AnswerBun.com

How can I understand these variograms?

Using grf function from R package geoR, I simulated 6 replicates (each with 1000 samples) of a Gaussian random field on [0, 1] x [0, 1], with zero mean, zero nugget, and an exponential spatial covariance $$gamma(h) = expleft(frac{h}{phi}right)sigma^2,$$ with $sigma^2 = 1$ and $phi = 0.5$. In the following figure, the blue solid curve gives the semi-variogram of this true model and the blue dashed line gives the sill (which is $sigma^2 = 1$). Then for each replicate, I

  1. compute an empirical semi-variogram with function variog (black solid dots);
  2. compute a monte-carlo envelope for the empirical semi-variogram (gray shaded polygon);
  3. fit an exponential spatial covariance model (black solid curve) to the empirical semi-variogram for $hat{sigma}^2$ and $hat{phi}$ (printed in the tile).

However, I find it difficult to understand the results / behavior as illustrated by the figure.

enter image description here

  1. I simulated data from a true model and fit that true model, why does the fitted semi-variogram differ from the true one so much? Or, why does $(hat{sigma}^2, hat{phi})$ differ from $(sigma^2, phi)$ so much? On some replicate the difference is drastic.
  2. How can the empirical semi-variogram (black solid dots), as well as the fitted variogram (black solid curve), breach sample variance (black dashed line)? Isn’t sample variance the sill?
  3. Why does the black dashed line so much away from the blue dashed line? (the answer is maybe as same as that to question 1)

Is a spatial process expected to behave like this? If I examine an AR(1) process by simulating a time series and fitting it using Yule-Walker method, I can get back to the truth, i.e., true autocorrelation coefficient and variance. This is not possible for a spatial process?


Reproducible R code

set.seed(0)

phi <- 0.5
sigmasq <- 1

## unconditional simulation of Gaussian random field with exponential covariance
EXP_GRF <- grf(1000, cov.pars = c(sigmasq, phi), cov.model = "exponential",
               nsim = 6)

## estimation of semi-variogram
semi_variog <- variog(EXP_GRF)

semi_variog_i <- semi_variog
semi_variog_i$v <- NULL

par(mfrow = c(2, 3), mar = c(2.5, 2.5, 1, 1))

for (i in 1:6) {

  ## the i-th empirocal semi-variogram
  semi_variog_i$v <- semi_variog$v[, i]

  ## mc envelops
  env <- variog.mc.env(coords = EXP_GRF$coords, data = EXP_GRF$data[, i],
                       obj.variog = semi_variog_i, nsim = 99, messages = FALSE)

  ## plotting range
  MAX <- max(env$v.lower, env$v.upper, semi_variog_i$v, sigmasq)
  with(semi_variog_i, plot(u, v, ylim = c(0, MAX), type = "n"))

  ## empirical semi-variogram
  polygon(c(semi_variog_i$u, rev(semi_variog_i$u)),
          c(env$v.lower, rev(env$v.upper)), col = 8, border = NA)
  with(semi_variog_i, points(u, v, pch = 19))

  ## variance of the data
  var_data <- semi_variog$var.mark[i]
  abline(h = var_data, lty = 2)

  ## semi-variogram of the true model
  lines.variomodel(EXP_GRF, col = 4, lwd = 2)
  abline(h = sigmasq, lwd = 2, col = 4, lty = 2)

  ## fit an variogram for parameter estimation
  model <- variofit(semi_variog_i, c(var_data, phi), "exponential",
                    fix.nugget = TRUE, nugget = 0, message = FALSE)

  ## fitted variogram
  lines.variomodel(model)

  ## add estimated `sigmasq` and `phi` as title
  title(sprintf("sigmasq = %.2f, phi = %.2f", model[[2]][1], model[[2]][2]))

  }

Cross Validated Asked by Zheyuan Li on January 7, 2022

1 Answers

One Answer

I didn't have a chance to look carefully into the code, but I can already guess what you are experiencing. It may have to due with the fact that you are simulating random fields on finite domains (in a computer).

Try to increase the domain to a larger box [0,10]x[0,10] or try to decrease the range of the variogram to something that is at least smaller than half of your domain size, say 0.1. Then plot the variograms up to a lag that is at most half of your domain. Everything that is separated by a lag that is equal or larger to half of your domain is not useful in empirical variogram calculations (too few samples, and hence spurious oscillations).

Geostatistics theory is more general than AR models, and there are theorems related to the issue you are experiencing, but I have no time at the moment to write down a more detailed answer.

Answered by juliohm on January 7, 2022

Add your own answers!

Related Questions

How to evaluate if year-over-year data is correlated and predictive

1  Asked on January 21, 2021 by basketballautomation

     

A/B Testing – Repeated Sampling

0  Asked on January 18, 2021 by thomas-moore

   

Clustering with Likert items and N/A option

1  Asked on January 17, 2021 by matthias

     

How to derive Bias of PCA estimator

0  Asked on January 16, 2021 by baz

     

Mean squared error of OLS smaller than Ridge?

5  Asked on January 15, 2021 by aristide-herve

       

Ask a Question

Get help from others!

© 2022 AnswerBun.com. All rights reserved.