Semivariograms
All forms of kriging assumes you know Cov (Z (s i ), Z (s j ))
Or, equivalently Cor (Z (s i ), Z (s j ))
Usually, need to estimate this
Non-spatial approach: data-based estimate
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Spatial data: two problems
1) only one observation at s i and one at s j
plot of Z (s i ) against Z (s j ) not very useful!
2) need vector of Cov (Z (s 0 ), Z (s i )) when haven’t observed Z (s 0 )
c Philip M. Dixon (Iowa State Univ.)
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Semivariograms
Need a model! How does Cov (Z (s i ), Z (s j )) depend on:
distance between s i and s j
direction from s i to s j
location of s i and s j in the study area
Use a model in the same way we use a regression model:
E Yi = β X i + εi .
Optimal prediction is E Y | X , mean of Y for that X
If have many Y at an X (more like ANOVA),
can calculate sample average
no model required
One Y per X (regression)
Use a model to predict Y at observed X or new X
use global characteristics to predict for a specific X
relies on assumptions being appropriate
Examine each Cov characteristic in turn
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Does Cov depend on location?
Two pairs of points, same direction, same distance,
different parts of study area
Same covariance?
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c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Stationary spatial processes
2nd order stationary
µ(s) constant across study area
Cov (Z (s), Z (s + h)) same across study area
h specifies a particular distance and direction
So in previous picture, the two pairs of points would have same Cov
above implies Var Z (s) constant across study area
instrinsic stationarity
Var (Z (s) − Z (s + h)) same everywhere
Slightly weaker assumption
Some really care about the difference. I don’t.
We’ll assume 2nd order stationarity
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
4 / 58
Does Cov depends on direction?
Isotropic spatial process
Cov (Z (s), Z (s + h)) same in all directions
Only depends on distance between two points, i.e. || h||
Anisotropic spatial process
Cov (Z (s), Z (s + h)) depends on direction
1.0
consider Cov as function of distance, but only for certain directions
0.8
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Spring 2016
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Geometric anisotropy
can redefine X, Y coordinates so rescaled process is isotropic
simple scaling & rotation → isotropy
Example: Swiss rain data, plots on next 3 slides
Original coordinates
rotated 45◦ counterclockwise = −pi/4 radians
stretched 5-fold in X direction
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Spring 2016
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Spatial Data Analysis - Part 4
Spring 2016
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Dealing with anisotropy
Think about the data
Can a covariate explain part of the pattern?
Swiss rain: elevation
But then need covariate values to make predictions
Geometric anisotropy
can transform coordinates to make isotropic
General anisotropy
can repeat what we’re about to do in different directions
more complications, more details, no change in concept
Discuss later
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
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Estimating covariances
We will assume isotropy and 2nd order stationarity
So, Cov (Z (s i ), Z (s j )) depends on ||s i − s j ||
i.e, the distance between s i and s j
smaller areas: Euclidean distance
larger areas: should use great circle distance
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Estimating covariances
Remember non-spatial data
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All X observations have same µ and σ 2 , ditto all Y
to estimate covariance / corr.:
If an obs has an above-average X , does it have an above-average Y ?
ditto below-average
calculate (Xi − µX )(Yi − µY ) for each obs, then average
Works because each pair (X , Y ) is an independent sample from the
population
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
13 / 58
Estimating spatial covariances
Spatial data:
Remember N = 1, even when 1000 locations
Data are one realization of a correlated spatial random vector
Concept: When 2nd order stationary
can use “replicates” in space instead of replicate realizations
called Ergodicity
What are spatial replicates?
Depends on assumptions:
pairs of points separated by same distance
pairs of points separated by same distance and same direction
We’ll assume isotropy, so only distance matters
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
14 / 58
Spatial covariance
Remember Cov = E (Z (s i ) − µi )) (Z (s j ) − µj )
2nd order stationary, so µ constant
Know h = the distance between s i and s j .
Find all pairs of locations separated by that same distance
Could plot Z (s i ) against Z (s j ). rarely done.
