University of California, Los Angeles Department of Statistics. Universal kriging

Transcription

1 University of California, Los Angeles Department of Statistics Statistics C173/C273 Instructor: Nicolas Christou Universal kriging The Ordinary Kriging (OK) that was discussed earlier is based on the constant mean model given by Z(s) = µ + δ(s) where δ( ) has mean zero and variogram 2γ( ) Many times this is a too simple model to use The mean can be a function of the coordinates X, Y, in some linear, quadratic, or higher form For example the value of Z at location s can be expressed now as Z(s i ) = β 0 + β 1 X i + β 2 Y i + δ(s i ), linear Or Z(s i ) = β 0 + β 1 X i + β 2 Y i + β 3 X 2 i + β 4 X i Y i + β 5 Y 2 i + δ(s i ), quadratic, etc If this is the case then we say that there is a trend of the polynomial type We need to take this into account when we find the kriging weights The predicted value Z(s 0 ) at location s 0 will be again a linear combination of the observed Z(s i ), i = 1,, n values: where Ẑ(s 0 ) = w 1 Z(s 1 ) + w 2 Z(s 2 ) + + w n Z(s n ) = w i Z i w 1 + w w n = 1 Suppose now that a trend of the linear form is present Then the value Ẑ(s 0) can be expressed as or Ẑ(s 0 ) = Ẑ(s 0 ) = w i Z i = w i β 0 + w i β 1 X i + w i β 2 Y i + w i δ(s i ) n n w i Z i = β 0 + β 1 w i X i + β 2 w i Y i + w i δ(s i ) (1) But also the value of Z(s 0 ) can be expressed (based on the linear trend) as Z(s 0 ) = β 0 + β 1 X 0 + β 2 Y 0 + δ(s 0 ) (2) Compare (1) and (2) In order to ensure that we have an unbiased estimator we will need the following conditions: 1

2 and w i X i = X 0 w i Y i = Y 0 w i = 1 As with ordinary kriging, to find the weights when a trend is present we need to minimize the mean square error (MSE) min ( ) 2 Z(s 0 ) w i Z(s i ) subject to the above constraints This minimization will be unconstrained if we incorporate the 3 constraints in the objective function The result is a system of n + 3 equations for the linear trend If the trend is quadratic we will need n + 6 equations, and n + 10 equations for a threedimentional quadratic trend, etc On the next page the system of equations for the linear trend example is presented in matrix form Note: We should try to understand why the trend exists based on the nature of our data, use a simple form of the trend if possible, and avoid extrapolation beyond the available data Once we decided about which trend to use, we subtract this trend from the observed data to obtain the residuals We then use the residuals to compute the sample variogram, fit a model variogram to it, predict the values at the unsampled locations ( kriged the residuals), and finally add the kriged residuals back to the trend 2

3 The Kriging System with linear trend as a function of the x, y coordinates γ(s1 s1) γ(s1 s2) γ(s1 s3) γ(s1 sn) x1 y1 1 γ(s2 s1) γ(s2 s2) γ(s2 s3) γ(s2 sn) x2 y2 1 w1 w2 = γ(sn s1) γ(sn s2) γ(sn s3) γ(sn sn) xn yn 1 wn x1 x2 xn λ1 y1 y2 yn λ2 λ0 γ(s0 s1) γ(s0 s2) γ(s0 sn) x0 y0 1 3

4 Example of universal kriging: Let s return to the simple example with the 7 observed points as shown below We use the exponential semivariogram model with c 0 = 0, c 1 = 10, α = 333, γ(h) = 10(1 e h 333 ) s i x y z(s i ) s s s s s s s s ??? First we calculate the distance matrix as shown below: Distance matrix = s 0 s 1 s 2 s 3 s 4 s 5 s 6 s 7 s s s s s s s s Using the universal kriging equations (assuming that there is a linear trend in our data) the weights, the estimate, and its variance are computed as follows: Universal kriging using the variogram: W = Γ 1 γ =

5 The answer is: W = w 1 w 2 w 3 w 4 w 5 w 6 w 7 λ 1 λ 2 λ 0 = The last 3 elements of the W vector are the Lagrange multipliers, λ 0 = 23, λ 1 = 0066, λ 2 = 0023 We can verify that the sum of the elements 1 through 7 is equal to 1, as it should be The estimate is equal to: 7 ẑ(s 0 ) = w i z(s i ) = 0220(477) (783) = The variance of the estimate is: Or σ 2 (s 0 ) = w i γ(s 0 s i ) + λ 0 + λ 1 x λ k x k0 σ 2 (s 0 ) = 0220(7384) (9823) (65) (137) =

