# 12.2 - Inverse Sampling for Capture-Recapture

12.2 - Inverse Sampling for Capture-Recapture

What 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 are 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:

$$\hat{V}ar(\hat{\tau})=\dfrac{X^2y(y-x)}{x^2(x+1)}$$

Note! here x is specified and we do not need to worry about the case x = 0.

## 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!

Estimate the total population size of eagles for the above example and find the variance of your estimate.

X = 200
x = 35, y = 100

$$\hat{\tau}=\dfrac{100}{35}\times 200=571.43$$

$$\hat{V}ar(\hat{\tau})=\dfrac{200^2\times 100(100-35)}{35^2(35+1)}=5895.69$$

$$\hat{S}D(\hat{\tau})=76.78$$

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