It seems that liker greatly overestimates the parameter sig1 in this
case. Actually, the use of this method relies on the hypothesis that the
animal is moving according to a Brownian motion between two successive
relocations. Depending on the speed of the animal, the bridge may be
made more or less narrow, the parameter sig1 is controlling this width.
When the parameter sig1 increases, the movement of the animal between
two successive relocations is supposed to be more "diffuse". Given the
fast speed supposed when setting a large parameter sig1, the time lag
between the two relocations is long enough to allow the animal to make
large moves, i.e. moving far from the straight line connecting the two
relocations. The relocations then are no longer "attractors" of the
animal during the movements supposed by the model, and the probability
under this model to find the animal far from these relocations is not
negligible when sig1 is large. Consider the following example (just copy
and paste in R):
par(mfrow=c(2,1))
plot(simm.bb(1:1000,end=c(10,10)), addp=FALSE)
plot(simm.bb(1:1000,end=c(1000,1000)), addp=FALSE)
The upper graph shows a Brownian bridge characterized by an end
relocation (in red) close to the the start point (in blue). This is
similar to a large value of sig1 (i.e. fast animal). And the lower graph
shows a Brownian bridge characterized by an end point far from the start
point. This is similar to a very small value of sig1. In the latter
case, the movement is close to the line segment separating the two
relocations. You can see on the first graph that the animal may move far
from the start and end relocations during the time lag separating the
relocations. And this may even be far from the other relocations, i.e.
far from the "true" home range. This is caused by the hypothesis that
the animal is moving according to a Brownian motion between successive
relocations. This Brownian bridge is a useful model when sig1 is not too
large, because it allows to take into account the fact that the animal
is moving between relocations. Taking a too large sig1 reduces the
usefulness of this model...
So, now the question is: why the liker function does return a so large
result? By curiosity, I compared on your data the results of the
function liker from adehabitat and of the program provided by Horne et
al. (2007, Ecology): the two functions give fairly similar results for
sig2=100:
adehabitat, function liker: sig1 = 10.16
Horne et al, visual basic program: sig1 = sqrt(103.735) = 10.18
Actually, as indicated by Horne et al., the estimation procedure of sig1
strongly relies on the assumption of the Brownian bridge model. This may
be problematic given your data. Indeed, the function kernelbb does not
consider the partitioning of the trajectory of an animal into "bursts".
Have a look at your data:
> tr
*********** List of class ltraj ***********
Type of the traject: Type II (time recorded)
Irregular traject. Variable time lag between two locs
Characteristics of the bursts:
id burst nb.reloc NAs date.begin date.end
1 1 11 1 0 2008-05-25 00:39:00 2008-05-25 00:39:00
2 1 110 6 0 2008-08-02 01:40:00 2008-08-02 05:56:00
3 1 111 5 0 2008-08-12 01:50:00 2008-08-12 04:58:00
4 1 112 10 0 2008-08-13 00:28:00 2008-08-13 07:02:00
...
The first "burst" is made of only one relocation. Then a time lag of
more than two months occur. Then a burst of 6 relocations collected
every hour do occur. Then, two days after the last relocation of this
burst, there is a new burst of relocations collected every hour, etc.
The function kernelbb assumes that the movement between any pair of
successives relocations is generated by the *same* Brownian bridge
model, whatever the time lag between the relocations. Therefore, even if
sig1 is set so that the movement between two relocations separated by an
hour is nearly straight, when two days are separating the relocations,
using the same sig1, the probability that the animal moves far from the
relocations under the assumed model is high. This is similar to the
following situation:
star <- c(simm.bb(1:1000,begin=c(0,0), end=c(1000,1000), burst="A0"),
simm.bb(round(seq(1001, 10001000, length=1000)),begin=c(1000,1000),
end=c(2000,2000)))
plot(star, addp=FALSE)
This graph presents three relocations (in red and blue) together with a
simulated Brownian bridge supposed between the relocations. The first
two relocations are separated by 1000 seconds and the last two by
10,000,000 seconds. You can see that the animal can move far from the
relocations under the supposed model.
Therefore, you have to think about the time scale at which you want to
have an estimate of the home-range. The function kernelbb does not (yet)
offer the possibility to estimate the home-range using pairs of
successive relocations at most separated by an interval of X seconds.
One possibility would be to estimate a brownian bridge for each burst
separately, and then to combine them together, giving them a weight
corresponding to the number of relocations used for the estimation. That
is, something like:
## remove the first burst, because only one relocation
tr2 <- tr[-1]
## estimate the Brownian bridge kernel for each burst separately
hrBB <- kernelbb(tr2, sig1 = 6, sig2 = 100, grid = 40, byburst=TRUE)
## then combine the UDs
liUD <- lapply(1:length(tr2), function(i) {
hrBB[[i]]$UD * (nrow(tr2[[i]]) / sum(sapply(tr2, nrow)))
})
UDend <- liUD[[1]]
for (i in 2:length(liUD))
UDend <- UDend + liUD[[i]]
## Plot the result
image(UDend)
contour(getvolumeUDs(UDend), level=95, col="red", lwd=2, add=TRUE)
And for the sig1 selection:
ii <- liker(tr3, c(2,100), 100, byburst=TRUE)
hist(sapply(ii, function(x) x$sig1), ncla=20, col="grey")
median(sapply(ii, function(x) x$sig1))
Note that a value sig1 = 6 seems reasonable.
Hope this helps
Best regards,
Clément Calenge
--
Clément CALENGE
Cellule d'appui à l'analyse de données
Office national de la chasse et de la faune sauvage
Saint Benoist - 78610 Auffargis
tel. (33) 01.30.46.54.14
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