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.

Minimizing the sum of squared residuals gives the normal equations for linear least squares, $X^{T}X\hat{\theta }=X^{T}Y$. Full column rank makes $X^{T}X$ positive definite and invertible, so the unique minimizer isBecause $\mathbb{E}Y=X\theta$ and $\mathrm{Cov}(Y)={\sigma }^2{I}_n$, one has $AX=I_p$ andFor the alternative linear estimator the corresponding general formulas areNo normality assumption on the errors is needed for these claims.