Past exam of the options I course of the University of Oxford / 2009 / O / O2

With immediate release and ideal independent uniform sampling of fish, each catch is a Bernoulli trial with tagged-success probability $p=k/N$. Stopping at the $r$th success makes the count of untagged catches a negative binomial random variable. A total of $x$ failures requires $r-1$ successes among the first $x+r-1$ trials and a success on the last trial, so the probability is $\left(\genfrac{}{}{0ex}{}{x+r-1}{r-1}\right)p^r(1-p{)}^x$. Separate days use disjoint independent sequences of trials, hence their counts form an independent and identically distributed sample. Equal catch probabilities alone do not mathematically imply independence; this is the usual repeated independent sampling assumption in the stated model.

Let $S={\sum }_i{x}_i$ and $\bar{x}=S/n$. Terms independent of $p$ can be dropped from the log-likelihood, leavingFor $S>0$, strict concavity gives the unique interior maximum likelihood estimatorIf $S=0$, the likelihood increases up to $p=1$: this is the boundary estimate on the closed parameter range, or a supremum if the model insists on $p<1$.

Using the given mean, the sample Fisher information isThe information per observation is $I_1(p)=r/[p^2(1-p)]$.

For a fixed true parameter $0<p<1$, the usual asymptotic normality of maximum likelihood estimators givesEquivalently, for a large sample, $\hat{p}$ is approximately normally distributed with mean $p$ and variance $p^2(1-p)/(nr)$. Replacing the unknown variance by its consistent plug-in estimate gives the Wald confidence intervalHere $z_{1-\alpha /2}$ is a standard-normal quantile. Intersect with the physical parameter range $(0,1)$ if needed. This is an interior, large-sample approximation; the degenerate estimate $\hat{p}=1$ from $S=0$ does not justify a zero-width finite-sample interval.

The transformation $N=k/p$ is strictly decreasing for $p>0$. Therefore an interval $[L,U]$ with $0<L<U\le 1$ transforms toThe event $L\le p\le U$ is exactly the event $k/U\le N\le k/L$, so the approximate coverage is inherited. If the lower endpoint for $p$ reaches zero, the transformed upper endpoint for $N$ is $+\infty$; if the upper endpoint reaches one, the lower endpoint is $k$. Finally intersect with integer values $N>k$, rounding the lower bound upward and the upper bound downward. This respects the original discrete population constraint.