Biology · Ecology

Simpson Index (Dominance and Diversity)

The Simpson index D measures the dominance in a community: the probability that two randomly drawn individuals belong to the same species. Its complement (1 − D) is the Simpson diversity index and rises with diversity.

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Formula

LaTeX: D = \sum_{i=1}^{S} p_i^{\,2}
D, 1 − D and pᵢ are dimensionless; pᵢ = nᵢ/N is a proportion (0…1)

Variables & units – Simpson Index (Dominance and Diversity)

SymbolMeaningUnit
DSimpson dominance index (probability of the same species)– (0…1)
pᵢrelative frequency of species i, that is nᵢ/N– (0…1)
nᵢnumber of individuals of species icount
Ntotal number of all individualscount
Snumber of species (species richness)count

Derivation & background – Simpson Index (Dominance and Diversity)

Edward H. Simpson introduced the measure in 1949. A large D (near 1) means that few species dominate, that is low diversity. Therefore one usually looks at the Simpson diversity index 1 − D (Gini-Simpson) or the reciprocal Simpson index 1/D, both of which increase with diversity. For finite samples one uses the unbiased form D = Σ nᵢ(nᵢ−1)/[N(N−1)]. Compared with the Shannon index, Simpson weights common species more strongly and reacts less sensitively to rare species.

Exam blueprint

Validity range

Applies to diversity and dominance comparisons of communities; the finite form D = Σ nᵢ(nᵢ−1)/[N(N−1)] is more accurate for small samples.

Derivation steps

The probability of drawing the same species twice is the sum of the squared species proportions.

  1. 1Compute the relative frequencies pᵢ = nᵢ/N of each species.
  2. 2Square the proportions and sum them to D = Σ pᵢ²; the diversity is 1 − D.

Rearrangements

Simpson diversity index

Increases with diversity; 0 for one species, near 1 for many equally frequent species.

Reciprocal Simpson index

Gives the effective number of species; equals exactly S under even distribution.

Unbiased form for samples

Draws without replacement; more accurate than the proportion form for small N.

Task variant

Three species with 10, 6, 4 individuals (N = 20). Compute D, 1 − D and 1/D.

p = 0.5; 0.3; 0.2. D = 0.25 + 0.09 + 0.04 = 0.38. Diversity 1 − D = 0.62, reciprocal 1/0.38 = 2.63.

Compare community A (50, 50) with B (90, 10), each N = 100. Which is more diverse?

A: D = 0.5² + 0.5² = 0.50, so 1 − D = 0.50. B: D = 0.9² + 0.1² = 0.82, so 1 − D = 0.18. A is clearly more diverse, B strongly dominated.

Common mistakes

Confusing D (dominance) with 1 − D (diversity).

Large D means high dominance and low diversity; the diversity is 1 − D.

Squaring counts instead of proportions.

In D = Σ pᵢ² the proportions pᵢ = nᵢ/N enter, not the raw numbers.

Mixing the proportion form and the finite form in the same task.

You commit to one form; the finite form uses nᵢ and N directly.

Treating Simpson and Shannon as identical.

Simpson weights common species more strongly, Shannon is more sensitive to rare species.

Exam context

  • Typical in ecology: comparing the diversity of two sites via 1 − D and contrasting it with the Shannon index.

These mistakes cost points in real exams. The set drills them until they stick.

Worked example

Community with 3 species, individuals 10, 6, 4 → N = 20, so p = 0.5; 0.3; 0.2. D = 0.5² + 0.3² + 0.2² = 0.25 + 0.09 + 0.04 = 0.38. The diversity index is 1 − D = 0.62, the reciprocal Simpson index 1/D = 1/0.38 = 2.63. The finite form gives D = (10·9 + 6·5 + 4·3)/(20·19) = 132/380 = 0.347.

Applications

Ecology (diversity comparison, dominance structures), environmental monitoring, nature conservation, comparing disturbed and undisturbed habitats

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Which formula describes Simpson Index (Dominance and Diversity)?

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How do you rearrange D = Σ pᵢ² for Simpson diversity index?

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Which common mistake happens with Simpson Index (Dominance and Diversity)?

