Talk:Mark and recapture

message: jollys method should be mentioned also.
Yes, the cormack jolly seber method should be described here. It is of extreme importance. 24.147.119.33 17:51, 20 March 2007 (UTC)
thread_title: Other methods

message: Some Sources, References, Links and related topics would be nice... Anyone with knowledge of the subject cares to do it?
--Lucas Gallindo 22:52, 12 July 2007 (UTC)
thread_title:  Sources 

message: Please see Talk:Lincoln Index to discuss relation of content here with that of the newer article. Melcombe (talk) 12:20, 2 December 2010 (UTC)
thread_title:  Lincoln Index 

message: I would like to propose that Tag and release is merged with Mark and recapture for two main reasons;
# Tag and release has very little content
# Tag and release is just another mark and recapture method
Jamesmcmahon0 (talk) 12:04, 2 May 2013 (UTC)
replies: text: Also, an example of an organization that refers to the practice as "tag and release" despite not being related to fishing: http://www.thevlm.org/turtle_tag_release.aspx 173.79.218.246 (talk) 09:36, 29 May 2013 (UTC)
thread_title:  Merge discussion 

message: 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
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. 
   b=.6000 LF~a=.11 10 4
   +/0>:2-/\b
 27
   *`%/a
 27.5
   *`%/&.:>:a
 25.4
   +/0 0.95 0.5<:/~(%{:)+/\b
 17 96 36
Bo Jacoby (talk) 06:14, 6 March 2014 (UTC)
thread_title:  Graphs 

message: 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.
:N\approx \mu \pm \sigma
where
:\mu =\sum_{N=K+n-k}^\infty  (N|k)N  
:\sigma^2+\mu^2 =\sum_{N=K+n-k}^\infty (N|k)N^2   
Bo Jacoby (talk) 14:10, 27 February 2014 (UTC)
thread_title:  Order of magnitude and statistical uncertainty 

message: A closed form for the above sums can be found using Gosper's algorithm. However Wolframalpha does not immediately do it . But the following detour does the trick.
Define the sums
:S_Q = {\sum_{N=K+n-k}^\infty \frac{\binom{N-K}{n-k}\binom{N}{Q}}{\binom{N}{n}}} for Q = 0, 1, 2
so
:\mu=\frac{S_1}{S_0}
and
:\frac{\sigma^2}{\mu}+\mu-1= 2\frac{S_2}{S_1} 
The sums are evaluated 
:\sum _{N=K+n-k}^{m-2} \frac{\binom{N-K}{n-k}\binom{N}{Q}}{\binom{N}{n}}=
\frac{\binom{n+K-k}{Q}}{\binom{n+K-k}{n}}A_Q
-\frac{\binom{m-1}{Q}}{\binom{m-1}{n}}\binom{m-K-1}{n-k}B_Q
where 
:A_Q=\,_2F_1(1+K-k,1+n-k;1+K+n-k-Q;1)
and
:B_Q=\,_3F_2(1,m-K,m-n;m-Q,m-(K+n-k);1)
are generalized hypergeometric functions.
The limiting case for m → ∞ is 
:S_Q=\frac{\binom{K+n-k}{Q}}{\binom{K+n-k}{n}}A_Q
and so
:Q \frac{S_Q}{S_{Q-1}}=(K+n-k-Q+1){A_Q\over A_{Q-1}}
Gauss's theorem
:_2F_1 (a,b;c;1)=\frac{(c-1)!}{(c-a-1)!} \frac{(c-a-b-1)!}{(c-b-1)!} 
gives the simplification
:A_Q=\frac{(K+n-k-Q)!}{(K-Q-1)!}\frac{(k-Q-2)!}{(n-Q-1)!}
so
:\frac{A_Q}{A_{Q-1}}=\frac{K-Q}{K+n-k-Q+1}\frac{n-Q}{k-Q-1}
and
:Q \frac{S_Q}{S_{Q-1}}=\frac{K-Q}1\frac{n-Q}{k-Q-1}
So μ and σ are given by
:\mu=\frac{K-1}1\frac{n-1}{k-2} for k≥3
and
:\frac{\sigma^2}{\mu}+\mu-1=\frac{K-2}1\frac{n-2}{k-3} for k≥4.
and the final result is 
:N\approx \frac{K-1}1\frac{n-1}{k-2}\pm\sqrt{\frac{K-1}1\frac{n-1}{k-2}\frac{K-k+1}{k-2}\frac{n-k+1}{k-3}}
Bo Jacoby (talk) 09:51, 25 July 2014 (UTC)
thread_title:  Summation 

message: 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)
thread_title:  Tagging may alert predators 

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thread_title:  External links modified 

message: The formula
K + n - k + \frac{(K-k+0.5)(n-k+0.5)}{(k+0.5)}\exp(\pm z_{\alpha/2}\hat{\sigma}_{0.5})
should I think be 
K + n - k -0.5 + \frac{(K-k+0.5)(n-k+0.5)}{(k+0.5)}\exp(\pm z_{\alpha/2}\hat{\sigma}_{0.5})
See (6) in 
I have made this change.
Nick Mulgan (talk) 01:53, 1 February 2019 (UTC)
thread_title:  Confidence Interval