12.2 - Inverse Sampling for Capture-Recapture
12.2 - Inverse Sampling for Capture-RecaptureWhat we covered already is the direct sampling of capture-recapture, i.e., the size of both the initial sample (capture) size and the second sample (recapture) size is pre-determined. When the second capture size is not pre-determined, then we have:
Inverse Sampling for Capture-Recapture
Again, assume that an initial sample of X individuals is captured, tagged, and released back into the population. Then, random sampling is conducted until x-tagged individuals are recaptured. If y denotes the second sample size, then:
\(\hat{\tau}=\dfrac{y}{x}X\)
Note that for inverse sampling, x is fixed but y is random. The estimated variance of \(\hat{\tau}\) is:
\(\hat{V}ar(\hat{\tau})=\dfrac{X^2y(y-x)}{x^2(x+1)}\)
Example 12-1: Number of Eagles
We want to estimate the total number of eagles in a wildlife preserve. A random sample of 200 eagles is trapped, tagged, and then released. In the same month, a second sample is drawn until 35 tagged eagles are recaptured. The sample size needed to get 35 tagged eagles is 100. (as opposed to having 100 eagles being recaptured to find 35 tagged ones in the direct capture-recapture).
Try it!
X = 200
x = 35, y = 100
\(\hat{\tau}=\dfrac{100}{35}\times 200=571.43\)
\(\hat{V}ar(\hat{\tau})=\dfrac{200^2\times 100(100-35)}{35^2(35+1)}=5895.69\)
\(\hat{S}D(\hat{\tau})=76.78\)