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\begin{document}


{\bf Question}

A population consists initially of $b$ individuals. At any
subsequent time an individual in the population has, during any
time interval of length $\delta t$, independent of previous
history and of other individuals,

\begin{description}
\item[(i)]
a probability $\alpha\delta t+o(\delta t)$ of producing a single
offspring,

\item[(ii)]
a probability $\beta\delta t+o(\delta t)$ of producing twins,

\item[(iii)]
a probability $o(\delta t)$ of producing more than two offspring,
as $\delta t \to 0$.

\end{description}
Let $p_n(t)$ denote the probability that the total population size
is $n$ at time $t$. Show that for $n=1,\ 2,\ \cdots$,

$$p_n'(t)=-(\alpha+\beta)np_n(t)+\alpha(n-1)p_{n-1}(t)+\beta(n-2)p_{n-2}(t).$$

Find the probability that, at time $t$, the population has not
changed from its original size.

Suppose that the mean number of individuals at time $t$ is

$$M(t)=\ds\sum_{n=0}^\infty np_n(t).$$

Show that $M'(t)=(\alpha+2\beta)M(t)$ and hence find $M(t)$.

\vspace{.25in}

{\bf Answer}

${}$

$P(X(t+\delta t)=n+1\ |\ X(t)=n)=\alpha n \delta t+o(\delta t)$

$P(X(t+\delta t)=n+2\ |\ X(t)=n)=\beta n \delta t+o(\delta t)$

$P(X(t+\delta t)=n\ |\ X(t)=n)=1-(\alpha+\beta)n \delta t+o(\delta
t)$

${}$

Now

$p_b(t+\delta t)=p_b(t)(1-(\alpha+\beta)b\delta t+o(\delta t))$

$p_{b+1}(t+\delta t)=p_{b+1}(t)(1-(\alpha+\beta)(b+1)\delta
t+o(\delta t)+p_b(t)(\alpha b \delta t+o(\delta t))$

For $n>b+1$

$\begin{array}{rcl} p_n(t+\delta t) & = &
p_n(t)(1-(\alpha+\beta)n\delta t+o(\delta t))\\ & & +
p_{n-1}(t)(\alpha(n-1)\delta t+o(\delta t))\\ & &
p_{n-2}(t)(\beta(n-2)\delta t+o(\delta t)) \end{array}$

In fact, since $p_n(t)=0$ for $n<b$, this equation encompasses the
first two.

Thus for $n=1,\ 2,\ \cdots$

$\begin{array}{rcl} \df{p_n(t+\delta t)-p_n(t)}{\delta t} & = &
-(\alpha+\beta)n p_n(t)+\alpha(n-1)p_{n-1}(t)\\ & & +
p(n-2)p_{n-2}(t)+o(\delta t) \end{array}$

Letting $\delta t \to 0$ gives

$p_n'(t0=-(\alpha+\beta)n p_n(t)+\alpha
(n-1)p_{n-1}(t)+\beta(n-2)p_{n-2}(t)$

${}$

Now $P_b(0)=1$ and $p_n(0)=0$ for $n \ne b$.

${}$

$P_b'(t)=-(\alpha+\beta)b p_b(t)$

so $p_b(t)=Ke^{-(\alpha+\beta)bt}$

and using $p_b(0)=1$ gives $K=1$.

${}$

Hence $p_b(t)=e^{-(\alpha+\beta)bt}$.

${}$

$n=b+1$

$\begin{array}{rcl}p_{b+1}(t+\delta t) & = &
p_{b+1}(t)(1-(\alpha+\beta)(b+1)\delta t)\\ & & +p_b(t)(\alpha b
\delta t)\\ \\ p_{b+1}'(t) & = &
-(\alpha+\beta)(b+1)p_{b+1}(t)+\alpha b p_b(t)\end{array}$

${}$

$n>b+1$

$\begin{array}{rcl} p_n(t+\delta t) & = & p_n(t)(1-(\alpha+\beta)n
\delta t)\\ & & + p_{n-1}(t)(\alpha(n-1)\delta
t)+p_{n-2}(t)(\beta(n-2)\delta t)\\ \\ p_n'(t) & = &
-(\alpha+\beta)n p_n(t)+\alpha(n-1)p_{n-1}(t)\\ & & +
\beta(n-2)p_{n-2}(t)
\end{array}$

${}$

Now $M(t)=\ds\sum_{n=0}^\infty n p_n(t)$

$\begin{array}{rcl} M'(t) & = & \ds\sum_{n=0}^\infty n p_n'(t)\\ &
= & \ds\sum_{n=0}^\infty n [-(\alpha+\beta)n
p_n(t)+\alpha(m-1)p_{n-1}(t)\\ & & +\beta(n-2)p_{n-2}(t)]\\ & = &
\ds\sum_{n=0}^\infty -n^2(\alpha+\beta)p_n(t)+\ds\sum_{n=0}^\infty
\alpha n(n-1) p_{n-1}(t)\\ & & +\ds\sum_{n=0}^\infty \beta
n(n-2)p_{n-2}(t)\\ & = & \ds\sum_{n=0}^\infty
-n^2(\alpha+\beta)p_n(t)+\ds\sum_{n=0}^\infty \alpha n(n+1)n
p_n(t)\\ & & +\ds\sum_{n=0}^\infty \beta(n+2) n p_n(t)\\ & = &
\ds\sum_{n=0}^\infty n
p_n(t)[-n(\alpha+\beta)+\alpha(n+1)+\beta(n+2)]\\ & = &
(\alpha+2\beta)M(t). \end{array}$

${}$

Hence $M(t)=He^{(\alpha+2\beta)t}$

${}$

Now $M(0)=b$, and so

$$M(t)=be^{(\alpha+2\beta)t}$$

\end{document}
