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Kriging — Examples

1 · Prerequisites

2 · Summary

Three finite cases isolate the assumptions in the covariance systems. Duplicated observations give a singular simple-kriging covariance and nonunique coefficients for one exact predictor. A quadratic trend observed at only two sites can make the design rank deficient and the universal unbiasedness constraint impossible even when observation covariance is nonsingular. With two independent unit-variance observations, simple kriging with known mean can have smaller error than the unbiased ordinary predictor.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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

Singular observation covariance in simple kriging

Example

Let $Ω = {- 1, 1}$ have equal probability and let $U (ω) = ω$ . Take three distinct sites $s_{0}, s_{1}, s_{2}$ and set $X (s_{0}) = X (s_{1}) = X (s_{2}) = U$ . The known mean is zero and $Z = (X (s_{1}), X (s_{2}))^{⊤} = (U, U)^{⊤}$ predicts $Y = X (s_{0}) = U$ . Then $Σ = (\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array})$ is singular, so the inverse formula for simple kriging is unavailable. Nevertheless, $E [(Y - λ^{⊤} Z)^{2}] = (1 - λ_{1} - λ_{2})^{2},$ so the mean-square error is zero exactly for weight pairs with $λ_{1} + λ_{2} = 1$ . Different coefficient pairs such as $(1, 0)$ and $(0, 1)$ give the same predictor $U$ .

Facts & Assumptions

Given: The equal-weight two-outcome probability space, its coordinate variable $U$ , and the field and sites in the example.

[F1]

Simple, ordinary, and universal kriging: the simple predictor is $m_{0} + λ^{⊤} (Z - m)$ , and its inverse formula requires a positive-definite observation covariance matrix.

[F2]

Finite probability spaces, outcome weights, events, and event probabilities and Real random variables on finite probability spaces and their finite distributions: a finite probability space assigns outcome weights summing to one, and a real random variable is a function on its outcomes.

[F3]

Expectation of a real random variable on a finite probability space: expectation is the finite weighted sum over outcomes.

[F4]

Variance, standard deviation, and covariance on a finite probability space: variance and covariance are the expectations of centered squares and centered products.

Verification

Given: The two equally weighted outcomes and $U (ω) = ω$ .

1.1givenF2F3F4algebra

Direct finite summation using [F2] and [F3] gives $E [U] = (- 1 + 1) / 2 = 0$ and $E [U^{2}] = 1$ , so [F4] gives $Var (U) = 1$ and $Cov (U, U) = 1$ . Thus the displayed observation covariance has determinant $1 - 1 = 0$ and is singular.

2.1F1F3F4step 1.1algebra∎

The simple predictor is $λ_{1} U + λ_{2} U$ because both known means are zero. Therefore $Y - λ^{⊤} Z = (1 - λ_{1} - λ_{2}) U$ , whose squared expectation is $(1 - λ_{1} - λ_{2})^{2} E [U^{2}] = (1 - λ_{1} - λ_{2})^{2}$ . It is zero exactly for $λ_{1} + λ_{2} = 1$ , and both $(1, 0)$ and $(0, 1)$ give the same random variable $U$ .

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

A rank-deficient universal-kriging design

Example

On the finite product space $Ω = {- 1, 1}^{3}$ with equal probability on each of its eight outcomes, let $ϵ_{s}$ be the coordinate indexed by $s \in {0, 1, 2}$ . For unrestricted $β \in R^{3}$ , define $X (s) = β_{0} + β_{1} s + β_{2} s^{2} + ϵ_{s}$ . Observe at $s_{1} = 0$ and $s_{2} = 1$ , and predict at $s_{0} = 2$ , using $p (s) = (1, s, s^{2})^{⊤}$ . The observation design matrix $P = (\begin{array}{ccc} 1 & 0 & 0 \\ 1 & 1 & 1 \end{array})$ has rank two, while $p (s_{0}) = (1, 2, 4)^{⊤}$ . The universal-unbiasedness constraint has no solution, although the observation covariance is $Σ = I_{2}$ .

Facts & Assumptions

Given: The equal-weight product space, the three coordinate variables, the quadratic trend, and the observation and target sites.

[F1]

Simple, ordinary, and universal kriging: for trend design matrix $P$ and target trend vector $p_{0}$ , universal-unbiased weights must satisfy $P^{⊤} λ = p_{0}$ .

[F2]

Finite probability spaces, outcome weights, events, and event probabilities and Real random variables on finite probability spaces and their finite distributions: the eight outcomes have weights $1 / 8$ , and each coordinate variable is a function on that finite space.

[F3]

Expectation of a real random variable on a finite probability space: expectations are finite weighted sums.

