Spatial Process Estimates - IMAGe

Spatial Process Estimates
SAMSI Summer School August, 2009
Douglas Nychka,
National Center for Atmospheric Research

Outline
• A spatial model and Kriging
• Kriging = Penalized least squares = splines
• Robust Kriging.
• Identifying a covariance function
• Inference
Supported by the National Science Foundation
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The additive model
Given n pairs of observations (x i , y i ), i = 1, . . . , n
ɛ i ’s are random errors.
y i = g(x i ) + ɛ i
Assume that g is a realization of a Gaussian process.
MN(0, σ 2 I)
and ɛ are
Formulating a statistical model for g
makes a very big difference in how we solve the problem.
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A Normal World
We assume that g(x) is a Gaussian process,
ρk(x, x ′ ) = COV (g(x), g(x ′ ))
For the moment assume that E(g(x)) = 0.
(A Gaussian process ≡ any subset of the field locations has a multivariate
normal distribution. )
We know what we need to do!
If we know k we know how to make a prediction at x!
ĝ(x) = E[g(x)|data]
i.e. Just use the conditional multivariate normal distribution.
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A review of the conditional normal
and
u ∼ N(0, Σ)
u =
(
u1
u 2
)
( )
Σ11 , Σ
Σ = 12
Σ 21 , Σ 22
[u 2 |u 1 ] = N(Σ 2,1 Σ −1
1,1 u 1, Σ 2,2 − Σ 2,1 Σ −1
1,1 Σ 1,2)
(distribution of u 2 given u 1 )
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Our application is
u 1 = y
and
(the Data)
u 2 = {g(x 1 ), ...g(x N )}
a vector of function values where we would like to predict.
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The Kriging weights
Conditional distribution of g given the data y is Gaussian.
Conditional mean
ĝ = COV (g, y) [COV (y)] −1 y = Ay
rows of A are the Kriging weights.
Conditional variance
COV (g, g) − COV (g, y) [COV (y)] −1 COV (y, g)
These two pieces characterize the entire conditional distribution
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Kriging as a smoother
Suppose the errors are uncorrelated Normals with variance
σ 2 .
ρK = COV (g, y) = COV (g, g)
and
COV (y) = (ρK + σ 2 I)
ĝ = ρK(ρK + σ 2 I) −1 y
= K(K + λI) −1 y = A(λ)y
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My geostatistics/BLUE overhead
For any covariance and any smoothing matrix (not just A above) we
can easily derive the prediction variance.
Question: Find the minimum of
over all choices of A.
E [ (g(x) − ĝ(x)) 2]
The answer: The Kriging weights ... or what we would do if
we used the Gaussian process and the conditional distribution.
Folklore and intuition: The spatial estimates are not very sensitive
if one uses suboptimal weights, especially if the observations
contain some measurement error.
It does matter for measures of uncertainty.
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Kriging with a fixed part
Adding a fixed component
g(x) = ∑ i
φ i (x)d i + h(x)
d is fixed
h is a mean zero process with covariance, k.
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BLUE/Universal Kriging
Find d by Generalized least squares
ˆd =
(
T T M −1 T ) −1
T T M −1 y
M = K + λI
”Krig” the residuals
ĥ = K(K + λI) −1 (y − T ˆd)
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In general:
ĝ(x) = ∑ i
φ i (x) ˆd i + ∑ j
k(x, x j )ĉ j
with
ĉ = (K + λI) −1 (y − T ˆd)
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The spline connection
Basis functions: determined by the covariance function
Penalty function: Ω = K is based on the covariance.
The minimization criteria:
min
d,c
n∑
i=1
(y − (T d + Kc) i ) 2 + λc T Kc
Kriging estimator is a spline with reproducing kernel k.
λ is proportional to the measurement (nugget) variance
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Robust Kriging
ρ(u) a robust function that grows slower than u 2
The robust minimization criteria:
min
d,c
n∑
i=1
ρ(y − (T d + Kc) i ) + λc T Kc
An example of ρ and its derviative
0 5 10 15
−5 0 5
−4 −2 0 2 4
−4 −2 0 2 4
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Identifying a covariance function
The Matern class of covariances:
φ(d) = ρψ ν (d/θ)) with ψ ν a Bessel function.
correlation
0.0 0.2 0.4 0.6 0.8 1.0
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2
1
0.5
0.0 0.5 1.0 1.5 2.0
θ a range parameter, ν smoothness
at 0.
• ψ ν is an exponential for ν =
1/2 as ν → ∞ Gaussian.
• m t h order thin plate spline in
R d ν = 2m − d.)
distance
Compactly support Wendland covariance
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What kind of processes are these?
Matern (.5,1.0,2.0) and Wendland (2.0)
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Correlations among ozone
In many cases spatial processes also have a temporal component.
Here we take the 89 days over the ”ozone season” and just find
sample correlations among stations.
correlation
−0.5 0.0 0.5 1.0
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0 100 200 300 400 500 600
miles
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Comparison to the variogram
Sample variogram for 89 days during summer 1987 :
Variance
0 100 300 500
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0 50 100 150 200 250 300
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Distance (miles)
”.” Sample variogram values
”–” fitted linear function Variogram ∼ distance/θ
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50
Sensitivity to the covariance
Exp (200) Matern (2, 200)
150
100
100
50
50
50
150
150
150
50
100
50
50
Thin plate spline Wendland (2, 180)
100
50
100
50
0
correlation
0.0 0.4 0.8
0 50 100 150 200 250 300
distance
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10
5
5
Sensitivity to the covariance (SE)
Exp (200) Matern (2, 200)
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20
20
10
20
20
10
5
10
10
20
20
10
20
10
10
5
10
10
20
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5
10
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10
20
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10
20
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20
20
10
10
5
10
10
10
10
10
10
60
50
40
30
20
5
20
10
10
5
10
10
0
Thin plate spline Wendland (2, 180)
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140
140
140
140
140
140
140
140
140
140
140
140
140
140
140
Uncertainty in exceeding 140 PPB
Mean field, four realizations of the conditional distribution
Contours from 10 cases
0 50 100 150 200
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Summary
A spatial process model leads to a penalized
least squares estimate
A spline = Kriging estimate= Bayesian posterior
mode
For spatial estimators the basis functions are
related to the covariance functions and can be
identified from data
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