Mark recapture with no knowledge of marked individuals

I am a math student working with a group of field biologists. In multiple experiments of mark-recapture of the same population, they claimed that if the number of observations (recaptures) is large enough, it is possible to infer both population size ($$N$$) and proportion ($$p=K/N$$) based only on the sample, without prior knowledge of the number of individuals that were originally marked ($$K$$).

Considering a sample of size $$n$$ with $$k$$ marked individuals is drawn from the population, a histogram of the observed proportions is created using jackknifes subsamples. Then a hypergeometric distribution is fitted to the histogram using minimum squares. This leads to a pair of estimated parameters $$\hat{p}$$ and $$\hat{N}$$.

I am convinced that under the assumption of $$K$$ unknown this estimator converges to the sample size ($$n$$), i.e $$\hat{N} \rightarrow n$$ (or at least to a value different than $$N$$). However, I haven't proved it yet.

They tested this estimator with simulated data. My conclusion is that in the simulations the standard deviation of $$\hat{N}$$ is large enough so that the values of $$N$$ and $$\hat{N}$$ sometimes coincide.

I am creating this question to verify whether I am correct. Also, I am not very good explaining myself, and need to create an argument to convince non-specialists that $$\hat{N} \rightarrow n$$. Thank all of you for your help.

• +1 Nice clear explanation. Do you know what estimators they're using to estimate N and p? Commented Jun 14 at 5:23
• They create a histogram of the observed proportions (they make many observations) and they fit a hypergeometric distribution trough minimum squares. Thanks Glen_b. Commented Jun 14 at 5:27
• Better said, they made 1 single (large) observation (size n) and they sub-sample (bootstrap or jackknifes, not entirely sure) this vector to create the histogram. Commented Jun 14 at 5:32
• Do you at least know if you've recaptured the same specimen, i.e. is the mark unique? If not you might be able to estimate $p$ but I don't see how that tells you anything about $N$. Commented Jun 14 at 6:09
• @Glen_b The population sizes are $N$ (total, unkown), $K$ (marked, unkown), and the sample sizes are $n$ (total sampled, known), $k$ (marked, known). Commented Jun 14 at 7:33

From first principles it is clear that the single observed hypergeometric sample of size $$n$$ out of which $$k$$ will be informative about little more than $$p=K/N$$. Both $$N$$ and $$K$$ will be nearly unidentifiable which can be seen from the fact that the likelihood tends towards a flat ridge along the line given by $$K/N=k/n$$ as $$N$$ and $$K$$ becomes large (this follows from the binomial as a limit of hypergeometic distribution). The likelihood ridge is illustrated below for the sample $$k=5$$ and $$n=10$$.

It is impossible that the subsampling method employed can produce any information about $$N$$ beyond the information provided by the above observed likelihood. Conditional on $$n$$ and $$k$$ (that is, conditional on the original sample), it is true that subsamples of size $$m$$ (assuming that these are also drawn without replacement from the orignal sample) will be independently hypergeometrically distributed but the parameters will be $$n,k,m$$ rather than $$N,K,m$$. So when estimating $$N$$ by fitting a hypergeomertic distribution to these subsamples, the field biologists are in reality indeed estimating $$n$$, not $$N$$. It is also true that the subsamples hypergeometrically distributed with parameters $$N,K,m$$ but this is only marginally (that is, when not conditioning on $$n$$ and $$k$$). And marginally, the subsamples are also not independent which makes fitting a hypergeometric distribution to the subsamples invalid when the aim is to estimate $$N$$.

N <- 1:100
K <- 1:100
n <- 10
k <- 5
contour(N, K,
outer(N, K, function(N,K) dhyper(k, K, N-K, n)),
xlab="N", ylab="K", nlevels = 50
)
#> Warning in dhyper(k, K, N - K, n): NaNs produced


Created on 2024-06-14 with reprex v2.1.0

• Sorry for my late response, I was waiting for further comments. Your answer gave me valuable insights. Thank you Jarle! Commented Jul 9 at 6:17