The covariogram is chosen to be a gaussian model. Assume \[ Y \sim \textrm{N}(10, \Sigma(\sigma^2, \phi))\] where \(\Sigma(\sigma^2, \phi)_{ij} = \sigma^2 \exp(-(d_{ij}/\phi)^2)\) with \(\sigma^2 = 2\) and \(\phi = 5\). The grid is a \(50 \times 30\) equally spaced grid on the US map.
The estimated mean parameter is 10.2652628.
Using initial values of \(\sigma^2_{(1)} = 2.5\) and \(\phi_{(1)} = 5.5\), the gaussian model is fitted.
The estimated parameter is \(\hat\sigma^2 = ``2.0681803``\) and \(\hat\phi = ``5.1847943``\). The parameter estimates are quite close. From the plot, we can see it fits the true semi-covariogram quite well.
The semivariogram on 4 directions is plotted. From the way we built the model, we expect it to be isotropic.
The empirical semi-variogram on 4 directions is estimated. From the plot we can see that when the distance is less than 6, the process seems to be isotropic (the four lines are almost indentical). But when distrance larger than 6, the difference becomes larger.