# Simpson's index

Simpson’s index is a diversity index which is the summation of the square of the probabilities of the species in a community. The value varies between 0-1.

here, pi=n/N where ni=number of individual of species i in the community and N=total no. of individuals.

S=total number of species.

It is the measure that equals the probability that two individuals taken at random from a community belongs to the same species.The probability decreases if the species richness and evenness increase. It is strongly influenced by the most abundant species.

From:

What is meant by 'It is strongly influenced by the most abundant species'?

• I'm assuming that term is meant as a critique/limitation or at least a point to note when taking the index as a measure of diversity. They just mean that the value for D in that equation is affected strongly by how abundant the most abundant species is. So, if you have one species that is very abundant, you might get a high value for D (=low diversity, strangely) even if there are hundreds of other, less abundant species, because adding 1 more species with low abundance does little to the measure. – Bryan Krause Jan 24 '17 at 20:37

I don't know whether this math will be clarifying forr you, but it might help you understand why "It is strongly influenced by the most abundant species." It is all about finding the best p's that does the minimization for $D$. You can skip to conclusion if you feel like skipping all the math.

You want to find the critical point (where the function attains its minimum and maximum) of:

$$D = p_{1}^2+p_{2}^2+...+p_{n}^2$$

given ${p_{1}+p_{2}+...+p_{n}=1}$, for $p_{i}$ of species.

Lagrange Multiplier method is one of the techniques to find the values given to $p_{1}, p_{2}, ...p_{n}$ to make $p_{1}^2+p_{2}^2+...+p_{n}^2$ go to maximum or minimum given some constraint on this.

Assume:

$$f = p_{1}^2+p_{2}^2+...+p_{n}^2$$ $$g = p_{1}+p_{2}+...p_{n}$$

According to Lagrange multiplier:

$$\partial_{p_{1},..p_{n}} f = \lambda \partial_{p_{1},..p_{n}} g$$

We have to find the value of $\lambda$.

So after partial fractions step, we will end up with these equations:

$$np_{1} = \lambda ; np_{2} = \lambda; ... np_{n} = \lambda$$

and we already know that $p_{1}+p_{2}+...+p_{n} = 1$

From above, we know that:

$$p_{1} = p_{2} = ... = p_{n} = \lambda/n$$ we get $\lambda = 2/n$ and from this the minimum values of the function:

$$f = D = p_{1}^2+p_{2}^2+...+p_{n}^2$$

is when all $p_{i} = 1/n$ for i is denoting each species in ${1,...,n}$

Conclusion: So $D$ value attains its minimum when all terms $p_{i}$ are equal. As any value goes away from $1/n$ ('n' is the total number of individuals of species you got)the value increases drastically as it is being squared in $D$.

Sample example will be like $0.5^2 + 0.5^2 = 0.25$ but a change from this say $0.6,0.4$ increases $0.5$ term by $0.11$ but decreases $0.4$ term by only $0.09$.

Hope this helps.