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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 $U$ .

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