I would like to use the function Kcross.inhom, but I'm confused about ways to
supply the lambda intensity vaules (arguments lambdaI and lambdaJ)
The help file says you can supply image files for these arguments, or if you
omit values for these arguments, then it will be estimated using a
'leave-one-out' kernel smoother through density.ppp.
My confusion is why you get different values if you use density.ppp to generate
an image file with either at="points" (creating a numeric vector of intensity
values at the points) or at="pixels" (creating an image showing intensities at
every location in a fine grid of the window).
The examples in the help file show this done with density.ppp using at=points,
yet the default value in this function is at="pixels".
An extension to the examples given in the help file (see below- method 3) show
that K1 and K2 are equivalent (estimate values ABOVE the theoretical Poisson
line), but K3 (using the image produced from density.ppp) is highly different
(with estimate values BELOW the theoretical Poisson line).
Can someone explain to me the difference?
Thank you!
Erika
# Lansing Woods data
data(lansing)
lansing <- lansing[seq(1,lansing$n, by=10)]
ma <- split(lansing)$maple
wh <- split(lansing)$whiteoak
# method (1): estimate intensities by nonparametric smoothing
lambdaM = density.ppp(ma, sigma=0.15, at="points")
lambdaW = density.ppp(wh, sigma=0.15, at="points")
K1 = Kcross.inhom(lansing, "whiteoak", "maple", lambdaW, lambdaM)
plot(K1)
# method (2): leave-one-out
K2 = Kcross.inhom(lansing, "whiteoak", "maple", sigma=0.15)
plot(K2)
# method (3): MY ADDITION
lambdaM.px = density.ppp(ma, sigma=0.15) #default at="pixels"
lambdaW.px = density.ppp(wh, sigma=0.15) #default at="pixels"
K3 = Kcross.inhom(lansing, "whiteoak", "maple", lambdaW.px, lambdaM.px)
plot(K3)
Erika Mudrak
Postdoctoral Researcher
Ecology, Evolution, and Organismal Biology
Bessey Hall 253
Iowa State University
mud...@iastate.edu
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