Half-tapering strategy for conditional simulation with large datasets

Abstract

Gaussian conditional realizations are routinely used for risk assessment and planning in a variety of Earth sciences applications. Assuming a Gaussian random field, conditional realizations can be obtained by first creating unconditional realizations that are then post-conditioned by kriging. Many efficient algorithms are available for the first step, so the bottleneck resides in the second step. Instead of doing the conditional simulations with the desired covariance (F approach) or with a tapered covariance (T approach), we propose to use the taper covariance only in the conditioning step (half-taper or HT approach). This enables to speed up the computations and to reduce memory requirements for the conditioning step but also to keep the right short scale variations in the realizations. A criterion based on mean square error of the simulation is derived to help anticipate the similarity of HT to F. Moreover, an index is used to predict the sparsity of the kriging matrix for the conditioning step. Some guides for the choice of the taper function are discussed. The distributions of a series of 1D, 2D and 3D scalar response functions are compared for F, T and HT approaches. The distributions obtained indicate a much better similarity to F with HT than with T.

Appendix: Proof of Proposition 2

We first establish two lemmas:

Lemma 1

(Adapted from Lu and Shiou 2002) Consider the symmetric block matrix

$$\left( \begin{array}{ll} {\mathbf{K}}_0^{-1} &{} \mathbf{D}_T \\ \mathbf{D}_T &{} {\mathbf{K}}_1\end{array}\right),$$

where $\mathbf{D}_T$ is diagonal and ${\mathbf{K}}_0$ and ${\mathbf{K}}_1$ are symmetric non singular matrices such that ${\mathbf{K}}_1 - \mathbf{D}_T {\mathbf{K}}_0 \mathbf{D}_T$ and ${\mathbf{K}}_0^{-1} - \mathbf{D}_T {\mathbf{K}}_1 \mathbf{D}_T$ are non singular. Then,

$$({\mathbf{K}}_0^{-1} - \mathbf{D}_T {\mathbf{K}}_1 \mathbf{D}_T)^{-1} = {\mathbf{K}}_0 + {\mathbf{K}}_0 \mathbf{D}_T ({\mathbf{K}}_1 - \mathbf{D}_T {\mathbf{K}}_0 \mathbf{D}_T)^{-1} \mathbf{D}_T {\mathbf{K}}_0.$$

Proof

This lemma is a direct consequence of Theorem 2 in Lu and Shiou (2002). $\square$

Lemma 2

Let ${\mathbf{x}}_1,\dots , {\mathbf{x}}_n$ be n sample points in finite domain D and let $C({\mathbf{h}})$ be a covariance function on D with $C(\mathbf{0})=1$. Let Z be a zero mean Gaussian random field with covariance function $C({\mathbf{h}})$ on D. Let further ${\mathbf{K}}$ be the $n \times n$ matrix with elements $[{\mathbf{K}}]_{ij} = C({\mathbf{x}}_i-{\mathbf{x}}_j)$, for $1 \le i,j \le n$ and ${\mathbf{k}}_s$ be the n vector with elements $[{\mathbf{k}}_{\mathbf{x}}]_i = C({\mathbf{x}}_i-{\mathbf{x}})$, for ${\mathbf{x}}\in D$ and $1 \le i \le n$. Then, the matrix

$${\mathbf{K}}- {\mathbf{k}}_{\mathbf{x}}{\mathbf{k}}^{\prime }_{\mathbf{x}}$$

is positive semi-definite for any ${\mathbf{x}}\in D$.

Proof

In order to show this, we need to show that for any vector $\pmb {\lambda }= (\lambda _1,\dots ,\lambda _n)^{\prime } \in {\mathfrak{R}}^n$, it holds that

$$Q = \sum _{i=1}^n \sum _{j=1}^n \lambda _i \lambda _j \left( [{\mathbf{K}}]_{ij} - [{\mathbf{k}}_{\mathbf{x}}]_{i} [{\mathbf{k}}_{\mathbf{x}}]_{j} \right) \ge 0.$$

(21)

Let us denote $S = \sum _{i=1}^n \lambda _i Z({\mathbf{x}}_i)$. Then, since $\hbox {Var}\{S\} = \sum _{i=1}^n \sum _{j=1}^n \lambda _i \lambda _j [{\mathbf{K}}]_{ij}$ and $\hbox {Cov}\{S,Z({\mathbf{x}})\} = \sum _{i=1}^n \lambda _i [{\mathbf{k}}_{\mathbf{x}}]_i$, Eq. (21) is equivalent to

$$Q = \hbox{Var}\{S\} -\hbox {Cov}\{S,Z({\mathbf{x}})\}^2.$$

Using $\hbox {Cov}\{S,Z({\mathbf{x}})\}^2 \le \hbox {Var}\{S\} \hbox {Var}\{Z({\mathbf{x}})\}$ and $\hbox {Var}\{Z({\mathbf{x}})\} = 1$, we thus get very easily that

