Universität StuttgartInstitut für Wasserbau, Lehrstuhl für Hydrologie und Geohydrologie Copulas...

Universität Stuttgart Institut für Wasserbau, Lehrstuhl für Hydrologie und Geohydrologie

Copulas (2)

András Bárdossy

Universität Stuttgart

Spatial problems

• Sampling only at a number of locations• What is between ?

– Estimate– Quality of estimation– Simulate realizations

• Geostatistics (Krige, Matheron)– Mining applications– Hydro and Environmental sciences

Geostatistics

• Z(x) Random function – Realisation z(xi)• Assumption – „uniform continuity“• No differences are known a-priori

• Independent of the location – depends only on h • (Semi)Variogramm Covariance function

2uZhuZEuZhuZVarh

Experimental Variogramm EC

0 500 1000 1500 2000 2500 3000

Point kriging

iii uZuZ

DuallformuZE

muZEuZE i

Unbiasedness

Estimation variance using the variogram

uZuZVaru

)()()(

Kriging equations using variogram

uuuu i

i=1,…,n

jjijK uuu

Problems

• Estimation variance is an index of spatial configuration– Does not depend on the local values– “Best” for Gaussian distribution– Symmetrical (high and low values not distinguished)

• Variogram estimation difficult– Squared differences – skewed distribution– Dominated by high values– Independence of the pairs not fulfilled

Strongly influenced by the marginal distribution

Symmetry

• Digital elevation models – water dominated regions– Maxima and minima

• Contaminations– Source vs Background concentrations

Known but unquantified deterministic processes lead to asymmetry and non-Gaussian dependence

Indicator Variables

Indicator variables

else 0

)( if0

)( if1)(

Indicator variogram

uIuIhN

h 2* ))()(()(2

Indicator variables

• Interpretation as probability• Interpolation of the indicators• Result pdf for each location • Simulation restricted to the observed range

Can copulas be used to overcome some of these problems?

Spatial copulas

• Assumption:– Multivariate copula exists for any number of points– The bi-variate marginal copulas corresponding to pairs

separated by a vector h are translation invariant

• How to find such copulas ?

vZFuZFPvu zz ))((,))((),,( xhxhCS

Empirical copulas

• Set of pdf pairs corresponding to points separated by the vector h

• Generalization of the variogram• Empirical density using kernel smoothing

hxxxxh iijnin ZFZFS |))(()),(()(

Empirical copula density chloride h=5000m

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

Empirical copula density nitrate h=5000m

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

Empirical copula density pH h=5000m

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

Cl Variogramm

0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

Chloride concentration (mg/l)

pH Variogramm

5.2 5.6 6 6.4 6.8 7.2 7.6 8 8.4 8.8 9.2 9.6

Conditional Entropy: Nitrate 3000 m and 30000m

0 5 10 15 20 25

Copulas and natural processes

• Natural processes influence high and low values differently– Erosion at high elevations– Pollution is spreading not the background– Weather relates the high discharges

• Copulas of digital elevation models:– Spain – eroded old landscape– Ecuador – younger but erroded– Mars – eroded and meteorites

Copula density of the pair C8 and C9

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

00.81.62.43.244.85.66.47.288.89.610.411.21212.813.614.415.21616.817.618.419.220

Copulas of daily rainfall

• 601 rainfall stations in the Rhein catchment Germany• Size = 100.000 km2• Days with important events with good spatial coverage

were selected (400 days of the period 1958-2003)• Spatial copulas (densities) for different distances were

calculated

Spatial dependence – 5 kmEvent 70

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Precip ita tion am ount location A

itatio

00.20.40.60.811.21.41.61.822.22.42.62.833.23.43.63.844.24.44.64.85

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

00.20.40.60.811.21.41.61.822.22.42.62.833.23.43.63.844.24.44.64.85

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

itatio

00.20.40.60.811.21.41.61.822.22.42.62.833.23.43.63.844.24.44.64.85

Radarniederschlag29. Dezember 2001 11:20-13:20

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

05101520253035404550556065707580859095100105110115120

Copula Radarniederschlag29. Dezember 2001 11:20-13:20

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

00.81.62.43.244.85.66.47.288.89.610.411.21212.813.614.415.21616.817.618.419.220

Requirements for a spatial copula

1. Stability of the multivariate marginals: which means that any multivariate marginal copula corresponding to a selected set of points should not depend on the set of other selected points used to define the multivariate copula.

2. Wide range of dependence: a geographically close set of points should have an arbitrarily strong dependence structure, while distant points should be independent.

3. Flexible parametrization: the multivariate copula should have a parametrization such that the dependence structure reflects the geometric position of the corresponding set of points.

• Definition of a copula from a multivariate distribution:

Possibilities

• Multivariate normal copula– Simple but symmetrical

• Derived multivariate copulas

• If g monotonic – no change of the copula• If g non monotonic one can get interesting copulas

)( ii YgX

Normal copula

• Correlation = 0.85

Chi square

• Non central chi-square distribution

mtmttg

tYtPtYPtG

)()()(

Chi square

• Multivariate case

)1(,...,)1(

,...,),...,(

tYttYtPttG

n-dimensional Chi-square copula

• Density

Chi-Square Copulas

Gauss – Chi-square

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

-4.5-4.1-3.7-3.3-2.9-2.5-2.1-1.7-1.3-0.9-0.5-0.10.30.71.11.51.92.32.73.13.53.94.3

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

-4.5-4.1-3.7-3.3-2.9-2.5-2.1-1.7-1.3-0.9-0.5-0.10.30.71.11.51.92.32.73.13.53.94.3

V-transformed copula

• Transformation function:

• Strong dependence of the extremes if shifted to one side and partly to the middle

• If k=1 then it is the chi square copula

else )(

if )()(

mxxmkxg

K=2, m=1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

Parameter estimation

• Non independent pairs – ML• Fit the rank correlation function and the asymmetry • Parametric form of the covariance of the original normal

• Further work needed

Interpolation

• For n+1 points the joint distribution is known– Calculate the conditional for the unobserved point

Full conditional distribution known – thus confidence intervals can be calculated

• Example:

4 points – corner of a unit square

A: two of them with F(x)=1 two with F(x)=0

B: all with F(x)=0.5

Example interpolation – conditional densities m=0, k=1

0 0.2 0.4 0.6 0.8 1

Example interpolation – conditional densities m=1, k=3

0 0.2 0.4 0.6 0.8 1

Example interpolation – conditional densities normal copula

0 0.2 0.4 0.6 0.8 1

Validation of the conditional densities

• Are the conditional densities OK ?• Cross validation

– Calculation of the frequencies of non exceedence for the observed values

– Comparison with the uniform

• V is much better then normal or Kriging

Thank you !

bardossy@iws.uni-stuttgart.de

Nitrat und Phosphat

Universität StuttgartInstitut für Wasserbau, Lehrstuhl für Hydrologie und Geohydrologie Copulas...

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