Cov is a theoretical average (E )
Estimate by
calculating (Z (s i ) − µ̂) (Z (s j ) − µ̂) for each of those pairs of obs.
and average those values
That average estimates Cov(h) for that distance h
repeat for other h to estimate the Cov(h) function.
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Spatial Data Analysis - Part 4
Spring 2016
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Covariogram cloud
Usually done for all distances simulaneously
calculate (Z (s i ) − µ̂) (Z (s j ) − µ̂) for each pair of obs.
calculate distances between each pairs of obs.
empirical cov vs. distance
Example: covariogram cloud for the Swiss rain data
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
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50000 100000
200000
300000
Distance
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
17 / 58
Semivariance
Need µ̂ to calculate empirical covariance
Can avoid calculating µ̂
Remember (a few weeks ago):
Var X = E (X − µ)2 = E (Xi − Xj )2 /2
Average squared diff. of pairs of observations estimates variance
Same idea works for covariances
2
Use [Z (σ i ) − Z (σ j )] /2 to evaluate the spatial dependence between a
pair of locations
Called the semivariance
Again, calculate for each pair, plot vs distance
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
120000
18 / 58
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c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
19 / 58
Semivariance
Why use semivariances instead of covariances?
When 2nd order stationary - can convert between the two
E
1
2
(Z (s i ) − Z (s j )) = σ 2 − E (Z (s i ) − µi ) (Z (s j ) − µj )
2
In rare situations, semivariance can be calculated but covariance
shouldn’t
difference between intrinsic and 2nd order stationarity
convenience: don’t have to estimate µ
Notation:
γ(h): theoretical semivariance at distance h
γ̂(h): empirical semivariance at distance h
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
20 / 58
Semivariogram
Single semivariance values are very variable
Smooth semivariogram cloud by averaging
“Classical” or Matheron estimator
γ̂(h) =
1
Σ
[Z (s i ) − Z (s j )]2
2 N(h) N(h)
Sum over pairs of points separated by distance h
N(h) is number of pairs at that distance
When locations on a grid, easy to do
many pairs of locations with same distance
When locations are irregular, have to create “distance bins”
Define a range of distances = a bin, e.g. 0-25000m
Calculate mean distance and mean semivariance for the bin
Repeat for rest of bins
Plot X = mean distance vs. Y= mean semivariance
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
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c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
22 / 58
Choice of bin size/number
How precise is the estimated semivariance?
Var γ̂(h) ≈
2γ(h)2
N(h)
How big should the bin be?
Want at least 30, preferably 50 or more, pairs per bin
Choice of binning matters
very common to make all bins equally wide
but, N(h) often small for large h
γ̂(h) not very precise, semivariogram is erratic
Solution: don’t calculate γ̂(h) for large distances
one recc.: calculate γ̂(h) to 1/2 max distance
Swiss rain: max distance is 291 km, so calculate to h = 150 km.
Notice default max distance in R is less
Would like to have 10-15 bins, but # pairs more important
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
23 / 58
Choice of bin size/number
When locations irregular, N(h) often small at short distances
less of a problem because γ(h) small, so Var γ̂(h) small
Alternative is to have equal # pairs per bin
→ wide bins for short distances and very long distances
concern is loss of information about γ̂(h) at small h
that info. is crucial for fitting models to semivariograms
I tend to use equi-distant bins without large lag distances
But, I really want a good estimate of γ(h) for h close to 0
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
24 / 58
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c Philip M. Dixon (Iowa State Univ.)
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Spatial Data Analysis - Part 4
Spring 2016
25 / 58
Cressie-Hawkins estimator
Classical estimator is very sensitive to outliers
One unusual value,
pairs.
