משתמש:צ'כלברה/ארגז חול

בהוכחה הבאה נוכיח שלכל חברה נתונה יהיה מדד שנון-ויבר מקסימלי אך ורק אם חלקו היחסי של כל מין בחברה (קרי מספר הפרטים של כל מין) יהיה שווה:

Expanding the index:

$H^{\prime }=-\sum _{i=1}^{S}{n_{i} \over N}\ln {n_{i} \over N}$

$NH^{\prime }=-\sum _{i=1}^{S}n_{i}\left(\ln n_{i}-\ln N\right)=-\sum _{i=1}^{S}n_{i}\ln n_{i}+\ln N\sum _{i=1}^{S}n_{i}$

$NH^{\prime }-N\ln N=-\sum _{i=1}^{S}n_{i}\ln n_{i}$

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 from 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.

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 each species 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:

${N-k \over S-p}={k \over p}={pN \over S}.$

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:

$H_{\max }=-\sum _{i=1}^{S}{1 \over S}\ln {1 \over S}=\ln S.$