[F4]

Variance, standard deviation, and covariance on a finite probability space: covariance is the expectation of a centered product.

Verification

Given: The eight equally weighted sign triples and the stated trend.

1.1givenF2F3F4algebra

For each coordinate $ϵ_{s}$ , four outcomes have value $1$ and four have value $- 1$ , so [F2] and [F3] give $E [ϵ_{s}] = 0$ and $E [ϵ_{s}^{2}] = 1$ . For distinct coordinates, each of the four sign pairs occurs twice, so [F4] gives $Cov (ϵ_{s}, ϵ_{t}) = 0$ . Hence the observation-error covariance is $I_{2}$ , which is positive definite.

2.1F1step 1.1algebra∎

The two rows of $P$ are linearly independent, but its last two columns are equal, so its rank is two rather than three. For every $λ = (λ_{1}, λ_{2})^{⊤}$ , $P^{⊤} λ = (λ_{1} + λ_{2}, λ_{2}, λ_{2})^{⊤}$ ; its last two coordinates cannot equal the target's $2$ and $4$ . Thus no weight vector satisfies $P^{⊤} λ = p (s_{0})$ , so no linear predictor is unbiased for every $β$ .

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

Ordinary kriging can have larger error than simple kriging

Example

Let $Ω = {- 1, 1}^{3}$ assign probability $1 / 8$ to each sign triple, with coordinate variables $U_{0}, U_{1}, U_{2}$ . For a common mean $θ \in R$ , let the target be $Y = θ + U_{0}$ and observations be $Z_{i} = θ + U_{i}$ for $i = 1, 2$ . With $θ$ known, simple kriging predicts $Y$ by $θ$ and has mean-square error $1$ . With $θ$ unknown, an ordinary predictor must use weights $λ_{1} + λ_{2} = 1$ ; its minimum mean-square error is $3 / 2$ , attained by $λ_{1} = λ_{2} = 1 / 2$ . This is a finite-model comparison, not a general inequality between ordinary and simple kriging.

Facts & Assumptions

Given: The equal-weight product space and the target and observation model in the example.

[F1]

Simple, ordinary, and universal kriging: a known-mean predictor uses the centered observations.

[F2]

Simple, ordinary, and universal kriging: ordinary-kriging weights are constrained to sum to one.

[F3]

Kriging systems and mean-square optimality: with positive-definite observation covariance, the simple minimizing weight is $Σ^{- 1} c_{0}$ .

[F4]

Kriging systems and mean-square optimality: under the full-rank constant-mean design, the ordinary system and error formula apply.

[F5]

Finite probability spaces, outcome weights, events, and event probabilities and Real random variables on finite probability spaces and their finite distributions: the eight sign triples have equal weight and the coordinates are real random variables on that space.

[F6]

Expectation of a real random variable on a finite probability space: expectations are finite weighted sums.

[F7]

Variance, standard deviation, and covariance on a finite probability space: variance and covariance are centered second moments.

Verification

Given: The three sign-coordinate variables on the equal-weight eight-outcome space.

1.1givenF5F6F7algebra

Each $U_{i}$ has mean zero and second moment one by [F5] and [F6]. For $i \neq j$ , each sign pair occurs equally often, so [F7] gives $Cov (U_{i}, U_{j}) = 0$ . Consequently, $Σ = I_{2}$ , $c_{0} = (0, 0)^{⊤}$ , and $v_{0} = 1$ .

2.1F1F2F3step 1.1

When $θ$ is known, [F3] and step 1.1 give simple weight $Σ^{- 1} c_{0} = 0$ , so the simple minimizing predictor is $θ$ . Its error is $U_{0}$ and its mean-square error is $E [U_{0}^{2}] = 1$ . For the ordinary model, the unbiasedness constraint is $λ_{1} + λ_{2} = 1$ by [F2].

3.1F2F4F7step 1.1step 2.1algebra∎

The ordinary system in [F4] applies because step 1.1 gives $Σ = I_{2}$ and the constant-mean design has full rank. Directly, the ordinary prediction error is $U_{0} - λ_{1} U_{1} - λ_{2} U_{2}$ . Expanding its square, all cross terms have expectation zero by step 1.1, so its mean-square error is $1 + λ_{1}^{2} + λ_{2}^{2}$ . Substituting $λ_{2} = 1 - λ_{1}$ gives $\frac{3}{2} + 2 (λ_{1} - \frac{1}{2})^{2}$ , whose unique minimum is $3 / 2$ at $λ_{1} = λ_{2} = 1 / 2$ . Since $3 / 2 > 1$ , ordinary kriging has the larger error in this example.