\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}


{\bf Question}

Explain what a branching Markov chain is. Suppose such a Markov
chain begins with single individual. Let $A(s)$ denote the
probability generating function for the number of offspring of any
individual. State how to use $A(s)$ to find the probability of
extinction. Prove that extinction occurs with probability 1 if and
only if the mean number of offspring per individual does not
exceed 1.

${}$

Let $F_n(s)$ denote the probability generating function for the
number of individuals in generation $n$. Assuming the relationship
$F_n(s)=F_{n-1}(s)(A(s))$, obtain expressions for the mean and
variance of the number of individuals in generation $n$, in terms
of the mean and variance of the number of offspring of any
individual.

${}$

An organism reproduces by multiple division. The number of
offspring of any individual has a Poisson distribution with
parameter $\lambda$. For what values of $\lambda$ is extinction
certain? Write down expressions for the mean and variance of the
number of individuals in generation $n$. Estimate the probability
of extinction correct to one decimal place when $\lambda=1.5$.



\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 are
identical.  Let $X_n$ denote the number of inderivdials in
generation n.  Then $(X_n)$ is called a branching Markov chain.

The probability of extiniction when the probability has size 1 is
given by the smallest positive root of $s=A(s)$.  Since $A(s)$ is
a power series with positive coefficients, it is concave upwards,
and so its graph meets $y=s$ at most twice, for $s>0$.  Since
$A(1)=1$ for a p.g.f. this gives 3 possibilities

\begin{description}
\item[(i)] PICTURE
\item[(ii)] PICTURE
\item[(iii)] PICTURE
\end{description}
So extinction happens with probability 1 if and only if $\mu \leq
1$

\begin{eqnarray*} F_n(s) & = & F_{n-1}(A(s)) \\ {\rm So\ \ }
F_n'(s) & = & F_{n-1}'(A(s))A'(s) \\ {\rm Putting\ s=1\ gives} \\
F_n'(1) & = & F_{n-1}'(1)A'(1) {\rm \ \ since\ A(1) = 1} \\ \mu_n
& = & \mu_{n-1}\cdot \mu \end{eqnarray*}

Since $\ds \mu_0 = 1$ we have $\mu_n = \mu^n$

Differentiating again gives $$F_n''(s) = F_n''(1)(A(s))(A'(s)) +
F_{n-1}'(1)A''(s)$$

Putting s=1 gives $$F_n''(1) =
f_{n-1}''(1)A'(1)^2+F_{n-1}'(1)A''(1)$$  \begin{eqnarray*}
\sigma_n^2 + \mu_n^2 - \mu_n & = & \mu^2(\sigma_{n-1}^2 +
\mu_{n-1}^2 - \mu_{n-1}) + \mu_{n-1}(\sigma^2 + \mu^2 - \mu) \\
\sigma_n^2 + \mu^{2n} = \mu^n & = & \mu^2(\sigma_{n-1}^2 +
\mu^{2n-2}-\nu^{n-1})+\mu^{n-1}(\sigma^2+\mu^2 - \mu) \\
\sigma_n^2 & = & \mu^2\sigma_{n-1}^2 + \mu^{n-1}\sigma^2 \\ & = &
\mu^2(\mu^2\sigma_{n-2}^2 + \mu^{n-2}\sigma^2)+\mu^{n-1}\sigma^2
\\ & = & \mu^4\sigma_{n-2}^2+\sigma(\mu^{n-1}+\mu^n) \\ & = &
\mu^6\sigma_{n-3}^2 + \sigma^2(\mu^{n-1} + \mu^n + \mu^{n+1} = ...
\\ & = & \mu^{2n-2}\sigma_1^2 + \sigma^2(\mu^{n-1} + ... +
\mu^{2n-3}) \\ & = & \sigma^2(\mu^{n-1} + \mu^n + .. + \mu^{2n-2}
\\ & = & \left\{\begin{array}{ll} \sigma^2\mu^{n-1}
\left(\frac{1-\mu^n}{1-\mu}\right) & \mu\not=0 \\ n\sigma^2 &
\mu=1 \end{array} \right. \end{eqnarray*}

The mean and variance of the Poisson distribution are both
$\lambda$

so $\ds \mu_n = \lambda^n$ and $\sigma_n^2 = \left\{
\begin{array}{ll} \lambda^n\left(\frac{1 - \lambda^n}{1 - \lambda}
\right) & \lambda \not=0 \\ n  & \lambda =1 \end{array} \right.$

The p.g.f. for the poisson $(\lambda)$ is $e^{\lambda(s-1)}$.  So
we have to estimate the solution for $e^{1.5(s-1)}=s$
\begin{eqnarray*} s & & e^{1.5(s-1)} \\ {\rm Inetial\ guess\ \ }
0.5 & > & 0.47... \\ 0.4 & <& 0.41... \\ 0.45 & > & 0.44...
\end{eqnarray*} Thus the extinction probability is 0.4 to 1 d.p.



\end{document}