A forest has $N$ (unknown) monkeys. A random sample of $n$ monkeys is selected from the forest, tagged and released back into the forest. After a few days, a random sample of $m$ monkeys is selected and $Y$ monkeys are found with the tags. Find and estimator of $N$.
I tried with hypergeometric distribution, while I did not think it was correct. If $Y$ is a hypergeometric random variable, $P(Y=y)=\frac{{n\choose y}{N-n\choose n-k}}{N\choose n}$. How can I get an estimate from here?
Any suggestions would be greatly appreciate!
This is the classical capture-recapture method of estimation, for which you can use the method of maximum likelihood. The random variable $Y$ denotes the number of tagged monkeys that were recaptured, so it is reasonable to assume a hypergeometric distribution for $Y$ in a sampling without replacement model.
The probability mass function of $Y$ is
$$P(Y=y)=\begin{cases}\dfrac{\binom{n}{y}\binom{N-n}{m-y}}{\binom{N}{m}}&,\text{ if }y=\max(0,m-N+n),\ldots,\min(m,n)\\\\ \qquad 0&,\text{ otherwise }\end{cases}$$
Therefore likelihood function of $N$ given the observation $y$ is $$L(N\mid y)=\frac{\binom{n}{y}\binom{N-n}{m-y}}{\binom{N}{m}}\,,\,N\in\{1,2,3,\ldots\}$$
To find the maximum likelihood estimator (MLE) of $N$, we consider the ratio
$$\frac{L(N\mid y)}{L(N-1\mid y)}=\frac{N-m}{N}\cdot\frac{N-n}{N-n-m+y}$$
Now studying when this ratio exceeds $1$, equals $1$ and is less than $1$, we can find out when $L(N\mid y)$ reaches its maximum. MLE of $N$ is the value of $N$ at which the likelihood is maximized.
Finally we arrive at the following conclusion:
If $\frac{mn}{y}$ is not an integer, the MLE is $\hat N(y)=\left\lfloor\frac{mn}{y}\right\rfloor$.
If $\frac{mn}{y}$ is an integer, MLE could be either of $\frac{mn}{y}$ and $\frac{mn}{y}-1$.