I saw the following in my notes, but I can’t quite remember why the argument holds... could someone help me please?
Suppose there are N number of fishes in a lake and we want to estimate N. We catch m=1000 fishes, mark them and release them back.
Then we catch n=1000 fishes and count X marked fishes. We know that X~hypergeo(N,m,n) and $Pr(X=x|N)=\frac{{m \choose x }{N-m \choose n-x}}{N \choose n}$.
Let $N’=arg \max_N Pr(X=x|N)$. To find $N’$, we can maximise
$\frac{Pr(X=x|N)}{Pr(X=x|N-1)}=\frac{N(N-n-m)+mn }{N(N-n-m)+Nx }$ $(\star)$
Then the argument said $(\star)$ is maximised when it is equal to 1, so we have $mn=N’x$. I don’t know why $(\star)$ is maximised when it a equal to 1?
The argument as you present it is wrong. $(\star)$ is not maximized when it is equal to $1$; rather, $\mathsf Pr(X=x\mid N)$ is maximized if the ratio $(\star)$ is equal to $1$. This is because $\mathsf Pr(X=x\mid N)$ is unimodal, and as long as $(\star)$ is greater than $1$, $\mathsf Pr(X=x\mid N)$ is increasing, and when $(\star)$ becomes less than $1$, $\mathsf Pr(X=x\mid N)$ starts decreasing.