Closed t-pollington closed 5 years ago
Dear @t-pollington,
You questions about of reason the accommodation of the kernel function in the estimators of the spatial and temporal mark variograms can be solve if you take time to check the classical literature of the spatial point patterns. I recommend you read the following books in order to understand the geometry of the counts behind the estimators and it relationship with the kernel function and bandwints:
If you are interested in more details or some kind of scientific cooperation please contact with me at the email frrodriguezc@unal.edu.co.
Regards!
Dear @frajaroco,
Thank you for the book references.
My second question is resolved! I realise that kernel functions can have negative arguments since they are symmetric.
I am still puzzled by the 0<epsilon<r condition in your paper and its purpose. If anything, for the Epanechnikov kernel, one needs epsilon<0.5*(ds[i]-ds[i-1]), assuming there are evenly spread ds[i]
, to prevent overlap of pairs selected for adjacent ds[i-1]
,ds[i]
,ds[i+1]
.
Yes let's discuss more. I'll be attending Spatial Statistics 2019 and hope to discuss more with you then on topics that interest us.
Kind regards, @t-pollington.
Dear @frajaroco,
Please could you help me in understanding the application of the kernel function _kepsilon regarding the spatial and temporal mark variograms as described in equations 7 and 8 (p129) of your Mark variograms for spatio-temporal point processes paper and how it relates to your code in
gspcore.f
as an example?This is my current understanding however it results in a contradiction.
1) Is the 0 < epsilon < r condition only for the edge-corrections and so only applies to the "
standardised
" approach, or does it also apply to the "simplified
" approach too? I think it may be the latter since online 79
you drop the first element ofds
which would be set tohs
. I think this is because you want to avoid the epsilon(hs
)=r(ds
) case.2) for the kernel function _k_epsilon(||x_i-xj|| - r) I presume that only an argument "
x
"(say) greater than or equal to r ( i.e. x >= r) is possible. But onlines 31-35
you divides(iu)-hij
byhs
(i.e. epsilon). But if you divide the argumentx
by something that has to be smaller than r (from 1 above as 0 < epsilon < r) then this will result in an argument larger than 1. So when the kernel function runs e.g.lines 90-104
then it will never run theif
condition, only theelse
.Kind regards,
@t-pollington.