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Following on from the basic branching processes introduction, we now calculate the expected number of individuals at the nth generation.
As in the above example, let Z_n be the number of individuals in generation n, and X be a random variable describing the number of offspring an individual has, with E[X]=\mu and Var[X]=\sigma^2
\begin{align*} \therefore E[Z_n] & = G'_n(1) \\ &= G'_{n-1} \Big (G(1) \Big)G'(1) \\ &= G'_{n-1}(1)G'(1) \\ &= E[Z_{n-1}] \mu \\ &= \mu^2 E[Z_{n-2}] =... \\ &= \mu^n E[Z_0] \\ &= \mu^n \end{align*}
Clearly the eventual population size is highly dependent on the value of \mu
So if each individual is expected to have more than one offspring, then the population will increase. If each individual is expected to have either one or no offspring, then the population will remain constant or decrease and eventually die out.
Question:
Why do elephants not die out, if the above comment on mean family size holds? What are the limitations of our model in representing the reproductive lifespan of an elephant?
By evaluating the mean, we see that ultimate extinction is certain only when the mean family size is \mu \leq 1.
To find this probability exactly, we let the probability of extinction at the nth generation be \theta_n=P(Z_n=0) . So the probability of ultimate extinction is \theta = lim_{n \to \infty} \theta_n=lim_{n \to \infty} P(X_n=0) .
\begin{align*} \theta_n & = G_n(0) \\ & = G_{n-1} \Big(G(0)\Big) \\ &= G\Big(G\big(...(s)...\big)\Big) \\ &= G\Big(G_{n-1}(s)\Big) \\ &= G(\theta_{n-1}) \end{align*}
So as n \rightarrow \infty , we have \theta_n\rightarrow \theta and G(\theta_{n-1}) \rightarrow G(\theta) . And so we can find \theta by solving \theta=G(\theta)
Now there may be other roots to this equation, so we show \theta is the smallest by supposing \alpha is also a root. Then \theta_1=G(0) \leq G(\alpha)=\alpha \Rightarrow \theta_2=G(\theta_1) \leq G(\alpha)=\alpha
And so proceeding by induction, \theta =lim_{n \to \infty} \theta_n \leq \alpha , which shows \theta is indeed the smallest non-negative root.
The dependence of \theta on the value of the mean family size, is shown in the diagrams below. The first case being \mu \leq 1 and the second case \mu > 1 .
Example:
In the previous elephant example, we now solve for \theta in the equation \begin{align*} \theta & = G(\theta) \\ &= (1-p^n)+p^n \theta\\ \theta(1-p^n) &= (1-p^n) \\ \therefore \theta &=1 \end{align*}
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