<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\df}{\ds\frac}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}


{\bf Question}

Explain what a branching Markov chain is, and state clearly the
Fundamental Theorem for branching chains.

Amoeba reproduce by dividing into two. Some amoeba will die
without dividing. Assume that amoeba reproduce identically and
independently and that an amoeba dies with probability $p$. Show
that if $p \geq \df{1}{2}$ then eventual extinction is effectively
certain, irrespective of the initial population size. Calculate
the probability of eventual extinction for $p &lt; \df{1}{2}$, for a
population of initial size $n$.

A related organism can reproduce by multiple division, giving $k$
offspring with probability $a_k$. The mean number of offspring per
organism is $m&gt;1$. Explain how you would calculate the probability
$P$ of eventual extinction if the equation of which $P$ is a root
cannot be solved algebraically. Explain in outline how your
approach can be justified.



\vspace{.25in}

{\bf Answer}

Suppose we have a population of individuals, each reproducing
independently of the others. Suppose the distributions of the
number of offspring of all individuals are identical.

Let $X_n$ be the number of individuals in the nth generation; then
$(X_n)$ is a branching Markov chain.

${}$

The Fundamental Theorem.

\begin{description}
\item[(i)]
The probability 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}

An amoeba produces no offspring with probability $p$ and 2
offspring with probability $(1-p)$.

The p.g.f is $A(s)=p+(1-p)s^2$

$\mu=2(1-p) \leq 1$ for $p \geq \df{1}{2}$

$x=A(x)$ gives $\begin{array}{rcl} (1-p)x^2-2+p &amp; = &amp; 0\\
(x-1)((1-p)x-p) &amp; = &amp; 0\end{array}$

so $x=\df{p}{1-p}$ is the smaller root for $p&lt;\df{1}{2}$.

By independence, if the population has initial size $n$, the
probability of extinction is

$$\left(\df{p}{1-p}\right)^n$$

Now let the p.g.f be $A(x)$. To find the probability of extinction
we have to solve $x=A(x)$. In proving the fundamental theorem it
is shown that $A(x)$ is convex downwards, and that it has at most
two roots in $[0,1]$, one of them being $x=1$.

The other root in this case is positive, for $x=0$ is not a root
as is easily seen. Since $\mu&gt;1$, the root is $&lt;1$, and so the
graph of $A(x)$ will be as follows

PICTURE \vspace{2in}

An iterative approach, with $0&lt;x_0&lt;1$, using $x_n=A(x_{n-1})$ will
therefore converge to $P$. In general it will be a 1st order
process.



\end{document}
</pre></body></html>