95% Home Range from sigma

[Elsa Bussiere]

Hello,

I am trying to use half-normal estimates of sigma to calculate 95% home ranges.
It seems the following formula: (95HR = 2.45^2 * Pi * sigma^2 ; roughly 19 * sigma^2) was mentioned during an SECR workshop in Montpellier (but I don't have any reference to use it). The HR estimates I get using this formula are slightly big but reasonable.

While searching for previous conversations on the subject, I found this question sent to the secrgroup a little bit more than a year ago:

It was highlighted that the sigma should not be directly compared between half-normal and negative exponential detection functions. Is there a simple way to estimate a 'home range radius' and a '95% home range area' from a negative exponential sigma? For the half-normal sigma I have generally used these formula (based on Ringler et al. 2014 and Noss et al. 2012):
HR_radius<-(qchisq(0.95,2)^0.5)*(sigma)  
HR95_area<-(pi*(3.36*sigma)^2)/10000

Looking at the Ringler et al. 2014 paper (attached), the underlying idea behind the formula seems to be the same (5% chi-squared test statistic), but the factor jumped from 19 to 35, which in my case will lead to HR estimates unreasonably big.

My question is: which formula should I use to estimate 95% Home Range from sigma and which reference must I provide?

Thanks very much
Kind regards,

Elsa

[Murray Efford]

Hi
Check the help for circular.r in 'secr'. The bivariate-normal multiplier is indeed around 19 (it also depends on whether you believe the hazard or the probability has a bivariate normal distribution). I think one of Beth Gardner's papers is a peer-reviewed source for SECR, and there are many others going back to Calhoun & Casby and Jennrich & Turner. I can't make sense of Ringler et al. They use a hazard-rate detection function for which the multiplier depends strongly on z, yet, as far as I can see on a quick browse, they do not acknowledge this or report their z-hat.

Whatever you do - remember this is really rough and depends on an implausible model for animal behaviour, so agreement with telemetry data is not expected.
Murray

[James Russell]

I followed up with David Ringler on this as its come up a few times on the list now:

 

We used the circular.r function to determine the sigma multiplier.
Assuming a hazard rate detection function and a z-estimate of 4 (which indeed is not in the paper), that gives :

circular.r(detectfn=1, detectpar=list(sigma=1, z=4), hazard=FALSE)
multiplier = 3.36

Also at the time we did that, integration of the circular.r function was not performed on the cumulative hazard rate. Using that new integration, the multiplier would be :

circular.r(detectfn=1, detectpar=list(sigma=1, z=4), hazard=TRUE)
multiplier = 1.8 (which is changing a bit our HR estimates)

James

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[Murray Efford]

Thanks very much for clarifying this.
Murray

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