\documentclass[a4paper,12pt]{article}
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\begin{document}


{\bf Question}

Explain what a branching Markov chain is, and state clearly the
fundamental theorem for branching chains. Suppose we begin with a
single individual. Let $F_n(s)$ denote the probability generating
function for the number of individuals in generation $n$, and let
$A(s)$ denote the probability generating function for the number
of offspring of any individual. Let $P$ denote the probability of
eventual extinction. Explain why

$$P+\lim_{n\to \infty} F_n(0)$$

Assuming the relationship $F_n(s)=A(F_{n-1}(s))$, prove the
fundamental theorem for branching chains.

${}$

The exquisite Lewis orchid reproduces as follows. Each plant
produces $0,\ 1,\ 2$ or 3 seeds, with equal probability. Then when
Autumn comes the plant dies. In the Spring a seed may be dead, or
it may produce a new plant, with equal probabilities. Each plant
and each seed behaves independently of all the others.

${}$

Find the probability of eventual extinction of this species of
orchid.



\vspace{.25in}

{\bf Answer}

Suppose we have a population of individuals each reproducing
independently of the others, and that the probability
distributions for the number of offspring of all individuals in
generation $n$. Then $X(n)$ is called a branching Markov chain.

The Fundamental Theorem states:

\begin{description}
\item[(i)]
The probabilities of extinction, when the population has size 1
initially, is the smallest positive root of the equation $x=A(x)$,
where $A$ is the p.g.f. of the number of offspring per individual.

\item[(ii)]
Extinction occurs with probability 1 if and only if $\mu \leq 1$,
where $\mu$ is the mean number of offspring per individual.
\end{description}

The probabilities of extinction at or before the nth generation if

$P(X_n=0)=F_n(0)$.

This is a non-decreasing function of $n$. The probability of
eventual extinction is the union of the increasing sequence of
events ($X_n=0$). So $P=\lim_{n\to\infty}F_n(0)$.

Since $F_n(0)=A(F_{n-1}(0))$, taking limits gives

$$P=A(P)$$

so $P$ is a root of $x=A(x)$, satisfying $0 \leq x \leq 1$.

${}$

Let $A(x)=a_0+a_1 x+a_2 x+\cdots$

If $a_0=0$ then each individual has at least one offspring, and
extinction is impossible. $\mu \geq 1$ in this case. If
$a_0+a_1=1$ then each individual has at most one offspring. If
$a_1<1$ then the probability that the population survives to
generation $n$ is $(a_1)^n \to 0$ as $n to\infty$, so extinction
is effectively certain.

${}$

Now $A'(x)=a_1+2a_2 x+3a_3 a^2+\cdots >0$ for $x>0$, for otherwise
all the $a$'s are zero, giving $a_0=1$ and extinction at the first
generation.

${}$

$A''(x)=2a_2+6a_3x+\cdots >0$ for all $x$, for otherwise
$a_0+a_1=1$, which we have already dealt with. So $A(x)$ is convex
and cuts the line $y=x$ in at most 2 places for $x>0$. Now
$A(1)=1$, so one root is $x=1$.

${}$

Let $P$ be the smaller of the two roots. We show by induction that
$F_n(0)<P$ for all $n$.

$F_0(0)=0<P$ since $F_0(s)=s$.

${}$

Suppose then that $F_n(0)<P$.

Then $F_{n+1}(0)=A(F_n(0))<A(P)=P$, since $A$ is strictly
increasing.

So $\lim_{n\to \infty} F_n(0)$ must be the smaller of the two
roots.

${}$

We consider 3 cases. Let $A(x)=x$ have roots $P_1,\ P_2$

\begin{enumerate}
\item
$P_1<P_2=1$ PICTURE \vspace{1in}
\item
$1=P_1<P_2$ PICTURE \vspace{1in}
\item
$P_1=P_2=1$ PICTURE \vspace{1in}
\end{enumerate}

Now $\mu=A'(1)$. In case 1 $\mu>1$. In case 2 $\mu<1$ and in case
3 $\mu=1$.

Extinction happens with probability 1 in cases 2 and 3, i.e. when
$\mu \leq 1$.

${}$

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So the probability distribution for the number of plants produced
from one plant is:

$p_0: \ds \frac{1}{4}+\ds \frac{1}{4}\cdot\ds \frac{1}{2}+\ds
\frac{1}{4}\cdot \ds \frac{1}{4}+\ds \frac{1}{4}\cdot\ds
\frac{1}{8}=\ds \frac{15}{32}$

$p_1: \ds \frac{1}{4}\cdot\ds \frac{1}{2}+\ds \frac{1}{4}\cdot \ds
\frac{1}{2}+\ds \frac{1}{4}\cdot\ds \frac{3}{8}=\ds \frac{11}{32}$

$p_2: \ds \frac{1}{4}\cdot \ds \frac{1}{4}+\ds \frac{1}{4}\cdot\ds
\frac{3}{8}=\ds \frac{5}{32}$

$p_3: \ds \frac{1}{4}\cdot\ds \frac{1}{8}=\ds \frac{1}{32}$

EITHER

$A(x)=\ds \frac{1}{32}(15+11x+5x^2+x^3)$

$x=A(x)$ gives $x^3+5x^2-21x+15=0$

i.e. $(x-1)(x^2+6x-15)=0$

The roots are $x=1,\ -3 \pm 2 \sqrt{6}$. Now $-3+2\sqrt{6}>1$ so
$P=1$.

OR

The mean number of offspring per individual is

$$1 \times \ds \frac{11}{32}+\ds \frac{2\ times 5}{32}+\ds \frac{3
\times 1}{32}=\ds \frac{24}{32}<1$$

so again $P=1$.



\end{document}