There is a formal relationship between effective sample area Ae, abundance and density. We feel that at the present time many ecologists perceive methods as being distinct based on whether they provide an estimate of Ae, or density, or N, or some combination (or something altogether different). However, any conceptually coherent procedure should imply a clear relationship between these quantities via the specified model.
To motivate the calculation of effective sample area, suppose that S was discrete, being composed of a large number of pixels of unit area sk(k = 1, 2,...). Suppose further that each pixel had associated with it a binary indicator variable zk such that 0k = Pr(zk = 1). Prior to sampling we flip a coin for each pixel, having success probabilities 0k, and each pixel for which zk = 1 is sampled; therefore, in this case, the sampled area (in units of pixels) is the outcome of a stochastic process, being the sum of the zk indicators:
Consider a 10 x 10 grid, flip a coin 100 times, and all of the grid cells that come up 'heads' are included in the sample. The sampled area is the sum of the area associated with those grid cells. Naturally we might estimate this quantity by its expected value
This is the expected or effective sample area. In the 10 x 10 grid example, if the 100 coins flips all have = 0.5, then the effective sample area during each sample is 50. That is, you expect to count individuals from 50 grid cells during the survey. An alternative way to frame this concept of effective sample area is to suppose that sk; (k = 1, 2,...) are the activity centers of hypothetical lizards and each lizard flips a coin with probability to determine whether it will be located within the sampled plot. Whether it is lizards that are flipping coins to decide their exposure to sampling, or biologists flipping coins to decide whether a pixel will be sampled, both situations yield the same conceptual formulation of the effective sample area. Note that 1—^k is interpretable as the temporary emigration probability (Kendall et al., 1997).
Now, suppose that the sampling will be conducted on more than one occasion, say J = 2 times. In this case, then a lizard will be exposed to sampling if zk = 1 in either (or both) samples. Thus, the lizard is exposed with probability ek = 1 — (1 — )2. In this case, the effective sample area is
We see that the effective sample area is not just a function of but also the number of times that the population is exposed to sampling. Effective sample area increases with J, the number of temporal samples.
This concept extends directly to the continuous space situation. That is, the effective sample area is the integral:
where e(s) = Pr(exposed to sampling|s). As before, 'exposed to sampling' is the event that x G X at least once during the J samples. Let ^(s) = Pr(x G X)|s); then if the J samples are independent,
We emphasize that the probability of exposure, and hence the effective sample area, both depend on the effort expended sampling - the number of samples, J. We do not believe that this is widely appreciated in spatial sampling problems. Note also that Pr(x G X|s) is a function of the parameter a, and so the effective sample area is also a function of the movement process of the species being sampled, as we would expect. An individual has more exposure to sampling as its movement distribution becomes concentrated in the surveyed region which is influenced by both a and s. For example, an individual having s right in the middle of the quadrat should have exposure nearly equal to 1.00, whereas an individual's exposure should
decrease to 0 as s moves away from the sampled area. Exposure to sampling is also related to the total sampling effort (J), such that more effort will increase exposure. Thus, with a = 0.187 we would not expect to capture an individual one-half unit from the sample plot in 1 or 2 samples, but with J = 14, we should have a reasonable chance of capturing the individual.
Figure 7.5 shows the effective sample area of the 9 ha quadrat under various values of a. The color of a pixel (let s index the pixel) was obtained by placing a normal density centered at s having prescribed a. Then, the normal pdf was evaluated at a fine grid of points and those points within the grid box were summed and divided by the overall total. The Ae shown in this figure is in units of ha, and so the estimate of Ae for a = 0.187 is 12.84 ha. We note that it would be easy to obtain a characterization of uncertainty associated with Ae within a Bayesian framework, as Ae is just a function of parameters for which we obtain posterior samples.
The quantity 0(s) can be computed easily, as the solution to the integral:
This is the overlap of a bivariate normal kernel with the region X, which can be computed using pnorm in R (if X is a simple polygon). In general, we find it efficient to apply a discrete approximation of the normal integral, by evaluating the kernel on a large grid of points and them summing them up within the box, and in total.
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