Past exam of the options I course of the University of Oxford / 2009 / O (Statistics)

By continuity of the cumulative distribution function and nonnegativity, $F(0)=0$; strict increase and $F(x)\to 1$ allow inversion for every $0<t<1$. The probability integral transform follows fromThus $F(X)$ is uniformly distributed on $(0,1)$.

The order statistics arrange the sample increasingly: $X_{(1)}\le \cdots \le X_{(n)}$. Strict monotonicity means $F(X_{(i)})$ is the $i$th order statistic of $n$ independent uniform variables. Its density isobtained by choosing the $i-1$ observations below $t$ and the $n-i$ above it. This is the beta distribution with parameters $i,n+1-i$, soFor the given log-logistic distribution, rearrangement givesConsequently plot $y_i=\log (i/(n+1-i))$ vertically against $x_i^{\ast }=\log X_{(i)}$ horizontally. Under a good fit, replacing $F(X_{(i)})$ by its plotting position $i/(n+1)$ gives an approximate straight line withThis is a diagnostic probability plot, rather than an assertion that the expectation of the nonlinear transformed order statistic equals that transformed plotting position. Positive observations occur almost surely under the model.

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.