'Detection functions' for polygon detectors

[Ben Stevenson]

Hi all,

I am trying to get my head around the following passage in the polygon detector vignette:

Although probability and cumulative hazard are mostly interchangeable for point detectors, it’s not so simple for polygon and transect detectors. The integration in equation 1 always uses the hazard form for h(u | x) (secr 3.0.0 and later), and only hazard-based detection functions are allowed (HHN, HHR, HEX, HAN, HCG). The default function is HHN.

I want to fit a model where the PDF of a cue's location is bivariate normal, centred on the activity centre. However, I'm struggling to figure out if this is what the default hazard halfnormal actually gives me, or what the non-hazard halfnormal would give me if secr let me use it.

For example, the hazard halfnormal detection function can have a long shoulder at 1 if lambda0 is large enough. So, if I use the hazard halfnormal here, is it the case that the PDF of a cue's location is not bivariate normal, but rather may have a circular, flat 'plateau' around the activity centre due to this shoulder? If secr did let me fit the non-hazard halfnormal detection function, how would this change the assumed PDF of a cue's location?

Many thanks,

Ben

[Murray Efford]

Hi Ben

I tried so hard to make that clear in the pdf, but I admit there is still room for the doubt you raise! As I remember/understand, the function h(u|x) should be interpreted directly as the spatial distribution (pdf) of cues - so HHN will give the bivariate normal you ask for. h(u|x) itself is integrated to get the expected number of (detected) cues. You will have read "The detection model is fundamentally different for polygon detectors..." Part of that is that at no point is h(u|x) converted to a probability via e.g. 1-exp(-h(u|x)). If one integrated 1-exp(-h(u|x)) then one would be modelling a different distribution, as you suggest, but 'secr' does not do that. It's somewhat arbitrary that 'secr' uses HHN rather than HN etc. for polygon detectors (I feel comfortable integrating hazards or probability densities, but not probabilities).

Hope this gives you some confidence...

Murray

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