Z (s ), can really mess up γ̂(h) because used in N-1
Cressie-Hawkins estimator is more robust to outliers
4
0.5 ΣN(h) | Z (s i ) − Z (s j ) |1/2
γ̂(h)CH =
0.494
0.045
0.457 + N(h) + N(h)2
Where does this come from?
| Z (s i ) − Z (s j ) |1/2 is not dominated by a single large squared
difference
That makes C-H more robust to outliers
When Z (s) ∼ N(0, 1),
i4
1 h
0.494
0.045
E
ΣN(h) | Z (s i ) − Z (s j ) |1/2 ≈ 2γ(h) 0.457 +
+
2
N(h)
N(h)
N(h)
⇒ C-H estimator (also called robust SV estimator)
25000
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
26 / 58
15000
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Cressie−Hawkins
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c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
27 / 58
Semivariogram parameters
With either estimator, plot of X=h vs. Y=γ̂(h) describes the pattern
of spatial correlation
Three important numbers that describe that plot:
nugget, sill, and practical range
Picture on next slide
Nugget: γ(h), for distances very very close to 0.
quantifies micro-scale variation
term from mining geology:
nuggets of metal in ore → micro-scale variation
Replicate observations at a single location assumed to be identical
(γ(0) = 0)
i.e., γ(h) not continuous - has a jump at 0
Later: will discuss measurement error kriging, in which replicate
observations are not identical
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
28 / 58
1.0
0.8
sill
0.4
0.6
practical
range
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0.2
nugget
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
29 / 58
Variogram parameters
Sill: γ(h), for large h.
If σ 2 constant, sill = σ 2
Sill is an asymptote, technically, γ(h) never equals sill.
Partial sill:
Magnitude of the “spatially associated” variability
partial sill = sill - nugget
Practical range: distance at which there 95% of the spatially
associated variability
i.e., γ(h)nugget = 0.95(sill − nugget)
obs. separated by more than the practical range are essentially
uncorrelated.
0.95 is traditional.
All above assumes one spatial process.
Can have multiple sills and practical ranges if multiple processes
Spatial Data Analysis - Part 4
Spring 2016
30 / 58
Spatial Data Analysis - Part 4
Spring 2016
31 / 58
0.0
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c Philip M. Dixon (Iowa State Univ.)
c Philip M. Dixon (Iowa State Univ.)
Semivariogram models
To use kriging, need Cov Z (s ), Z (s ) for all locations with observed
data and all prediction locations
If field is second order stationary,
γ(h) = σ 2 − Cov (h) = Cov (0) − Cov (h), so
Cov (h) = Cov (0) − γ(h)
So, need γ(h) for any h
Fit a model to the variogram or variogram cloud.
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
32 / 58
Semivariogram models
Many different models
Three very commonly discussed models:
"
#
Spherical:
1 h 3
3h
−
, for h ≤ α
γ(h) = σ 2
2α 2 α
Cov exactly 0 for h > α
Exponential:
γ(h) = σ 2 [1 − exp(−3h/α)]
Gaussian:
γ(h) = σ 2 [1 − exp(−3
2
h
)]
α
In all, α is the practical range, σ 2 is the sill
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
33 / 58
Some variogram models
10000
0
5000
Semivariance
15000
Matern, 0.25
Exponential
Whittle
Gaussian
0
20
40
60
80
100
120
Distance (km)
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
34 / 58
Semivariogram models
Exponential and Gaussian are two specific examples of models in the
Matern class, a large family of semivariogram models
#
"
k
1
θh
γ(h) = σ 2 1 −
2Kk (θh)
Γ(k) 2
Γ(k) and K (θh) are math special functions (Gamma and modified
Bessel fn of 2nd kind, order k)
θ controls the range, = 1/α, σ 2 controls the sill
k controls the shape of the variogram
Gaussian is k = 2, exponential is k = 0.5.
k = 1 is the Whittle model. I have found it quite useful
h
h
γ(h) = σ 2 1 − K1
α
α
The Gaussian is one of the historically traditional models,
but modern opinion is to avoid it
Wackernagel, 2003. The Gaussian model is “pathological”.
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
35 / 58
More variogram models
linear: γ(h) = θh
no sill!, unless you force one.