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Scientific sources

Common notations & search queries

D = sum p_i^2D=Σpi²Simpson-IndexSimpson Diversitätsindexsimpson index1-D Simpsonreziproker Simpson-IndexDominanzindex ÖkologieBiodiversität Simpson

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Frequently asked questions about Simpson Index (Dominance and Diversity)

How do you calculate the Simpson index?+

You determine the relative frequency pᵢ of each species, that is the species count divided by the total. Then you square each proportion and sum the squares to the dominance index D = Σ pᵢ². Example: three species with 10, 6 and 4 individuals give the proportions 0.5, 0.3 and 0.2 at N = 20, so D = 0.25 + 0.09 + 0.04 = 0.38. D is the probability of randomly drawing the same species twice. As a diversity measure one usually uses 1 − D = 0.62 or the reciprocal index 1/D = 2.63. For small samples one uses the more accurate finite form D = Σ nᵢ(nᵢ−1)/[N(N−1)], which draws without replacement. For the same data it gives D = 132/380 = 0.347.

What is the difference between D and 1 − D?+

The pure Simpson index D = Σ pᵢ² measures dominance: the probability that two randomly drawn individuals belong to the same species. A large D means few species dominate the community, that is low diversity. This is unintuitive, because a high value stands for low diversity. Therefore one usually forms the Simpson diversity index 1 − D, also called the Gini-Simpson index. It is the probability of drawing two different species and increases with diversity: 0 for one species only, near 1 for many equally frequent species. Example: D = 0.38 means 1 − D = 0.62. A common mistake is to confuse D and 1 − D and thus reverse the statement. You should always state which form is meant.

What is the reciprocal Simpson index?+

The reciprocal Simpson index is the reciprocal of the dominance, that is 1/D. Its great advantage is the intuitive interpretation: it gives the effective number of species, that is the number of equally frequent species that would produce the same dominance. Example: at D = 0.38, 1/D = 2.63, so the community behaves like about 2.6 equally frequent species. If S equally frequent species occurred, 1/D would equal exactly S, because then each pᵢ = 1/S and D = S·(1/S)² = 1/S. The reciprocal index therefore always lies between 1 and the actual number of species. It is especially useful for comparing diversity across sites of different richness, because its unit "effective species" is easy to understand and allows linear comparisons.

When do you use the finite form of the Simpson index?+

The finite or unbiased form D = Σ nᵢ(nᵢ−1)/[N(N−1)] is used when you draw a sample from a population and the individual counts are small. It models drawing without replacement: once you have drawn an individual of a species, one fewer of that species remains, which the factor (nᵢ − 1) and the denominator N(N − 1) account for. For large individual counts this form approaches the simple proportion form D = Σ pᵢ², because the difference between nᵢ and nᵢ − 1 then becomes negligible. For school and rough comparisons the proportion form usually suffices. Scientifically, especially with small samples, the finite form is more correct. It is important to always use the same form within a comparison so that the values stay consistent.

Why is the Simpson index less sensitive to rare species?+

The Simpson index squares the species proportions. As a result, common species with large pᵢ contribute disproportionately strongly, while rare species with small pᵢ almost vanish after squaring. A species with proportion 0.5 contributes 0.25, a rare species with proportion 0.02 only 0.0004, that is practically nothing. Therefore few dominant species determine the value, and the addition or loss of rare species barely changes D. The Shannon index, by contrast, weights rare species much more strongly via the logarithm and reacts more sensitively to them. Which measure fits better depends on the question: if the dominance structure and common species matter, Simpson is suitable; if rare species are ecologically important, for example in conservation, the Shannon index or a comparison of both measures is more sensible.

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How do you calculate with Simpson Index (Dominance and Diversity)?

Here is how to work through a typical Simpson Index (Dominance and Diversity) (D = Σ pᵢ²) task step by step:

  1. 1

    Task

    Three species with 10, 6, 4 individuals (N = 20). Compute D, 1 − D and 1/D.

    Solution path

    p = 0.5; 0.3; 0.2. D = 0.25 + 0.09 + 0.04 = 0.38. Diversity 1 − D = 0.62, reciprocal 1/0.38 = 2.63.

  2. 2

    Task

    Compare community A (50, 50) with B (90, 10), each N = 100. Which is more diverse?

    Solution path

    A: D = 0.5² + 0.5² = 0.50, so 1 − D = 0.50. B: D = 0.9² + 0.1² = 0.82, so 1 − D = 0.18. A is clearly more diverse, B strongly dominated.