Lecture 10

DispersionDispersionandandthetheNegativeNegativeBinomialBinomialDistributionDistribution

DispersionDispersion

Remark

There are three ways in which objects (events, individuals,...) can be dispersed in space and/ or time: regularly, randomly, and contaguously. In Ecology these three types of dispersion are also known as uniform, random and aggregated dispersion, respectively. We will illustrate the three types of dispersion with pictures in two dimensional space. The dots represent the objects.

The first picture shows regular dispersion.

The second picture shows random dispersion.
The third picture shows contiguous dispersion.
Note that new pictures will be generated if you click the reload button, giving you an unlimites supply of examples.

Remark

The counts of objects in a given area are modeled with probability distributions.

Regularly dispersed objects have less variation in the counts compared with randomly dispersed objects. Regular dispersion counts are modeled with the binomial distribution. This is because for the binomial distribution, the coefficient of dispersion is less than 1:$$CD=\frac{\sigma^2}{\mu}=\frac{np(1-p)}{np}=1-p<1$$On the other hand, the coefficient of dispersion for randomly dispersed objects is close to 1. The Poisson distribution is used to model the counts of randomly dispersed objects since its coefficient of dispersion is 1:$$CD=\frac{\sigma^2}{\mu}=\frac{\mu}{\mu}=1$$To model counts of contiguously dispersed objects, we need a distribution with a coefficient of dispersion greater than 1. The Negative Binomial distribution which we will discuss next is used for this purpose.

NegatiaveNegatiaveBinomialBinomialDistributionDistribution

Formula

In a series of independent binary trials the negative binomial formula returns the probability that there will be $x$ failures before the $r^{th}$ success. Let $p$ be the probability for success. Then
$$P(X=x)=_{x+r-1}C_{r-1} \cdot p^r\cdot (1-p)^{x} $$The mean and variance of the negative binomial distribution are$$ \mu=\frac{(1-p) r}{p} \; ; \quad \sigma^2=\frac{(1-p) r}{p^2} $$

Note

Observe that$$CD=\frac{\sigma^2}{\mu}=\frac{1}{p}>1$$The negative binomial distribution is used to model the counts of contiguously dispersed objects.

Remark

To use the negative binomial distribution to model the counts of contiguously dispersed objects, we need to estimate $\mu$ and $\sigma^2$ from sample data. The formulas are as follows:$\mu \sim \bar{x}\quad;\quad \sigma^2 \sim s^2$.$$ \begin{aligned} \bar{x}&=\frac{1}{n} \sum f_i\, x_i \\ &\\ s^2&=\frac{1}{n-1} \sum_i f_i\, \left(x_i-\bar{x}\right)^2 \end{aligned}$$The formulas for $p$ and $r$ are$$ p=\frac{\mu}{\sigma^2} \sim \frac{\bar{x}}{s^2}\quad;\quad r=\frac{\mu^2}{\sigma^2-\mu} \sim \frac{\bar{x}^2}{s^2-\bar{x}} $$

Example

The number of aquatic invertebrates on the lake floor in $400$ quadrats is given in the table below.

$$ \begin{array}{l|{6*c}} \text { # of invertebrates } & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \text { frequency } & 213 & 128 & 37 & 18 & 3 & 1 \end{array} $$

$$\mu \sim \bar{x}=\frac{\sum f_i x_i}{n}=\frac{1}{400}[(213)(0)+\dots+(5)(1)]=0.6825 $$$$\begin{aligned} \sigma^2 \sim s^2 =\frac{1}{(n-1)} \sum f_i\left(x_i-\bar{x}\right)^2&=\frac{1}{399}\left[213(0-0.6825)^2+\cdots+1(5.0 .6825)^2\right] \\ & =0.8137 \end{aligned}$$$$CD= \frac{\sigma^2}{\mu} \approx \frac{0.8137}{0.6825}=1.192>1 \quad \Rightarrow \quad$$ The counts are contiguously dispersed. The Negative Binomial distribution might be a good fit. Let us estimate $p$ and $r$ from sample data:$$ p \approx \frac{\bar{x}}{s^2}=0.8388 \quad;\quad r \approx \frac{\overline{x^2}}{s^2-\bar{x}}=3.55$$We are ready to use the model to make predictions. However the negative binomial formula cannot be evaluated on common calculators. We will use Excel:$$ \begin{array}{l|{6*c}} x & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline p(x) & 0.59 & 0.29 & 0.09 & 0.02 & 0.006 & 0.001 \end{array} $$E.g. the probability of finding 2 invertebrates in a random quadrat is $0.09$.

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Lecture 9Lecture 11