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Oppose – Tag and release can include Mark and recapture, but it can also be quite different. For example small archival tags can be attached to marine animals like fish. These archival tags can be equipped with a camera or sensors that monitor and log things like salinity, temperature, depth, acceleration, and pitch and roll. They are designed to detach at a later date and float to the surface where some method can be used to retrieve the logged data. This has nothing to do with "marking" or recapture. Tag and release is often linked with catch and release, and is a term widely used, particularly in fisheries and by recreational fishermen. --Epipelagic (talk) 04:21, 3 May 2013 (UTC)[reply]
Support – In school, we learned about Mark and recapture as Tag and release. I actually have never heard of mark and recapture and "tag and release" is what I always knew "mark and recapture" as. Mark and recapture is definitely known as tag and release in many textbooks and the MCPS curriculum. 173.79.218.246 (talk) 09:32, 29 May 2013 (UTC)[reply]
Oppose - Despite Tag and release being a small page, it doesn't mention anything about the purpose of estimating population size. - Paul2520 (talk) 09:33, 21 October 2015 (UTC)[reply]
Support - The current article related to methods of tagging, releasing, and the recapturing fish, so does fall within the scope of Mark and recapture. Note that other forms of tracking and datacollection are aready discussed on GPS wildlife tracking, now linked from the Mark and recapture page. Klbrain (talk) 15:55, 21 April 2016 (UTC)[reply]
I was requested here to improve this article. However my expertise is on Bayesian statistics and so any contribution of mine on Mark and Recapture might be considered original research.
Assume that K animals out of a population of the unknown size N have been marked. Later n animals are captured out of which k animals turned out to be marked.
N−K animals are unmarked. N−n animals are uncaptured. n−k captured animals are unmarked. K−k marked animals are uncaptured. N−K−n+k uncaptured animals are unmarked. As all these numbers are non-negative, the following inequalities result: ( N ≥ K+n−k) and (n ≥ k) and (K ≥ k) and (k ≥ 0).
Knowing K and n and k the problem is to estimate N.
Probability distribution
The conditional probability (k|N) of observing k knowing N (and n and K), is the hypergeometric distribution.
But we were interested in estimating N knowing k.
Credibility distribution
When there is no prior knowledge regarding N and k, the credibility distribution (N|k) is proportional to the likelihood function (k|N).
Inserting the expression for (k|N) and cancelling the common factor:
The denominator series is convergent for k ≥ 2.
Graphs
load 'plot' NB. plotting software
LF =: 4 : 0 NB. Likelihood Function
'K n k'=:x
(N>:K+n-k)*((n-k)!N-K)%n!N=:i.y
)
g =: [: 'dot; labels 1 0 ; pensize 4' & plot LF
11 10 0 g 701
11 10 1 g 501
11 10 2 g 301
11 10 3 g 101
11 10 4 g 71
Likelihood for Mark and Recapture. K=11, n=10, k=0Likelihood for Mark and Recapture. K=11, n=10, k=1Likelihood for Mark and Recapture. K=11, n=10, k=2Likelihood for Mark and Recapture. K=11, n=10, k=3Likelihood for Mark and Recapture. K=11, n=10, k=4
This J programming created the 5 graphs to the right showing likelihood functions for the total number of animals, N, for K=11 marked animals and n=10 captured animals, and for k= 0, 1, 2, 3 and 4 recaptured marked animals.
When k=0 we didn't recapture any marked animals and the physical limit on how many animals there are around is not reduced by the mark and recapture observation. The likelihood function has no maximum, and so the maximum likelihood estimate is infinite.
When k=1 we recaptured a single marked animal, and the likelihood function has a maximum at N=110, but the median is infinite.
When k=2 we recaptured two marked animals. The maximum likelihood is at N=55, and the median is at N=137. A 95% confidence interval is 19≤N≤1367, but the mean value is infinite.
When k=3 we recaptured three marked animals. The maximum likelihood is at N=36, and the median is at N=57. A 95% confidence interval is 18≤N≤236. The mean value is finite but the standard deviation is infinite.
When k=4 the maximum likelihood is at N=27. The median is at N=36. The 95% confidence interval is 17≤N≤ 96.
The frequentist formulas from the article gives the estimates N ≈ Kn/k = 27.5 and N ≈ (K+1)(n+1)/(k+1)−1 = 25.4.
Knowing the credibility distribution function, (N|k), one can compute the order of magnitude, μ, and the statistical uncertainty, σ, of the unknown number N.
A closed form for the above sums can be found using Gosper's algorithm. However Wolframalpha does not immediately do it [1]. But the following detour does the trick.
Thanks for this, I think I mostly followed what you have here, the stuff I've been looking at is using frequentist MLE approximations. Seeing it done from a Bayesian perspective is very interesting though. Have you seen any references or is it purely yourself? Jamesmcmahon0 (talk) 11:18, 4 March 2014 (UTC)[reply]
Reference to 2013 paper in The American Statistician (arXiv) for a Bayesian analysis, which demonstrates the above. — Preceding unsigned comment added by 194.81.223.66 (talk) 13:11, 15 December 2014 (UTC)[reply]
Just read this the linked article will probably have contributions for this article. I don't have time for the foreseeable future unfortunately. 86.179.58.222 (talk) 18:54, 20 November 2014 (UTC)[reply]
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The formula
should I think be
See (6) in [1]
I have made this change.
Nick Mulgan (talk) 01:53, 1 February 2019 (UTC)[reply]
^Sadinle, Mauricio (2009-10-01). "Transformed Logit Confidence Intervals for Small Populations in Single Capture–Recapture Estimation". Communications in Statistics - Simulation and Computation. 38 (9): 1909–1924. doi:10.1080/03610910903168595. ISSN0361-0918.