Not second order stationary, C (0) = σ 2 doesn’t exist unless you force a
sill
But is intrinsic stationary:
Var (Z (s i ) − Z (s j )) is constant for any distance h
So semivariogram is well defined
My experience is that a linear trend semivariogram usually indicates
trend across the study area
I suggest modeling the trend, then computing a semivariogram from
the residuals from that trend
wave or hole-effect:
h
h
γ(h) = σ 2 1 −
sin
α
α
models periodic spatial patterns
picture on next slide
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
36 / 58
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More Semivariogram models
0
Spherical
Linear w/sill
Hole
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c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
37 / 58
Nuggets
Can add to any model
e.g. Exponential with nugget
γ(h) = σ02 + σ 2 [1 − exp(−3h/α)]
σ 2 is the partial sill, the spatially-associated variation
σ02 is the micro-scale variation
In practice, very few pairs of observations are really close to each
other, i.e. with h ≈ 0
So, the estimated nugget is an extrapolation down to h = 0
nugget can be very sensitive to the choice of semivariogram model.
Even if there is a nugget, γ(0) is still defined as 0
The Kriging prediction at an obs. location is the obs. value
But, γ() = nugget, where is a very small number
Which means that the prediction of Z (s) can be quite different a very
small distance away from that observed location
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
38 / 58
Estimating a SV model
Given an empirical average SV or SV cloud and a model, how do you
estimate the SV model parameters (e.g. σ 2 , α, and perhaps σ02 )?
All SV models are non-linear functions of their parameters
Not X1i β1 + X2i β2
Use non-linear least-squares
Could fit to cloud, but tradition is to fit to binned averages
Given γ̂(hi ) for a set of bins and a specific model
Find θ = (σ 2 , α, σ02 ) for which Σ(γ(hi , θ) − γ̂(hi ))2 is as small as
possible
Same criterion as linear regression
No closed-form solution - need iterative algorithm
must provide starting values
bad start → big trouble
generally well-behaved if start is reasonable
I (and everyone else) uses eyeball guess as starting values
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
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c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
40 / 58
Fitting SV models
One issue:
LS assumes all obs equally important
But, if Var Y2 > Var Y1 , want to give more attention to fitting Y1
more closely
2
Remember Var (γ̂(h) ≈ 2 γ(h)
#pts
more pts →↓ Var
↑ γ̂(h) →↑ Var
Used weighted LS to put more emphasis on obs with lower variance
Minimize Σwi [γ(hi , θ) − γ(hi )]2
statistically optimal weights are wi = 1/Var γ(hi )
Problem: Var depends on γ(h), which is what we are trying to
estimate!
Solution (not the only one): use wi = #pts
h2 as the weight
assumes γ(h) ≈ h
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
41 / 58
Spring 2016
42 / 58
Which SV model to use?
Which model to use? Two approaches:
1) which model best fits semivariogram
use wt SS as measure of fit. For Swiss rainfall:
Exponential: SS = 1.608
Spherical: SS=1.184
Gaussian: SS=1.258
Matern, k=1: 1.237
Matern, k=5: 1.153
Matern, k=4: 1.140
Suggests Spherical or Matern with k=4
2) which model gives most accurate predictions?
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Which SV model to use?
One small problem with assessing predictions
Using data twice:
Once to fit SV model
Again, to assess precision of predictions
known to be overly optimistic and favor models with more parameters
leads to overfitting data at hand
Three solutions, all often used:
1) Training/test set
fit model to subset of data (training set)
evaluate on remaining data (test set)
requires arbitrary division into training and test
both are subsets of full data set
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
43 / 58
Cross-validation
2) cross-validation
remove 1st obs, using N − 1 obs. to fit SV and predict Z (s 1 )
h
i2
calculate squared error for 1st obs: Z (s 1 ) − Ẑ (s 1 )
Ẑ (s 1 ) NOT based on Z (s 1 ), so valid “out-of-sample” prediction
return 1st obs. to data set, remove 2nd obs., repeat above
repeat for all obs., average squared errors
h
i2
gives mean squared error of prediction: N1 (Z (s i ) − Ẑ (s i )
exact same concept as PRESS statistic in linear regression
3) k-fold CV
If many points, leave-one-out can be slow
Divide data into k “folds”=sets of obs., k=5 and k=10 are common
remove entire set of obs., fit model on 90% (for k=10) of data
predicte omitted points, calculate rMSEP
repeat for all other parts
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
44 / 58
Cross-validation
For Swiss rainfall data:
Exponentional: MSEP = 3615
Spherical: MSEP = 3447
Gaussian: MSEP = 3725
Matern, k=1: 3561
Matern, k=5: 3610
Matern, k=4: 3591
Suggests Spherical (one of two with good fit)
But, none substantially better than any other
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
45 / 58
Cross-validation
CV also provides a way to assess model in various ways
plots on next pages
plot predicted values vs residuals
should be flat sausage, just like for linear regression
spatial (or bubble) plot of residuals
should be no big clusters
if there are, don’t have a good spatial model
outliers are more apparent
very large residuals, usually surrounded by residuals with opposite sign
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
46 / 58
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Spatial Data Analysis - Part 4
Spring 2016
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c Philip M. Dixon (Iowa State Univ.)
