O5 (source code)
= O5
{c}
{scope}
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 $\chi^2_{k-2}$ 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 $\chi^2_{k-1}$. Reject for a value above the chosen upper <quantile>.
For the ungrouped <logarithmic distribution>[logarithmic-series] observations write $n=1188$ and $S=\sum_rX_r=6968$. The log <likelihood> is
$$
\ell(\theta)=-n\log[-\log(1-\theta)]+S\log\theta-\sum_r\log X_r,\qquad0<\theta<1.
$$
Setting its <derivative> to zero gives
$$
\boxed{\frac{-\hat\theta_{\rm full}}{(1-\hat\theta_{\rm full})\log(1-\hat\theta_{\rm full})}=\frac{6968}{1188}.}
$$
Bisection of this monotone mean equation, or <Newton root-finding iteration>[Newton's method] with a valid bracket, obtains $\hat\theta_{\rm full}\approx0.9442280$.
The supplied table instead uses ten grouped cells, with the last probability $\pi_{\ge10}(\theta)=1-\sum_{i=1}^9\pi_i(\theta)$. Its <likelihood> is determined by the cell counts and is not the ungrouped <likelihood> just differentiated. From its listed values, the <local minimum>[minimum] deviance occurs near $\boxed{\hat\theta_{\rm grouped}\approx0.93}$; 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.
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