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\begin{document}


{\bf Question}

Explain what a branching Markov chain is. Suppose a population is
descended from a single individual (generation 0). Let $A(s)$ be
the probability generating function for the number of offspring of
any individual. Let $X_n$ be the number of individuals in general
$n$, with probability generating function $F_n(s)$.


Prove that $F_n(s)=F_{n-1}(A(s))$ and deduce that
$F_n(s)=A(F_{n-1}(s))$.

Suppose that the probability distribution of the number $Z$ of
offspring of any individual is given by

$$P(Z=k)=qp^k\ \rm{for}\ k=0,\ 1,\ 2, \cdots$$

where $0<p<1,\ q=1-p$ and $p \ne q$. Obtain the probability
generating function $A(s)$ in this case, and verify that for
$n=1,\ 2,\ \cdots\ ,$

$$F_n(s)=\df{q(p^n-q^n-(p^{n-1}-q^{n-1})ps)}{p^{n+1}-q^{n+1}-(p^n-q^n)ps}$$

Find the probability of eventual extinction of the population.



\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.

Suppose $P(z=k)=a_k$ and  $A(s)=\ds\sum_{k=0}^\infty a_ks^k$.

Now $P(X_n=l\ |\ X_{n-1}=j)=P(z_1+\cdots+z_j=1)=$ coeff. of $s^l$
in $[A(s)]^j$ as the $z_i$ are i.i.d.

$P(x_n=l)=\ds\sum_{j=0}^\infty P(X_n=1\ |\ X_{n-1}=j)P(X_{n-1}=j)$

so

$\begin{array}{rcl} F_n(s) & = & \ds\sum_{l=0}^\infty
\ds\sum_{j=0}^\infty\ (\mbox{coeff. of $s^l$ in
$[A(s)]^j$})P(X_{n-1}=j)s^l\\ & = & \ds\sum_{l=0}^\infty
\left(\ds\sum_{j=0}^\infty\ (\mbox{coeff. of $s^l$ in}\
[A(s)]^j)s^l\right) P(X_{n-1}=j)\\ & = & \ds\sum_{j=0}^\infty
P(X_{n-1}=j)[A(s)]^j=F_{n-1}(A(s)) \end{array}$

Now

$P(X_0=1)=1$ so $F_0(s)=s$.

${}$

$F_1(s)=F_0(A(s))=A(s)$

$F_2(s)=F_1(A(s))=A(A(s))$

$\ \vdots$

$F_n(s)=\undb{A(A(\ \cdots\ (A}(s)))\cdots=A(F_{n-1}(s))$

\hspace{.8in} $n$ times

Now when $a_k=qp^k$

$$A(s)=\ds\sum_{k=0}^\infty qp^ks^k=\df{q}{1-p^s}$$

Now $F_1(s)=A(s)$ - which fits the given formula for $n=1$. Assume
the formula is true for $n$

$\begin{array}{rcl} & & F_{n+1}(s)\\ & = & A(F_n(s))\\ & = &
\df{q}{1-\df{pq[p^n-q^n-(p^{n-1}-q^{n-1})ps]}{p^{n+1}-q^{n+1}-(p^n-q^n)ps}}\\
& = & \df{q[p^{n+1}-q^{n+1}-(p^n-q^n)ps]}
{p^{n+1}-q^{n+1}-(p^n-q^n)ps-pq[p^n-q^n-(p^{n-1}-q^{n-1})ps]}\\ &
= & \df{q[p^{n+1}-q^{n+1}-(p^n-q^n)ps]}
{p^{n+1}(1-q)-q^{n+1}(1-p)-[p^n(1-q)-q^n(1-p)]ps}\\ & = &
\df{q[p^{n+1}-q^{n+1}-(p^n-q^n)ps]}{p^{n+2}-q^{n+2}-(p^{n+1}-q^{n+1})ps}
\end{array}$

as $p+q=1$.

Hence the result by induction.

The probability of extinction is the smallest positive root of the
equation $A(s)=s$, and so is given by

$$\df{q}{1-ps}=s$$

i.e. $ps^2-s+q=0\ \ (ps-q)(s-1)=0$ as $p+q=1$

$s=\df{q}{p},\ s=1$

So the extinction probability is 1 if $q \geq p$ and $\df{q}{p}$
if $q<p$.




\end{document}
