The sizes of animal populations are often estimated by using a capture–tag– reca

Question

The sizes of animal populations are often estimated by using a capture–tag– recapture method. In this method r animals are captured, tagged, and then released into the population. Sometime later n animals are captured, and Y, the number of tagged animals among the n, is noted. The probabilities associated with Y are a function of N, the number of animals in the population, so the observed value of Y contains information on this unknown N. Suppose that r=4 animals are tagged and then released. A sample of n=3 animals is then selected at random from the same population. Find P(Y=1) as a function of N. What value of N will maximize P(Y=1)?

Explanation / Answer

We are given that 3 animals are randomly chosen from N animals.
Also, of these N animals, 4 are tagged and N - 4 are untagged.

The total number of equally likely outcomes is N choose 3.

Y = 1 means that exactly 1 of the 3 chosen animals is tagged. For this to occur, 1 tagged animal must be chosen from the 4, and 2 untagged animals must be chosen from the other N - 4.
So the number of favorable equally likely outcomes is (4 choose 1)((N - 4) choose 2)).
Also note that for Y = 1 to be possible, N must be at least 6, so we will assume N >= 6.

Thus P(Y = 1) = (4 choose 1)((N - 4) choose 2)) / (N choose 3)
= 4(N - 4)(N - 5)/2! / [N(N - 1)(N - 2)/3!]
= 12(N - 4)(N - 5) / [N(N - 1)(N - 2)]

Now let f(N) = P(Y = 1) = 12(N - 4)(N - 5) / [N(N - 1)(N - 2)].

Then f(N + 1)/f(N)
= {12(N - 3)(N - 4) / [(N + 1)(N)(N - 1)]} / {12(N - 4)(N - 5) / [(N)(N - 1)(N - 2)]}
= [(N - 3) / (N - 5)] / [(N + 1) / (N - 2)]
= (N - 3)(N - 2) / [(N - 5)(N + 1)]
= (N^2 - 5N + 6) / (N^2 - 4N - 5)

Assuming N >= 6, note that the ratio f(N + 1)/f(N) is equal to 1 if N = 11, less than 1 if N > 11, and greater than 1 if N < 11.

It follows that f(N) = P(Y = 1) is maximum both at N = 11 and at N = 12

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