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Shannon-wiener Index

The Shannon-Wiener Index (also called the Shannon Index or the Shannon-Weaver Index)

H^{\prime}

is one of several measurements of biodiversity. The advantage of this index is that it takes into account the number of species and the evenness of the species. The index is increased either by having more unique species, or by having a greater #evenness. To compute the index, you need to know the number of animals in each population in the area you are computing the Shannon-Wiener index for.

Definitions

$n_i$ The number of individuals in each species.
$S$ The number of species. Also called Species Richness.
$N$ The total number of all all individuals: $\sum_{i=1}^S n_i$
$p_i$ The proportion of the individual species to the total: $n_i\over N$

Computing the Index

H^\prime = -\sum_{i=1}^S p_i \ln p_i

By applying calculus, it can be shown that for any given number of species, there is a maximum possible

H^\prime

,

H_{max}=\ln S

which occurs when all species are present in equal numbers.

Proof that Maximum Evenness Maximizes the Index

The following will prove that any given population will have a maximum Shannon-Wiener Index if and only if each species represented is composed of the same number of individuals. Expanding the index:

H^\prime = -\sum_{i=1}^S {n_i\over N} \ln {n_i\over N}

N H^\prime = -\sum_{i=1}^S n_i \left ( \ln n_i - \ln N \right ) = -\sum_{i=1}^S n_i \ln n_i + \ln N \sum_{i=1}^S n_i

N H^\prime - N \ln N = -\sum_{i=1}^S n_i \ln n_i

Now, let's define

H_s = -\sum_{i=1}^S n_i \ln n_i

Clearly, since

N

is a positive constant for a given population size, and

N\ln N

is also a constant, then maximizing

H_s

is equivalent to maximizing

H^\prime

.

Strategy

Let's split an arbitrarily sized population into two groups, with each group receiving an arbitrary number of individuals and an arbitrary number of species. Now, within each group, each species has the same number of individuals as any other species in that group, but the number of individuals per species in the first group may be different than the number of individuals per species in the second group. Now, if it can be proven that

H_s

is maximized when the number of individuals per species in the first group matches the number of individuals per species in the second group, then it has been proved that the population has a maximum index only when each species in the population is evenly represented.

H_s

doesn't depend on the total population. So

H_s

may built by simply adding the indices of two sub-populations. Since the population size is arbitrary, this proves that if you have two species (the smallest number that can be considered two groups), their index is maximized if they are present in equal numbers. So the rules of mathematical induction have been satisfied.

Actual Proof

Now, divide the species into two groups. Within each group, the population is evenly distributed among the species present.

$k$ The number of individuals in the second group.
$p$ The number of species in the second group.
$n_{i2} = k/p$ Number of individuals in each species in the second group.
$N-k$ The number of individuals in the first group.
$S-p$ The species in the first group.
$n_{i1} = {N-k \over S-p}$ The individuals in the first group.

H_s = -\sum_{i=1}^{S-p} {N-k \over S-p} \ln {N-k \over S-p}

     - \sum_{i=1}^p {k\over p} \ln {k \over p}     = -\left ( N-k \right ) \ln  {N-k \over S-p}       - k \ln {k\over p}

To find out which value of

k

will maximize

H_s

, we must find the value of

k

which satisfies the equation:

{d\over dk}\, H_s=0

Differentiating...

\ln { N-k \over S-p} + (N-k){1 \over N-k} - \ln {k\over p} - k {1 \over k} = 0

\ln {N-k\over S-p} = \ln {k \over p}

Exponentiating:

{N-k\over S-p} = {k \over p}

So

= {pN \over S}

Now by applying the definitions of

N_{i1}

and

N_{i2}

, we get

N_{i1} = N_{i2} = {N\over S}

.

Result

Now we have accomplished the proof that the Shannon-Wiener index is maximized when each species is present in equal numbers (see #strategy). But what is the index is that case? Well,

n_i = {N\over S}

, so

p_i = {1\over S}

Therefore:

H_{max} = - \sum_{i=1}^S {1\over S} \ln {1\over S} = \ln S

Evenness

Evenness is how equal the populations are numerically. So if there are 40 foxes, and 1000 dogs, the population isn't very even. But if there are 40 foxes and 42 dogs, the population is quite even. The evenness of a population can be represented by

E={ H^\prime \over H_{max} }

which is constrained between 0 and 1. The less variation in populations between the species, the higher

E

is.

See Also

*Information theory

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