 O5

The multinomial likelihood is a parameter-independent factor times ${\prod }_i{\pi }_i^{n_i}$. Maximize its logarithm subject to $\sum {\pi }_i=1$: the multiplier equations give $n_i/{\pi }_i=c$, and summation makes $c=n$. Thus ${\hat{\pi }}_i=n_i/n$. Positivity of all counts makes this the unique interior maximum by strict concavity.

The likelihood ratio compares the fitted null likelihood with this saturated likelihood. Taking minus twice its logarithm gives $\Lambda (\hat{\theta })=2{\sum }_i{n}_i\log [n_i/(n{\pi }_i(\hat{\theta }))]$. Under regularity and large expected counts, compare it to ${\chi }_{k-2}^2$ for one fitted parameter: the saturated probabilities have $k-1$ free coordinates and the null has one. If the parameter is fixed at ${\theta }_0$, use $\Lambda ({\theta }_0)$ and ${\chi }_{k-1}^2$. Reject for a value above the chosen upper quantile.

For the ungrouped logarithmic-series observations write $n=1188$ and $S={\sum }_r{X}_r=6968$. The log likelihood isSetting its derivative to zero givesBisection of this monotone mean equation, or Newton's method with a valid bracket, obtains ${\hat{\theta }}_{\mathrm{full}}\approx 0.9442280$.

The supplied table instead uses ten grouped cells, with the last probability ${\pi }_{\ge 10}(\theta )=1-{\sum }_{i=1}^9{\pi }_i(\theta )$. Its likelihood is determined by the cell counts and is not the ungrouped likelihood just differentiated. From its listed values, the minimum deviance occurs near $\boxed{{\hat{\theta }}_{\mathrm{grouped}}\approx 0.93}$; direct minimization of this grouped deviance gives about $0.9333714$. The supplied two forms of the estimate therefore concern different likelihoods. They must not be identified: the full-data score equation gives about $0.94423$, while the grouped table points near $0.93$. The following requested tests use the table's grouped likelihood, with ten cells. The source values and totals are preserved unchanged in TeX.