^{2} \geq 0$ . Thus the kernel is positive semidefinite, and [A1, F0] supplies the stated canonical prior. At $X = (a, b)$ , $K = I_{2}$ , so $A_{X} = K + R = 2 I_{2} ≻ 0$ and the inverse required in [F2] exists by [F3].

2.1A1F2step 1.1algebra

Since $m_{X} = 0$ and $K_{* X} = K_{X X} = I_{2}$ , [F2] gives posterior mean $(2 I_{2})^{- 1} y = \frac{1}{2} (2, - 1)^{T} = (1, - \frac{1}{2})^{T}$ . It gives posterior covariance $I_{2} - I_{2} (2 I_{2})^{- 1} I_{2} = \frac{1}{2} I_{2}$ .

3.1F1F2step 2.1algebra∎

By the Example's hypothesis, $η_{*}$ is independent of the pair $(f, ε)$ , so [F2]'s test-noise clause applies: the conditional mean is unchanged and $I_{2}$ is added to the covariance, yielding $\frac{1}{2} I_{2} + I_{2} = \frac{3}{2} I_{2}$ .

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

Positive observation noise with a singular rank-one prediction

Example

Assume AC. Let $T = {a, b}$ with distinct points, $ϕ (a) = 1$ , $ϕ (b) = 2$ , $m = 0$ , and $k (s, t) = ϕ (s) ϕ (t)$ . Use the canonical cylinder-space Gaussian-process prior with this mean and kernel. For $X = X_{*} = (a, b)$ , set $v = (1, 2)^{T}$ , $K_{X X} = K_{* *} = v v^{T}$ , and $R = I_{2}$ , so $A_{X} = K_{X X} + R = I_{2} + v v^{T} ≻ 0$ . With data $y = (1, 0)^{T}$ , the latent posterior mean is $v / 6$ and the latent posterior covariance is $K_{X X} / 6$ , which is singular.

Example references

Rasmussen and Williams, GPML, Chapter 2 §2.2, printed pp. 13–18

Bishop, PRML, §6.4.2, printed pp. 306–309

Facts & Assumptions

Given: AC, the two-point set, feature values $ϕ (a) = 1, ϕ (b) = 2$ , kernel $k (s, t) = ϕ (s) ϕ (t)$ , zero mean, ordered lists $X = X_{*} = (a, b)$ , noise covariance $R = I_{2}$ , and $y = (1, 0)^{T}$ .

[A1]

AC is the axiom used by the canonical GP construction and finite conditioning interface. The Axiom of Choice

[F0]

[F1]

The finite observation model defines Gram matrices, the observation vector, and noise covariance. Finite Gaussian-process regression observation model

[F2]

For positive-definite observed covariance, the posterior latent mean and covariance are $m_{*} + K_{* X} A_{X}^{- 1} (y - m_{X})$ and $K_{* *} - K_{* X} A_{X}^{- 1} K_{X *}$ . Finite Gaussian-process regression posterior and predictive law

[F3]

The observed matrix of PSD Gram and noise matrices is invertible exactly on the positive-definite branch. Positive definiteness is exactly the inverse condition for the observed GP covariance

Verification

Given: The finite model in the Example.

1.1A1F0F1F3givenconstructalgebra

For any finite list $t_{1}, \dots, t_{r} \in T$ and real $c_{i}$ , the Gram form for $k (s, t) = ϕ (s) ϕ (t)$ is $(\sum_{i} c_{i} ϕ (t_{i}))^{2} \geq 0$ . Thus [A1, F0] supplies the canonical prior. At $X = (a, b)$ this gives $K = v v^{T} = (\begin{array}{cc} 1 & 2 \\ 2 & 4 \end{array})$ . Therefore $A_{X} = I_{2} + v v^{T}$ satisfies $c^{T} A_{X} c = ∥ c ∥^{2} + (v^{T} c)^{2} > 0$ for every nonzero $c$ ; [F3] confirms the inverse branch applies.

2.1F2step 1.1algebra

The identity $(I_{2} + v v^{T}) v = (1 + v^{T} v) v = 6 v$ gives $A_{X}^{- 1} v = v / 6$ , and hence $A_{X}^{- 1} y = A_{X}^{- 1} (1, 0)^{T} = (5 / 6, - 1 / 3)^{T}$ . Since $K = v v^{T}$ , $K A_{X}^{- 1} y = v (v^{T} A_{X}^{- 1} y) = v / 6$ . Formula [F2] therefore gives the stated posterior mean.

3.1F2step 1.1algebra∎

Also $K A_{X}^{- 1} K = v (v^{T} A_{X}^{- 1} v) v^{T} = (5 / 6) v v^{T}$ , because $v^{T} A_{X}^{- 1} v = 5 / 6$ . Thus the posterior covariance is $K - (5 / 6) K = K / 6$ . The nonzero vector $(2, - 1)^{T}$ is annihilated by $K / 6$ , while $K / 6 \neq 0$ , so this covariance is singular.

Gaussian-process regression and prediction — Examples

2 · Logical flowchart

3 · Definitions, theorems and proofs

4 · Examples, counterexamples and false statements

Noisy interpolation on two independent coordinates

Example

Example references

Facts & Assumptions

Verification

Positive observation noise with a singular rank-one prediction

Example

Example references

Facts & Assumptions

Verification

Sources