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Kriging systems and mean-square optimality

Statement

Use the notation of Simple, ordinary, and universal kriging. For simple kriging, if $Σ$ is positive definite, the unique mean-square minimizing weight vector is $λ_{S K} = Σ^{- 1} c_{0}$ , and its mean-square prediction error is $v_{0} - c_{0}^{⊤} Σ^{- 1} c_{0} .$

For an unknown trend with design matrix $P$ and target vector $p_{0}$ , $λ^{⊤} Z$ is unbiased for every $β \in R^{q}$ if and only if $P^{⊤} λ = p_{0}$ . Suppose $Σ$ is positive definite and $P$ has full column rank (that is, $P y = 0$ implies $y = 0$ for $y \in R^{q}$ ). Then the unique unbiased mean-square minimizing weights $λ \in R^{n}$ and Lagrange multiplier $ν \in R^{q}$ solve $(\begin{array}{cc} Σ & P \\ P^{⊤} & 0 \end{array}) (\begin{array}{c} λ \\ ν \end{array}) = (\begin{array}{c} c_{0} \\ p_{0} \end{array}) .$ The augmented matrix is invertible under these hypotheses. The resulting mean-square prediction error is $v_{0} - λ^{⊤} c_{0} - ν^{⊤} p_{0} .$ For ordinary kriging, take $P = 1_{n}$ and $p_{0} = 1$ , giving $(\begin{array}{cc} Σ & 1_{n} \\ 1_{n}^{⊤} & 0 \end{array}) (\begin{array}{c} λ \\ ν \end{array}) = (\begin{array}{c} c_{0} \\ 1 \end{array}) .$

Facts & Assumptions

Given: A square-integrable target $Y$ and observation vector $Z$ from the definition, with covariance matrix $Σ$ , covariance vector $c_{0}$ , and target variance $v_{0}$ .

[F1]

Simple, ordinary, and universal kriging: simple prediction centers $Z$ at its known mean and the target at its known mean.

[F2]

Simple, ordinary, and universal kriging: in a finite trend model, unbiasedness for every $β$ is constrained by $P^{⊤} λ = p_{0}$ .

[F3]

Normal equations for best affine $L^{2}$ prediction: the best affine $L^{2}$ predictor exists as a unique almost-sure class, and its centered coefficients satisfy the covariance normal equations.

[F4]

Moments, variance, and covariance on a probability space and Variance and covariance identities for random variables: variance is the expectation of a centered square, and covariance is symmetric and bilinear on finite linear combinations of square-integrable variables.

Proof

Given: The finite second-moment model, $Σ ≻ 0$ , and, for the unknown-trend case, full column rank of $P$ .

1.1givenF1F4algebra

For a centered error vector $ε = Z - E Z$ and centered target error $ε_{0} = Y - E Y$ , every unbiased linear predictor has error $ε_{0} - λ^{⊤} ε$ . Expanding its squared expectation by [F4] gives the common objective $Q (λ) = v_{0} - 2 λ^{⊤} c_{0} + λ^{⊤} Σ λ$ . The centered predictor in the simple case has this same error expression.

1.2F2algebra

In the trend model, $E [λ^{⊤} Z - Y] = (P^{⊤} λ - p_{0})^{⊤} β$ . This expression is zero for every $β \in R^{q}$ if $P^{⊤} λ = p_{0}$ . Conversely, if it is zero for every $β$ , taking $β$ successively to be each coordinate vector shows every coordinate of $P^{⊤} λ - p_{0}$ is zero. Thus the unbiased weights are exactly the affine constraint $P^{⊤} λ = p_{0}$ .

1.3givenalgebra

Let $M = (\begin{array}{cc} Σ & P \\ P^{⊤} & 0 \end{array})$ and suppose $M (\binom{x}{y}) = 0$ . The lower block gives $P^{⊤} x = 0$ . Taking the inner product of $x$ with the upper block gives $x^{⊤} Σ x + x^{⊤} P y = 0$ ; the second term is $(P^{⊤} x)^{⊤} y = 0$ , so positive definiteness gives $x = 0$ . The upper block now gives $P y = 0$ , and full column rank gives $y = 0$ . Hence the finite square matrix $M$ is injective and therefore invertible.

2.1givenF3step 1.1algebra

In the simple case, [F3] gives the covariance normal equations $Σ λ = c_{0}$ for the minimizing centered affine predictor. Positive definiteness makes the solution unique, $λ = Σ^{- 1} c_{0}$ ; substituting it into step 1.1 gives $Q (λ) = v_{0} - c_{0}^{⊤} Σ^{- 1} c_{0}$ .

2.2step 1.1step 1.2step 1.3algebra

Invertibility gives a unique solution $(λ, ν)$ of the displayed augmented system, and its lower block makes $λ$ feasible. For any other feasible vector $λ + h$ , one has $P^{⊤} h = 0$ . Using the upper block $Σ λ + P ν = c_{0}$ in the quadratic from step 1.1 gives $Q (λ + h) - Q (λ) = 2 h^{⊤} (Σ λ - c_{0}) + h^{⊤} Σ h = h^{⊤} Σ h$ . This is positive for nonzero $h$ and zero for $h = 0$ , so $λ$ is the unique constrained minimizer.

3.1step 1.1step 1.2step 2.2algebra∎

Multiplying $Σ λ + P ν = c_{0}$ on the left by $λ^{⊤}$ and using $P^{⊤} λ = p_{0}$ yields $λ^{⊤} Σ λ + ν^{⊤} p_{0} = λ^{⊤} c_{0}$ . Substitution in $Q (λ)$ from step 1.1 gives $Q (λ) = v_{0} - λ^{⊤} c_{0} - ν^{⊤} p_{0}$ , as claimed.

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Sources

Evgeny Spodarev, Spatial Statistics: Lecture Notes (2024), Sections 3.1–3.2 (standard reference, not scraped)
Dennis Sun, Stats 253 Lecture 5: Prediction and Kriging (Stanford University) (standard reference, not scraped)