6 Universal kriging using the covariance: The same example is solved using the covariance function: C(h) = 10e h 333 Reminder: The corresponding semivariogram function was γ(h) = 10(1 e h 333 ) W 1 = C 1 1 c 1 = The answer is: W 1 = w 1 w 2 w 3 w 4 w 5 w 6 w 7 λ 1 λ 2 λ = The last 3 elements of the W 1 vector are the Lagrange multipliers, λ 0 = 23, λ 1 = 0067, λ 2 = 0023 We can verify that the sum of the elements 1 through 7 is equal to 1, as it should be The estimate is equal to: 7 ẑ(s 0 ) = w i z(s i ) = 0220(477) (783) = The variance of the estimate is: Or σ 2 (s 0 ) = C(0) w i C(s 0 s i ) + λ 0 + λ 1 x λ k x k0 σ 2 (s 0 ) = 10 [0220(2612) (0176)] (65) (137) =

7 Universal kriging using geor > a <- readtable(" kriging_11txt", header=true) > b <- asgeodata(a) Using the ksline function: > univ_kr <- ksline(b, covmodel="exp", covpars=c(10,333), nugget=0, locations=c(65,137), m0="kt", trend=1) ksline: kriging location: 1 out of 1 Kriging performed using global neighbourhood Warning message: assuming that there is only 1 prediction point in: checklocations(locations) And here is the estimate and its variance: > univ_kr$predict [1] > univ_kr$krigevar [1] Note: The argument trend=1 is required if the argument m0="kt" is used Using the krigeconv function: The locations argument in the krigeconv function must be a data frame You can create it as dd <- dataframe(65,137) Or simply use dataframe(65,137) as shown below: > kc <- krigeconv(b, locations=dd,krige=krigecontrol(typekrige="ok", covmodel="exp", covpars=c(10, 333), nugget=0, trendl="1st", trendd="1st")) kc <- krigeconv(b, locations=dataframe(65,137),krige=krigecontrol(typekrige="ok", covmodel="exp", covpars=c(10, 333), nugget=0, trendl="1st", trendd="1st")) Note: The arguments trendl, trendd specify the trend of the locations to be estimated and the trend of the data They must be the same A complete example on how to use universal kriging in geor and gstat is presented on the next page using the Wolfcamp data set 7

8 #Read the data: site=" a <- readtable(file=site, header=true) #Load the geor and gstat packages: library(geor) library(gstat) #Convert the data into a geodata object (for geor): b <- asgeodata(a) #Create the grid for spatial predictions: xrange <- asinteger(range(a[,1])) yrange <- asinteger(range(a[,2])) grd <- expandgrid(x=seq(from=xrange[1], to=xrange[2], by=2), y=seq(from=yrange[1], to=yrange[2], by=2)) #Perform universal kriging (geor using the function ksline): krig_trend <- ksline(b, covmodel="sph", covpars=c(30000, 60), nugget=10000, m0="kt", trend=1, locations=grd) #Perform universal kriging (geor using the function krigeconv): kc <- krigeconv(b, locations=grd,krige=krigecontrol(typekrige="ok", covmodel="sph", covpars=c(30000, 60), nugget=10000, trendl="1st", trendd="1st")) #Perform universal kriging (gstat - krige): pr_univk <- krige(id="level", level~x+y, locations=~x+y, model=vgm(30000, "Sph", 60, 10000), data=a, newdata=grd) #Compare the three methods: pred <- cbind(pr_univk$levelpred, krig_trend$pred, kc$predict) predvar <- cbind(pr_univk$levelvar, krig_trend$krigevar, kc$krigevar) #Display the first 5 rows: > pred[1:5,] [,1] [,2] [,3] [1,] [2,] [3,] [4,] [5,]

9 > predvar[1:5,] [,1] [,2] [,3] [1,] [2,] [3,] [4,] [5,] #Load the lattice package: library(lattice) #Construct a raster map using the kriged values: levelplot(pr_univk$levelpred~x+y, pr_univk, aspect ="iso", main="universal kriging predictions") #Construct a raster map using the variances of the kriged values: levelplot(pr_univk$levelvar~x+y, pr_univk, aspect ="iso", main="universal kriging variances") Here are the plots: Universal kriging predictions y x Universal kriging variances y x 9

10 Using the imageorig function: #Collapse the predicted values into a matrix: qqq <- matrix(pr_univk$levelpred, length(seq(from=xrange[1], to=xrange[2], by=2)), length(seq(from=yrange[1], to=yrange[2], by=2))) #Construct the raster map of the predicted values: imageorig(seq(from=xrange[1], to=xrange[2], by=2), seq(from=yrange[1],to=yrange[2], by=2), qqq, xlab="west to East",ylab="South to North", main="raster map of the predicted values") #Collapse the variances into a matrix: qqq1 <- matrix(pr_univk$levelvar, length(seq(from=xrange[1], to=xrange[2], by=2)), length(seq(from=yrange[1], to=yrange[2], by=2))) #Construct the raster map of the variances: imageorig(seq(from=xrange[1], to=xrange[2], by=2), seq(from=yrange[1],to=yrange[2], by=2), qqq1, xlab="west to East",ylab="South to North", main="raster map of the variances") Here are the plots: Raster map of the predicted values South to North West to East Raster map of the variances South to North West to East 10