[AniMov] summing BB outputs

Clément Calenge clement.calenge at oncfs.gouv.fr
Fri May 20 12:44:35 CEST 2011


Hello,

> I am looking for some advice. I’ve been struggling with tyring to 
> combine Brownian Bridge UD’s of the class estUD for work similar to 
> that carried out by Sawyer et al (Ecological Applications 2009, 
> 19:2016-2025). My goal is to get some ideas of common space use 
> (similar to migration corridors) for my animals by summing the 
> individual UD’s and rescaling to a single probability surface.
>
> I have created the BB probability surfaces using the function kernelbb 
> which creates a class estUD. What I am having trouble with is 
> determining how to sum the different surfaces so that I can then 
> rescale the cumulative values back to 1 and then create a Volume 
> surface using getvolumeUD. Any suggestions on how I would go about 
> summing the different UD’s would be extremely helpful.
>

The UD of a given animal i corresponds to the probability to find an 
animal at a given point (x,y), given that this animal is this animal i, 
i.e. P(x,y|i). What you want to estimate is the probability to find *any 
animal of the population* at a given place x,y, i.e. P(x,y). Simple 
probability calculation gives:

P(x,y) = Sum_i P(x,y | i) P(i)

if there are N exchangeable animals in the finite population of animals, 
then P(i) = 1/N, and

P(x,y) = Sum_i P(x,y | i)/N

Where the sum is over all the animals *in the population*. Therefore, 
the probability to find an animal at a given place is simply the average 
of the UDs of *all* animals in the population. You want to estimate the 
population mean P(x,y) from the UDs of the animals in a sample of n 
animals. This is a problem of estimation, and the way you can estimate 
P(x,y) from a sample of animals depends on your sampling design (i.e. 
the way you selected monitored animals).

One way to deal with this problem is to consider it from the point of 
view of the sampling theory for finite populations (e.g. Cochran 1977, 
Sampling techniques, John Wiley and Sons). Note that I suppose here a 
finite population. The way the animals have been sampled defines your 
sampling design. For example, you may consider that your sampling design 
is a random sample without replacement of n animals in the population of 
N animals (this is just an example), you can estimate the population 
mean of UDs P(x,y) by the sample mean of UDs P(x,y |s) without bias. In 
other words, the estimator of P(x,y) for simple random sampling is:

P(x, y | s ) = Sum_i P(x,y | i)/n

where the sum is over the n animals in the sample s. So, *in the case of 
simple random sampling*, the sample mean of the UDs is an unbiased 
estimate of the population mean of the UDs, i.e. P(x,y).

However, you should be aware that any estimation is associated with a 
sampling variability: if you repeated your study, the sample of 
monitored animals would probably have been different (the animals 
monitored would not have been the same). In other words, the value of 
P(x, y | s ) would also have been different. For a given location (x,y), 
the set of values of P(x, y | s ) obtained from all possible samples s 
of n animals in the population defines a statistical distribution with a 
given variability. When you use P(x, y | s) as an estimate of P(x,y), 
you do not take into account this variability (i.e. there is no standard 
error, nor confidence interval associated with this estimation). For 
finite populations, you cannot estimate this variability without knowing 
the size of the population N (which is usually unknown).

Therefore:

* You should be aware that when the proportion of animals in the 
population that have been monitored with GPS is small, this variability 
is very large, and the estimates P(x,y | s) obtained from the sample of 
monitored animals can be very different from the actual value of P(x,y) 
(that is highy variable estimation, although unbiased, i.e. if we 
consider all possible samples drawn with this design, the mean of the 
distribution of the P(x,y|s) is equal to P(x,y)).

* Here, I took the example of simple random sampling without 
replacement. You have to consider whether this sampling design is a 
correct approximation for the way your animals have been sampled. To 
make this assumption likely, your capture sites should be distributed 
uniformly over all the area where there is a chance to capture an animal 
(and should not be more dense in some places), the spatial distribution 
of the capture effort should be uniform. The capture effort should not 
be focused on one class of animals (male/female, young/adults, etc.), etc.
Note that other sampling designs (stratified, two-phase, etc.) can be 
supposed when these assumptions are unlikely, but the estimator of 
P(x,y) will not be the same. See Cochran (1977) for further information 
about the sampling theory.

Now, from a purely technical point of view, using for example the 
dataset puechcirc as an example, to estimate the sample mean of the UDs 
with adehabitatHR:

## Load the data
data(puechcirc)

## Field studies have shown that the mean standard deviation (relocations
## as a sample of the actual position of the animal) is equal to 58
## meters on these data (Maillard, 1996, p. 63). Therefore
sig2 <- 58

## Find the maximum likelihood estimation of the parameter sig1
## First, try to find it between 10 and 100.
liker(puechcirc, sig2 = 58, rangesig1 = c(1, 10))

## We define sig1 = 5.5
## Now, estimate the brownian bridge with the same grid for all
tata <- kernelbb(puechcirc, sig1 = 5.5, sig2 = 58, grid = 100, 
same4all=TRUE)

## convert the result as a SpatialPixelsDataFrame:
sdf <- estUDm2spixdf(tata)

## calculate the sample mean for each pixel:
smea <- apply(sdf at data,1,mean)

## converts to spatialpixelsdataframe
df <- data.frame(smea)
coordinates(df) <- coordinates(sdf)
gridded(df) <- TRUE

## the mean is in df
image(df)

HTH,


Clément Calenge

-- 
Clément CALENGE
Cellule d'appui à l'analyse de données
Direction des Etudes et de la Recherche
Office national de la chasse et de la faune sauvage
Saint Benoist - 78610 Auffargis
tel. (33) 01.30.46.54.14



More information about the AniMov mailing list