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{\bf Question}

The Albanian singing flea is hermaphrodite, and at the beginning
of the breeding season the population consists of b fleas.  At any
subsequent time during the breeding season an individual flea has,
during any time interval of length $\delta t,$ independent of
previous history and other fleas,
\begin{description}
\item[(i)] a probability $\alpha \delta t + o(\delta t)$ of
producing one new flea,
\item[(ii)] a probability $\beta \delta t + o(\delta t)$ of
producing two new fleas,
\item[(iii)] a probability $o(\delta t)$ of producing more then two new
fleas,or dying, as $\delta t \to 0$
\end{description}

${}$

With the first winter frost a few fleas manage to hibernate to
serve as the starting population for the next season, and the
remainder perish.

${}$

Let$p_n(t)$ denote the probability the the total population size
is n at time t (for values of t with in the breeding season). Show
that for any n = 1, 2, ... $$p_n'(t) = -(\alpha+\beta)np_n(t) +
\alpha(n-1)p_{n-1}(t) + \beta(n-2)p_{n-2}(t).$$ Find the
probability that, at time t, the population has not changed from
its original from its original size.

${}$

Suppose that the mean number of fleas at time t is $$M(t) =
\sum_{n=0}^\infty np_n(t)$$  Show that $\ds M'(t) = (\alpha +
2\beta)M(t)$ and hence find $M(t)$.


\vspace{.25in}

{\bf Answer}

\begin{eqnarray*} 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) \end{eqnarray*}

Arguing conditionally gives:\begin{eqnarray*} 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{eqnarray*}

(Note that for $k \leq b$ $p_k(t)=0$, so we don't need special
equations for $k \leq b+2$)

Thus we have: \begin{eqnarray*} \frac{p_n(t+\delta t) -
p_n(t)}{\delta t} & = & -(\alpha+\beta)n p_n(t) +
\alpha(n-1)p_{n-1}(t) \\ & & + \beta(n-2)p_{n-2}(t) + o(\delta t)
\end{eqnarray*}

Letting $\delta t \to 0$ gives $$p_n'(t) = -(\alpha+\beta)np_n(t)
+ \alpha(n-1)p_{n-1}(t) + \beta(n-2)p_{n-2}(t)$$

Now $p_b(0)=1$ and $p_n(0) = 0$  for $n \not= b$

So $p_b'(t) = -(\alpha+\beta)bp_b(y)$ and $p_b(0)=1$

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

${}$

Now $$ M(t) = \sum_{n=0}^\infty n p_n(t) \hspace{.3in} M'(t) =
\sum_{n=0}^\infty np'_n(t)$$ \begin{eqnarray*}  & = &
\sum_{n=0}^\infty -n^2(\alpha+\beta) p_n(t) + \sum_{n=0}^\infty
\alpha(n-1)np_{n-1}(t) + \sum_{n=0}^\infty \beta(n-2)np_{n-2}(t)
\\ & = & \sum_{n=0}^\infty -n^2(\alpha+\beta) p_n(t) + \sum_{n=0}^\infty
\alpha(n+1)np_n(t) + \sum_{n=0}^\infty \beta(n+2)np_n(t) \\ & = &
\sum n p_n(t) [-n^2(\alpha+\beta) + \alpha(n+1) +\beta(n+2)] \\ &
= & (\alpha + 2\beta)M(t) \end{eqnarray*} Now $M(0) = b$ and so
$$M(t) = be^{(\alpha+2\beta)t}$$



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