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Cross-validation
Compare range of residuals to range of obs. values
obs. values: 0 to 493
residuals: -240 to 160
these residuals are pretty variable
the problem for this data set is (I think) the anisotropy
process is more constant ⇒ smooth more in one direction than the
other
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
49 / 58
Other ways to fit SV models
REML (REstricted/REsidual Maximum Likelihood)
likelihood-based, after removing fixed effects
avoids binning
most commonly used with linear models for trend / treatment effects
more esoteric
Composite likelihood and generalized estimating equations
not commonly used
Also even more robust ways to estimate semivariances
Genton’s estimator (instead of Cressie-Hawkins)
available in geoR, not gstat
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
50 / 58
Anisotropy
Up to now: assume isotropy. Only distance matters
Anisotropy:
Spatial dependence dependence on direction as well as distance
How to assess?
What you know about the data. Is isotropy reasonable?
Directional semivariograms
Calculate semivariogram using only points in one direction
repeat for other directions
looking for similar curves in all directions
Variogram map
Plot of average semivariance for distance / direction combinations
looking for “bulls eye” patterns.
No single best tool
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
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Spatial Data Analysis - Part 4
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Spatial Data Analysis - Part 4
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c Philip M. Dixon (Iowa State Univ.)
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Spatial Data Analysis - Part 4
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Anisotropy
How to deal with different types of anisotropy
Geometric anisotropy:
Range longer in one direction than another
Nugget and (partial) sill same in all directions
Rotate coordinate system so longer direction aligned with an axis
Then rescale that axis
Do this by:
cos θ − sin θ
s ∗ = 10 λ0
s
sin θ
cos θ
Where θ is the angle of the major axis and λ is the ratio of length of
minor to length of major axis
krige() will do this for you
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
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Zonal anisotropy
variogram sills vary with direction
The direction with the shorter range also has the shorter sill.
model by a combining an isotropic model and a model which depends
“only on the lag-distance in the direction θ of the greater sill”
(Schabenberger and Gotway, 2005, p 152).
γ(h ) = γ1 (|| h ||) + γ2 (hθ )
Chiles and Delfiner (1999, p. 96) warn against axis-specific models,
e.g.: γ(h ) = γ1 (hx ) + γ2 (hy )
Under certain circumstances they can lead to Var Z (s) = 0, which not
good.
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
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Kriging with anisotropy
Geometric anisotropy
Do not have to transform and back transform coordinates
krige() can do this for you
in the vgm() specification of the variogram model,
add anis=c(θ, λ)
θ is the direction of the longest range (highest correlation)
in degrees, measured clockwise from the + vertical axis (N)
λ is the ratio of minor range to major range (a value from 0 to 1)
so anis=c(60,0.2) specifies a major axis at 2 o’clock with a length 5
times the minor axis.
Zonal anisotropy
“fake it” by setting up a geometrically anisotropic component with a
major axis much longer than the minor axis (e.g. λ = 0.00001.
Only distances only along the major axis contribute to γ(h )
c Philip M. Dixon (Iowa State Univ.)
Spatial Data Analysis - Part 4
Spring 2016
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