Combining counts and incidence data: an efficient approach for estimating the log-normal species abundance distribution and diversity indices

Abstract

Obtaining accurate estimates of diversity indices is difficult because the number of species encountered in a sample increases with sampling intensity. We introduce a novel method that requires that the presence of species in a sample to be assessed while the counts of the number of individuals per species are only required for just a small part of the sample. To account for species included as incidence data in the species abundance distribution, we modify the likelihood function of the classical Poisson log-normal distribution. Using simulated community assemblages, we contrast diversity estimates based on a community sample, a subsample randomly extracted from the community sample, and a mixture sample where incidence data are added to a subsample. We show that the mixture sampling approach provides more accurate estimates than the subsample and at little extra cost. Diversity indices estimated from a freshwater zooplankton community sampled using the mixture approach show the same pattern of results as the simulation study. Our method efficiently increases the accuracy of diversity estimates and comprehension of the left tail of the species abundance distribution. We show how to choose the scale of sample size needed for a compromise between information gained, accuracy of the estimates and cost expended when assessing biological diversity. The sample size estimates are obtained from key community characteristics, such as the expected number of species in the community, the expected number of individuals in a sample and the evenness of the community.

Appendix

For the simulation, three parameters are fixed: the expected total number of individuals in a sample E(N), the variance σ², and the number of species s in the community. The parameter μ is deduced from these three parameters as follows: if the observed number of individuals for species i is $N_i \sim \text{Poisson}(e^{X_{i}})$ and X ∼ N(μ, σ²), then:

$$ \begin{aligned} E(N_i)&=E\left(E\left(N_i\mid X\right)\right)=E\left(e^X\right)=e^{\mu+\frac{1}{2}\sigma^2},\\ E(N)&=E\left(\sum_{i} N_i\right)=s e^{\mu+\frac{1}{2}\sigma^2},\\ \ln(E(N))&=\ln s+\mu+\frac{1}{2} \sigma^2, \end{aligned} $$

giving:

$$ \mu=\ln \left(\frac{E(N)}{s}\right)-\frac{1}{2} \sigma^2. $$