Past exam of the options II course of the University of Oxford / 2009 / O / O5

For unrestricted category probabilities, the log-likelihood is ${\sum }_i{n}_i\log {\pi }_i$ subject to ${\sum }_i{\pi }_i=1$. A multiplier gives $n_i/{\pi }_i$ constant, hence the unique maximum, because all counts are positive and the log-likelihood is strictly concave:The likelihood ratio test compares this saturated maximum with the maximum under the parametric model. Twice the log ratio isFor $k$ bins and one regular fitted parameter, its usual large-sample reference distribution is ${\chi }_{k-2}^2$; reject for a sufficiently large value. Under a fully specified parameter ${\theta }_0$, use $\Lambda ({\theta }_0)$ and ${\chi }_{k-1}^2$. This relies on an identifiable interior fit and adequate expected bin counts.

For individual ungrouped abundances, let $S$ be the number of species and $T$ the total number of individuals. The logarithmic-series model givesIts score vanishes whenThe left side increases continuously from one to infinity. One way to see strict increase is to set $t=-\log (1-\theta )>0$, making it $(e^t-1)/t$, whose derivative is positive because $e^t(t-1)+1$ increases from zero. The score has the sign of $T/S$ minus this increasing function, so this is the unique maximum. Bisection on $(0,1)$ is a robust numerical method; Newton's method is another option.

There are two fits implicit in the supplied presentation. Using the ungrouped sufficient statistics gives ${\hat{\theta }}_{\mathrm{full}}\approx 0.944228$. By contrast the displayed ten-bin deviance is minimized near $0.93$, so that is the table-based estimate ${\hat{\theta }}_{\mathrm{grouped}}$. Grouping the upper tail discards its detailed abundance total; minimizing the grouped deviance maximizes the grouped likelihood, not the ungrouped likelihood used in the preceding score equation. The estimates therefore need not agree. The following table-based decisions use its grouped-fit convention.

The displayed table has ten categories, including one pooled upper-tail category. With the parameter fixed, no model parameter is fitted and the appropriate reference degrees of freedom are $10-1=9$. The supplied statistic is $15.59$, below the stated five-percent critical value $16.92$. Hence do not reject the specified log-series model at the five-percent level. Using eight degrees of freedom here would incorrectly charge a fitted parameter to a fully specified null.

For the one-parameter fit to ten grouped categories, the usual reference degrees of freedom are $10-1-1=8$. The table minimum is about $8.29$, far below the supplied critical value $15.51$, so do not reject the fitted log-series model at the five-percent level.

This standard grouped likelihood-ratio calibration uses the grouped maximum, approximately $0.93$, as the preceding answer explains. If the ungrouped estimate from the total abundance is inserted into a grouped statistic instead, the usual automatic loss of exactly one chi-squared degree of freedom is not justified by that same grouped-likelihood argument. The problem's table-based decision should therefore be kept distinct from that alternative procedure.