# Spatial Process Estimates - IMAGe 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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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.

4

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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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

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

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

0 50 100 150 200 250 300

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

20

10

5

10

10

10

20

10

10

20

20

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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