Using the example data provided, when I derive a spatial range estimate from
a semivariogram based on the spatial organization of the ruf values I am
looking at initial values in the region of 2-4 (code below), yet the
estimate provided in the example is 0.2, and the ML estimate of spatial
range in
Hi John and thanks for the response - I posted the question for the sake of
other users, but also because I had written to the developer, Mark Handcock,
in the past with no response.
Since the example of ruf.fit in the ruf package of R uses data from one
individual jay (d412), and since as you say
Two parameters are required in order to fit a Resource Utilization Function
to spatial attribute data using ruf.fit, 1) *spatial range* and 2)
smoothness (theta). With regard to the spatial range parameter, rho,
Marzluff et al. (2004) state this to be the range of spatial dependence in *
metres*.