$$Q \ge \hbox {Var}\{S\} -\hbox {Var}\{S\} \hbox {Var}\{Z({\mathbf{x}}_0)\} = 0,$$

which finishes the proof. $\square$

We are now ready to provide the proof of Proposition 2. We must show that $\sigma ^2_{k,C_1}({\mathbf{x}}) \ge \sigma ^2_{k,C_0}({\mathbf{x}})$ for all ${\mathbf{x}}\in D$. As usual, we drop the dependency on ${\mathbf{x}}$ for sake of conciseness. Since $\sigma ^2_{k,C_1} = \sigma ^2_0 - {\mathbf{k}}^{\prime }_1 {\mathbf{K}}_1^{-1} {\mathbf{k}}_1$ and $\sigma ^2_{k,C_0} = \sigma ^2_0 - {\mathbf{k}}^{\prime }_0 {\mathbf{K}}_0^{-1} {\mathbf{k}}_0$, we need to prove that:

$${\mathbf{k}}^{\prime }_0 {\mathbf{K}}_0^{-1} {\mathbf{k}}_0 - {\mathbf{k}}^{\prime }_1 {\mathbf{K}}_1^{-1} {\mathbf{k}}_1 \ge 0.$$

(22)

Since ${\mathbf{k}}_1 = {\mathbf{k}}_0 {\mathbf{k}}_T$ and ${\mathbf{K}}_1 = {\mathbf{K}}_0 \odot {\mathbf{K}}_T$, Eq. (22) is equivalent to

$$\sum _{i=1}^n \sum _{j=1}^n [{\mathbf{k}}_0]_i \left( [{\mathbf{K}}_0^{-1}]_{ij} - [{\mathbf{k}}_T]_i \, [\{{\mathbf{K}}_0 \odot {\mathbf{K}}_T\}^{-1}]_{ij} \, [{\mathbf{k}}_T]_j\right) [{\mathbf{k}}_0]_j \ge 0.$$

(23)

To show that this expression is always nonnegative, we will show that the matrix ${\mathbf{M}}$ with elements $[{\mathbf{M}}]_{ij} = [{\mathbf{K}}_0^{-1}]_{ij} - [{\mathbf{k}}_T]_i \, [\{{\mathbf{K}}_0 \odot {\mathbf{K}}_T\}^{-1}]_{ij} \, [{\mathbf{k}}_T]_j$, for $1 \le i,j \le n$ is positive definite (p.d.) except for the trivial case ${\mathbf{K}}_0={\mathbf{K}}_1, {\mathbf{k}}_0={\mathbf{k}}_1$, corresponding to a taper with infinite range, where ${\mathbf{M}}={\mathbf{0}}$ and Eq. (22) equals zero. Introducing the diagonal matrix $\mathbf{D}_T = \hbox {diag}({\mathbf{k}}_T)$, this matrix can also be written

$${\mathbf{M}}={\mathbf{K}}_0^{-1} - \mathbf{D}_T\, \{{\mathbf{K}}_0 \odot {\mathbf{K}}_T\}^{-1} \mathbf{D}_T.$$

(24)

Since ${\mathbf{M}}$ is invertible, it is p.d. if and only if ${\mathbf{M}}^{-1}$ is p.d. Using Lemma 1, its inverse is

$${\mathbf{M}}^{-1} = {\mathbf{K}}_0 + {\mathbf{K}}_0 \mathbf{D}_T\, \{{\mathbf{K}}_0 \odot {\mathbf{K}}_T - \mathbf{D}_T {\mathbf{K}}_0 \mathbf{D}_T\}^{-1} \mathbf{D}_T {\mathbf{K}}_0.$$

(25)

Using Lemma 2, one has that ${\mathbf{K}}_T - {\mathbf{k}}_T {\mathbf{k}}_T^{\prime }$ is p.d. Hence, using Schur’s product theorem, ${\mathbf{K}}_0 \odot ({\mathbf{K}}_T - {\mathbf{k}}_T {\mathbf{k}}_T^{\prime }) = {\mathbf{K}}_0 \odot {\mathbf{K}}_T - {\mathbf{K}}_0 \odot {\mathbf{k}}_T {\mathbf{k}}_T^{\prime }= {\mathbf{K}}_0 \odot {\mathbf{K}}_T - \mathbf{D}_T {\mathbf{K}}_0 \mathbf{D}_T$ is p.d. and so is its inverse. As sums and products of p.d. matrices are p.d., we can conclude that ${\mathbf{M}}^{-1}$ in Eq. (25) is also p.d., which completes the proof. $\square$