Kriging and Radial Basis Functions

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SIAM OP05 Kriging and RBFs 7• Unified approacha ∈ R m . a = Qa R + Na N , a R = Q T a, a N = N T aAssume that y = Fβ +z, z = Nz NE [ z z T] = σ 2 NAN T , A = D˜ΦD, D = diag(˜Φ −1/2ii )A is symmetric, positive definite, and A ii =1. Valid correlation matrix.A = DC T CDBLUE:min ‖v‖ 2 2 s.t. Fβ +NDC T v = y Solution: C T v =(−1) µ D −1 N T yv,gbv ∈ R m−q has variance σ 2 I. Estimate: σ 2 = ‖v‖ 2 2/(m − q)}Assume Gaussian process: θ ∗ = argmin θ{Γ(θ) ≡ det A(θ) · σ 2(m−q) (θ)SIAM OP05 Kriging and RBFs 8Theorem: For φ(r) ∈{r, r 3 ,r 2 log r}, ω∈R + :s(ωθ,x) = s(θ,x),Γ(ωθ) = Γ(θ)Remarks:“Scalar” θ has no effect.“Full” θ : Finding θ ∗ reduces to a problem in R d−1+ instead of R d +
SIAM OP05 Kriging and RBFs 9• ExamplesY (x) =d∏e kx kcos(2kx k )k=1Educated guess: θ ∗ =(1,2,...,d) T2Test regions T of sizes 2 2 , 2 3 , 0.5 40−2Successively insert new designpoint atargmax x∈T|s(x) − Y (x)|−4−60−0.5−1−1.50.511.52SIAM OP05 Kriging and RBFs 1010 0 No scaling: θ = (1, 1)10 0 With scaling10 −210 −4Thin plate splineMultiquadricGaussian0 10 20 30 40 50 60 70 80 90 1002-D10 −210 −40 10 20 30 40 50 60 70 80 90 10010 0 No scaling: θ = (1, 1, 1)10 0 With scaling10 −1Thin plate splineMultiquadricGaussian10 −20 10 20 30 40 50 60 70 80 90 1003-D10 −110 −20 10 20 30 40 50 60 70 80 90 10010 0 No scaling: θ = (1, 1, 1, 1)10 0 With scaling10 −1Thin plate splineMultiquadricGaussian10 −210 20 30 40 50 60 70 80 90 1004-D10 −110 −210 20 30 40 50 60 70 80 90 100
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Kriging and Radial Basis Functions

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