exp.image.cov: Exponential,Gaussian and "power" covariance family for 2-d gridded locations

Description

Given two sets of locations on a 2-d grid efficiently multiplies a cross covariance with a vector.

Usage

exp.image.cov(ind1, ind2, Y, cov.obj = NULL, setup = FALSE, grid, ...)

Arguments

ind1

Matrix of indices for first set of locations this is a two column matrix where each row is the row/column index of the image element. If missing the default is to use all grid locations.

ind2

Matrix of indices for second set of locations. If missing this is taken to be ind2. If ind1 is missing ind2 is coerced to be all grid locations.

Vector to multiply by the cross covariance matrix. Y must be the same locations as those referred to by ind2.

cov.obj

A list with the information needed to do the multiplication by convolutions. This is usually found by using the returned list when setup=T.

setup

If true do not do the multiplication but just return the covariance object required by this function.

grid

A grid list giving the X and Y grids for the image. (See example below.) This is only required if setup is true.

...

Any arguments ( e.g. theta and p) to pass to the exp.cov in setting up the covariance object. This is only required if setup is true.

Value

A vector that is the multiplication of the cross covariance matrix with the vector Y.

Details

This function was provided to do fast computations for large numbers of spatial locations and supports the conjugate gradient solution in krig.image. In doing so the observations can be irregular spaced but their coordinates must be 2-dimensional and be restricted to grid points. (The function as.image will take irregular, continuous coordinates and overlay a grid on them.)

Returned value: If ind1 and ind2 are matrices where nrow(ind1)=m and nrow(ind2)=n then the cross covariance matrix, Sigma is an mXn matrix (i,j) element is the covariance between the grid locations indexed at ind1[i,] and ind2[j,]. The returned result is Sigma%*%Y. Note that one can always recover the coordinates themselves by evaluating the grid list at the indices. e.g. cbind( grid$x[ ind1[,1]], grid$y[ind1[,2])) will give the coordinates associated with ind1. Clearly it is better just to work with ind1! Functional Form: Following the same form as exp.cov for irregular locations, the covariance is defined as exp( -(D.ij **p)) where D.ij is the Euclidean distance between x1[i,] and x2[j,] but having first been scaled by theta. Specifically,

D.ij = sqrt( sum.k (( x1[i,k] - x2[j,k]) /theta[k])**2 ).

Note that if theta is a scalar then this defines an isotropic covariance function.

Implementation: This function does the multiplication on the full grid efficiently by a 2-d FFT. The irregular pattern in Y is handled by padding with zeroes and once that multiplication is done only the appropriate subset is returned.

As an example assume that the grid is 100X100 let big.Sigma denote the big covariance matrix among all grid points ( If the parent grid is 100x100 then big.Sigma is 10K by 10K !) Here are the computing steps:

temp<- matrix( 0, 100,100)

temp[ ind2] <- Y

temp2<- big.Sigma%*% temp

temp2[ind1]

Notice how much we pad with zeroes or at the end throw away! Here the matrix multiplication is effected through convolution/FFT tricks to avoid creating and multiplying big.Sigma explicitly. It is often faster to multiply the regular grid and throw away the parts we do not need then to deal directly with the irregular set of locations.

Confusion: In this entire discussion Y is treated as vector. However is one has complete data then Y can also be interpreted as a image matrix conformed to correspond to spatial locations. See the last example for this distinction.

Examples

Run this code

# generate a grid and set of indices based on discretizing the locations
# in the precip dataset
data(precip)
out<-as.image( precip$y, x= precip$x)
ind1<- out$ind
grid<- list( x= out$x, y=out$y)
# there are 750 gridded locations 

# setup to create cov.obj for exponential covariance with range= .25
cov.obj<- exp.image.cov( setup=TRUE, grid=grid, theta=.25) 

result<- exp.image.cov( ind1, ind1, Y= rnorm(750),cov.obj=cov.obj)

# The brute force way would be  
# result<- exp.cov( precip$x, precip$x,theta=.25, C=rnorm(750))

# evaluate the covariance between all grid points and the center of the
#grid

Y<- matrix(0,64,64)
Y[32,32]<- 1
result<- exp.image.cov( Y= Y,cov.obj=cov.obj)
image.plot( matrix( result, 64,64 ))